Using the constant acceleration formula v^2 = u^2 + 2as, we can figure out that it would take a distance of 193.21m to reach 27.8m/s
The mechanical work done by the sprinter during this time will be 4537.5 J , the average power the sprinter must generate will be 907.5 W and if the sprinter converts food energy to mechanical energy with an efficiency of 25% then he will be burning calories at 54.20 calories per second.
Work in physics is the energy that is transferred to or from an item when a force is applied along a displacement. It is frequently described in its most basic form as the result of force and displacement.
The quantity of energy moved or transformed per unit of time is known as power in physics. The watt, or one joule per second, is the unit of power in the International System of Units.. A scalar quantity is power.
Given 75-kg sprinter accelerates from rest to a speed of 11.0 m/s in 5.0 s.
So let,
m = 75 kg
v = 11.0 m/s
t = 5.0 s
So the mechanical work done by the sprinter during this time will be as follow:
W = 0.5 mv²
W = 0.5 (75)(11)²
W = 4537.5 J
The average power the sprinter must generate will be as follow:
Power(P) = W / t
P = 4537.5/5
P = 907.5 W
Only 25% is absorbed. So, the sprinter only absorbed 226.875 J per second which is equal to 54.20 calories per second.
Hence mechanical work done by the sprinter during this time will be 4537.5 J , the average power the sprinter must generate will be 907.5 W and if the sprinter converts food energy to mechanical energy with an efficiency of 25% then he will be burning calories at 54.20 calories per second.
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Answer:
a) v, v
b) 2mv^2
c) Elastic collion
Explanation:
(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction
Here x, y represent direction.They are not variable. 1 and 2 represent before and after.
2vm=v1xm+v2xm, we find v1x=v.
From momentum conservation in y-direction
0 =v1ym+v2ym, we findv1y=v.
(b) By energy conservation principle
Before: K=1/2m(2v)^2=2mv^2.
After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2
(c) The collision is elastic
Answer:
It is a force in the direction of the motion that allows the box to move
Explanation:
Answer:
<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>
Explanation:
The Non conservative force is defined as a force which do not store energy or get he energy dissipate the energy from the system as the system progress with the motion.
Given are
<em> mass of the student 73 kg</em>
<em> height of water glide 11.8 m</em>
<em> work done as -5.5*10³ J</em>
Have to find speed at which the student goes down the glide.
According to<em> Law of Conservation of energy</em>,
K.E =P.E+Work Done
mv²/2=mgh +W
Rearranging the above eqn for v
v = √2(gh+W/m)
Substituting values,
V = 12.48 m/s.
<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>
