Answer:
C 80 m
Explanation:
Given:
v₀ = 30 m/s
a = -10 m/s²
t = 8 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (30 m/s) (8 s) + ½ (-10 m/s²) (8 s)²
Δy = -80 m
The ball lands 80 m below where it started. So the height of the cliff is 80 m.
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
Distance in the y direction.
Buddy, I think you need to evaluate your question and fix it. Because it's not making any sense, whatsoever.
Answer:
v = √[gR (sin θ - μcos θ)]
Explanation:
The free body diagram for the car is presented in the attached image to this answer.
The forces acting on the car include the weight of the car, the normal reaction of the plane on the car, the frictional force on the car and the net force on the car which is the centripetal force on the car keeping it in circular motion without slipping.
Resolving the weight into the axis parallel and perpendicular to the inclined plane,
N = mg cos θ
And the component parallel to the inclined plane that slides the body down the plane at rest = mg sin θ
Frictional force = Fr = μN = μmg cos θ
Centripetal force responsible for keeping the car in circular motion = (mv²/R)
So, a force balance in the plane parallel to the inclined plane shows that
Centripetal force = (mg sin θ - Fr) (since the car slides down the plane at rest, (mg sin θ) is greater than the frictional force)
(mv²/R) = (mg sin θ - μmg cos θ)
v² = R(g sin θ - μg cos θ)
v² = gR (sin θ - μcos θ)
v = √[gR (sin θ - μcos θ)]
Hope this Helps!!!