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3 years ago

How to find instantaneous velocity physics?

1 answer:

8 0

You should have the velocity as a function of time either given explicitly or implicitly (a graph)

v = ds/dt (differentiating the position vector)

integrating the acceleration.

you can use impulse or work and energy principle and also newton law of motion to find acceleration then velocity

NOT SURE IF THAT WHAT YOU WANT.

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A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used

zloy xaker [14]

Answer:

The compression is $\sqrt{2} \ d$ .

Explanation:

A Hooke's law spring compressed has a potential energy

$E_{potential} = \frac{1}{2} k (\Delta x)^2$

where k is the spring constant and $\Delta x$ the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

$E_{kinetic} = \frac{1}{2} m v^2$ .

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity $v_1$ . Knowing that the energy is constant.

$\frac{1}{2} m v_1^2 = \frac{1}{2} k d^2$

If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:

$2 * (\frac{1}{2} m v_1^2) = \frac{1}{2} k D^2$

But, in the left side we can use the previous equation to obtain:

$2 * (\frac{1}{2} k d^2) = \frac{1}{2} k D^2$

$D^2 = \frac{2 \ (\frac{1}{2} k d^2)}{\frac{1}{2} k}$

$D^2 = 2 \ d^2$

$D = \sqrt{2 \ d^2}$

$D = \sqrt{2} \ d$

And this is the compression we are looking for

3 0

4 years ago

Read 2 more answers

What happens when two continental plates meet ?

dalvyx [7]

Answer:

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Explanation:

8 0

3 years ago

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What happens when a wave reflects?

solong [7]

They bounce off just like sun rays when they heat the atmosphere they bounce off reflection

4 0

4 years ago

Please answer and show work:

Vinvika [58]

The answer is c
Distance/speed=time
Therefore 4000/500=8

6 0

3 years ago

HELP PLEASE NOWWWW !!!! ( 1 ) Jill and Scott both road their bikes for 30 minutes. Jill traveled 5 kilometers and Scott trave

katovenus [111]

Jill and Scott both moves for 30 minutes

now if Jill cover 5 km distance and Scott cover 10 km distance

now we know that the formula of speed is given as

$speed = \frac{distance}{time}$

now we will have

speed of Jill

$v = \frac{5}{0.5} = 10 km/h$

speed of Scott

$v = \frac{10}{0.5} = 20 km/h$

so correct answer here is

Scott had the faster speed since he rode at 20 k/h while Jill only traveled 10 km/h.

#2

distance travelled by each car is given as

$d = 400 miles$

now here it is given that

time taken by green car

$t_1 = 10 hours$

time taken by yellow car

$t_2 = 8 hours$

now we can find the speed of two cars

$speed = \frac{distance}{time}$

speed of green car

$v = \frac{400}{10} = 40 mph$

speed of yellow car

$v = \frac{400}{8} = 50 mph$

so correct answer will be

The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

6 0

3 years ago

Read 2 more answers