Answer:
The magnitude of magnetic flux is 0.849 Wb.
Explanation:
Given that,
Number of turns in a coil, N = 290
Radius of the coil, r = 11.5 cm
Magnetic field, B = 0.0705 T
The magnetic field is directed perpendicular to the plane of the coil. The magnetic flux is given by :
So, the magnitude of magnetic flux is 0.849 Wb.
The momentum of the brick is 15.84 Kg m/s.
We have a 1.6 kg brick parachuting straight downward.
We have to determine its Momentum.
<h3>What is Linear Momentum ?</h3>
The linear momentum of a body of mass M moving with velocity v is -
p = mv.
According to the question -
Mass = 1.6 Kg
Velocity = 9.9 m/s
Therefore -
p = 1.6 x 9.9 = 15.84 Kg m/s
Hence, the momentum of the brick is 15.84 Kg m/s.
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K= (1/2) m v^2
so 0.5*0.145*<span>1281.64=92.92J of KE</span>
Answer:
3.0 m/s north
Explanation:
Take north to be positive.
Initial momentum = final momentum
(1576 kg) (-32 m/s) + (1837 kg) (33 m/s) = (1576 kg + 1837 kg) v
v = 3.0 m/s
As the spring returns to it's equilibrium position, it performs
1/2 (4975 N/m) (0.097 m)² ≈ 23 J
while the gravitational force (opposing the block's upward motion) performs
-(0.244 kg) g<em> </em>(0.097 m) ≈ -2.3 J
of work on the block. By the work energy theorem, the total work done on the block is equal to the change in its kinetic energy:
23 J - 2.3 J = 1/2 (0.244 kg) v² - 0
where v is the speed of the block at the moment it returns to the equilibrium position. Solve for v :
v² = (23 J - 2.3 J) / (1/2 (0.244 kg))
v = √((23 J - 2.3 J) / (1/2 (0.244 kg)))
v ≈ 44 m/s
After leaving the spring, block is in free fall, and at its maximum height h it has zero vertical velocity.
0² - (44 m/s)² = 2 (-g) h
Solve for h :
h = (44 m/s)² / (2g)
h ≈ 2.3 m