The specific heat of other liquid is 2.09 J/g°C.
Using the formula Q = mcΔT, where Q is amount of heat lost, m is the mass of liquid, c is specific heat of liquid, and ΔT is the change in temperature, we can set up two equations for each liquid.
For water, Q = (50 + 30)g × 4.18 J/g°C × (30 - 25)°C = 1045 J
For other liquid, Q = 100g × c × (30 - 25)°C = 500c J
Since both liquids lost same amount of heat, took same amount of time to cool, we can set these two equations equal to each other and solve for c: 1045 J = 500c J, c = 2.09 J/g°C
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--The correct Question is, A mass of 50 g of a certain metal at 150 degree C is immersed in 100 g of water at 11 degree C. The final temperature is 20 degree C. Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is 4.2Jg −1 K−1 .--
at what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm?
The first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm is approximately 6.2°.
The angle of the first-order maximum refers to the angle at which the brightest interference pattern appears on a screen placed behind two closely spaced slits when illuminated with the blue light of 450-nm wavelength.
The angle is determined by the equation:
theta_m = (m*lambda)/d
where m is the order, lambda is the wavelength, and d is the slit separation.
theta_m = (1*450E-9 m)/0.0500 mm
theta_m = 6.2°
Thus, the first-order maximum for double slits of 0.0500 mm at 450 nm λ blue light is around 6.2°.
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once body density is determined as with hydrostatic weighing and air displacement plethysmography, percent body fat can be calculated using
Once body density is determined as with hydrostatic weighing and air displacement plethysmography, percent body fat can be calculated using the Siri equation. Body density refers to the measurement of an individual's body mass. It is the mass of an individual's body divided by the volume of their body.
It is expressed in kilograms per cubic meter in SI units. Body density can be used to calculate body fat percentage. Body fat percentage, also known as adiposity index, is the amount of body fat present in an individual's body divided by their total body mass. Body fat is essential for proper functioning of the body, but it needs to be maintained in the right amount for overall health and well-being.
Percent body fat calculation using the Siri equation Once the body density is determined, percent body fat can be calculated using the Siri equation. The Siri equation is expressed as: Percent body fat = [(4.95/Body Density) - 4.50] x 100The Siri equation is an accurate way of calculating percent body fat. It uses body density as its basis for measurement.
Body density is determined by measuring the mass and volume of the individual's body. Hydrostatic weighing and air displacement plethysmography are the two most common methods for determining body density. These methods are accurate and reliable for body density measurement.
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a certain truck has twice the mass of a car. both are moving at the same speed. if the kinetic energy of the truck is k, what is the kinetic energy of the car?
The kinetic energy of the car is also (m_car) (v_car)², which is half the kinetic energy of the truck.
What is kinetic energy?
The kinetic energy of an object is given by the equation:
KE = (1/2)mv²
where KE is the kinetic energy, m is the mass of the object, and v is the speed of the object.
Given that the truck has twice the mass of the car and both are moving at the same speed, we can write:
m_truck = 2m_car
v_truck = v_car
The kinetic energy of the truck is given as k. Therefore, we can write:
k = (1/2)(m_truck)(v_truck)²
Substituting the values of m_truck and v_truck, we get:
k = (1/2)(2m_car)(v_car)²
k = (m_car)(v_car)²
Therefore, the kinetic energy of the car is also (m_car)(v_car)², which is half the kinetic energy of the truck.
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explain how apostivley charged object can be used to leave another metalic boject with a net negative charge
When an object acquires an electric charge, it gains either a positive or a negative charge. Positively charged objects can be used to leave another metallic object with a net negative charge. The process is known as charging by conduction.
The process of charging a conductor without touching it to the charging body but by bringing the charged body near the uncharged conductor is known as charging by conduction. A positively charged object can be used to leave another metallic object with a net negative charge by charging through conduction. It is possible because of the conservation of charges.
Charging through conduction involves the following steps: Bring a positively charged object closer to an uncharged metallic object. The positive object will transfer some of its electrons to the uncharged object because of its close proximity. The uncharged metallic object, after gaining electrons from the positively charged object, becomes negatively charged. The positive object loses electrons in the process and becomes less positively charged or sometimes negatively charged, depending on the situation.
When a positively charged object is brought near an uncharged metallic object, it gains electrons from the positive object, which creates a net negative charge in the metallic object. As a result, the metallic object is negatively charged due to the transfer of electrons from the positive object.
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use the impulse-momentum theorem to find how long a falling object takes to increase its speed from 4.23 m/s to 10.47 m/s?
The time it takes the object to fall through the change in speed using the impulse-momentum theorem is 0.62 seconds.
What is impilse-momentum theorem?
The impulse-momentum theorem states that the change in momentum of an object is equal to the impulse exerted on it.
To calculate the time it takes the object to increase it speed using the impulse-momentum theorem, we use the formula below.
Formula:
Ft = m(v-u)Ft/m = (v-u)Recall that F/m = acceleration. Therefore,
at = v-ua = (v-u)/t.......................... Equation 1Where:
a = Acceleration due to gravityv = Final velocityu = Initial velocityt = TimeFrom the question,
Given:
v = 10.47 m/su = 4.23 m/sg = 9.8 m/s²Substitute these values into equation 1 and solve for t
9.8 = (10.27-4.23)/tt = (10.27-4.23)/9.8t = 6.04/9.8t = 0.62 secondsHence, the time it takes the object to fall is 0.62 seconds.
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the time it takes a planet to complete one full orbital revolution is commonly known as its question 25 options: period frequency acceleration velocity
The time it takes a planet to complete one full orbital revolution is commonly known as its period. Option a is the correct choice.
The period of a planet refers to the time it takes for the planet to complete one full orbit around its star or sun. This time period is determined by the distance between the planet and the star, as well as the planet's velocity. The period is an important concept in astronomy and is used to calculate a planet's orbital speed, distance, and other orbital parameters. By studying the periods of planets, astronomers can make predictions about their behavior and gain insights into the workings of the solar system and the universe as a whole. Therefore, option a is correct.
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10. a kangaroo is capable of jumping to 250 cm height. determine time taken in jump and takeoff speed of the kangaroo.
The time taken and the takeoff speed of the kangaroo jumping is 0.71 seconds and 7 m/s, alternately. The result is obtained by using the equations for uniformly accelerated motion.
Uniformly Accelerated Straight MotionA uniformly accelerated straight motion is a motion with acceleration or deceleration in a straight line. The equations apply in vertical dimension are
v₁ = v₀ + gtv₁² = v₀² + 2ghh = v₀t + ½ gt²The maximum height of kangaroo jumping is 250 m.
Determine the time taken in jump and takeoff speed of the kangaroo!
At the maximum height, the velocity is zero. So, using the equations for uniformly accelerated motion, the takeoff speed (initial speed) of the kangaroo is
v₁² = v₀² + 2gh
0 = v₀² + 2(-9.81)(2,5)
v₀² = 2(9.81)(2,5)
v₀ = √49
v₀ = 7 m/s
The time taken for the kangaroo jumping is
v₁ = v₀ + gt
0 = 7 + (-9.81)t
t = 7/9.81
t = 0.71 seconds
Hence, the time taken is 0.71 seconds and the initial speed of the kangaroo jumping is 7 m/s.
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a long, straight wire carries a current of 8.60 a. an electron is traveling in the vicinity of the wire. at the instant when the electron is 4.50 cm from the wire and traveling at a speed of 6.00 * 104 m>s directly toward the wire, what are the magnitude and direction (relative to the direction of the current) of the force that the magnetic field of the current exerts on the electron?
The magnitude and direction of the force that the magnetic field of the current exerts on the electron in a a long, straight wire is 1.96 x 10⁻¹⁸ N and direction of the force is opposite to the direction of the current.
The magnetic field of the current exerts a force on the electron of magnitude 6.072 x 10⁻¹³ N in a direction that is opposite to the direction of the current.
where
Current, I = 8.60 A
Distance of electron from wire, r = 4.50 cm = 0.045 m
Velocity of electron, v = 6.00 x 10^4 m/s
The force on the electron due to magnetic field of current-carrying wire is given by:
F = (μ * I * q) / (2 * π * r)
where μ is the magnetic permeability of free space and is equal to 4π x 10⁻⁷ Tm/A,
q is the charge of electron and is equal to -1.6 x 10⁻¹⁹ C, and
r is the distance between the electron and the wire.
Substituting the values, we get:
F = (4π x 10⁻⁷ Tm/A) * (8.60 A) * (-1.6 x 10⁻¹⁹ C) / (2 * π * 0.045 m)
F = -1.96 x 10⁻¹⁸ N.
The negative sign indicates that the direction of force is opposite to the direction of the current.
So, the magnitude of the force exerted by the magnetic field on the electron is 1.96 x 10⁻¹⁸ N, and the direction of the force is opposite to the direction of the current.
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when 171 v is applied across a wire that is 11 m longa nd has a 0.44 mm radius, the magnitude of the current density is 1.3 x 10^4
When 171 V is applied across a wire that is 11 m long and has a 0.44 mm radius, the magnitude of the current density is 1.3 x 10^4 A/m2.
solution:
This current density can be calculated using the following equation:
Current Density = Voltage / (Length x Resistance)
Therefore: Current Density = 171 V / (11 m x 1.05 Ω)
Current Density = 1.3 x 10^4 A/m2
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example 16-3: sound intensity on the street. at a busy street corner, the sound level is 75 db. what is the intensity of sound there?
The intensity of sound at the busy sound corner is 3.162 × 10⁻² W/m².
The sound intensity, represented by I, is defined as the power conveyed by a sound wave per unit area. Watts per square metre (W/m2) are the units of measurement.
waves are a type of energy propagation through a medium by means of adiabatic lading and unloading. Important amounts for describing aural swells are aural pressure, flyspeck haste, flyspeck relegation and aural intensity.
The formula for determining sound intensity from decibel level is as follows:
I = I₀ × 10^(L/10)
where I0 is the reference intensity and L is the decibel level.
Plugging in the values from the issue yields:
I = (1×10⁻¹² W/m²) × 10^(75/10) = 3.162 × 10⁻² W/m²
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each resistor is rated at 0.50 w w (maximum without overheating), what is the maximum voltage that can be applied across the whole network?
The maximum voltage that can be applied across the whole network is 1.28 V.
To calculate the maximum voltage that can be applied across the whole network, you need to apply Ohm's Law and power formula.
The equation for power is P = V²/R,
where P is power, V is voltage, and R is resistance.
Therefore, V = sqrt(P * R).
Given that each resistor is rated at 0.50 W, the power of the network is 2 x 0.50 W = 1 W.
Since the resistors are in parallel, the equivalent resistance can be calculated as follows:
1/R = 1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5R = 1/(1/3 + 1/3 + 1/6 + 1/8 + 1/16)R = 1.6375 Ω
Therefore, the maximum voltage that can be applied across the whole network is V = sqrt(P * R) = sqrt(1 * 1.6375) = 1.28 V (approx).
Therefore, the maximum voltage that can be applied across the whole network is 1.28 V.
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over the course of a half of a year the relative position of the sample star, as seen from earth, is seen to change by 0.400''. what is the parallax angle (p) in this case?\
Over the course of half of a year the relative position of the sample star, as seen from earth, is seen to change by 0.400''. The parallax angle in this case is: 0.400''
Given that the relative position of the sample star as seen from earth is seen to change by 0.400'' over the course of half of a year. We are to determine the parallax angle in this case. Parallax angle (p) can be defined as the angle between the baseline and the line of sight to the star. It is the angle between two lines drawn from the star to the Earth, separated by six months, and viewed at a right angle to the baseline.
It is measured in seconds of arc (or arcseconds), and it is usually too small to measure directly. The parallax angle can be calculated using the formula below: parallax angle (p) = (d/b)
where d is the distance from the Earth to the star and b is the baseline, which is half of the distance that the Earth moves in its orbit over six months, which is equal to 1 astronomical unit (AU).
Thus, using the given values, we can calculate the parallax angle as follows: [tex]p = (d/b) = (0.400/1) = 0.400''[/tex]
Thus, the parallax angle, in this case, is 0.400'' (arcseconds). Therefore, the relative position of a star as seen from Earth changes with the change in the Earth's position. The change in position helps to determine the distance from the Earth to the star using the parallax angle.
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which satellite channel measures the temperature of the underlying surfaces (i.e., clouds, ocean, land)? group of answer choices visible infrared water vapor
Visible Infrared (IR) satellite channels measure the temperature of underlying surfaces. This includes clouds, oceans, and land.
IR channels work by detecting the amount of infrared radiation emitted from the Earth's surface. The intensity of the radiation is then converted into a digital number, which is displayed as a color on a satellite image. The higher the digital number, the warmer the surface temperature. This data can then be used to track changes in temperatures over time. The satellite channel that measures the temperature of the underlying surfaces is visible infrared. The surface temperature measurement is made possible by the difference in temperatures of objects in the infrared spectrum. An object's temperature and the level of radiation it emits have a direct correlation, and this is what visible infrared satellites use to take the temperature of the underlying surfaces. The visible infrared (VI) channel is used to estimate cloud cover and surface temperature. Infrared radiation from the surface of the earth is detected in this channel. The temperature of clouds, oceans, and land can be estimated using the visible infrared (VI) channel. It also provides data on how temperature changes with latitude and over time. Furthermore, the VI channel aids in the identification of cold and hot surfaces. Water vapor (WV) is another channel utilized in satellite imagery to observe the atmosphere's water vapor content. It enables meteorologists to forecast the occurrence of rainfall and other weather patterns. In general, satellite measurements assist in understanding Earth's weather and its impact on humans and the environment. These satellites help scientists to forecast severe weather, monitor weather changes over time, and analyze natural disasters. In addition, they assist in tracking the effects of climate change on the planet.
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the pilot of an airplane notes that the compass indicates a heading due west. the airplane's speed relative to the air is 100 km/h. the air is moving in a wind at 31.0 km/h toward the north. find the velocity of the airplane relative to the ground.
The pilot of an airplane notes that the compass indicates a heading due west. The airplane's speed relative to the air is 100 km/h. The air is moving in the wind at 31.0 km/h toward the north. The velocity of the airplane relative to the ground is: 104 km/h
The airplane's velocity relative to the ground is calculated by adding the velocity of the airplane relative to the air with the velocity of the air relative to the ground.
The velocity of the airplane relative to the ground is obtained by vector addition of the airplane's velocity relative to the air and the air's velocity relative to the ground. Given that the compass indicates a heading due west, the airplane's velocity relative to the air is 100 km/h towards the west.
The air is moving towards the north at 31.0 km/h, therefore the velocity of the air relative to the ground will be towards the north. The velocity of the air relative to the ground will be equal to 31.0 km/h towards the north.
To find the velocity of the airplane relative to the ground, we need to add the velocity of the airplane relative to the air to the velocity of the air relative to the ground.
Hence, we get the velocity of the airplane relative to ground = velocity of the airplane relative to air + velocity of air relative to ground. The velocity of the airplane relative to the ground = (100 km/h)2 + (31.0 km/h)2 = 104 km/h.
The velocity of the airplane relative to the ground is 104 km/h.
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two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. the electric potential of the inner conductor, with respect to the outer conductor, is 600 v. what is the maximum electric field magnitude between the cylinders? ( k
We can use the formula for electric field between two cylindrical conductors to calculate the maximum electric field magnitude between the cylinders:
E = (V ln(b/a))/d
where V is the potential difference between the conductors, ln is the natural logarithm, b and a are the radii of the outer and inner conductors, respectively, and d is the distance between the conductors.
Given:
V = 600 V
a = 20 mm = 0.02 m
b = 80 mm = 0.08 m
The distance between the conductors is the difference in their radii:
d = b - a = 0.08 m - 0.02 m = 0.06 m
The electric constant, k, is also needed:
k = 8.98755 × 10^9 N·m^2/C^2
Substituting these values into the formula, we get:
E = (V ln(b/a))/d
E = (600 V ln(0.08/0.02))/0.06
E = 3.5983 × 10^8 V/m or approximately 3.60 × 10^8 V/m
Therefore, the maximum electric field magnitude between the cylindrical conductors is approximately 3.60 × 10^8 V/m.
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Which segment of copper wire has the highest resistance at room
temperature?
(1) 1.0 m length, 1.0 × 10-6 m² cross-sectional area
(2) 2.0 m length, 1.0 × 10-6 m² cross-sectional area
(3) 1.0 m length, 3.0 x 10-6 m² cross-sectional area
(4) 2.0 m length, 3.0 x 10-6 m² cross-sectional area
The segment of copper wire with the highest resistance at room temperature is segment (2), which is 2.0 m in length and has a cross-sectional area of 1.0 x [tex]10^{-6}[/tex] m².
What is the resistance?
The resistance of a conductor is given by the formula:
R = (ρL) / A
where R is the resistance, ρ is the resistivity of the material, L is the length of the conductor, and A is the cross-sectional area of the conductor.
Assuming that the resistivity of copper is constant, we can compare the resistance of the different segments of copper wire using the above formula.
We can calculate the resistance of each segment of copper wire as follows:
(1) R = (1.68 x [tex]10^{-8}[/tex] Ωm x 1.0 m) / (1.0 x [tex]10^{-6}[/tex] m²) = 0.017 Ω
(2) R = (1.68 x [tex]10^{-8}[/tex] Ωm x 2.0 m) / (1.0 x [tex]10^{-6}[/tex] m²) = 0.034 Ω
(3) R = (1.68 x [tex]10^{-8}[/tex] Ωm x 1.0 m) / (3.0 x [tex]10^{-6}[/tex] m²) = 0.0056 Ω
(4) R = (1.68 x [tex]10^{-8}[/tex] Ωm x 2.0 m) / (3.0 x [tex]10^{-6}[/tex] m²) = 0.0112 Ω
Therefore, the segment of copper wire with the highest resistance at room temperature is segment (2), which is 2.0 m in length and has a cross-sectional area of 1.0 x [tex]10^{-6}[/tex] m².
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Complete question is: The segment of copper wire with the highest resistance at room temperature is segment (2), which is 2.0 m in length and has a cross-sectional area of 1.0 x [tex]10^{-6}[/tex] m².
speed camera uses electromagnetic radiation, with a wavelength 26 mm, to identify cars moving at 60 km/h or faster away from the camera. the cars act as the use of radiation, reflecting the radiation from the camera. what range of frequency decrease identifies cars above the speed limit?
The range of frequency decrease that identifies cars above the speed limit is determined by the wavelength of the electromagnetic radiation used by the high-speed camera. Since the wavelength of the radiation is 26 mm, the frequency of the radiation must be 11.54 GHz.
Thus, any frequency decrease below 11.54 GHz will identify cars moving at 60 km/h or higher. This works because, as the car moves away from the camera, it reflects some of the radiation instead of all of it.
This reduces the frequency of the radiation, and any reduction below 11.54 GHz indicates that the car is moving faster than the speed limit. This ability of the camera to identify cars moving at or above the speed limit is essential for safety and enforcement of traffic laws.
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when a knowledgeable amateur astronomer tells you that she has a 14-inch telescope, what does the number 14 refer to?
When a knowledgeable amateur astronomer tells you that she has a 14-inch telescope, the number 14 refers to the diameter of the telescope’s objective lens.
A telescope is a device used to view distant objects by utilizing electromagnetic radiation to either magnify their apparent size or gather more light than the human eye can. Telescopes are used for scientific, commercial, and amateur purposes. The telescope comprises an objective lens or mirror and an eyepiece to magnify the images created by the objective. Most telescopes have a viewfinder to make it simpler to aim the telescope precisely at the object of interest. They may also have a motorized mount to track celestial objects as they move across the sky.
Telescopes come in a variety of sizes, designs, and shapes and they range from large observatory telescopes to handheld amateur models. They are often classified into two types, reflecting and refracting telescopes and the size of a telescope is determined by the diameter of its objective lens or mirror. The bigger the diameter, the more light the telescope can collect, and the clearer the image. The diameter of the objective is the most significant aspect of a telescope when it comes to observing the heavens. For instance, a 14-inch telescope has an objective lens with a diameter of 14 inches, this large lens is capable of collecting a lot of light and providing clear images, making it a perfect tool for viewing the night sky.
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a box is given a push so that it slides across the floor. how far will it go, given that the coefficient of kinetic friction is 0.11 and the push imparts an initial speed of 3.8 m/s ?
The box will slide a distance of 6.96 m before coming to a stop due to the force of kinetic friction.
To determine how far the box will slide on the floor after it is given a push with an initial speed of 3.8 m/s, we need to use the equations of motion for constant acceleration. The force of kinetic friction acting on the box will cause it to decelerate, eventually coming to a stop.
The distance traveled by the box can be found using the equation:
d = [tex](v_i^2 - v_f^2) / (2 * a)[/tex]
where d is the distance traveled, v_i is the initial speed, v_f is the final speed (which is zero since the box comes to a stop), and a is the deceleration caused by the force of kinetic friction.
The deceleration can be found using the equation:
a = -F[tex]_friction / m[/tex]
where Ffriction is the force of kinetic friction and m is the mass of the box.
Assuming a mass of 5 kg for the box and a coefficient of kinetic friction of 0.11, the force of kinetic friction can be found using the equation:
F_friction = friction coefficient * F_normal
where F_normal is the normal force (equal to the weight of the box) and the friction coefficient is a dimensionless quantity that depends on the nature of the contact surface.
The weight of the box is:
Fweight = m * g
where g is the acceleration due to gravity (9.81 m/s²).
Therefore, the force of kinetic friction is:
F_friction = (0.11) * (5 kg * 9.81 m/s²) = 5.40 N
Using the equation for deceleration, we get:
a = -Ffriction / m = -(5.40 N) / (5 kg) = -1.08 m/s²
Finally, we can use the equation for distance traveled to find the distance the box will slide:
d = [tex](v_i^2 - v_f^2) / (2 * a)[/tex] =[tex](3.8 m/s)^2 / (2 * 1.08 m/s^2)[/tex] = 6.96 m
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a standing wave experiment is performed to determine the speed of waves in a rope. the rope makes 36 complete vibrational cycles in exactly one minute. if the wavelength is 3 m, what is the speed (in m/s) of the wave?
The speed of the wave is 1.8 m/s.
The speed of a wave in a rope is equal to the wavelength divided by the time it takes for a single cycle. In this experiment, the wavelength is 3 m and the time for a single cycle is 1/36 min, so the speed is:
Speed = \frac{3 \text{m}}{\frac{1 \text{min}}{36}} = \frac{3 \times 36 \text{m}}{1 \text{min}} = 108 \text{m/s}
A standing wave experiment is performed to determine the speed of waves in a rope. The rope makes 36 complete vibrational cycles in exactly one minute. If the wavelength is 3 m, The formula for wave speed (v) is given by v = λfWhere,v = Wave speedλ = Wavelength f = Frequency. Since the rope makes 36 complete vibrational cycles in exactly one minute or 60 seconds, its frequency is give by f = Number of cycles/time= 36/60= 0.6 Hz. Substituting the values of wavelength and frequency, we get
v = λf= 3 m × 0.6 Hz= 1.8 m/s
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a pen placed 13.2 cm from a concave spherical mirror produces a real image 14.0 cm from the mirror. a) what is the focal length of the mirror? answer in units of cm.
The focal length of the mirror is 8.57 cm.
The focal length is the distance between the mirror and the focal point.
For a convex lens the focal point is that point at which parallel rays will be focused after passing thru the lens.
For a convex lens, which is thicker at the center and thinner at the edges, the focal point is the point where parallel rays of light that pass through the lens converge.
When parallel rays of light pass through a convex lens, they are refracted, or bent, towards the center of the lens due to the lens's shape and refractive index.
As a result, these rays of light converge at a point on the opposite side of the lens from where the light entered, and this point is known as the focal point of the lens.
The distance between the convex lens and its focal point is called the focal length. It is usually denoted by the symbol 'f' and is an important parameter in lens design and optical systems.
The focal length of a convex lens determines how much the lens will bend or refract light and how much the light will converge at the focal point.
A lens with a shorter focal length will bend light more and converge it at a closer focal point, while a lens with a longer focal length will bend light less and converge it at a farther focal point.
The focal length of the concave spherical mirror can be calculated using the formula: 1/f = (1/p) + (1/q),
where p is the distance from the object to the mirror (13.2 cm) and q is the distance from the image to the mirror (14.0 cm).
Therefore, the focal length of the mirror is 8.57 cm.
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mountain bike tires have large, knob-like treads. these tires are useful on steep slopes because they responses increase friction. increase friction. carry heavy weights. carry heavy weights. increase the stopping distance. increase the stopping distance. prevent braking of the bicycle.
The correct option for the given statement is the first option i.e., they increase friction.
Mountain bike tires have large, knob-like treads. These tires are useful on steep slopes because they increase friction. Friction is a force that opposes motion between two surfaces that are in contact, and this force can be helpful when trying to stop or slow down the bike.
The treads help the tire to grip the surface better, which increases friction and makes it easier to control the bike. Additionally, mountain bike tires are wider than road bike tires, which also increases their contact area with the ground and thus, the friction.
They are also designed to withstand more abuse than road bike tires, as they are meant to handle rougher terrain, so they are less likely to puncture or wear down quickly. Hence, it can be concluded that mountain bike tires are useful on steep slopes because they increase friction.
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find the tension in an elevator cable if the 1 500-kg elevator is descending with an acceleration of 2.8 m/s2, downward.
The tension in an elevator cable if the 1 500-kg elevator is descending with an acceleration of 2.8 m/s² is 18,900 N.
The tension in the elevator cable, for net force is :
[tex]F_{net} = ma[/tex]
where [tex]F_{net}[/tex] is the net force,
m is the mass of the elevator, and
a is the acceleration of the elevator.
Since the elevator is descending, we can take the upward direction as positive.
The forces acting on the elevator are the force of gravity (mg) and
the tension in the cable (T), where T is in the upward direction.
Therefore, the net force acting on the elevator is:
[tex]F_{net}= T - mg[/tex]
where g is the acceleration due to gravity (9.8 m/s²).
Substituting the given values into the equation:
[tex]F_{net} = T - mg[/tex]
[tex]ma = T - mg[/tex]
Rearranging the equation, we get:
[tex]T = ma + mg[/tex]
where T is the tension in the cable,
m is the mass of the elevator,
a is the acceleration of the elevator, and
g is the acceleration due to gravity.
Also Substituting the given values:
T = (1500 kg) × (2.8 m/s²) + (1500 kg) × (9.8 m/s²)
T = 4200 N + 14700 N
T = 18900 N
Therefore, the tension in the elevator cable is 18,900 N when the 1,500-kg elevator is descending with an acceleration of 2.8 m/s², downward.
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skateboarder begins down a ramp at a speed of 1.0 m/s. after 3 seconds, her speed has increased to 4.0 m/s. calculate her acceleration
The acceleration of the skateboarder while going down the ramp is found to be 1m/s².
The skateboarder began to go down the ramp and that at a speed of 1.0m/s. After 3 seconds it is found that the speed of the skater is increased to 4.0m/s.
We can use the equation,
V = U+at, where, V is final speed, a is acceleration, t is time and U is initial speed.
Putting all the values,
4 = 1 +a(3)
a = 3/3
a = 1m/s²
The acceleration of the skateboarder is 1m/s².
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a bowling ball (2.6kg) is going down a lane at 5m/s to the right attempting to stike a lone stationary pin (0.3kg). the ball bounces back at a velocity of 1 ,/s at an angle of 30 below the horizontal. what is teh final velocity and direction of the pin
The final velocity and direction of the pin is 10m/s, at an angle of 30 degrees below the horizontal.
The final velocity and direction of the pin can be calculated by using the law of conservation of momentum. Momentum (P) is equal to the mass (M) multiplied by the velocity (V). The momentum of the system before the collision is the sum of the momentum of the ball and the pin, which can be expressed as follows:
P initial = (M ball * V ball) + (M pin * V pin)
Since the pin was initially stationary, V pin = 0. Therefore:
P initial = (2.6kg * 5m/s) + (0.3kg * 0)
P initial = 13 kgm/s
After the collision, the momentum of the system must remain constant. Therefore:
P final = (M ball * V ball) + (M pin * V pin)
P final = (2.6kg * 1m/s) + (0.3kg * V pin)
Pfinal = 13 kgm/s
Solving for V pin, we get:
V pin = 10m/s
The final velocity of the pin is 10m/s, at an angle of 30 degrees below the horizontal.
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through what potential difference should electrons be accelerated so that their speed is 1.2 % % of the speed of light when they hit the target?
The potential difference of the electron is 370V.
To determine the potential difference required to accelerate electrons to a speed of 1.2% of the speed of light, we can use the following equation:
v = √[(2qV)/m]
where:
v is the velocity of the electron
q is the charge of the electron
V is the potential difference
m is the mass of the electron
Since we are given the desired velocity of the electrons, we can rearrange the equation to solve for V:
V = (mv^2)/(2q)
We know the mass of an electron, which is approximately 9.11 × 10^-31 kg. We also know the charge of an electron, which is -1.6 × 10^-19 C.
So, plugging in the values, we get:
V = [(9.11 × 10^-31 kg) × (0.012c)^2] / (2 × -1.6 × 10^-19 C)
where "c" is the speed of light.
Simplifying and solving for V, we get:
V = 370 V
Therefore, electrons should be accelerated through a potential difference of 370 V so that their speed is 1.2% of the speed of light when they hit the target.
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a projectile is launched over a horizontal surface in such a manner that its maximum height is 4/5 of its horizontal range. determine the launch angle.
The launch angle a horizontal range maximum height is 4/5 is 77.47°.
Given, the maximum height of a projectile is 4/5 of its horizontal range. Let us assume that the maximum height and horizontal range be h and R respectively.
Let the initial velocity of the projectile be v₀ and the angle of projection be θ.
Since the projectile is launched over a horizontal surface, the initial vertical velocity of the projectile is 0.
Using the formulae of motion under constant acceleration, we can write, h = v₀sinθ)²/2g R = v₀²sin2θ/g
Where g is the acceleration due to gravity.
Substituting the value of v₀ from the first equation into the second equation, we get,
R = h tanθ/2 = 4R/5 tanθ/2
On simplification, we get,
tanθ/2 = 8/5
tanθ = 16/5
tan⁻¹16/5 = 77.47°
So, the launch angle is 77.47°.
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what do astronomers mean when they talk about the seeing conditions at a potential observatory site?
When astronomers talk about the seeing conditions at a potential observatory site, they are referring to the atmospheric turbulence and how it affects the quality of images obtained from telescopes at that location.The seeing conditions can have a significant impact on the image quality as well as the scientific output of an observatory.
Turbulent air creates a blurring effect on the images which is known as atmospheric distortion. This limits the telescope’s ability to resolve fine details in the observed objects.The quality of the seeing conditions at a potential observatory site depends on various factors such as the altitude, climate, and topography.
Astronomers evaluate the seeing conditions by monitoring the atmospheric turbulence at the site. They use a device called a seeing monitor that measures the fluctuations in the air density and temperature.The seeing conditions are critical for the success of an observatory.
Astronomers prefer sites with stable atmospheric conditions, low turbulence, and dry climate. These conditions help to minimize the effects of atmospheric distortion on the images and enable astronomers to study celestial objects in greater detail.
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why does the pressure rise as the volume of a cylinder filled with a gas is decreased by a piston? multiple choice question. gas particles move faster in a smaller volume. collisions with the walls are more frequent. collisions of gas particles with each other are more frequent.
When the volume of a cylinder filled with a gas is decreased by a piston, the pressure inside rises because of the increased frequency of collisions between gas particles.
This is due to the fact that when the available space is reduced, the particles are forced to move faster in order to maintain their average kinetic energy. Furthermore, the number of collisions between gas particles and the walls of the container increases, resulting in a higher pressure.
Additionally, as the volume decreases, the number of collisions between gas particles and each other increases, which also contributes to the rise in pressure. Therefore, when the volume of a cylinder filled with a gas is decreased by a piston, the pressure inside will rise due to the increased frequency of collisions between gas particles.
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what does pluto most resemble? what does pluto most resemble? a terrestrial planet a jovian planet a comet
Pluto most closely resembles a terrestrial planet, like the other planets in our solar system. Terrestrial planets are composed mostly of rock and metal, and have solid surfaces. Pluto is believed to have a rocky core surrounded by a mantle of ice, which makes it a terrestrial planet.
Pluto most resembles a terrestrial planet. What are terrestrial planets? Terrestrial planets are planets composed primarily of silicate rocks or metals, which are relatively near to the Sun. They are named after the Earth, as they share many common features. Venus, Earth, and Mars are the three most well-known planets in this group. Pluto is similar to a terrestrial planet since it is composed of rocky material like the Earth. Despite being a dwarf planet, it shares many characteristics with the terrestrial planets. Pluto is a small, icy world that orbits the Sun, is believed to be covered in water ice and various kinds of frozen gases, and has an atmosphere that is primarily composed of nitrogen. Pluto was originally identified as the ninth planet in our solar system but was later reclassified as a dwarf planet because it failed to meet the International Astronomical Union's criteria for being considered a planet. Although Pluto is no longer classified as a planet, it remains one of the most interesting objects in the outer solar system, and the study of Pluto is essential to our understanding of the development of our solar system.
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