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zimovet [89]
3 years ago
5

Use the velocity vs time graph to analyze the motion of the object.

Physics
1 answer:
arsen [322]3 years ago
7 0

Explanation:

the object has constant velocity for 2 seconds and it get a constant accelration (2ms-2)

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What real-world examples show no work being done? Can you think of examples other than resisting the force of gravity?
GalinKa [24]

-- pushing on a brick wall

-- standing on your little brother's back so that he can't get up

-- taking a nap while on the job

-- squeezing anything that doesn't yield to your squeeze, such as a glass bottle or your girl friend

-- watching TV

-- solving math problems in your head

-- making pictures out of clouds in the sky

8 0
3 years ago
A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally
Alex787 [66]

Answer:

47 ms

well the time it takes to fall 100m is the same time it takes to travel 65m horizontally.

The time to fall vertically, t is sqrt(2d/g). [this comes from d = 1/2at^2]

so t = sqrt(2*100/9.8) = 4.52s.

The vertical speed at that time is g*t = 9.8*4.52 = 44.3m/s

The horizontal speed is the horizontal distance over the same 4.52s, = 65/4.52 = 14.4m/s.

so the final velocity is = sqrt(44.3^2 + 14.4^2) = 47.ms

4 0
3 years ago
What type of relationship is shown in the graph
fenix001 [56]

Answer:

A linear, positive relationship

Explanation:

The graph is a line, so it is linear and the slope is positive, so it is positive...

6 0
3 years ago
3. Three forces are acting on an object which is in
kirza4 [7]

Answer:

hard

Explanation:

7 0
3 years ago
A snowball is rolling down a hill at 4.5 m/s and accumulating snow as it goes. Its diameter begins at 0.50 m and ends at the bot
Reil [10]
To find the change in centripetal acceleration, you should first look for the centripetal acceleration at the top of the hill and at the bottom of the hill.

The formula for centripetal acceleration is:
Centripetal Acceleration = v squared divided by r

where:
v = velocity, m/s
r= radium, m

assuming the velocity does not change:

at the top of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 0.25 m
                                      = 81 m/s^2

at the bottom of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 1.25 m
                                      = 16.2 m/s^2

to find the change in centripetal acceleration, take the difference of the two.
change in centripetal acceleration = centripetal acceleration at the top of the hill - centripetal acceleration at the bottom of the hill

= 81 m/s^2 - 16.2 m/s^2
= 64.8 m/s^2 or 65 m/s^2
6 0
4 years ago
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