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3 years ago

If a ball is rolling in a straight line and you push it to the right its velocity will to the right

1 answer:

4 0

The answer would be "velocity will stay in the same direction." Since the ball is going in the same the ball is still going straight but just right. Many people get confused over velocity and speed, speed it the average amount on how fast a object is going and velocity is the amount of force a object has when it has speed. So, in this case since how the ball is going straight but just right it would stay in the same direction.

Hope this helps!

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a physical science textbook has a mass of 2.2 kg. If the textbook weighs 19.6 newtons on Venus, what is the strength of gravity

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3 years ago

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3 years ago

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serious [3.7K]

Answer:

The answer to your question is:

a) t = 3.81 s

b) vf = 37.4 m/s

Explanation:

Data

height = 71.3 m = 234 feet

t = 0 m/s

vf = ?

vo = 0 m/s

Formula

h = vot + 1/2gt²

vf = vo + gt

Process

h = vot + 1/2gt²

71.3 = 0t + 1/2(9.81)t²

2(71.3) = 9,81t²

t² = 2(71.3)/9.81

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3 0

3 years ago

An object thrown vertically upward from the surface of a celestial body at a velocity of 36 m/s reaches a height of sequalsminu

Anarel [89]

Answer:

v = -1.8t+36

20 seconds

360 m

40 seconds

36 m/s

The object speed will increase when it is coming down from its highest height.

Explanation:

$s=-0.9t^2+36t$

Differentiating with respect to time we get

$\frac{ds}{dt}=-1.8t+36\\\Rightarrow v=-1.8t+36$

a) Velocity of the object after t seconds is v = -1.8t+36

At the highest point v will be 0

$0=-1.8t+36\\\Rightarrow t=\frac{-36}{-1.8}\\\Rightarrow t=20\ s$

b) The object will reach the highest point after 20 seconds

$s=-0.9t^2+36t\\\Rightarrow s=-0.9\times 20^2+36\times 20\\\Rightarrow s=360\ m$

c) Highest point the object will reach is 360 m

$s=-0.9t^2+36t\\\Rightarrow 360=-0.9t^2+36t\\\Rightarrow -0.9t^2+36t-360=0\\\Rightarrow -9t^2+360t-3600=0$

$\frac{-360\pm \sqrt{0}}{2\left(-9\right)}\\\Rightarrow t=20\ s$

d) Time taken to strike the ground would be 20+20 = 40 seconds

$[tex]v=u+at\\\Rightarrow v=0+0.9\times 2\times 20\\\Rightarrow v=36\ m/s$

Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of $s=ut+\frac{1}{2}at^2$

e) The velocity with which the object strikes the ground will be 36 m/s

f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.

4 0

2 years ago