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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:
send the wagon down a higher hill
Answer:
no.
Explanation:
because the mass of an object never changes.
Answer:
Explanation:
This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:
![[m_av_a+m_ev_e]_b=[m_av_a+m_ev_e]_a](https://tex.z-dn.net/?f=%5Bm_av_a%2Bm_ev_e%5D_b%3D%5Bm_av_a%2Bm_ev_e%5D_a)
which, in words, is
The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:
![[(90.0)(0)+(.50)(0)]_b=[(90.0)(v)+(.50)(-4.0)]_a](https://tex.z-dn.net/?f=%5B%2890.0%29%280%29%2B%28.50%29%280%29%5D_b%3D%5B%2890.0%29%28v%29%2B%28.50%29%28-4.0%29%5D_a)
Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).
0 = 90.0v - 2.0 and
2.0 = 90.0v so
v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.
12 = .022t and
t = 550 seconds, which is the same thing as 9.2 minutes
Answer:
The work done on the sled by friction, W = - 4593.75 J
Explanation:
Given data,
The combined mass of sled and the boy, m = 75 kg
The displacement of the boy, S = 25 m
The coefficient of the friction, u = 0.25
The frictional force acting on the boy,
<em>F = u η</em>
Where,
η - is the normal force acting on the boy (mg)
Substituting the values,
F = 0.25 x 75 x 9.8
= 183.75 N
Since the direction of the frictional force is against the direction of motion
F = - 183.75 N
The work done on the sled by friction,
W = F x S
= - 183.75 x 25
= - 4593.75 J
Hence, the work done on the sled by friction, W = - 4593.75 J
