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hammer [34]
2 years ago
5

A wheel of radius 0.5 m rotates with a constant angular speed about an axis perpendicular to its center. A point on the wheel th

at is 0.2 m from the center has a tangential speed of 2 m/s. Determine the tangential acceleration of the point that is 0.2 m from the center.
Physics
1 answer:
Masja [62]2 years ago
8 0

Answer:

The tangential acceleration is 0 m/s².

Explanation:

Given:

Radius of the wheel = 0.5 m

The point of observation for calculating tangential acceleration = 0.2 m from center.

Tangential speed at the point of observation = 2 m/s

The angular speed of the wheel is a constant.

In order to determine the tangential acceleration, we make use of the following formula:

Tangential acceleration at a point = Angular acceleration × Distance of the point from center

Or, a_t=\alpha \times r

Now, angular acceleration is defined as the rate of change of angular speed.

Here, the angular speed of the wheel is a constant. So, the change of angular speed is 0. Therefore, the angular acceleration is also 0 rad/s².

Now, from the above formula, as angular acceleration is 0, the magnitude of tangential acceleration at a point that is 0.2 m from the center of the wheel is also 0 m/s².

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A 24 kg crate is moving at a constant speed because it is being pushed with a force of 53n. What would be the coefficient of kin
____ [38]

Answer:

The coefficient of kinetic friction between the crate and the floor can be calculated using the formula μ = Ff / N, where Ff is the frictional force, N is the normal force, and μ is the coefficient of kinetic friction.

In this case, the normal force is equal to the weight of the crate, which is 24 kg * 9.8 m/s2 = 235.2 N. The frictional force can be calculated using the formula Ff = μ * N, where μ is the coefficient of kinetic friction and N is the normal force.

If we substitute the values for N and Ff into the formula for the coefficient of kinetic friction, we get:μ = 53 N / 235.2 N = 0.225

Therefore, the coefficient of kinetic friction between the crate and the floor is 0.225.

6 0
1 year ago
A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held b
Elina [12.6K]

Answer:

0.3 m

Explanation:

Initially, the package has both gravitational potential energy and kinetic energy.  The spring has elastic energy.  After the package is brought to rest, all the energy is stored in the spring.

Initial energy = final energy

mgh + ½ mv² + ½ kx₁² = ½ kx₂²

Given:

m = 50 kg

g = 9.8 m/s²

h = 8 sin 20º m

v = 2 m/s

k = 30000 N/m

x₁ = 0.05 m

(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²

x₂ ≈ 0.314 m

So the spring is compressed 0.314 m from it's natural length.  However, we're asked to find the additional deformation from the original 50mm.

x₂ − x₁

0.314 m − 0.05 m

0.264 m

Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.

8 0
3 years ago
What morbid structure traditionally has thirteen steps?
tatiyna

Answer:

A gallow

Explanation:

6 0
3 years ago
According to the second law of thermodynamics, it is impossible for ____________. According to the second law of thermodynamics,
Tomtit [17]

Answer:

It's impossible for an ideal heat engine to have non-zero power.

Explanation:

Option A is incomplete and so it's possible.

Option B is possible

Option D is related to the first lae and has nothing to do with the second law.

Hence, the correct option is C.

The ideal engine follows a reversible cycle albeit an infinitely slow one. If the work is being done at this infinitely slow rate, the power of such an engine is zero.

We can also stat the second law of thermodynamics in this manner;

It is impossible to construct a cyclical heat engine whose sole effect is the continuous transfer of heat energy from a colder object to a hotter one.

This statement is known as second form or Clausius statement of the second law.

Thus, it is possible to construct a machine in which a heat flow from a colder to a hotter object is accompanied by another process, such as work input.

7 0
3 years ago
2.A chef leaves a copper spoon, a wooden spoon, and a steel spoon in a pot of boiling soup for several minutes. Which spoon shou
Shkiper50 [21]

Answer:

Option B

Explanation:

The correct answer is Option B

After leaving the spoon of copper , steel and wooden in the soup for several minute chef can grab Wooden spoon with bare hand.

Wooden material are insulator of heat means heat does not transfer through the wooden material.

where as heat transfer take place in steel as well as copper and chef will not be able to lift the spoon bare hand.

5 0
2 years ago
Read 2 more answers
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