Answer:
98.13m
Explanation:
Complete question
Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building
CHECK THE ATTACHMENT
From the figure, using trigonometry
Tan(θ ) = opposite/adjacent
Where Angle (θ )= 63°
Opposite= X = height of the building
Adjacent= 50 m
Then substitute the values we have
Tan(63)= X/50
1.9626= X/50
X= 1.9626 × 50
X= 98.13m
Hence, the height of the building is 98.13m
F = G*((m sub 1*m sub 2)/r^2)
Answer:
it is B
Explanation:
Because I agree with her and also got 100
Answer:
Explanation:
Acceleration of particle A is 7.3 times the acceleration of particle B.
Let the acceleration of particle B is a, then the acceleration of particle A is
7.3 a.
Let the period of particle A is T and the period of particle B is 2.5 T.
Let the radius of particle A is RA and the radius of particle B is RB.
Use the formula for the centripetal force
So, 
The ratio of radius of A to the radius of B is given by
RA : RB = 1.17
The work done on a 50 Kg student by the elevator in 2 seconds if the elevator is accelerating upwards from rest at a rate of 2 ms⁻² would be
<h3>What is work done?</h3>
The total amount of energy transferred when a force is applied to move an object through some distance.
The work done is the multiplication of applied force with displacement.
Work Done = Force × Displacement
As given in the problem we have to find the work done on a 50 Kg student by the elevator in 2 seconds if the elevator is accelerating upwards from rest at a rate of 2 ms⁻²,
Lets us first calculate the displacement of the elevator in 2 seconds if it is accelerating upwards from rest at a rate of 2 ms⁻².
s = ut + 0.5at²
= 0 + 0.5×2×2²
= 4 meters
The work done on the elevator = mgh
=50×9.81×4
=1962 joules
Thus, the work done on the elevator would be 1962 joules.
To learn more about work done, refer to the link;
brainly.com/question/13662169
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