300N/25 kg= divide them for the answer
Answer:
0 N
Explanation:
Applying,
F = qvBsin∅................. Equation 1
Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.
From the question,
Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°
Substitute these values into equation 1
F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)
Since sin0° = 0,
Therefore,
F = 0 N
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The new gravitation force at the new location is 40 N
Explanation:
The weight of the astronaut is given by the equation
(1)
where
m is the mass of the astronaut
g is the acceleration of gravity
The acceleration of gravity at a certain distance
from the centre of the Earth is given by

where G is the gravitational constant and M is the Earth's mass. So we can rewrite eq.(1) as

When the astronaut is on the Earth's surface,
(where R is the Earth's radius), so his weight is

Later, he moves to another location where his distance from the Earth's surface is 3 times the previous distance, so the new distance from the Earth's centre is

Therefore, the new weight is

Which means that his weight has decreased by a factor 16: therefore, the new weight is

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Answer:
5.95 m
Explanation:
Given that the biggest loop is 40.0 m high. Suppose the speed at the top is 10.8 m/s and the corresponding centripetal acceleration is 2g
For the car to stick to the loop without falling down, at the top of the ride, the centripetal force must be equal to the weight of the car. That is,
(MV^2) / r = mg
V^2/ r = centripetal acceleration which is equal to 2g
2 × 9.8 = 10.8^2 / r
r = 116.64 /19.6
r = 5.95 m