When I divided by 6, I got W(moon) = mg(moon) = 10 x 10/6 = 16.7N (the moon's gravity is only 1/6 that of Earth in this case). 100N is equal to W(Earth) = mg(Earth) = 10 x 10. On the moon and the Earth, m = 10 kg.
How can you determine the strength of Earth's gravitational field at the moon?
M = 6 10 24 K g is the mass of the earth, and g = - G M R 2 is the gravitational field of the earth at the point of the moon. The distance between the earth and moon is 84 10 8 meters. field of gravitation.
How do you determine the distance between Earth and the moon where the gravitational field is at its weakest?
You only discover points where the forces from all of the different local bodies are in balance, or equal to each other, because there is no place in the universe where the gravitational field intensity is zero. All bodies, not just the Earth and Moon, must be taken into account in the computation to arrive at these positions.
To know more about gravitational field strength at moon visit;
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The question is incomplete. Here is the entire question.
A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?
Answer: Δx = - 42m
Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.
For this "type" of motion, displacement (Δx) can be determined by:
is the initial velocity
a is acceleration and can be positive or negative, according to the referential.
For Referential, let's assume rightward is positive.
Calculating displacement:
= - 42
Displacement of the boat for t=6.0s interval is = - 42m, i.e., 42 m to the left.
Answer:
Coefficient of dynamic friction= md= 0.09931
Explanation:
To determine the coefficient of dynamic friction we must first match the friction force that is permendicular to the normal force of the block and opposite to the drag force, to the component of the drag force in this same direction. This component on the X axis of the drag force will be:
F= 90N × cos(30°) = 77.9423N
This component on the X axis of the drag force must be equal to the dynamic friction force that is equal to the coefficient of dynamic friction by the normal force of the block weight:
F= md × m × g= 77.9423N
m= mass of the block
md= coefficient of dynamic friction
g= gravity acceleration
F= md × 80kg× 9.81 (m/s²)= 77.9423(kg×m/s²)
md= (77.9423(kg×m/s²) / 784.8 (kg×m/s²)) = 0.09931
Answer:
Work is done by friction = -165 J
Explanation:
Given:
Mass of block (m) = 15 kg
Ramp inclined = 28°
Friction force (f) = 30 N
Distance (d) = 5.5 m
Find:
Work is done by friction.
Computation:
Work is done by friction = -Fd
Work is done by friction = -(30)(5.5)
Work is done by friction = -165 J
Answer:
0.5 m.
Explanation:
From the question given above, the following data were obtained:
Force (F) = 5 N
Work done (W) = 2.5 J
Distance (s) =?
Workdone is simply defined as the product of force (F) and distance (s) moved in the direction of the force. Mathematically, it is expressed as:
Work done (W) = Force (F) × Distance (s)
W = F × s
With the above formula, we can obtain the distance to which the book has moved as follow:
Force (F) = 5 N
Work done (W) = 2.5 J
Distance (s) =?
W = F × s
2.5 = 5 × s
Divide both side by 5
s = 2.5/5
s = 0.5 m
Therefore, the book moved 0.5 m when the force was applied.
