Answer:
ΔE = 37.8 x 10^9 J
Explanation:
The energy required will increased the potential energy and increase the kinetic energy.
As the altitude change is fairly small compared to the earth radius, we can ASSUME that the average gravity will be a good representative
Gravity acceleration at altitude would be 9.8(6400²/8000²) = 6.272 m/s²
G(avg) = (9.8 + 6.272)/2 = 8.036 m/s²
ΔPE = mG(avg)Δh = 1000(8.036)(8e6 - 6.4e6) = 12.857e9 J
The centripetal force at orbit must be equal to the gravity force
mv²/R = mg'
v²/8.0e6 = 6.272
v² = (6.272(8.0e6)) = 50.2e6 m²/s²
The maximum velocity when resting on earth at the equator is about 460 m/s.
The change in kinetic energy is
ΔKE = ½m(vf² - vi²)(1000)
ΔKE = ½(1000)(50.2e6 - 460²) = 25e9 J
Total energy increase is
25e9 + 12.857e9 = 37.8e9 J
Answer:
<h3>displacement can be zero...because it only care about initial and final positions</h3>
Answer:
The force acting in x direction is 34.44 N
Explanation:
We have,
Force acting on the box is 65 N at an angle of 58 degrees to the horizontal the surface of frictional.
It is required to find the net force in x- direction.
The net force acting in the x- direction is given by :

So, the force acting in x direction is 34.44 N.
Answer:
4.08 kg
Explanation:
We can apply the Newton's second law of motion to find the mass of this object:
,
where
is mass, and
is acceleration, which is substituted by
here. Now, plugging given numbers in the equation, we have
.
Solving for mass, we have

Answer:
The change in the magnetic flux in the ring is 10.99 Wb.
Explanation:
Given that,
An MRI technician moves his hand from a region of very low (B=0) magnetic field strength into an MRI scanner’s 3.50 T field with his fingers pointing in the direction of the field.
Diameter of the ring, d = 2 cm
Radius, r = 1 cm
It takes 0.4 s to move it into the field. We need to find the change in magnetic flux in the ring. Magnetic flux is given by :

So, the change in the magnetic flux in the ring is 10.99 Wb. Hence, this is the required solution.