<h2>
<em><u>⇒</u></em>Answer:</h2>
In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)
Step-by-Step Solution:
Solution 35PE
This question discusses about the increased range. So, we shall assume that the angle of jumping will be as the horizontal range is maximum at this angle.
Step 1 of 3<
/p>
The legs have an extension of 0.600 m in the crouch position.
So, m
The person is at rest initially, so the initial velocity will be zero.
The acceleration is m/s2
Acceleration m/s2
Let the final velocity be .
Step 2 of 3<
/p>
Substitute the above given values in the kinematic equation ,
m/s
Therefore, the final velocity or jumping speed is m/s
Explanation:
Answer:
B) 16.67
Explanation:
If the dimension of one lumber is 2" × 6", the total area of one lumber will be 12inch²
If the total board feet of lumber there is 200in, therefore the total board of lumber that will be needed is 200/12 which gives 16.67 lumbers
Answer:
692.31 N
Explanation:
Applying,
F = ma............... Equation 1
Where F = Average force required to stop the player, m = mass of the player, a = acceleration of the player
But,
a = (v-u)/t............ Equation 2
Where v = final velocity, u = initial velocity, t = time.
Substitute equation 2 into equation 1
F = m(v-u)/t............ Equation 3
From the question,
Given: m = 75 kg, u = 6.0 m/s, v = 0 m/s (to stop), t = 0.65 s
Substitute these values into equation 3
F = 75(0-6)/0.65
F = -692.31 N
Hence the average force required to stop the player is 692.31 N
