the launch speed of the projectile is approximately 12.3 km/s that rises vertically above the Earth to an altitude equal to 11 REarth before coming to rest momentarily
We can use the conservation of energy principle to solve this problem. At the highest point, the projectile has zero kinetic energy and a certain potential energy, given by:
PE = mgh
where m is the mass of the projectile, g is the acceleration due to gravity, and h is the altitude above the Earth's surface. The potential energy can also be expressed in terms of the gravitational potential energy per unit mass, which is given by:
U = GM/r
where M is the mass of the Earth, r is the distance from the center of the Earth to the projectile, and G is the gravitational constant. At the highest point, the projectile has a distance of 12 REarth from the center of the Earth (11 REarth above the surface plus the radius of the Earth).
Setting these two expressions for potential energy equal to each other, we have:
mgh = GMm/r
where we can cancel the mass m from both sides to obtain:
gh = GM/r
Solving for the speed v at the launch point, we have:
v = sqrt(2gh)
where we can substitute:
h = 11 REarth
r = 12 REarth
g = GM/(12 REarth)2
Substituting these values and using the fact that REarth = 6.37×106 m, we get:
v = sqrt(2×(11+1)REarth×GM/(12REarth)) = 12.3 km/s
where we used the fact that the total distance traveled by the projectile is 2 REarth, so the factor of (11+1) appears in the expression for h.
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A rectangle with a base labeled uppercase L and a height labeled lowercase w. A vertical, dashed line runs along the leftside of the rectangle's height. A circular arrow around the vertical, dashed line indictates that the rectangle rotates about its left edge.
Calculate the moment of inertia if the plate has a length of 9.00 cm, a width
of 7.00 cm, and a uniform mass density of 2.50 g/cm^2
A = wL = 63 cm²
m = (2.5 g/cm²)(63 cm²)
m = 157.5 g = 0.1575kg
[tex]I=\frac{1}{3} mL^2\\\\I = \frac{1}{3}(0.1575kg)(0.09m)^2[/tex]
I = 4.2525×10⁻⁴ kg/m²
A 1.3 KG blocks flies along a frictionless surface at 1.0 M/S.a2 block sliding at a faster 5.0 M/S collides with the first from behind and sticks to it. The final velocity of the combine blocks is 2.0 M/S. What was the mass of the second block?
the initial momentum of the system of block m1 and block m2 is
Pi= m1v1 + m2v2
the final momentum of the combine blocks is
Pf= (m1+m2)V
according to the law of convervation of momentum
Pi = Pf
m1v1 + m2v2 = (m1+m2)V
1.3 × 1 + 5m2 = 1.3 × 2 + 2m2
m2= 1.3/3 kg
A truck accelerates uniformly from rest to 18.5 m/s in 5.7 s along a level stretch of road. Determine the average power (in W) required to accelerate the truck for the following values of the weight (ignore friction). (a)
The average power required to accelerate the truck with a weight of 20,000 N is approximately 116,930.68 W.
The average power required to accelerate an object is given by the formula:
Power = Work / Time
where Work is the change in kinetic energy of the object and Time is the time interval over which the change in kinetic energy occurs.
The change in kinetic energy of the truck is given by:
ΔK = 1/2 * m *[tex]v^2[/tex]
where m is the mass of the truck and v is its final velocity.
Since the truck starts from rest, its initial kinetic energy is zero, so the work done on the truck is equal to its change in kinetic energy. Therefore, the average power required to accelerate the truck is:
Power = Work / Time = ΔK / Time
Substituting the given values, we get:
(a) For weight w = 10,000 N (approximately 1,020 kg):
m = w / g = 10000 N / [tex]9.81 m/s^2[/tex] = 1019.3 kg
ΔK =[tex]1/2 * m * v^2[/tex] = 1/2 * 1019.3 kg * [tex](18.5 m/s)^2[/tex]= 333,036.6 J
Power = ΔK / Time = 333036.6 J / 5.7 s ≈ 58,426.84 W
Therefore, the average power required to accelerate the truck with a weight of 10,000 N is approximately 58,426.84 W.
Note: g is the acceleration due to gravity, which is approximately 9.81 [tex]m/s^2.[/tex]
(b) For weight w = 20,000 N (approximately 2,040 kg):
m = w / g = 20000 N / 9.81 m/s^2 = 2041.2 kg
ΔK = 1/2 * m *[tex]v^2[/tex] = 1/2 * 2041.2 kg * [tex](18.5 m/s)^2[/tex]= 666,073.3 J
Power = ΔK / Time = 666073.3 J / 5.7 s ≈ 116,930.68 W.
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Find the position of the center of mass of the system of the sun and Jupiter. (Since Jupiter is more massive than the rest of the planets combined, this is essentially the position of the center of mass of the solar system.)
it's only beyond the surface of the sun! There is a barycenter throughout our entire solar system. All of the planets in the solar system, including the sun, center of mass.
Is the Sun more massive than the solar system as a whole?The sun is significantly more massive than Earth and has a radius that is likewise much greater. The mass of the sun is more than 333,000 times more than the mass of the Earth and makes up nearly all of the solar system's mass (99.8%).
How much more does the Sun compare to Jupiter?The Sun is 1000 times more than Jupiter, the solar system's most planet , but Jupiter is still 1000 times less than it.
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What is the charge of an electric field 0.40 m away from a
source charge of 3.00 x 10-5C?
The electric field 0.40 m away from a source charge of 3.00 x 10^-5 C is 1.69 x 10^6 N/C, directed away from the source charge.
How to find the chargeThe electric field created by a point charge at a distance r from the charge is given by the equation:
E = kQ/r^2
where
E is the electric field in N/C,
k is Coulomb's constant (9.0 x 10^9 N m^2/C^2),
Q is the source charge in Coulombs, and
r is the distance from the source charge in meters.
Substituting the given values into the equation, we get:
E = (9.0 x 10^9 ) x (3.0 x 10^-5) / (0.40)^2
E = 1.69 x 10^6 N/C
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A physics class conducting a research project on projectile motion construct a device that can launch a cricket ball.the launching device is designed so that the ball can be launch at ground level with an initial velocity of 28m/s at an angle of 30 degrees to the horizontal.
Calculate the horizontal of the velocity of the all:
a) initially
B) after 1.0 seconds
C) after 2.0 seconds
A projectile motion is any object thrown into space upon which the only acting force is gravity. The primary force acting on a projectile is gravity.The velocity's horizontal component is 24.25 m/s at time t = 2 seconds.The velocity's horizontal component is 24.25 m/s at time t = 3 seconds.
This doesn’t necessarily mean that other forces do not act on it, just that their effect is minimal compared to gravity.The particle is moving vertically (downwards) along the y-axis due to uniform acceleration.A particle's vertical and horizontal projectile motions can both accelerate: The only force acting on a particle when it is launched into the air at some speed is the acceleration brought on by gravity (g). The downward motion of this acceleration is vertical.
347u = 28 m/s for the starting velocity
projecting at a 30° angle
The horizontal component of velocity is constant since there is no acceleration in the horizontal direction.
Vertical component of speed, u cos = 28 x Cos 30 = 24.25 m/s
The velocity's horizontal component is 24.25 m/s at time t = 2 seconds.
The velocity's horizontal component is 24.25 m/s at time t = 3 seconds
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MS-PS2-1, MS-PS2-2: Newton...
16
Distance (m)
Distance vs. Time
B
120
Time (sec.)
A
B
Which car is faster, A or B? Choose the statement that BEST fits the graph.
Car A is faster than B because A's speed is 120 m/s.
Car B is faster than A because B's speed is 90 m/s.
Car A is faster than B because A's speed is 20 m/s and B is 15 m/s.
D) Car B is faster than A because B's motion is accelerating and A is constant speed.
The statement that best fits the graph is: D) Car B is faster than A because B's motion is accelerating and A is constant speed.
Which car is faster, A or B?From the graph, we can see that the distance traveled by both cars increases over time. However, the slope of the distance-time graph for car B is steeper than that of car A. This means that car B covers more distance in the same amount of time as car A, indicating that car B is traveling faster than car A.
Furthermore, we can see that the speed-time graph for car B is a straight line with a positive slope, indicating that its speed is increasing over time. On the other hand, the speed-time graph for car A is a horizontal line, indicating that its speed is constant.
Therefore, we can conclude that car B is faster than car A because car B's motion is accelerating, while car A is moving at a constant speed.
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a person that just start driving a car is called
Answer:
A person who drives a car for themselves, generally taking themselves back and forth to work or errands, is generally referred to as a driver.
The Earth surface temperature is around 270 K and emissivity of 0.8, while space has temperature of around 2K. What is the net power radiated by the Earth in free space?
The net power radiated by the Earth in free space is approximately
1.2 x 10^ 17 W watts.
How to find the power radiatedTo calculate the net power radiated by the Earth in free space, we can use the Stefan-Boltzmann Law:
Power = σ * A * ε * (T^4 - T0^4)
where:
σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4)
A is the surface area of the Earth (4πR^2, where R is the radius of the Earth)
ε is the emissivity of the Earth (0.8)
T is the temperature of the Earth surface (270 K)
T0 is the temperature of space (2 K)
Plugging in the values, we get:
Power = 5.67 x 10^-8 * 4πR^2 * 0.8 * (270^4 - 2^4)
The radius of the Earth is approximately 6.37 x 10^6 m, so:
Power = 5.67 x 10^-8 * 4π(6.37 x 10^6)^2 * 0.8 * (270^4 - 2^4)
Power ≈ 1.193 x 10^ 17 W
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what is the mechanica advantage of alever then can lift 100 Newton with an input force of 20 Newton
Answer:
Explanation:
To calculate the mechanical advantage of a lever, we use the formula:
Mechanical Advantage = Output Force / Input Force
In this case, the output force is 100 Newton and the input force is 20 Newton, so:
Mechanical Advantage = 100 N / 20 N
Mechanical Advantage = 5
Therefore, the mechanical advantage of the lever is 5. This means that for every 1 Newton of input force applied to the lever, the lever will produce 5 Newtons of output force. So, in this case, an input force of 20 Newtons applied to the lever would produce an output force of 100 Newtons.
A car slows down from -27.7m/s to -10.9m/s while undergoing a displacement of -105. What was the acceleration? (Unit=m/s^2)
The acceleration of the car was 6.88 [tex]m/s^2[/tex]. The negative sign indicates that the car was decelerating, or slowing down.
To find the acceleration of the car, we can use the following formula:
acceleration = (final velocity - initial velocity) / time
However, we are not given the time it took for the car to undergo the displacement. To find the time, we can use the following formula:
displacement = (final velocity + initial velocity) / 2 * time
Solving for time, we get:
time = displacement / ((final velocity + initial velocity) / 2)
Plugging in the given values, we get:
time = [tex]-105 / ((-10.9 - 27.7) / 2) = 2.29 s[/tex]
Now that we have the time, we can use the first formula to find the acceleration:
acceleration = [tex](-10.9 - (-27.7)) / 2.29 = 6.88 m/s^2[/tex]
Therefore, the acceleration of the car was [tex]6.88 m/s^2[/tex]. The negative sign indicates that the car was decelerating, or slowing down.
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QUICK ANSWER Fiber optic cables utilize internal reflection to transmit signals. TRUE OR FALSE
PLS MRK ME BRAINLIEST
Answer:
True
Explanation:
optical fibre consists of core and cladding. The signal is converted to light using transducers. The light travels across the cable undergoing multiple internal reflections. At the other end the light is converted back to Signal using transducers.
Answer: True
Explanation: Optical fiber uses the optical principle of "total internal reflection" to capture the light transmitted in an optical fiber and confine the light to the core of the fiber.
If I get this wrong im sorry
QUICK ANSWER Rainbows are caused by the refraction of visible wavelengths of light. TRUE OR FALSE
Answer:
true it is refraction of visible
Which statements are true about the amount of gravitational potential energy an object has?
The amount increases as the object is lifted higher.
The amount varies according to the material of the object.
The amount varies according to the mass of the object.
The amount increases the more quickly the object is lifted.
Answer:
The statement "The amount increases as the object is lifted higher" is true about the amount of gravitational potential energy an object has. The statement "The amount varies according to the material of the object" is false. The amount of gravitational potential energy only depends on the mass of the object and its elevation from the reference point. The statement "The amount varies according to the mass of the object" is true. The more massive an object is, the more gravitational potential energy it has. The statement "The amount increases the more quickly the object is lifted" is false. The amount of gravitational potential energy only depends on the object's height above the reference point, not the speed at which it is lifted.
Answer: (A)
Explanation: The amount increases as the object is lifted higher.
Remember the steps that the American Heart Association recommends for saying no to cigarettes. Which one of these will you use if you are faced with peer pressure to smoke? Which one of these will you use if you are faced with peer pressure to drink or take drugs? Explain the method you will use and why you chose it.
Answer: just say no and never talk to that person again.
Explanation:
because that is what i would do to prevent drugs and/or nicotine to enter my body.
Consider a converging lens with focal length 9.56 cm. The distance between an object and
a real image of the object created by the lens is 59.6 cm. Find the distance between the
object and the lens if the lens is closer to the object than it is to the image. Answer in cm
The object's distance from the lens is 11.4 cm.
Calculation-The thin lens equation can be used to determine how a converging lens's object distance (p), image distance (q), and focal length (f) relate to one another:
1/p + 1/q = 1/f
where p denotes the distance to the object, q is the distance to the picture, and f is the focal length.
Let's solve for the object distance using the thin lens equation:
1/p + 1/59.6 = 1/9.56
Combining both sides with p59.69.56 results in:
59.69.56 + p9.56 = p*59.6
Adding and subtracting:
570.176 + 9.56p = 59.6p
50.04p = 570.176
p = 11.4 cm
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The current in a circuit is 0.50 A. The circuit has two resistors connected in series: one is 110 Ω and the other is 130 Ω. What is the voltage in the circuit?
Answer:
120 V
Explanation:
Given:
I (total) = 0,5 A
R1 = 110 Ω
R2 = 130 Ω
Find: V (total) - ?
The resistors are connected in series
That means, the current in the circuit is the same for every resistor:
I (total) = I1 = I2
Now, we need to find the voltage in each resistor:
V1 = I × R1
V1 = 0,5 × 110 = 55 V
V2 = I × R2
V2 = 0,5 × 130 = 65 V
Now, since the connection is in series, in order to find the total voltage in the circuit, we have to add the voltages of the resistors:
V (total) = V1 + V2
V (total) = 55 + 65 = 120 V
The following conversion equivalents are given:
1 m 100 cm 1 in = 2.54 cm 1 ft = 12 in
A bin has a volume of 1.5 m3. The volume of the bin, in ft3, is closest to:
Answer:52.972ft^3
Explanation:
It is unit conversion based and for a volume of a bin from cubic meter to cubic foot as we know 1 meter =3.281 foot.
where volume=1.5m^3
multiply 1.5*(3.281)^3 ft^3
v=52.972 ft^3
Jonathan and Jane are sitting in a sleigh that is at rest on frictionless ice. Jonathan's weight is 800 N
, Jane's weight is 600 N
, and that of the sleigh is 1000 N
. They see a poisonous spider on the floor of the sleigh and immediately jump off. Jonathan jumps to the left with a velocity (relative to the ice) of 5.00 m/s
at 30.0∘
above the horizontal, and Jane jumps to the right at 7.00 m/s
at 36.9∘
above the horizontal (relative to the ice).
1) Calculate the magnitude of the sleigh's horizontal velocity after they jump out?
2) What is the direction of the sleigh's horizontal velocity after they jump out?
Since the positive x-direction was established to be to the right, the sleigh's velocity is in this direction (the positive x-direction).
Calculation-The whole horizontal momentum is conserved since there is no external force operating on the system (the sleigh plus Jonathan and Jane) in the horizontal direction.
Let's say that the positive x-direction is to the right and the positive y-direction is up. Then, we may divide Jonathan and Jane's velocities into their x- and y-components as follows:
vx1 = 5.00 cos 30.0° = 4.33 m/s and vy1 = 5.00 sin 30.0° = 2.50 m/s are the speeds of Jonathan.
Jane's speed is given by the formulas vx2 = 7.00 cos 36.9° = 5.61 m/s and vy2 = 7.00 sin 36.9° = 4.16 m/s.
The sleigh is initially at rest, hence there is no initial total momentum in the x-direction. The total x-direction momentum after Jonathan and Jane jumps out is:
px = (−800 N)(−4.33 m/s) + (600 N)(5.61 m/s) = 6430 N·s
The final momentum in the x-direction is:
p'x = (1000 N + 800 N + 600 N)vx'
where vx' is the final velocity of the sleigh in the x-direction.
Therefore, we can solve for xv:
vx' = px / (1000 N + 800 N + 600 N) = 6430 N·s / 2400 N = 2.68 m/s
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Please help I need to make sure my answer is correct
The measure that helps your body to stretch is flexibility, so the correct answer is D.
Flexibility for Body Stretching.Flexibility is the ability of your body to move your joints and muscles through their full range of motion. It is an essential component of physical fitness and can help improve posture, balance, and coordination. Regular stretching exercises can increase flexibility and reduce muscle stiffness, which can lead to improved physical performance, decreased risk of injury, and better overall health.
Muscular strength and muscular endurance are related to the ability of your muscles to generate force or sustain effort over time, respectively. While they are important for overall fitness, they are not directly related to flexibility.
In summary, flexibility is the key measure that helps your body to stretch, and regular stretching exercises can improve your flexibility and overall physical fitness.
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A 10,000 kg freight car is rolling along a track at 3.00m.s.Calculate the time needed for a force of 1000N for stop the car
It takes 0.3 seconds for a force of 1000 N to halt a 10,000 kg goods car moving at 3.00 m/s along a track.
Calculate the time needed for a force of 1000N for stop the carWe must first establish the car's starting kinetic energy in order to calculate the time required to stop the vehicle:
Kinetic Energy (KE) is equal to half of mass times speed, or 10,000 kg times 3.00 m/s.
KE = 45,000 J
Then, we may use the designed with the intent, which asserts that an object's change in kinetic energy equals the jobs performed by an external force: Work equals Force x Distance x Change in KE.
The gain in kinetic energy is equal to the starting kinetic energy because the car is coming to a stop: KE Change = -45,000 J
As a result, the external force's work is: Work equals force times distance, or -45,000 J.
When we solve for distance, we obtain: Work / Force = -45,000 J / 1000 N Distance
Location = -45 m
Because the force is against the direction of the car's motion, you'll see that the range is negative.
Finally, we can calculate the travel time using the kine model for uniformly accelerated motion: Distance is equal to 1/2*acceleration*time2. Time is calculated as sqrt(2 * Distance / Acceleration) as well as sqrt(2 * 45 m / (1000 N / 10,000 kg)).
time equals sqrt(0.09 s2/kg).
time = 0.3 s
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A metal ball bearing with mass 5.0 g falls out of a factory machine and drops to the concrete floor 3.0 m below. It bounces back up to its starting point. Find the changes in the bearing's potential and kinetic energies as it a) travels from the machine down to the floor, and b) travels up from the floor back to its starting point.
a) When the metal ball bearing falls from the machine to the concrete floor, its potential energy decreases while its kinetic energy increases.
The potential energy lost by the ball bearing can be calculated using the formula PE = mgh, where m is the mass of the ball bearing (0.005 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height the ball bearing falls (3.0 m). Thus, the potential energy lost by the ball bearing is:
PE = (0.005 kg)(9.8 m/s²)(3.0 m) = 0.147 J
At the same time, the ball bearing's kinetic energy increases by an amount equal to the potential energy lost. Therefore, the ball bearing's initial kinetic energy is zero, and its final kinetic energy is:
KE = 0.147 J
b) As the ball bearing bounces back up from the concrete floor to its starting point, its kinetic energy decreases while its potential energy increases. The kinetic energy lost by the ball bearing can be calculated as the same value as before:
KE = 0.147 J
The ball bearing's potential energy at its starting point is equal to the potential energy it lost on the way down, which is:
PE = 0.147 J
Therefore, the ball bearing's change in potential energy is 0.147 J (from down to up), and its change in kinetic energy is -0.147 J (from up to down).
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Answer: a) When the ball bearing falls from the machine to the floor, there is a change in its potential and kinetic energies. The potential energy of an object at a height h above the ground is given by mgh, where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s^2), and h is the height above the ground. Initially, the ball bearing is at rest on the machine, so its kinetic energy is zero. Therefore, the initial energy of the ball bearing is purely potential energy, given by:
PEi = mgh = (0.005 kg)(9.8 m/s^2)(3.0 m) = 0.147 J
When the ball bearing hits the ground, its potential energy is zero and its kinetic energy is at a maximum. The velocity of the ball bearing just before it hits the ground can be found using the equation:
v^2 = u^2 + 2as
where u is the initial velocity (which is zero), a is the acceleration due to gravity (-9.8 m/s^2), s is the distance fallen (3.0 m), and v is the final velocity just before hitting the ground. Solving for v, we get:
v = sqrt(2as) = sqrt(2*(-9.8 m/s^2)*(3.0 m)) = 7.67 m/s
The kinetic energy of the ball bearing just before it hits the ground is given by:
KEf = (1/2)mv^2 = (1/2)(0.005 kg)(7.67 m/s)^2 = 0.145 J
Therefore, the change in potential energy is:
ΔPE = PEf - PEi = 0 - 0.147 J = -0.147 J
And the change in kinetic energy is:
ΔKE = KEf - KEi = 0.145 J - 0 J = 0.145 J
b) When the ball bearing bounces back up to its starting point, there is another change in its potential and kinetic energies. Just before it reaches its highest point, the ball bearing's velocity is zero, so its kinetic energy is also zero. Therefore, its energy is purely potential energy, given by:
PEf = mgh = (0.005 kg)(9.8 m/s^2)(3.0 m) = 0.147 J
The ball bearing reaches its highest point when all of its initial kinetic energy has been converted to potential energy. At this point, its potential energy is at a maximum and its kinetic energy is at a minimum. The ball bearing then starts to fall back down towards the ground, and its potential energy starts to decrease while its kinetic energy increases. Just before it hits the ground, its kinetic energy is at a maximum and its potential energy is at a minimum, as we saw in part (a). When it bounces back up, the process repeats.
Therefore, the change in potential energy as the ball bearing travels from the floor back up to its starting point is:
ΔPE = PEf - PEi = 0.147 J - 0 J = 0.147 J
And the change in kinetic energy is:
ΔKE = KEf - KEi = 0 J - 0 J = 0 J
Note that the change in kinetic energy is zero because the ball bearing starts and ends at rest.
Explanation: can i get brainliest
31. IP Point charges +4.1 µC and -2.2 μC are placed on the x axis at (11 m, 0) and (-11 m, 0), respectively. (a) Sketch the electric potential on the x axis for this system. (b) Your sketch should show one point on the x axis between the two charges where the potential vanishes. Is this point closer to the +4.1-C charge or closer to the -2.2-µC charge? Explain. (c) Find the point referred to in part (b).
Explanation:
let at a distance x from the c
Suppose an object is weighed with a spring balance, first in air and then whilst totally immersed in water. The readings on the balance are 0.48N and 0.36N respectively. Calculate the density of the object. (2)
The density of the object is is 4000 kg/m³
What is density?Density is the ratio of mass to volume of a body.
To calculate the density of the obeject, we use the formula below
Formula:
D = D'[W/(W-W')]........................ Equation 1Where:
D = Density of the objectD' = Density of waterW = Weight of the object in airW' = Weight of the object in waterFrom the question,
Given:
W = 0.48 NW' = 0.36 ND' = Constant = 1000 kg/m³Susbtitute these values into equation 1
D = 1000[0.48/(0.48-0.36)]D = 1000(0.48/0.12)D = 4000 kg/m³Hence, the density is 4000 kg/m³
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WHAT IS QUANTUM PHYSICS
hi!
Answer:
Quantum physics is the study of matter and energy at the most fundamental level. It aims to uncover the properties and behaviors of the very building blocks of nature.
The lightweight wheel on a road bike has a moment of inertia of 0.097 kg⋅m2
. A mechanic, checking the alignment of the wheel, gives it a quick spin; it completes 5 rotations in 2.2 s. To bring the wheel to rest, the mechanic gently applies the disk brakes, which squeeze pads against a metal disk connected to the wheel. The pads touch the disk 7.1 cm from the axle, and the wheel slows down and stops in 1.2 s.
What is the magnitude of the friction force on the disk?
the magnitude of the friction force on the disk is approximately 1.16 N.
We can use the conservation of energy to find the friction force on the disk. The initial kinetic energy of the wheel is equal to the work done by the friction force on the disk:
K_i = W_friction
The initial kinetic energy of the wheel can be found from its moment of inertia and angular velocity:
K_i = (1/2) I [tex]ω^2[/tex]
where I is the moment of inertia, ω is the angular velocity, and the factor of 1/2 comes from the rotational kinetic energy formula.
The final kinetic energy of the wheel is zero, since it comes to a stop. The work done by the friction force can be found from the distance over which it acts:
W_friction = F_friction d
where F_friction is the friction force and d is the distance over which the pads act on the disk. We can find the distance from the angular displacement of the wheel:
θ = ω t
where θ is the angle through which the wheel rotates, t is the time for the wheel to come to a stop, and the factor of 1/2π converts from rotations to radians. The distance over which the pads act is then:
d = r θ = 0.071 m × (5/2π) ≈ 0.562 m
Now we can put everything together:
K_i = W_friction
(1/2) I [tex]ω^2[/tex] = F_friction d
We can solve for the friction force:
F_friction = (1/2) I [tex]ω^2[/tex] / d
Plugging in the given values:
F_friction = (1/2) ×[tex]0.097 kg⋅m^2[/tex] × [tex](5/2.2π rad/s)^2[/tex] / 0.562 m ≈ 1.16 N
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If astronomers were to find they have made a mistake and our solar system is actually 7.2 (rather than 8.2) kpc from the center of the galaxy, but the orbital velocity of the sun is still 240 km/s, what is the minimum mass of the galaxy?
Assuming a circular orbit for the Sun, we can use the equation:
v^2 = GM/r
where v is the orbital velocity of the Sun, r is the distance from the center of the galaxy, G is the gravitational constant, and M is the mass of the galaxy.
We can solve for M:
M = v^2 * r / G
Using the given values of v = 240 km/s and r = 7.2 kpc = 7.2 * 3.086e+19 m, and G = 6.6743e-11 N m^2/kg^2, we get:
M = (240000 m/s)^2 * 7.2 * 3.086e+19 m / 6.6743e-11 N m^2/kg^2
M = 1.47e+42 kg
Therefore, the minimum mass of the galaxy, if the distance of the solar system from the center is actually 7.2 kpc, is approximately 1.47 x 10^42 kg.
A metal weigh 20N in air, 12N in water, and 14N in Kerosene. Cal the the Relative density of (a) Metal (b) kerosene
The relative density of kerosene is 0.006, or 6 kg/m³.
A metal block that weighs 60 tonnes in the air and 40 tonnes beneath water, what is its density?1 Response. Employ the Archimedes' Principle to your advantage: the buoyant force of a fluid is equal to the apparent loss of weight in water and is also equal to the weight of the displaced fluid. Weight of the displaced water equals 60 - 40 N, or the apparent loss in weight of the metal block. Water displacement mass is calculated as 20 / 9.8 = 2.04 kg.
Buoyant force = Weight of the displaced water = 20 N - 12 N = 8 N
Volume of water displaced = Buoyant force / Density of water = 8 N / 9.8 m/s² = 0.8163 kg
Density of metal = Weight of metal in air / Volume of metal = 20 N / (density of air x volume of metal)
Density of metal = 20 N / 0.8163 kg = 24.5 kg/m³
Buoyant force = Weight of the displaced kerosene = 20 N - 14 N = 6 N
Volume of kerosene displaced = Buoyant force / Density of kerosene = 6 N / 9.8 m/s² = 0.6122 kg
Density of kerosene = Mass of kerosene / Volume of kerosene displaced = 14 N / 0.6122 kg = 22.86 kg/m³
Relative density of kerosene = Density of kerosene / Density of water = 22.86 kg/m³ / 1000 kg/m³ = 0.006, or 6 kg/m³.
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Two +1 C charges are separated by 300m. What is the magnitude of the electric force between them
Answer:
1000 N
Explanation:
The electric force between two charges is given as
F = kq'q/r²......................... Equation 1
Where F = electric force between the charges, q' = magnitude of the first charge, q = magnitude of the second charge, r = distance between the charges, k = coulombs constant
Given: q' = q = 1 C, r = 3000 m, k = 9×10⁹ Nm²/kg²
Substitute into equation 2
F = 1²×(9×10⁹)/3000²
F = 1000 N.
Hence the magnitude of the electric force between them = 1000 N
Two friends, Burt and Ernie, are standing at opposite ends of a uniform log that is floating in a lake. The log is 4.0 m
long and has mass 250 kg
. Burt has mass 30.0 kg
and Ernie has mass 39.0 kg
. Initially the log and the two friends are at rest relative to the shore. Burt then offers Ernie a cookie, and Ernie walks to Burt's end of the log to get it.
a) Relative to the shore, what distance has the log moved by the time Ernie reaches Burt? Neglect any horizontal force that the water exerts on the log and assume that neither Burt nor Ernie falls off the log.
The log moves a distance of 2.0 m relative to the shore when Ernie reaches Burt.
Since the log is uniform, we can treat it as a system of three point masses: one at each end representing Burt and Ernie, and one at the center representing the center of mass of the log. We can use conservation of momentum to solve this problem.
Initially, the momentum of the system is zero since everything is at rest. When Ernie walks to Burt's end of the log, he exerts a force on the log, which in turn exerts an equal and opposite force on Ernie. This force is internal to the system and does not change the momentum of the system. Therefore, the momentum of the system is still zero after Ernie reaches Burt.
Total momentum = 0 (since the center of mass is at rest)
Burt's momentum = 0 (since he is at rest)
Ernie's momentum = 0 (since he is at rest)
After Ernie walks:
Total momentum = 0 (since the center of mass remains at rest)
Burt's momentum = 0 (since he remains at rest)
Ernie's momentum = (39.0 kg) * v, where v is the velocity of Ernie and the log in the opposite direction
Since the total momentum is conserved, we can equate the momentum before and after Ernie walks:
0 = (39.0 kg) * v
v = 0 m/s
the distance that the log moves when Ernie reaches Burt is simply the distance between the initial and final positions of Ernie, which is half the length of the log:
distance = (1/2) * (4.0 m) = 2.0 m
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