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3 years ago

14

While in empty space, an astronaut throws a ball at a velocity of 9 m/s. What will the velocity of the ball be after it has trav

eled 6 meters?

1 answer:

Snezhnost [94]3 years ago

5 0

Answer:

1.5

Explanation:

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The left side of the lever was forced down 10 inches in order to raise the rock 7 inches. The ideal mechanical advantage is

Yanka [14]

Answer:

1.42

Explanation:

<em> got it right on my homework </em>

6 0

3 years ago

What three variable factors determine the force of gravity between any two objects?

brilliants [131]

The force of gravity between two objects is:
F = G*m1*m2/r^2

So, it is dependent of the two masses and the distance between their centers of mass.

5 0

3 years ago

For general projectile motion, the horizontal component of a projectile's velocity aSelect one:O a. continuously increases.O b.

liubo4ka [24]

ANSWER:

d. remains a non-zero constant.

STEP-BY-STEP EXPLANATION:

If we consider that there is no air resistance and that the horizontal component would be at x, the velocity remains a non-zero constant

3 0

1 year ago

Thought Experiment: A monkey escapes from a zoo and climbs a tree. After failing to entice the monkey down, a zookeeper fires a

Andreyy89

Explanation:

When bullet is shot towards the monkey then let say the distance of monkey from the bullet is "d"

so we can find the time to reach the bullet to the monkey

$t = \frac{d}{vcos\theta}$

Now similarly we can find the vertical displacement of the bullet in the same time

$\Delta y = vsin\theta t - \frac{1}{2}gt^2$

$\Delta y = v sin\theta (\frac{d}{vcos\theta}) - \frac{1}{2}gt^2$

so it is given as

$\Delta y = d tan\theta - \frac{1}{2}gt^2$

here if the monkey is initially at height H above the ground at given angle then we can say

$H = dtan\theta$

so we can say that

$\Delta y = H - \frac{1}{2}gt^2$

So if at the same time monkey will fall down then the height of monkey from ground after time "t" is given as

$\Delta y = H - \frac{1}{2}gt^2$

so here bullet will hit the monkey as both monkey and bullet are at same position.

3 0

3 years ago

A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between

SashulF [63]

To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:

$\Delta W = \Delta KE$

$\Delta W = \frac{1}{2} mv^2$

Here,

m = mass

v = Velocity

Our values are given as,

$m = 79.7kg$

$v = 4.77m/s$

Replacing,

$\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2$

$\Delta W = 907J$

Therefore the mechanical energy lost due to friction acting on the runner is 907J

6 0

3 years ago