Hi I’m having some difficulty with this question could really use some help!

a 14 kg rock starting from rest free falls through a distance of 5.0 m with no air resistance. find the momentum change of the r

Alchen [17]

Use the equation p = mv

P - Momentum : We need to find out

M - Mass : 14 kg

V - Velocity : ?

Sorry this wasn't much help, you should probably report me, but i tried my best to help you out :/

3 0

4 years ago

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If the pressure of a gas is increased , then the volume will decrease because the particles of the gas will be forced more close

sergey [27]

Answer:

true

Explanation:

4 0

4 years ago

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The acceleration of a bus is given by ax(t)=αt, where α = 1.28 m/s3 is a constant.

xz_007 [3.2K]

The text looks incomplete. Here the complete question found on google:

<em>"The acceleration of a bus is given by ax(t)=αt, where α = 1.28m/s3 is a constant. Part A If the bus's velocity at time t1 = 1.13s is 5.09m/s , what is its velocity at time t2 = 2.02s ? If the bus's position at time t1 = 1.13s is 5.92m , what is its position at time t2 = 2.02s ?"</em>

A) 6.88 m/s

The velocity of the bus can be found by integrating the acceleration. Therefore:

$v(t) = \int a(t) dt = \int (\alpha t)dt=\frac{1}{2}\alpha t^2+C$ (1)

where

$\alpha = 1.28 m/s^3$

C is a constant term

We know that at $t_1 = 1.13 s$ , the velocity is $v=5.09 m/s$ . Substituting these values into (1), we can find the exact value of C:

$C=v(t) - \frac{1}{2}\alpha t^2 = 5.09 - \frac{1}{2}(1.28)(1.13)^2=4.27 m/s$

So now we can find the velocity at time $t_2 = 2.02 s$ :

$v(2.02)=\frac{1}{2}(1.28)(2.02)^2+4.27=6.88 m/s$

B) 11.2 m

To find the position, we need to integrate the velocity:

$x(t) = \int v(t) dt = \int (\frac{1}{2}\alpha t^2 + C) dt = \frac{1}{3}\alpha t^3 + Ct +D$ (2)

where D is another constant term.

We know that at $t_1 = 1.13 s$ , the position is $v=5.92 m$ . Substituting these values into (2), we can find the exact value of D:

$D = x(t) - \frac{1}{6}\alpha t^3 -Ct = 5.92-\frac{1}{6}(1.28)(1.13)^3 - (4.27)(1.13)=0.79 m$

And so now we can find the position at time $t_2 = 2.02 s$ using eq.(2):

$x(2.02)=\frac{1}{6}(1.28)(2.02)^3 + (4.27)(2.02) +0.79=11.2 m$

5 0

4 years ago

3 The components of a 15-meters-per-second

sammy [17]

First draw a picture to help make the problem easier. (see attached)

Next you can use sin and cos to find the x and y components of the velocity.

First lets find the vertical component:
$sin = \frac{opp}{hyp}$
$sin(60) = \frac{y}{15m/s}$
Multiply both sides by 15m/s
$15m/s * sin(60) = y$
y ≈ 13 m/s

Now you can find the horizontal component:
$cos = \frac{adj}{hyp}$
$cos(60) = \frac{x}{15m/s}$
Multiply both sides by 15m/s
$15m/s * cos(60) - x$
x = 7.5

Therefore the answer is 2, the vertical (y) component is 13m/s and the horizontal (x) component is 7.5m/s.

8 0

3 years ago

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An example of an atom that has no change is one that has A. 1 proton, 2 electrons, and 3 neutrons B. 3 protons,2 electrons, and

Vikki [24]

Your question asks which is an example of an atom that has no change, or charge.

Your answer would be C). 2 protons, 2 electrons, and 1 neutron

These atoms would be known as a neutral atom.

In order for an atom to be neutral, it must have the same amount of protons and electrons.

We know that Protons are positive

We also know that Electrons are negative

In order for the atom to be neutral, they must have an equal amount of protons and neutrons to "cancel out"

In answer choice C, you would see that there are 2 protons and 2 electrons.

2- 0 = 0, in which allows this atom to be neutral.

Therefore, answer choice C. 2 protons, 2 electrons, and 1 neutron would be your answer.

4 0

3 years ago

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