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1 month ago

A uniform thin circular rubber band of mass M and spring constant k has an original radius R?

1 answer:

6 0

Yes. A rubber band can be characterized by its mass, M, spring constant K, and its rest radius, which is the radius of the circular rubber band when no force is applied. Thus, indeed, a uniform, thin circular rubber band of mass M and spring constant K has an original radius R.

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A 1000-kg car comes to a stop without skidding. The car's brakes do 50,000 J of work to stop the car. Which of the following was

inna [3103]

Answer:

10m/s

Explanation:

7 0

1 month ago

A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by

serg [3582]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

The car's position over time t can be described by

$x(t)=\alpha t^2 - \beta t^3$

where

$\alpha = 1.50 m/s^2$

$\beta = 0.05 m/s^3$

To find the average velocity, we divide the displacement by the elapsed time:

$v=\frac{\Delta x}{\Delta t}$

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 2.00 s, the position is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

This leads us to the displacement of

$\Delta x = x(2)-x(0)=5.6-0=5.6 m$

The duration for this interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

Therefore, the average velocity during this period is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 4.00 s, the position is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

Thus, the displacement is

$\Delta x = x(4)-x(0)=20.8-0=20.8 m$

The time interval is

$\Delta t = 4.0 - 0 = 4.0 s$

This yields an average velocity of

$v=\frac{20.8}{4.0}=5.2 m/s$

The position at t = 2 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

And at t = 4 s it is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

This gives us a displacement of

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

$\Delta t = 4.0 - 2.0 = 2.0 s$

So the resulting average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

Find out more about average velocity:

6 0

3 months ago

Kai takes a hamburger off of the grill and puts a piece of cheese on it. Explain how and why the cheese melts when Kai puts it o

Keith_Richards [3271]

When Kai removes the burger from the grill and adds cheese, the cheese melts due to the heat retained by the burger from cooking. The warmth remains in the burger for a while, which explains the melting of the cheese.

4 0

1 month ago

Read 2 more answers

A ball rolls up a slope. At the end of three seconds its velocity is 20 cm/s; at the end of eight seconds its velocity is 0. Wha

serg [3582]

Answer:

$a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

Explanation:

Data provided

initial velocity v₀=20 cm/s at time t=3s

final velocity vf=0 at time t=8 s

Required

Average Acceleration for the interval from 3s to 8s

Solution

Acceleration can be defined as the first derivative of velocity concerning time

$a_{acceleriation} =\frac{dv_{velocity}}{dt_{time}}\\a_{acceleriation} =\frac{v_{f}-v_{o} }{dt}\\ a_{acceleriation} =\frac{0-20cm/s }{8s-3s}\\ a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

8 0

2 months ago