The magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.
<h3>emf induced in the loop</h3>
The magnitude of e.m.f induced in the loop is calculated as follows;
emf = dФ/dt
Ф = 6t² + 7t
dФ/dt = 12t + 7
at t = 2 seconds
emf = dФ/dt = 12(2) + 7 = 31 V
Thus, the magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.
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Answer:
V is approximately = 23m/s
Explanation:
Kinetic energy = ½ mv²
Where m= mass = 0.450kg
V= velocity =?
K. E = 119J
Therefore
K. E = ½ mv²
Input values given
119= ½ × 0.450 × v²
Multiply both sides by 2
119 ×2 = 2 × 1/2 × 0.450 × v²
238= 0.450v²
Divide both sides by 0.450
238/0.450 = 0.450v²/0.450
v² = 528.89
Square root both sides
Sq rt v² = sq rt 528.89
V = 22.998m/s
V is approximately = 23m/s
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Answer:
Net force on the block is 32 N.
Acceleration of the object is 6.4 m/s².
Explanation:
Let the acceleration of the object be 
m/s².
Given:
Mass of the block is, 
Force of pull is, 
Frictional force on the block is, 
The free body diagram of the object is shown below.
From the figure, the net force in the forward direction is given as:
Now, from Newton's second law of motion, net force is equal to the product of mass and acceleration. So,
Therefore, the acceleration of the object in the forward direction is 6.4 m/s².
Answer:
4.61 seconds
Explanation:
Given data
Initial velocity= 12m/s
acceleration= -2.6m/s^2
From the given data
we can find the time t
we know that
Acceleration= velocity/time
time= velocity/acceleration
time= 12/2.6
time= 4.61 seconds
