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3 years ago

15 POINTS!

2 answers:

8 0

The first law of inertia says that an object’s speed will not change unless something makes it change.

Newton's First Law of Motion

Place the hard boiled egg on its side and spin it. Put your finger on it gently while it is still spinning in order to stop it. Remove your finger when it stops.

Place the raw egg on its side and spin it. Place your finger gently on the egg until it stops. Once you remove your finger, the egg should start to spin again. The liquid inside the egg has not stopped so it will continue to spin until enough force is applied.

The second law: the strength of the force equals the mass of the object times the resulting acceleration.

Newton's Second Law of Motion

Drop a rock or marble and a wadded-up piece of paper at the same time. They fall at the same rate of speed, but the rock's mass is greater so it hits with greater force.

Finally, the third law says that for every action there is a reaction.

Newton's Third Law of Motion

Pull one ball or swing back and let it go.

It will swing into the other balls making the ball at the other end swing.

adelina 88 [10]3 years ago

6 0

1. The Egg Drop.

2. The Car down a ramp.

3. Maybe an Airplane race with things on it.

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What happens during

alukav5142 [94]

Answer:

Explanation:

8 0

3 years ago

Read 2 more answers

What is the difference between a steep and shallow slope

Airida [17]

Whether a line is "steep" or "shallow" depends on the slope of the line. In the equation y=mx+b, m is the slope. In your first equation, 1/2x+4, the slope would be 1/2 which means there is 1 increase in the vertical, or y, value for each 2 increases in the horizontal, or x value. This line would be shallow.

4 0

4 years ago

Read 2 more answers

A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how

GalinKa [24]

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=48.1 m/s$ is the cannonball's initial velocity

$\theta=0$ because we are told the cannonball is shot horizontally

$t$ is the time since the cannonball is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the cannonball

$y=0$ is the final height of the cannonball (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it $X_{max}$ and this occurs when $y=0$ .

So, firstly we will find the time with (2):

$0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2$ (3)

Rearranging the equation:

$0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m$ (4)

$-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0$ (5)

This is a <u>quadratic equation</u> (also called <u>equation of the second degree</u>) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (6)

Where:

$a=-4.9 m/s^{2}$

$b=48.1 m/s$

$c=1.5 m$

Substituting the known values:

$t=\frac{-48.1 \pm \sqrt{48.1^{2}-4(-4.9)(1.5)}}{2(-4.9)}$ (7)

Solving (7) we find the positive result is:

$t=9.847 s$ (8)

Substituting this value in (1):

$x=(48.1 m/s)cos(0\°) (9.847 s)$ (9)

$x=473.640 m$ This is the horizontal distance the cannonball traveled before it landed on the ground.

3 0

3 years ago

a ball is thrown upward with initial velocity of 20 m/s. (a) how long is the ball in the air? (b) what is the greatest height re

Lena [83]

a) t = 4.08 s

b) Response = 20.39m

c) t = 0.991 and 3.087 seconds.

Initially, the vertical speed was 20 m/s.

(a) how long is the ball in the air?

The ball is tossed upward, rises to its highest point, and then hits the ground again.

Consequently, the end height will be equal to the starting height.

h = h o = Om

Use the equation s= ut + -at.

2764 – An = 14

Put a negative sign for g because the ball is being thrown upwards and g is acting downwards.

Om = t - 0.5 * 9.81 m/s * 20 m/s

0 = 20 *t - 0.5 * 9.81 *t

20 *t=0.5*9.81 *t

20 = 0.5 * 9.81 *t

20 0.5*9.81

Answer (a): t = 4.08 s

(b) what is the greatest height reached by the ball?

Vertical velocity for the ball decreases to zero when it reaches its highest point.

Use the equation 2-u=295

-2ghmas for - u

(20m/s - (Om/s)2) = -2 * 9.81 m/s2 *

02 – 20= -2*9.81 *

-20% = -2 *9.81 *

202 = 2 * 9.81 *

b) RESPONSE: H = 20.39m

(c) when is the ball 15 m above the ground?

The ball will be 15 meters above the earth on two separate occasions.

1. When climbing

2. When descending

Use the equation s= ut + -at.

2,264 – fn = 4

0.5 * 9.81 m/s * 15m = 20 m/s*t

15 = 20 *t - 0.5 * 9.81 *t

15 = 20t – 4.9057

4.905 – 20t + 15 = 0

Calculator-based quadratic equation solution

(c) t = 0.991 and 3.087 seconds.

Hence the answers are,

a) t = 4.08 s

b) RESPONSE: Hmar = 20.39m

c) t = 0.991 and 3.087 seconds.

To learn more about initial velocity, click brainly.com/question/14154244

#SPJ4

4 0

1 year ago

How does the height the ball is released affect the amount of potential energy

aleksley [76]

The potential energy is directly proportional to the height.
If you double the height you double the potential energy.

5 0

3 years ago