Answer:
692.31 N
Explanation:
Applying,
F = ma............... Equation 1
Where F = Average force required to stop the player, m = mass of the player, a = acceleration of the player
But,
a = (v-u)/t............ Equation 2
Where v = final velocity, u = initial velocity, t = time.
Substitute equation 2 into equation 1
F = m(v-u)/t............ Equation 3
From the question,
Given: m = 75 kg, u = 6.0 m/s, v = 0 m/s (to stop), t = 0.65 s
Substitute these values into equation 3
F = 75(0-6)/0.65
F = -692.31 N
Hence the average force required to stop the player is 692.31 N
Answer:
The magnitude of the magnetic force acting on the wire is zero, because the magnetic field is parallel to the wire.
In fact, the magnetic force exerted by the magnetic field on the wire is
where I is the current in the wire, L the length of the wire, B the magnetic field intensity and the angle between the direction of B and the wire. In our problem, B and the wire are parallel, so the angle is and so , therefore the magnetic force is zero: F=0.
Answer:
Vi = 0.055 m³ = 55 L
Explanation:
From first Law of Thermodynamics, we know that:
ΔQ = ΔU + W
where,
ΔQ = Heat absorbed by the system = 52.5 J
ΔU = Change in Internal Energy = -102.5 J (negative sign shows decrease in internal energy of the system)
W = Work Done in Expansion by the system = ?
Therefore,
52.5 J = - 102.5 J + W
W = 52.5 J + 102.5 J
W = 155 J
Now, the work done in a constant pressure condition is given by:
W = PΔV
W = P(Vf - Vi)
where,
P = Constant Pressure = (0.5 atm)(101325 Pa/1 atm) = 50662.5 Pa
Vf = Final Volume of System = (58 L)(0.001 m³/1 L) = 0.058 m³
Vi = Initial Volume of System = ?
Therefore,
155 J = (50662.5 Pa)(0.058 m³ - Vi)
Vi = 0.058 m³ - 155 J/50662.5 Pa
Vi = 0.058 m³ - 0.003 m³
<u>Vi = 0.055 m³ = 55 L</u>
Answer:
force pushed the object = 10 × 5 = 50N