The maximum force that the athlete exerts on the bag is equal to 1,500 N and in the opposite direction as the force that the bag exerts on the athlete.
<h3>
Newton's third law of motion</h3>
Newton's third law of motion states that action and reaction are equal and opposite.
Fa = -Fb
The force exerted by the athlete on the bag is equal to the force the bag exerted on the athlete but in opposite direction.
Thus, the maximum force that the athlete exerts on the bag is equal to 1,500 newtons and in the opposite direction as the force that the bag exerts on the athlete.
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Answer:
15.5 m/s.
Explanation:
Potential energy of the balloon has been converted to kinetic energy.
potential energy = kinetic energy.
mgh = ½mv².
10* 10* 12= ½ *10 *v²
1200 = 5v²
v²=1200÷5
v=√240
v= 15.49~15.5 m/s.
Answer:
Explanation:
Given a square side loop of length 10cm
L=10cm=0.1m
Then, Area=L²
Area=0.1²
Area=0.01m²
Given that, frequency=60Hz
And magnetic field B=0.8T
a. Flux Φ
Flux is given as
Φ=BA Sin(wt)
w=2πf
Φ=BA Sin(2πft)
Φ=0.8×0.01 Sin(2×π×60t)
Φ=0.008Sin(120πt) Weber
b. EMF in loop
Emf is given as
EMF= -N dΦ/dt
Where N is number of turns
Φ=0.008Sin(120πt)
dΦ/dt= 0.008×120Cos(120πt)
dΦ/dt= 0.96Cos(120πt)
Emf=-NdΦ/dt
Emf=-0.96NCos(120πt). Volts
c. Current induced for a resistance of 1ohms
From ohms law, V=iR
Therefore, Emf=iR
i=EMF/R
i=-0.96NCos(120πt) / 1
i=-0.96NCos(120πt) Ampere
d. Power delivered to the loop
Power is given as
P=IV
P=-0.96NCos(120πt)•-0.96NCos(120πt)
P=0.92N²Cos²(120πt) Watt
e. Torque
Torque is given as
τ=iL²B
τ=-0.96NCos(120πt)•0.1²×0.8
τ=-0.00768NCos(120πt) Nm
If the runner is running in a circular track then yes when something or someone is moving in a circular motion at a constant speed they are indeed accelerating. They’re accelerating because the direction of the velocity vector is changing
Your question: The strong nuclear force felt by a single proton in a large nucleus _______________________.
Answer: is about the same as that felt by a single proton in a small nucleus.
