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Pie
3 years ago
5

If the 50 kg boy were in a spacecraft 5.0r from the center of the earth, what would his weight be?

Physics
1 answer:
Lady_Fox [76]3 years ago
7 0
·The acceleration of gravity is proportional to

                       1 / (the square of the distance from the center) .

When we're on the surface, we're 1 radius from the center of the Earth,
and the acceleration of gravity is  9.8 m/s² .

The boy's weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)

                                                             =      490 newtons .

At the distance of 5 radii from the center (4 radii altitude from the surface),
the acceleration of gravity is

                 (9.8 m/s²) · (1/5)²  =  0.39 m/s² .

The boy's weight is    (mass) · (gravity)  =  (50kg) · (0.39 m/s²)

                                                                  =     19.6 newtons .

Just as we expected, his weight at that distance is

         (19.6 / 490) = 0.04  =  1/25  =  1/5²  of his weight on the surface.
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A ball is projected horizontally from the top of a cliff. At the same moment, a second identical ball is dropped from rest from
almond37 [142]

Answer:3

Explanation:

First ball is thrown with horizontal velocity while other ball is dropped from cliff such that both have zero vertical velocity. So both balls have to cover a distance equal to the height of cliff with same initial velocity.

time taken is given by t=\sqrt{\frac{2h}{g}}

where h=height of cliff

g=acceleration due to gravity

horizontal velocity to first ball will make the ball to travel more horizontal distance as compared to second ball.

Option 3 is correct

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3 years ago
Suppose a bicycle was coasting on a level surface, and there was no friction. What would happen to the bicycle?
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3 0
3 years ago
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Calculate the watts of power expended when a force 400 N travels 12 meters over 10 seconds.
Pachacha [2.7K]

Answer:

Explanation:

Example

7 0
3 years ago
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons
Virty [35]

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

H_m=1.65m

H_E=1.16307m

Explanation:

From the question we are told that

Mass of ball M=2kg

Length of string L= 2m

Wind force F=13.2N

Generally the equation for \angle \theta is mathematically given as

tan\theta=\frac{F}{mg}

\theta=tan^-^1\frac{F}{mg}

\theta=tan^-^1\frac{13.2}{2*2}

\theta=73.14\textdegree

Max angle =2*\theta= 2*73.14=>146.28\textdegree

Generally the equation for max Height H_m is mathematically given as

H_m=L(1-cos146.28)

H_m=0.9(1+0.8318)

H_m=1.65m

Generally the equation for Equilibrium Height H_E is mathematically given as

H_E=L(1-cos73.14)

H_E=0.9(1+0.2923)

H_E=1.16307m

8 0
2 years ago
Point charge q1 of 30 nC is separated by 50 cm from point charge q2 of -45 nC. As shown in the diagram, point a is located 30 cm
Angelina_Jolie [31]

Answer:

E1 =  2996.667N/C E2 = 11237.5N/C

Explanation:

E1 = kQ1/r^2

  =8.99 x 10^9 x 30 x 10^-9/(30x10^-2)^2

  = 2996.667N/C

E2 = kQ2/r^2

      = 8.99 x 10^9 x 50 x 10^-9/(20x10^-2)^2

      = 11237.5N/C

The direction are towards the point a

6 0
3 years ago
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