A horizontal pull A pulls two wagons over a horizontal frictionless floor, the first wagon is 500N, the second is 2000 N. The te
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Answer:
The pitching speed of your friend is 33.20 m/s
Explanation:
<em>Lets explain how to solve the problem</em>
Your friend throw the ball horizontally that means the vertical initial
component of velocity is zero ().
The ball is thrown from a height 4 meters above the ground.
The height
<u><em>Remember:</em></u> the height is negative value because its below the point of
thrown (initial position)
h = -4 m , and g = -9.8 m/s²(downward)
<em>Substitute these values in the rule above</em>
⇒
⇒ -4 = -4.9t² (multiply both sides by -1)
⇒ 4 = 4.9t² (divide both sides by 4.9)
⇒ 0.81633 = t² (take √ for both sides)
⇒ <em>t = 0.9035</em>
Then the time of the ball to land on the ground is 0.9035 seconds
The range of the ball on the ground is 30 m
The range , where
is the horizontal
component of the initial velocity
R = 30 meters and t = 0.9035
⇒ (divide both sides by 0.9035)
⇒ m/s
<em>The pitching speed of your friend is 33.20 m/s </em>
Answer:
Explanation:
1+7 = 6+2 =8 -protons
1+15 = 12+4 = 16 - protons +neutrons