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il63 [147K]
3 years ago
10

Need help with 4 &5 please help

Physics
1 answer:
DochEvi [55]3 years ago
6 0
I can help you with 5. 5 is a displacement-time graph.
At A it is moving in a positive direction and speed is constant.
At B it is stationary or at rest.
At C, movement is negative and it is decreasing or decelerating
At D, it’s a general downwards movement in the negative direction that can be described as overall constant
You might be interested in
An unbalanced force of 15 N is applied to a 13 kg mass. What is the acceleration of the mass?
Alex787 [66]

Answer:

1.15m/s2

Explanation:

accerelation =force÷mass

4 0
3 years ago
Please fast answerrr thank you​
Vedmedyk [2.9K]

Answer:

50J

Explanation:

At the top you have(A)

KE_a = O

PE_a = 100J

KE + PE = 100J

At the bottom you have (C)

KE_c= 100J

PE_c=0J

KE+PE = 100J

At point C:

You are at half the height.

We know that at H, PE =100J

PE_c = mgH

At C,

PE_c= mg (H/2) *at half the height

*m and g stay the same

Intuitively, the higher you are, the more potential energy you have.

If you decrease the height by a half, your PE will also decrease

At A:

PE_a / (mg) = H

At B:

PE_b / (mg) = H/2

to also get H on the right hand side, multiply by 2

2 (PE_b/ (mg))= H

2PE_b / (mg) = H

Ok, now that we have set up 2 equations (where H is isolated), find PE at B

AT A = AT B *This way you are saying that H = H (you compare both equations)

PE_a / (mg) = 2x PE_b / (mg)

*mg are the same for both cancel them (you can do that because of the = sign)

PE_a =  2PE_b

We know that PE_a = 100J

100J/2 = PE_b

PE at b = 50J

**FIND KE at b

We know that

KE_b + PE_b is always 100J

100J = 50J + KE_b

KE_b = 50J

4 0
2 years ago
Terra tosses a 0.20 kg volleyball straight up at 10.0 m/s. how high does it go?
Sliva [168]

Answer:

5.1 meters

Explanation:

Terra tosses a 0.20kg volleyball up at at a speed of 10 m/s

The height can be calculated as follows

= v^2/2g

= 10^2/2×9.8

= 100/19.6

= 5.1 meters

Hence the height is 5.1 meters

5 0
2 years ago
 Why does a child in a wagon seem to fall backward when you give the wagon a sharp pull forward? 15. What force is needed to acc
slavikrds [6]

Answer:

Force, F = 77 N

Explanation:

A child in a wagon seem to fall backward when you give the wagon a sharp pull forward. It is due to Newton's third law of motion. The forward pull on wagon is called action force and the backward force is called reaction force. These two forces are equal in magnitude but they acts in opposite direction.

We need to calculate the force is needed to accelerate a sled. It can be calculated using the formula as :

F = m × a

Where

m = mass = 55 kg

a = acceleration = 1.4 m/s²

F=55\ kg\times 1.4\ m/s^2

F = 77 N

So, the force needed to accelerate a sled is 77 N. Hence, this is the required solution.

3 0
3 years ago
The amp is the unit for _________.
adell [148]
B.) <span>The amp is the unit for "Current"

Hope this helps!</span>
7 0
3 years ago
Read 2 more answers
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