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kaheart [24]
4 years ago
8

If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, A)compression B)w

ork C)radiation is being done.
Physics
2 answers:
Gnesinka [82]4 years ago
7 0
Salutations!

If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy,  _______________ is being done.

<span>If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, work is being done. Energy being transferred and the object begins to move is called work.

Thus, your answer is option B.

Hope I helped (:

Have a great day!</span>
Mademuasel [1]4 years ago
5 0

B) work is being done

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high frequency sound waves have a shorter wavelength and a higher A. amplitude B. pitch C. wavelength? than low frequency sound
gizmo_the_mogwai [7]
Pitch of the sound increases as frequency increases. 
choose B pitch 
8 0
3 years ago
Read 2 more answers
A 129-kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 65.
LUCKY_DIMON [66]

Answer:

Moment of inertia of the system is 289.088 kg.m^2

Explanation:

Given:

Mass of the platform which is a uniform disk = 129 kg

Radius of the disk rotating about vertical axis = 1.61 m

Mass of the person  standing on platform = 65.7 kg

Distance from the center of platform = 1.07 m

Mass of the dog on the platform = 27.3 kg

Distance from center of platform = 1.31 m

We have to calculate the moment of inertia.

Formula:

MOI of disk = \frac{MR^2}{2}

Moment of inertia of the person and the dog will be mr^2.

Where m and r are different for both the bodies.

So,

Moment of inertia (I_y_y )  of the system with respect to the axis yy.

⇒ I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g

⇒ I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2

⇒ I_y_y=\frac{129(1.61)^2}{2} +65.7(1.07)^2+27.2(1.31)^2

⇒ I_y_y=289.088\ kg.m^2

The moment of inertia of the system is 289.088 kg.m^2

7 0
3 years ago
A car has a speed of 23 m/s and a momentum of 31,050 kg·m/s. What is the mass of the car?
Ilya [14]

Answer:

Mass = 1350kg

Explanation:

Given the following data;

Velocity = 23m/s

Momentum = 31,050 kg·m/s.

To find the mass of car;

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

Momentum = mass * velocity

Substituting into the equation, we have

31,050 = mass*23

Mass = 31,050/23

Mass = 1350kg

Therefore, the mass of the car is 1350 kilograms.

3 0
3 years ago
If a 3-kg rock is thrown at a speed of 2 m/s in a gravity-free environment (presuming one could be found), then an unbalanced fo
Katarina [22]

Answer:

a. False

Explanation:

For an object to be moving at a constant velocity, a net force of 0 N would be required.

Newton's 1st Law of Motion states that an object will remain at rest unless acted upon by an unbalanced force, and an object will remain in motion unless acted upon by an unbalanced force.

Therefore, the unbalanced force of 6 N would not allow the rock to maintain its constant speed.

The answer to this question is A) False.

3 0
3 years ago
Read 2 more answers
A car traveling 75 km/h slows down at a constant 0.50 m/s2 just by "letting up on the gas." calculate (a) the distance the car c
Serjik [45]

Solution:

At 1st convert km/h to m/s. 1 km = 1000 m, 1 h = 3600 s, 1 km/m = 1000/3600 = 5/18 m/s  

Initial velocity = 75 * 5/18 = 20.8 m/s  

The car’s velocity decreases from 20.8 m/s to 0 m/s at the rate of 0.5 m/s each second. We have the final velocity, initial velocity, and the acceleration.  

Now according to the equation determine the distance.  

vf^2 = vi^2 + 2 * a * d  

a = -0.5 m/s^2  

0 = 20.8^2 + 2 * -0.5 * d  

so d = 431.64 m  

since we have the final velocity, initial velocity, and the acceleration. Use the following equation to determine time.  

vf = vi + a * t  

0 = 20.8 – 0.5 * t  

Solve for t = 41 seconds  

(c) the distance travels by it during the first and fifth second are.  

d = vi * t + ½ * a * t^2  

d1 = 20.8 * 1 – ½ * 0.5 * 1^2 = 20.55 m  

The easiest way to the distance for the 5th second is:

d = vi * t + ½ * a * t^2, a = -0.5  

d5 = 20.8 * 5 – ½ * 0.5 * 5^2 = 91.5 m  

d6 = 20.8 * 6 – ½ * 0.5 * 6^2 =  106.8m

d6 – d5 = 15.3 m  

this is the required solution.


3 0
3 years ago
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