Answer:
B = 0.8 T
Explanation:
It is given that,
Radius of circular loop, r = 0.75 m
Current in the loop, I = 3 A
The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.
When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.
We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :
B is magnetic field
So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.
Answer:
D
friction acts in the opposite direction of motion but does not affect the motion of the object
Answer:
659.01W
Explanation:
The cab has a mass of 1250 kg, the weight of the cab represented by Wc will be
Wc = mass of the cab × acceleration due to gravity in m/s²
Wc = 1250 × 9.81 = 12262.5 N
but the counter weight of the elevator represented by We = mass × acceleration due to gravity = 995 × 9.81 = 9760.95 N
Net weight = weight of the cab - counter weight of the elevator = Wc - We = 12262.5 - 9760.95 = 2501.55 N
the motor of the elevator will have to provide this in form of work
work done by the elevator to lift the cab to height of 49 m = net weight × distance (height) = 2501.55 × 49m
power provided by the motor of the elevator = workdone by the motor / time in seconds
Power = (2501.55 × 49) ÷ ( 3.1 × 60 seconds) = 659.01 W
Answer:
k = 9.6 x 10^5 N/m or 9.6 kN/m
Explanation:
First, we need to use the expression to calculate the spring constant which is:
w² = k/m
Solving for k:
k = w²*m
To get the angular velocity:
w = 2πf
The problem is giving the linear velocity of the car which is 5.7 m/s. With this we can calculate the frequency of the car:
f = V/x
f = 5.7 / 4.9 = 1.16 Hz
Now the angular velocity:
w = 2π*1.16
w = 7.29 rad/s
Finally, solving for k:
k = (7.29)² * 1800
k = 95,659.38 N/m
In two significant figures it'll ve 9.6 kN/m
C. Populations.
Hope that's right.
