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erastovalidia [21]
3 years ago
10

2. A solid sphere and a solid cylinder, both uniform and of the same mass and radius, roll without slipping at the same forward

speed. It is correct to say that the total kinetic energy of the solid sphere is A) more than the total kinetic energy of the cylinder. B) less than the total kinetic energy of the cylinder. C) equal to the total kinetic energy of the cylinder.
Physics
1 answer:
Kruka [31]3 years ago
7 0
The correct answer is B) less then the total kinetic energy of the cylinder,
Please mark brainliest
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What is the term for the matter through which a mechanical wave travels?
pav-90 [236]
The answer is “ The medium”
7 0
3 years ago
Every surface has different ____________________ of friction.
dedylja [7]

Answer:

amounts

Explanation:

every surface has different amounts of friction

hope this helps :) plz brainliest?

3 0
3 years ago
You place a 10 kg block on a ramp with an angle of 20 degrees. You push the block up the ramp giving it an initial velocity of 1
Ainat [17]

Answer:

L = 15.97 m

Explanation:

Given:-

- The mass of the block, m = 10 kg

- The inclination of ramp, θ = 20°

- The initial speed, Vi = 15 m/s

- The coefficient of friction u = 0.4

Find:-

find the total distance the block travels before it turns around and slides back down the ramp.

Solution:-

- The total distance travelled by the block up the ramp is defined when all the kinetic energy is converted into potential energy and work is done against the friction. Final velocity V2 = 0.

- Develop a free body diagram of the block. Resolve the weight "W" of the block normal to the surface of ramp. Then apply equilibrium condition for the block in the direction normal to the surface:

                                N - W*cos( θ ) = 0

Where, N : The contact force between block and ramp.

                                N = m*g*cos ( θ )

- The friction force (Ff) is defined as:

                               Ff = u*N

                               Ff = u*m*g*cos ( θ )

- Apply the work-energy principle for the block which travels a distance of "L" up the ramp:

                               K.E i = P.E f + Work done against friction

Where,  K.E i = 0.5*m*Vi^2

             P.E f = m*g*L*sin( θ )

             Work done = Ff*L

- Evaluate "L":

                        0.5*m*Vi^2 = m*g*L*sin( θ ) + u*m*g*cos ( θ )*L

                        0.5*Vi^2 = g*L*sin( θ ) + u*g*cos ( θ )*L

                        0.5*Vi^2 = L [ g*sin( θ ) + u*g*cos ( θ ) ]

                        L = 0.5*Vi^2 / [ g*sin( θ ) + u*g*cos ( θ ) ]

                        L = 0.5*15^2 / [ 9.81*sin( 20 ) + 0.4*9.81*cos ( 20 ) ]

                        L = 15.97 m

7 0
3 years ago
A 2000 kg car moves along a horizontal road at speed vo
cluponka [151]

Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a (Eq. 1)

\Sigma F_{y} = N-m\cdot g = 0 (Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

a = -\mu_{k}\cdot g (Eq. 3)

Where:

a - Deceleration of the car, measured in meters per square second.

\mu_{k} - Kinetic coefficient of friction, dimensionless.

g - Gravitational acceleration, measured in meters per square second.

If we know that \mu_{k} = 0.0735 and g = 9.807\,\frac{m}{s^{2}}, then deceleration of the car is:

a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )

a = -0.721\,\frac{m}{s^{2}}

The stopping distance of the car (\Delta s), measured in meters, is determined from the following kinematic expression:

\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a} (Eq. 4)

Where:

v_{o} - Initial speed of the car, measured in meters per second.

v - Final speed of the car, measured in meters per second.

If we know that v_{o} = 15.9\,\frac{m}{s}, v = 0\,\frac{m}{s} and a = -0.721\,\frac{m}{s^{2}}, stopping distance of the car is:

\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}

\Delta s = 175.319\,m

The shortest possible stopping distance of the car is 175.319 meters.

8 0
4 years ago
Mary cycled at an average speed of 8 km/h. How far has she traveled if she rides for 4 hours?
Sati [7]

Answer:

Which sentence from the passage shows that the function of the river depicted here has carried through to modern times?

Explanation:

Which sentence from the passage shows that the function of the river depicted here has carried through to modern times?

4 0
3 years ago
Read 2 more answers
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