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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<
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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<
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Substitute the above given values in the kinematic equation ,
m/s
Therefore, the final velocity or jumping speed is m/s
Explanation:
Answer: A projectile is any object in which the only force is gravity
Explanation: Equations on how to calculate projectile velocity is stated below:
The initial velocity Vo being a vector quantity, has two componentsVox and Voy
V0x = V0 cos(θ)
V0y = V0 sin(θ)
The acceleration A is a also a vector with two components Axand Ay given
Ax = 0 and Ay = - g = - 9.8 m/s2
Along the x axis the acceleration is equal to 0 and therefore the velocity Vx is constant
Vx = Vocos(θ)
Along the y axis, the acceleration is uniform and equal to - g and the velocity at time t is g
Vy = Vo sin(θ) - g t
Along the x axis the velocity Vx is constant and therefore the component x of the displacement is
x = Vocos(θ) t
Along the y axis, the motion is of uniform acceleration and the y component of the displacement is
y = Vo sin(θ) t - (1/2) g t2
Answer:
Orbital period, T = 1.00074 years
Explanation:
It is given that,
Orbital radius of a solar system planet, 
The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :
M is the mass of the sun
T = 31559467.6761 s
T = 1.00074 years
So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.
The velocity of the submarine immediately after firing the missile is 0.0104 m/s
Explanation:
Mass of the submarine M=50 tonne=
Mass of the missile m=40 kg
velocity of the missile v= 13m/s
we have to calculate the velocity of the submarine after firing
This is the recoil velocity and its expression is derived from the law of conservation of momentum
recoil velocity of the submarine
