For any mass m:
a = F/m
v = √2*F/m*s = √2F/sm = k/√m
Momentum = mv = k√m
Energy = 1/ mv² = 1/2 m.k²/m = 1/2k²
SO
Both will have same energy
The larger mass will have greater momentum
Answer:
0.94 m/s^2 downwards
Explanation:
m = 70 kg, m g = 70 x 9.8 = 686 N
R = 620 N
Let the acceleration be a, as the apparent weight decreases so the elevator is moving downwards with an acceleration a.
mg - R = ma
686 - 620 = 70 x a
a = 0.94 m/s^2
The initial speed of car A is 15.18 m/s.
Momentum is defined as mass in motion. If there are two objects (the two objects in motion or only one object in motion and the other in stationary) that collide and no other forces work in the system, the law of momentum conservation applies in the system.
p=p'
pa+pb = pa'+pb'
(ma×va) + (mb×vb) = (ma×va') + (mb×vb')
- ma = mass of object A (kg) = 1,783 kg
- mb = mass of object B (kg) = 1,600 kg
- va = speed of object A before collides (m/s)
- va' = speed of object A after collides (m/s) = 8 m/s
- vb = speed of object B before collides (m/s) = 0 m/s
- vb' = speed of object B after collides (m/s) = 8 m/s
- p = momentum before collision (Ns)
- p' = momentum after collision (Ns)
(ma×va) + (mb×vb) = (ma×va') + (mb×vb')
(1,783×va) + (1,600×0) = (1,783×8) + (1,600×8)
(1,783×va) + 0 = 14,264+12,800
(1,783×va) = 27,064

va = 15.18 m/s
Learn more about The law of momentum conservation here: brainly.com/question/7538238
#SPJ4
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