<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<
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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<
/p>
Substitute the above given values in the kinematic equation ,
m/s
Therefore, the final velocity or jumping speed is m/s
Explanation:
Answer:11686.5 joules
Explanation:
elastic constant(k)=53N/m
extension(e)=21m
Elastic potential energy=(k x e^2)/2
Elastic potential energy=(53 x (21)^2)/2
Elastic potential energy=(53x21x21)/2
Elastic potential energy=23373/2
Elastic potential energy=11686.5
Elastic potential energy is 11686.5 joules
Answer:
W_apparent = 93.1 kg
Explanation:
The apparent weight of a body is the weight due to the gravitational attraction minus the thrust due to the fluid where it will be found.
W_apparent = W - B
The push is given by the expression of Archimeas
B = ρ_fluide g V
ρ_al = m / V
m = ρ_al V
we substitute
W_apparent = ρ_al V g - ρ_fluide g V
W_apparent = g V (ρ_al - ρ_fluide)
we calculate
W_apparent = 980 50 (2.7 - 0.8)
W_apparent = 93100 g
W_apparent = 93.1 kg
Answer:
b. Relates the electric field at points on a closed surface to the net charge enclosed by that surface
Explanation:
Gauss's law states that the flux of certain fields through a closed surface is proportional to the magnitude of the sources of that field within the same surface. The electric flux expresses the measure of the electric field that crosses a certain surface. Therefore, the electric field on a closed surface is proportional to the net charge enclosed by that surface.
<span>Mass of the ball is m = 0.10kg
Initial speed of the Ball v = 15m/s
a. When the ball is at maximum height the velocity is 0
Momentum of ball = mass x velocity
Momentum = 0.10kg x 0 = 0
b. Getting the maximum height,
Using the conservation of energy equation KEinitial = mgh
1/2mVin^2 = mgh => h = v^2/2g
h = 15^2/2x9.8 = 11.48m => Half Height h = 5.96m
Applying the conservation of energy equation at halfway V^2 = 2gh
V = square root of (2x9.8x5.96) => V = square root of (116.816)
So the velocity at the half way V = 10.81 m/s
Momentum M = m x V => M = 0.10 x 10.81 => M = 1.081kg-m/s</span>
