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2 years ago

T/-15=-3. what is t?

Physics

1 answer:

sweet-ann [11.9K]2 years ago

4 0

Answer:

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When you jump upward, your hang time is the time your feet are off the ground. Does hang time depend on the vertical component o

hjlf

Answer:

It only depends on the vertical component

Explanation:

Hello!

The horizontal component will tell you how much you travel in that direction.

You could have a large horizontal velocity, but if the vertical velocity is zero, you will never be out of the ground. Similarly, you could have a zero horizontal velocity, but if you have a non-zero vertical velocity you will be some time off the ground. This time can be calculated by two means, one is using the equation of motion (position as a function of time) and the other using the velocity as a fucntion of time.

For the former you must find the time when the position is zero.

Lets consider the origin of teh coordinate system at your feet

y(t) = vt - (1/2)gt^2

We are looking for a time t' for which y(t')=0

0 = vt' - (1/2)gt'^2

vt' = (1/2)gt'^2

The trivial solution is when t'=0 which is the initial position, however we are looking for t'≠0, therefore we can divide teh last equation by t'

v = (1/2)gt'

Solving for t'

t' = (2v/g)

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3 years ago

A screw is a type of simple machine. If we look closely at a screw, we see that it is really a _________ wrapped around a centra

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4 years ago

Read 2 more answers

Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as

Dmitriy789 [7]

Answer:

a) $v=5.6725\,m.s^{-1}$

b) $h= 1.6420\,m$

Explanation:

Given:

mass of the body, $M=20\,kg$

mass of the tyre, $m=10\,kg$

length of hanging of tyre, $l=3.5m$

distance run by the body, $d=10m$

acceleration of the body, $a=3.62m.s^{-2}$

(a)

Using the equation of motion :

$v^2=u^2+2a.d$ ..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

$u=0m.s^{-1}$

putting the values in eq. (1)

$v^2=0^2+2\times 3.62 \times 10$

$v=8.5088\,m.s^{-1}$

Now, the momentum of the body just before the jump onto the tyre will be:

$p=M.v$

$p=20\times 8.5088$

$p=170.1764\,kg.m.s^{-1}$

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

$(M+m)\times v'=p$

$(20+10)\times v'=170.1764$

$v'=5.6725\,m.s^{-1}$

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

$\frac{1}{2} (M+m).v'^2=(M+m).g.h$

$\frac{1}{2} (20+10)\times 5.6725^2=(20+10)\times 9.8\times h$

$h\approx 1.6420\,m$

above the normal hanging position.

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Explanation:

This is in line with the left hand rule

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