The distance from the base of the table that the second object lands will be the same as the distance from the base of the table that the first object lands.
This is because the initial kinetic energy and spring potential energy that the objects possess is the same in both cases. The only difference between the two scenarios is the mass of the objects, which does not affect the distance traveled. This is because the time taken by the objects to travel the same distance is inversely proportional to their masses, so the total time taken by both objects to travel the same distance is the same.
This means that the distance traveled by both objects is the same, and hence the distance from the base of the table that the second object lands will be the same as the distance from the base of the table that the first object lands.
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An astronaut weighs 8.00 × 102
newtons on the
surface of Earth. What is the weight of the astronaut
6.37 × 106
meters above the surface of Earth?
The weight of the astronaut 6.37 × 106 meters above the surface of the Earth would be 160 N.
Weight of an astronautThe weight of an object depends on its mass and the gravitational field it is in. Near the surface of the Earth, the acceleration due to gravity is approximately 9.81 m/s².
We can use the formula F = mg to calculate the weight of the astronaut at the surface of the Earth:
The gravitational field weakens as the distance from the center of the Earth increases, thus, we can use the formula for gravitational acceleration at a distance r from the center of the Earth:
g' = g (R / (R + h))²g' = 9.81 m/s² x (6.37 × 106 m / (6.37 × 106 m + 6.37 × 106 m))²g' = 1.96 m/s²Now we can calculate the weight of the astronaut at this height:
F' = mg'F' = mass x (1.96 m/s²)We don't know the mass of the astronaut, but we can use the weight at the surface of the Earth to find it:
F = mgm = F / g= 8.00 × 102 N / 9.81 m/s²
= 81.63 kg
F' = (81.63 kg) x (1.96 m/s²)
F' = 160 N
Therefore, the weight of the astronaut 6.37 × 106 meters above the surface of the Earth is approximately 160 N.
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(1]2] Which row links both the photoelectric effect and electron diffraction to the properties of
waves and particles?
[1 mark]
Photoelectric effect | Electron diffraction
a Particle property Particle property
8 | Wave property Wave property
Particle property Wave property
| Wave property Particle property
The row that links both the photoelectric effect and electron diffraction to the properties of waves and particles is the first row, which includes the terms "Particle property" and "Wave property".
The photoelectric effect refers to the phenomenon where electrons are emitted from a material when light shines on it, while electron diffraction refers to the scattering of electrons by a crystal lattice.
Both of these phenomena can be explained using the wave-particle duality of matter, which suggests that matter can exhibit both particle-like and wave-like properties. The photoelectric effect can be explained by treating light as a particle (photon) that transfers energy to an electron, while electron diffraction can be explained by treating electrons as waves that interfere with each other.
Understanding the properties of waves and particles is essential in understanding these phenomena and many other fundamental concepts in physics. The study of wave-particle duality has also led to the development of quantum mechanics, which is a cornerstone of modern physics. The correct option is "Particle property" and "Wave property".
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you pull a sled with a package on it across a snow covered flat lawn. if u apply a force of 46.8 N to the sled, it accelerates at 0.75 m/s. what is the combined mass of the package and sled
The combined mass of the package and the sled can be found to be 62.4 kg.
How to find the combined mass ?We can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F net = ma), to solve this problem.
F net = ma
46.8 N = (m sled + m package) × 0.75 m/s²
To find the combined mass of the sled and package, we need to add their individual masses together. Therefore, we can rewrite the equation as:
46.8 N = (m sled + m package) × 0.75 m/s²
46.8 N / 0.75 m/s² = m sled + m package
62.4 kg = m sled + m package
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When a wind-up toy is set in motion, elastic potential energy that was stored in a compressed spring is converted into the __________ of the toy’s moving parts
Answer:When a wind-up toy is set in motion, elastic potential energy that was stored in a compressed spring is converted into the kinetic energy of the toy's moving parts.
Explanation:
A landscaper uses 15. 00 newtons of force to push a lawn mower. How much work, in joules, does the landscaper use to move the lawn mower?
The landscaper uses 75.00 joules of work to move the lawn mower.
Work is the product of force and displacement, in the direction of the force.
Given that the landscaper uses a force of 15.00 N to push a lawn mower, the amount of work done depends on the distance the mower is pushed.
If we assume that the mower is pushed a distance of 5 meters, the work done can be calculated as follows:
Work = force x distance x cos(theta)
where theta is the angle between the force and the direction of displacement, which we assume to be zero degrees in this case. Therefore, the work done can be calculated as:
Work = 15.00 N x 5 m x cos(0) = 75.00 J
Therefore, the landscaper uses 75.00 joules of work to move the lawn mower.
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Katie and her dad gather ingredients to make pancakes. Then they follow the following six steps. Mix the ingredients together in a bowl. Add a spoon of pancake mix to a hot pan. See that bubbles start to form and the pancake starts to harden. Wait until the bubbles stop forming. Flip the pancake so the other side can darken and harden. Serve the pancake on a plate. Which three steps in the process best show that new substances are created when making pancakes?
A. Steps 1, 2, and 5
B. Steps 3, 4, and 5
C. Steps 1, 4, and 6
D. Steps 2, 3, and 6
In the process of making pancakes, the three steps that best show that new substances are created are: Steps 3, 4, and 5. The correct option is B.
These steps involve seeing bubbles form as the pancake starts to harden, waiting until the bubbles stop forming, and flipping the pancake so the other side can darken and harden.
During Step 3, the formation of bubbles indicates a chemical reaction taking place, as the heat causes the pancake batter to release carbon dioxide gas. This leads to the creation of a new substance: the cooked pancake. Step 4, waiting for the bubbles to stop forming, further demonstrates the chemical changes occurring as the batter continues to cook and transform.
Lastly, in Step 5, flipping the pancake allows the other side to darken and harden, completing the cooking process and solidifying the new substance: a fully cooked pancake. The correct option is B.
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A baseball is hit straight up at an initial velocity of 30m/s if the ball has a negative acceleration of about 10m/s2 how long does the ball take to reach the too of its path
The ball takes about 3.06 seconds to reach the top of its path.
When a ball is thrown or hit straight up, it will reach its maximum height at the point where its vertical velocity becomes zero.
At this point, the ball's acceleration will be equal to the acceleration due to gravity, which is -9.8 m/s².
Using the equation of motion for an object with constant acceleration, we can find the time it takes for the ball to reach the top of its path:
vf = vi + at
where vf is the final velocity (which is zero when the ball reaches its maximum height), vi is the initial velocity (30 m/s), a is the acceleration (-9.8 m/s²), and t is the time we're looking for.
Rearranging the equation, we get:
t = (vf - vi) / a
Since the final velocity is zero, we have:
t = -vi / a
Substituting the values, we get:
t = -30 m/s / (-9.8 m/s²)
t = 3.06 seconds
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A light bulb carries a current i. the power dissipated in the light bulb is p. what is the power dissipated if the same light bulb carries a current of 3i
The power dissipated in the light bulb if it carries a current of 3i is 9p.
The power dissipated by a light bulb is given by the equation P = I²R, where I is the current flowing through the bulb and R is its resistance. Since the same light bulb is being used, its resistance remains constant.
When the current flowing through the bulb is increased to 3i, the power dissipated is given by P' = (3i)²R = 9i²R = 9P,
where P is the power dissipated when the current was i.
Therefore, the power dissipated in the light bulb is multiplied by a factor of 9 when the current is increased to 3i.
Assuming the resistance of the light bulb remains constant, we can use Ohm's law to find the new current when the current is tripled. Ohm's law states that the current flowing through a conductor is directly proportional to the voltage applied across it and inversely proportional to its resistance. Therefore, when the current is tripled, the voltage across the bulb will also triple since the resistance remains constant.
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Is it possible to play the lowest string with your finger on any of the frets shown and hear the same frequency as the highest string?.
No, it is not possible to play the lowest string with your finger on any of the frets shown and hear the same frequency as the highest string.
The frets on a stringed instrument, such as a guitar, are placed in specific positions along the neck to produce different pitches or frequencies when the strings are pressed against them.
Each fret represents a specific note, and when you press a string against a particular fret, you effectively shorten the vibrating length of the string, which increases the frequency and raises the pitch of the sound produced.
As you move your finger along the fretboard, the pitch of the note played changes.
The lowest string on a guitar, typically the thickest string, has the lowest pitch or frequency when played open (without pressing any frets). As you press down on higher frets, you increase the pitch of the note.
The highest string on a guitar, typically the thinnest string, has the highest pitch or frequency when played open.
Therefore, pressing a fret on the lowest string will never produce the same frequency as the open (unfretted) highest string because the length and tension of the strings are different, resulting in different natural frequencies for each string.
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what is the apparant position of an object bellw a 6cm thick rectangular block of glass if a 4 cm water is on top of glass
note:in my book it took mew of glass independently .. (I mean with air but there is water is top of it, will it affect mew ?) (a pic is attached check it)
Yes, the presence of water on top of the glass block will affect the apparent position of the object.
Total apparent depth of the block and water is 8 cm.
Why does water affect apparent position?This is because the light rays passing through the water will refract or bend as they enter the glass block, and then bend again as they exit the glass and enter the air above.
To determine the apparent position of the object, you will need to know the refractive indices of water and glass. The refractive index of water is 1.33, and the refractive index of glass is typically around 1.5.
Assuming the light rays are traveling perpendicular to the surfaces of the block, the apparent depth of the block as seen from above the water line will be:
apparent depth = actual depth / refractive index
For the water, the apparent depth is simply its actual depth, since the light rays are not refracted when passing from air to water.
So, for the glass block:
apparent depth = 6 cm / 1.5 = 4 cm
And for the water:
apparent depth = 4 cm
Therefore, the total apparent depth of the block and water is 4 + 4 = 8 cm. If an object is placed below the water line but above the top surface of the block, its apparent position will appear to be shifted upward by this amount.
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Find the temperature of an ideal gas being quasi-statically compressed by 700 j of work to one-third of its initial volume
The temperature of an ideal gas being quasi-statically compressed by 700 j of work to one-third of its initial volume can be found using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the work done on the system plus the heat added to the system.
Since the gas is considered ideal, the heat added is assumed to be zero. Therefore, the change in internal energy is simply equal to the work done, which is 700 j. Since internal energy is proportional to temperature, the temperature of the gas can be found by dividing the work done by the gas's specific heat capacity.
The temperature will increase as the gas is compressed, and the final temperature can be determined by multiplying the initial temperature by the ratio of the final volume to the initial volume. In this case, the final temperature would be three times the initial temperature.
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A capacitor of capacitance 8. 1x10-6 F is discharging through a 1. 3 M Ω resistor. At what time will the energy stored in the capacitor be half of its initial value?
Answer in seconds and upto 1 decimal place
TimeTime when energy stored in the capacitor will be half of its initial value=7.3s
To find the time at which the energy stored in the capacitor will be half of its initial value, we can use the formula for the time constant (τ) of an RC circuit and the fact that energy is halved when the voltage across the capacitor is reduced to 1/√2 of its initial value.
The time constant (τ) of the RC circuit is given by τ = R * C, where R is the resistance and C is the capacitance.
τ = (1.3 * 10^6 Ω) * (8.1 * 10^-6 F) = 10.53 seconds
Now, we can use the formula for discharging a capacitor: V(t) = V_initial * e^(-t/τ)
We need to find the time (t) when V(t) = V_initial / √2. So:
V_initial / √2 = V_initial * e^(-t/10.53)
Divide both sides by V_initial:
1 / √2 = e^(-t/10.53)
Take the natural logarithm of both sides:
-ln(√2) = -t / 10.53
Now, solve for t:
t = 10.53 * ln(√2) ≈ 7.3 seconds
So, the energy stored in the capacitor will be half of its initial value at approximately 7.3 seconds.
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Which latitude receives the most direct rays of the sun year-round?.
The latitude that receives the most direct rays of the sun year-round is the equator, which has a latitude of 0 degrees.
Due to the Earth's axial tilt, the sun's rays strike the Earth at different angles at various latitudes throughout the year. Near the equator, the sun's rays are nearly perpendicular to the Earth's surface, resulting in a more direct and intense sunlight.
At the equator, the sun is positioned directly overhead at least once a year during the equinoxes (around March 21st and September 21st). This means that the equator receives the most direct and concentrated sunlight throughout the year compared to other latitudes.
As one moves away from the equator towards higher latitudes, the angle at which the sun's rays hit the Earth becomes progressively steeper, resulting in less direct and more diffuse sunlight. This is why regions closer to the poles experience more significant variations in daylight and seasonal changes in sunlight intensity.
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At point a is the interference between the two sources constructive or destructive?.
At point A, the interference between the two sources is constructive. This is because the two waves are in phase at this point, meaning the peaks and troughs of the two waves line up.
When this happens, the amplitude of the combined wave is greater than the amplitude of either individual wave. This increased amplitude results in a stronger wave, which is an example of constructive interference.
Constructive interference can also be caused when two waves have a phase difference of 0, 180, or multiples of 180. In this case, the two waves have a phase difference of 0, resulting in constructive interference.
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A speeding car traveling at 41 m/s passes a parked police car. One second after getting passed, the police car begins pursuit. The police car accelerates at a rate of 7.5 m/s/s. The police car catches up after 12.8 seconds and the police car travels 527 meters.
What is the velocity of the police car when it catches up to the speeding car?
Answer:
To solve this problem, we can use the equation:
distance = initial velocity x time + 1/2 x acceleration x time^2
First, we need to find the initial distance between the two cars. The speeding car travels for 1 second before the police car begins pursuit, so its initial distance from the parked police car is:
initial distance = 41 m/s x 1 s = 41 m
Now we can use the equation to find the time it takes for the police car to catch up to the speeding car:
distance = initial velocity x time + 1/2 x acceleration x time^2
527 m = 0 m/s x t + 1/2 x 7.5 m/s^2 x t^2
Simplifying:
t = sqrt((2 x 527 m) / 7.5 m/s^2) = 12.92 s
So the police car catches up to the speeding car after 12.92 seconds. Now we can use the equation:
final velocity = initial velocity + acceleration x time
to find the velocity of the police car when it catches up to the speeding car:
final velocity = 0 m/s + 7.5 m/s^2 x 12.92 s = 96.9 m/s
Therefore, the velocity of the police car when it catches up to the speeding car is 96.9 m/s.
Explanation:
Which class of fires consists of flammable liquids including.
Answer:
class b
Explanation:
flamabal liquids , gas alcohol or oil
what energy refers to the kinetic energy of moving particles of matter
Answer:
Thermal Energy
Explanation:
A 5.10 kg cast-iron skillet is heated on the stove from 295 k to 450 k. how much heat had to be transferred to the iron (specific heat of iron is 450j/kg k)?
The amount of heat transferred to the cast-iron skillet is approximately 351,450 J.
To calculate the amount of heat transferred to the cast-iron skillet, we can use the formula:
Q = m * c * ΔT
where:
Q is the heat transferred,
m is the mass of the skillet,
c is the specific heat capacity of iron, and
ΔT is the change in temperature.
Given:
m = 5.10 kg (mass of the skillet)
c = 450 J/(kg*K) (specific heat capacity of iron)
ΔT = 450 K - 295 K (change in temperature)
Let's calculate the heat transferred:
Q = (5.10 kg) * (450 J/(kg*K)) * (450 K - 295 K)
Q = 5.10 kg * 450 J/(kg*K) * 155 K
Q ≈ 351,450 J
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You find some limestone rock in southern Indiana and notice that it has fossil trilobites in it. Later you find the same fossil trilobites in a limestone in Colorado. From this you determine that the two rock types were deposited during the same time period using what concept or principle?
The concept or principle used to determine that the two rock types were deposited during the same time period is the principle of faunal succession.
This principle states that fossils of similar organisms found in rocks from different locations were deposited during the same time period, as the distribution of fossils in the rock layers is related to the relative ages of the rocks.
By finding the same fossil trilobites in both the Indiana and Colorado limestone rocks, it can be inferred that the rocks were deposited during the same time period and were likely part of the same geologic formation.
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A bar of length L = 0. 36m is free to slide without friction on horizontal rails. A uniform magnetic field B = 2. 4T is directed into the plane. At one end of the rails there is a battery with emf = 12V and a Switch S. The bar has the mass 0. 90kg and resistance 5. 0ohm. Ignore all the other resistance in the circuit. The switch is closed at time t = 0. A) Just after the switch is closed, what is the acceleration of the bar? b)what is the acceleration of. The bar when its speed is 2. 0m/s? c) what is the bar's terminal speed?
The acceleration of the bar just after the switch is closed is [tex]6.91 m/s^2[/tex]. When the bar's speed is 2.0 m/s, its acceleration is zero. The terminal speed of the bar is 1.49 m/s.
To solve this problem, we will use the equation of motion for an object under the influence of a force and the equation for the current in a circuit under the influence of an emf and resistance.
a) Just after the switch is closed, the current in the circuit will be given by Ohm's Law:
I = emf / R = 12 V / 5.0 Ω = 2.4 A
The bar will experience a magnetic force due to the magnetic field that is perpendicular to its motion. The magnetic force can be calculated using the formula:
F = BIL
where B is the magnetic field, I is the current, and L is the length of the bar. The bar will experience a force in the direction opposite to its motion. Therefore, the acceleration of the bar can be calculated using Newton's second law:
a = F / m = (BIL) / m
Substituting the given values, we get:
a = (2.4 T)(2.4 A)(0.36 m) / 0.90 kg = [tex]6.91 m/s^2[/tex]
Therefore, the acceleration of the bar just after the switch is closed is [tex]6.91 m/s^2[/tex].
b) To calculate the acceleration of the bar when its speed is 2.0 m/s, we need to use the equation of motion:
v = u + at
where v is the final velocity, u is the initial velocity (which is zero in this case), a is the acceleration, and t is the time.
We can rearrange this equation to solve for time:
t = (v - u) / a = v / a
Substituting the given values, we get:
t = 2.0/ 6.91 = 0.289 s
Now we can use the equation of motion again to calculate the distance traveled by the bar during this time:
[tex]$s = ut + \frac{1}{2}at^2 = \frac{1}{2}at^2$[/tex]
Substituting the given values, we get:
[tex]$s = \frac{1}{2}(6.91 , \mathrm{m/s^2})(0.289 , \mathrm{s})^2 = 0.115 , \mathrm{m}$[/tex]
Therefore, the distance traveled by the bar when its speed is 2.0 m/s is 0.115 m. To calculate the acceleration, we can use the formula:
a = F / m = (BIL) / m
Substituting the given values and using the fact that the bar is now moving at a constant speed (i.e., the net force on the bar is zero), we get:
a = 0
Therefore, the acceleration of the bar when its speed is 2.0 m/s is zero.
c) The terminal speed of the bar can be calculated using the formula:
[tex]$v_{\text{terminal}} = \frac{\text{emf}}{\text{BRL}} \cdot \left(1 - e^{-\frac{\text{BRL}}{\text{m}}}\right)$[/tex]
where emf is the emf of the battery, B is the magnetic field, R is the resistance of the bar, L is the length of the bar, and m is the mass of the bar.
Substituting the given values, we get:
[tex]$v_{\text{terminal}} = \frac{12 , \mathrm{V}}{(2.4 , \mathrm{T})(5.0 , \Omega)(0.36 , \mathrm{m})} \cdot \left(1 - e^{-\frac{(2.4 , \mathrm{T})(5.0 , \Omega)(0.36 , \mathrm{m})}{0.90 , \mathrm{kg}}}\right)$[/tex]
Simplifying this expression, we get:
v_terminal = 1.49 m/s
Therefore, the terminal speed of the bar is 1.49 m/s.
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What would the RPM be if we were turning a 1. 00" diameter work piece made out of mild
steel, using HSS cutting tool?
The recommended RPM for turning a 1.00" diameter mild steel workpiece using an HSS cutting tool would be approximately 400 RPM.
To calculate the RPM for turning a 1.00" diameter workpiece made of mild steel using an HSS (high-speed steel) cutting tool, you can use the following formula:
RPM = (Cutting Speed x 4) / Workpiece Diameter
For mild steel, the recommended cutting speed with HSS tools is approximately 100 surface feet per minute (SFM). Using this value and the given workpiece diameter, we can calculate the RPM:
RPM = (100 SFM x 4) / 1.00" Diameter
RPM = 400 / 1.00
RPM ≈ 400
So, the recommended RPM for turning a 1.00" diameter mild steel workpiece using an HSS cutting tool would be approximately 400 RPM.
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A cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 70.0 m above ground level, and the ball is fired with initial horizontal speed v0 . Assume acceleration due to gravity to be g = 9.80 m/s2 .
A)Assume that the cannon is fired at time t=0 and that the cannonball hits the ground at time tg . What is the y position of the cannonball at the time tg/2 ? Answer numerically in units of meters.
The vertical position of the cannonball at the time tg/2 is 87.5 meters above ground level.
What is the vertical position of the cannonball?The horizontal motion of the cannonball is independent of its vertical motion. Since the cannonball is fired horizontally, its initial vertical velocity is zero, and it only experiences a downward acceleration due to gravity.
We can use the following kinematic equation to determine the time it takes for the cannonball to hit the ground:
h = v₀_y * t + (1/2) * g * t²,
where;
h is the initial height of the cannonball, v₀_y is the initial vertical velocity of the cannonball (which is zero), and t is the time it takes for the cannonball to hit the ground.Solving for t, we get:
t = √(2*h/g)
Plugging in the given values, we get:
t = √(2*70/9.8) = 3.78 s
Therefore, the cannonball hits the ground at time t = 3.78 s.
Now, let's consider the vertical motion of the cannonball. At the time tg/2, the time elapsed since the cannon was fired is tg/2.
The vertical position of the cannonball at this time can be calculated using the following kinematic equation:
y = h + v₀_y * t + (1/2) * g * t²,
Since v₀_y is zero, we have:
y = h + (1/2) * g * (tg/2)²
Plugging in the given values, we get:
y = 70 + (1/2) * 9.8 * (3.78/2)² = 87.5 m (rounded to one decimal place)
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ii. how long it takes to travel 294 m below the point of projection.
It takes 10 seconds for the stone to travel 294 m below the point of projection. That's how long it take to travel.
How do we calculate how long it take to travel 294 m below the point of projection.?The equation to use to find how long it take to travel 294m below the point of projection is:
4.9t² - 19.6t - 294 = 0
We need the quadratic equation
t = [-b ± √(b² - 4ac)] / (2a)]
a = 4.9, b = -19.6, and c = -294.
t = [19.6 ± √((-19.6)² - 44.9(-294))] / (2×4.9)
t = [19.6 ±√384.16 + 5745.6)] / 9.8
t = [19.6 ± √(6129.76)] / 9.8
t = [19.6 ± 78.3] / 9.8
The possible answers are;
t1 = (19.6 + 78.3) / 9.8 = 10 seconds
t2 = (19.6 - 78.3) / 9.8 = -6 seconds
considerinf that the answer cannot be in the negative, therefore t₁ is the answer.
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A motorcyclist traveling due north at 50km/h. the wind appears to come from north west at 60km/h . what is the true velocity of the wind
The true velocity of the wind is approximately 43.10 km/h.
To find the true velocity of the wind when a motorcyclist is traveling due north at 50 km/h, and the wind appears to come from the northwest at 60 km/h, we can use vector addition.
Step 1: Break the wind's apparent velocity into its north and west components. Since the wind is coming from the northwest, the north and west components will be equal.
Using the Pythagorean theorem (a² + b² = c²) to find the components:
North component:
a = 60 * cos(45°)
= 60 * 0.707
= 42.43 km/h
West component:
b = 60 * sin(45°)
= 60 * 0.707
= 42.43 km/h
Step 2: Subtract the motorcyclist's northward velocity from the north component of the wind's apparent velocity:
True north component of the wind:
42.43 - 50 = -7.57 km/h (southward)
Step 3: Combine the true north and west components of the wind's velocity using the Pythagorean theorem:
True wind velocity = √((-7.57)² + (42.43)²)
= √(57.36 + 1800.06)
= √1857.42
≈ 43.10 km/h
The true velocity of the wind is approximately 43.10 km/h.
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A wheel of diameter 40. 0 cm starts from rest and rotates with a constant angular acceleration of 4rpm. At the instant the wheel has computed its second revolution, compute the radial acceleration of a point on the rim in two ways: (a) using the relationship a_rad = w^2r and (b) from the relationship a_rad = v^2/r
(a) Using a_rad = [tex]\omega^{2r[/tex]: Approximately 100.53 m/s²
(b) Using a_rad = [tex]v^{2/r[/tex]: Approximately 31.42 m/s²
To solve the problem, let's first convert the angular acceleration from revolutions per minute (rpm) to radians per second squared (rad/s²):
Given:
Diameter of the wheel (D) = 40.0 cm
Radius of the wheel (r) = D/2 = 20.0 cm = 0.20 m
Angular acceleration (α) = 4 rpm
(a) Using the relationship a_rad = [tex]\omega^{2r[/tex]:
The angular acceleration (α) can be converted to angular velocity (ω) using the formula:
ω = αt, where t is the time taken to complete two revolutions.
Since the wheel starts from rest, the time taken to complete two revolutions is given by:
t = (2 rev) / (4 rpm) = 0.5 min = 30 s
Now we can calculate the angular velocity (ω):
ω = αt = (4 rpm) × (2π rad/1 min) × (1 min/60 s) × (30 s) = 4π rad/s
Using the relationship a_rad = [tex]\omega^{2r[/tex], we can calculate the radial acceleration:
a_rad = [tex]\omega^{2r[/tex] = (4π rad/s)² × 0.20 m
a_rad = 16π² × 0.20 m ≈ 100.53 m/s²
Therefore, the radial acceleration of a point on the rim, calculated using a_rad = [tex]\omega^{2r[/tex], is approximately 100.53 m/s².
(b) Using the relationship a_rad = [tex]v^{2/r[/tex]:
The wheel starts from rest, so its initial linear velocity (v) is zero.
The final linear velocity (v) can be calculated using the formula:
v = ωr
The time taken to complete two revolutions is already calculated as 30 seconds, so we can find the final angular velocity (ω) as follows:
ω = αt = 4π rad/s (same as before)
Now we can calculate the final linear velocity (v):
v = ωr = (4π rad/s) × 0.20 m ≈ 2.513 m/s
Using the relationship a_rad = [tex]v^{2/r[/tex], we can calculate the radial acceleration:
a_rad = [tex]v^{2/r[/tex] = (2.513 m/s)² / 0.20 m
a_rad ≈ 31.42 m/s²
Therefore, the radial acceleration of a point on the rim, calculated using a_rad = [tex]v^{2/r[/tex], is approximately 31.42 m/s².
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If a galaxy is receeding from the earth at 560,000 mi/hr, how far away is it from earth?
The distance of the galaxy from Earth is approximately 34 Mpc or 111 million light-years away.
The distance of the galaxy from Earth can be calculated using Hubble's law, which states that the recessional velocity of a galaxy is proportional to its distance from Earth.
The proportionality constant is known as the Hubble constant, denoted by H. The current estimated value of the Hubble constant is approximately 73.3 km/s/Mpc.
To convert the given velocity from miles per hour to kilometers per second, we need to divide it by 2.237 × 10^5. Thus, the recessional velocity of the galaxy in km/s is approximately 2510 km/s.
Using Hubble's law, we can calculate the distance of the galaxy from Earth by dividing its velocity by the Hubble constant.
Therefore, the distance of the galaxy from Earth is approximately 34 Mpc or 111 million light-years away.
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If the force between two charges is initially 1800 N then what will it be if one of the charges is moved 3x farther away?
When one of the charges is moved 3 times farther away, the force between the two charges will be 200 N.
The force between two charges is described by Coulomb's Law, which states that the force (F) is proportional to the product of the charges (q1 and q2) and inversely proportional to the square of the distance (r) between them:
F = k × (q1 × q2) / r²
Here, k is Coulomb's constant.
Initially, the force between the two charges is 1800 N. Let's assume the initial distance between the charges is r. Now, one of the charges is moved 3 times farther away, making the new distance between the charges 3r.
To find the new force, we can apply Coulomb's Law again:
F_new = k × (q1 × q2) / (3r)²
Notice that k × (q1 × q2) / r² = 1800 N (initial force). To make calculations easier, we can replace the expression with 1800 N:
F_new = 1800 N / 3²
F_new = 1800 N / 9
F_new = 200 N
So, when one of the charges is moved 3 times farther away, the force between the two charges will be 200 N. This demonstrates the inverse-square relationship between the force and the distance in Coulomb's Law.
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Estimate how long dr.mann went without human contact while on the ice planet:
the lazarus mission was ___ years ago. it takes ____ years to get to saturn from earth. copper and dr. brand were on miller’s planet for approximately _____ years. this means that the total time dr. mann went without human contact is: _____ years.
The Lazarus mission was mentioned to be around 10 years ago in the movie Interstellar. It takes approximately 6 years to travel from Earth to Saturn. Copper and Dr. Brand were on Miller's planet for approximately 23 years. This means that the total time Dr. Mann went without human contact is estimated to be around 33 years.
It is stated in the movie that it takes approximately 6 years to travel from Earth to Saturn, where the wormhole that leads to other galaxies was located. This indicates that the Lazarus mission likely took 6 years to reach Saturn from Earth.
During the Lazarus mission, two of its members, Dr. Mann and Dr. Brand, were sent to explore separate planets that were potentially suitable for human colonization. Dr. Mann was sent to a planet named Mann's planet, while Dr. Brand was sent to Miller's planet.
On Miller's planet, time was significantly dilated due to its proximity to a massive black hole, known as Gargantua. For every hour that passed on the planet, approximately 7 years passed outside the planet's gravitational influence.
This means that the time Dr. Brand and the other crew members spent on Miller's planet felt like 23 years had passed for the rest of the universe.
Considering the 6-year journey to Saturn, the time spent on Miller's planet, and the 10 years that had already passed since the Lazarus mission, the total time that Dr. Mann went without human contact can be estimated to be around 33 years.
This is derived from adding the 6-year journey, the 23 years on Miller's planet, and the 10 years prior to the Lazarus mission.
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How far do you have to lift a 10kg bag of salt to do 250j of work?
You have to lift the 10kg bag of salt approximately 2.55 meters to do 250J of work.
To determine how far you have to lift a 10kg bag of salt to do 250J of work, we need to use the work-energy theorem and the formula for gravitational potential energy. The work-energy theorem states that the work done on an object is equal to the change in its potential energy. The formula for gravitational potential energy is:
PE = m * g * h
where PE is the potential energy, m is the mass (10kg), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height the object is lifted.
Since the work done is 250J, we can set the potential energy equal to the work done:
250J = 10kg * 9.8 m/s² * h
Now, we need to solve for h:
250J = 98 kg*m/s² * h
h = 250J / 98 kg*m/s²
h ≈ 2.55 meters
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in a football game wanting kicks off to the other at the moment the receiver catches the ball he is 40 m from the nearest Tucker the receiving runs left to right at a speed of 10 m/s the toddler runs right to left at a speed of 6 m/s
Answer: yes
Explanation: good luck