It would be c. As it can’t accelerate faster thus not having a faster velocity so it’s inertia
Answer:
The second knife-edge must be placed 46.2 cm from the zero mark of the rod.
Explanation:
From the law of equilibrium, ΣF = 0 and ΣM = 0.
Let R be the reaction at the knife edge. Since the weight of the rod and zinc load act downward, and we take downward position as negative
-32 N - 2 N + R = 0
-34 N = -R
R = 34 N
Also, let us assume the knife-edge is x cm from the zero mark. Taking moments about the weight and assuming the knife-edge is right of the weight of the rod. Taking clockwise moments as positive and anti-clockwise moments as negative,
-(45 - 25)2 + (x - 45)R = 0
-(20)2 + (x - 45)34 = 0
-40 = -(x - 45)34
x - 45 = 40/34
x - 45 = 1.18
x = 45 + 1.18
x = 46.18 cm
x ≅ 46.2 cm
The second knife-edge must be placed 46.2 cm from the zero mark of the rod.
Answer:
V=20cm/s
Explanation:
The average speed is the distance total divided the time total:
First stage:
T1=5s
But, 
(decelerates to rest)
then: 
on the other hand:
X1=75cm
Second stage:
T2=5s
X2=125cm
Finally:
X=X1+X2=200cm
T=T1+T2=10s
V=X/T=20cm/s
The magnetic field strength in a coil is directly proportional to the number of turns, or loops, in the coil.
Therefore, when there are four loops instead of one, the magnetic field strength has increased four times, making it harder to push the magnet in.
Answer:
The induced current is 0.084 A
Explanation:
the area given by the exercise is
A = 200 cm^2 = 200x10^-4 m^2
R = 5 Ω
N = 7 turns
The formula of the emf induced according to Faraday's law is equal to:
ε = (-N * dφ)/dt = (N*(b2-b1)*A)/dt
Replacing values:
ε = (7*(38 - 14) * (200x10^-4))/8x10^-3 = 0.42 V
the induced current is equal to:
I = ε /R = 0.42/5 = 0.084 A
