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mixas84 [53]
3 years ago
11

Two masses of size m and 4m are connected by a massless thread and are strung over a frictionless pulley of radius R and moment-

of-inertia I. The two masses are held motionless at the same height and then released from rest. After the mass 4m has fallen a distance L and the mass m has risen the same distance L, what is the total kinetic energy of the entire system
Physics
1 answer:
tekilochka [14]3 years ago
6 0

Answer:

Total kinetic energy of entire system is 3 mgl

Explanation:

Given two masses: m and 4m.

Since the pulley is frictionless and the thread is massless, the energy here is linked to the two masses.

Total kinetic energy of entire system = decrease in gravitational potential energy of the system.

Therefore, we have :

ΔKE = Δp

ΔKE = 4mgl - mgl

= 3 mgl

Total kinetic energy of entire system is 3 mgl

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A 500 kg car traveling at 20 m/s rear ends another car of 600 kg at rest. The collision is great enough that the two cars stick
aksik [14]

Answer: its 50

Explanation:

im waffling does anybody have syrup

4 0
3 years ago
Two cars are heading towards one another. Car A is moving with an acceleration of aA = 4 m/s2. Car B is moving with an accelerat
Paladinen [302]

Answer:

Car B reaches car A in 19.7 s.

Explanation:

Hi there!

The equation of the position of an object moving in a straight line at constant acceleration is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration

When both cars meet, their positions are the same. At the meeting point:

position of car A = position of car B

xA = xB

x0A + v0A · t + 1/2 · aA · t² = x0B + v0B · t + 1/2 · aB · t²

Let´s place the origin of the frame of reference at the point where A is located. In that case x0A = 0 and x0B = 2900 m. Since both cars are initially at rest, v0A and v0B = 0. So, the equation gets reduced to this:

1/2 · aA · t² = x0B + 1/2 · aB · t²  

If we replace with the data we have and solve for t:

1/2 · 4 m/s² · t² = 2900 m - 1/2 · 11 m/s² · t²

2 m/s² · t² =  2900 m - 5.5 m/s² · t²

5.5 m/s² · t² + 2 m/s² · t² = 2900 m

7.5 m/s² · t² = 2900 m

t² = 2900 m / 7.5 m/s²

t = 19.7 s

Car B reaches car A in 19.7 s.

4 0
3 years ago
The blade of metal cutter is shorter than that scissors of tailors<br><br>​
Naily [24]

Explanation:it is beause they are sharper and also have less surface area and therefore more pressure

8 0
3 years ago
The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the magnitude of the
PIT_PIT [208]

The drag force is directly proportional to the square of the velocity of motion of the object. So as the speed is doubled, the magnitude of drag force will get quadrupled.

<u>Explanation: </u>

Drag force is the opposing or resisting force acting on any object by the medium in which it is moving. So in this case, you are riding a bicycle, thus the medium can be considered as air.

The formula for calculating drag force is as below:

               \text {Drag force }=\frac{1}{2} \rho v^{2} C_{D} A

Here, ρ is the density of the air molecules, v is the velocity or speed of the bicycle, CD is the drag coefficient and A is the area of the bicycle.

In the above equation, only the term velocity will be a varying quantity with respect to time and other quantities will remain constant throughout the single situation of riding of bicycle.

So, the equation can be,

             \text { Drag force }=k v^{2}

Where ,  

     k=\frac{1}{2} \rho C_{D} A (constant for this whole condition)

Now given the speed of bicycle increased from v to 2v, so the initial drag force will be

                   N_{i}=k v^{2}

After increase in speed, the final drag force will be  

                   N_{f}=k(2 v)^{2}

                   N_{f}=4 k v^{2}=4 N_{i}

Thus, if the speed of the bicycle is doubled, the drag force will get increased by four times.

4 0
3 years ago
Given two vectors A= 4.00i + 7.00jand B= 5.00i-2.00ja.Find the magnitude of each vector.b.Find the vector differenceC= A –B, giv
Tpy6a [65]

Answer:

444.99

Explanation:

8 0
3 years ago
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