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3 years ago

This is a change in the position of a body with respect to time relative to a reference point.

Physics

2 answers:

dusya [7]3 years ago

5 0

Answer: MOTION

Explanation:

motion is defined as the displacement of an object with respect to time relative to a stationary object (reference point). A good example of an object that can serve as a reference point includes: a tree or a building. The movement of a body at constant speed towards a particular direction at regular intervals of time can be determined and it's called uniform motion.

There are different types of motion, these includes: simple harmonic motion,

linear motion,

circular motion,

Brownian motion,

Rotatory motion

malfutka [58]3 years ago

5 0

Answer:rest

Explanation:

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Say you dropped a cannonball from the 17.0–meter mast of a ship sailing at 2.0 meters/second. How far from the base of the mast

Mice21 [21]

The correct choice is A. 0 meters.

If you simply dropped the cannonball and didn't throw it horizontally,

then it'll fall straight down the mast and land on the deck right next to

the mast.

While you were up there holding it, before you dropped it, the cannonball

was moving horizontally, at 2.0 meters/second, along with the rest of the

ship and everything else aboard. It continued doing that after it dropped,

and from the point of view (in the reference frame) of the mast and everyone

on the ship, it fell straight down, parallel to the mast.

Now that we have that question answered, we can proceed to the more-

important ones. I answered the easy one, but YOU'll have to answer these:

==> WHY did you climb the mast carrying a cannonball ?

Have you been drinking sea water or bad rum ?

==> WHY did you drop it, and never even yell "LOOK OUT BELOW !" ?

==> How many formerly-able-bodied souls were injured by the

plummeting cannonball ?

==> What did everybody ELSE yell after the impact ?

==> What did they do to you after they brought you down ?

3 0

3 years ago

Read 2 more answers

on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker

Aleksandr-060686 [28]

B4 the tackle:

<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>

<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>

<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>

<span>The vector triangle is right angled: </span>

<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>

<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>

<span>v(f) = 5.6 m/s (to 2 sig figs) </span>

<span>direction of v(f) is the same as the direction of the final momentum </span>

<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>

<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>

<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>

4 0

3 years ago

Why is it impossible to build a machine that produces more energy than it uses?

Irina18 [472]

Because then it could mess up the machine with to much energy

8 0

3 years ago

Why don't atoms get too close?

Lera25 [3.4K]

Answer:

I would have to go with A, or maybe....yea A

3 0

2 years ago

Read 2 more answers

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

Evgen [1.6K]

Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :