Answer:
Radius of the solenoid is 0.93 meters.
Explanation:
It is given that,
The magnetic field strength within the solenoid is given by the equation,
, t is time in seconds

The induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, x = 2 m
The electric field due to changing magnetic field is given by :

x is the distance from the axis of the solenoid
, r is the radius of the solenoid


r = 0.93 meters
So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.
Answer: a. F doubled
b. F reduced by one-quarter i.e
1/4*(F)
c. 1/9*(F)
d. F increased by a factor of 4 i.e 4*F
e. F reduces 3/4*(F)
Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is
F = k*(qA*qB)/d²
a. If qA is doubled therefore the force is doubled since they are directly proportional.
b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4
Which means the product of charge is divided by 4 so the force would be divided by 4 too since they are directly proportional.
c. If d is tripped that is multiplied by 3. From the formula new d would be (3*d)²=9d² but force is inversely proportional to d² so instead of multiplying by 9 the force will be divided by 9
d. If d is cut into half that is divided by 2. The new d would be (d/2)²=d²/4. So d² is divided by 4 so the force would be multiplied by 4
e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,
3/4*F
The average speed can be easily calculated by taking the
ratio of distance and time. That is:
average speed = distance / time
so calculating:
average speed = 4875 ft / 6.85 minutes
<span>average speed = 711.68 ft / min</span>
Two significant figures, the 6 and the 9
Answer:
Answered
Explanation:
v= 1 m/s
A= 1 m^2
m= 100 kg
y= 1 mm
μ = ?
ζ= viscosity of SAE 20 crankcase oil of 15° C= 0.3075 N sec/m^2
forces acting on the block are
F_s ← ↓ →F_f
mg
N= mg
F_s= shear force = ζAv/y F_f= friction force = μN
now in x- direction F_s= F_f
ζAv/y = μN
0.3075×1×1×1/1×10^{-3} = μ×100
⇒μ=0.313 (coefficient of sliding friction for the block)
Now, as the velocity is increased shear force also increases and due to this frictional force also increases.
Now, to compensate this frictional force friction coefficient must increase
as v∝μ