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<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:
α=0.625rad/s^2
v=340m/s
w=10rad/s
θ=320rad
Explanation:
Constant angular acceleration = ∆w/∆t
angular acceleration = 20/32
α=0.625rad/s^2
Linear velocity v=wr
v = 20×17= 340m/s
Average angular velocity
w0+w1/2
w= 0+20/2
w= 20/2
w=10rad/s
What angle did it rotate with
θ=wt
θ= 10×32
=320rad
Answer:
-2.8 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity The S. I unit of acceleration is m/s²
Using the equation of motion,
v² = u² + 2as................... Equation 1
Where v = Final velocity, u = initial velocity, a = acceleration, s = distance,
Given: v = 6.0 m/s, u = 8.0 m/s, s = 5.0 m.
Substituting into equation 1
6² = 8²+2(a)5
36 = 64 + 10a
10a = 36-64
10a = -28
10a/10 = -28/10
a = -2.8 m/s²
Note: a is negative because because the skater decelerate on the rough ice
Hence the magnitude of her acceleration is = -2.8 m/s²
Explanation:
Formula to represent thrust is as follows.
F = 
= 
or, p = 

F = 
= 
= 201.67 N
Thus, we can conclude that the thrust is 201.67 N.
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