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Inessa05 [86]
3 years ago
12

An 880n box is pushed across a level floor for a distance of 5.0m with a force of 440n. how much work was done on the box

Physics
1 answer:
kupik [55]3 years ago
5 0
W=Fd, F=440N, d=5m. 5x440=2,200J of work
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A motor exerts a force of 12,00N to lift an elevator 8.0m in 7.0secs what is the power produced by the motor?
SashulF [63]

Answer:

1371.4watt

Explanation:

from power=energy/time

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2 years ago
Which of Newton's Three Laws Applies? Law 1, 2, or 3?
frozen [14]

Answer:

1. Newton's first law

2.Newton's second law

3.Newton's third law

Explanation:

1. Newton's first law stated, In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force... this is base of the concept of inertia.

2. Newton's second law stated, In an inertial frame of reference, the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration a of the object: F = ma, or in easier words, F is directly proportional to a.

3. Newton's third law stated, When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body., In this case, the Normal Are opposite with gravititional force.

7 0
3 years ago
A ball is thrown horizontally from the top of a 55 m building and lands 150 m from the base of the building. Ignore air resistan
PtichkaEL [24]

Answer:

a) t =3.349 s

b) V_x,i = 44.8 m/s

c) V_y,f = 32.85 m/s

d)  V = 55.55 m/s

Explanation:

Given:

- Total throw in x direction x(f) = 150 m

- Total distance traveled down y(f) = 55 m

Find:

a) How long is the rock in the air in seconds.  

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

                                 y(f) = y(0) + V_y,i*t + 0.5*g*t^2

- V_y,i = 0 (horizontal throw)

                                 55 = 0 + 0 + 0.5*(9.81)*t^2

                                 t = sqrt ( 55 * 2 / 9.81 )

                                 t =3.349 s

- Use the second equation of motion in x direction:

                                 x(f) = x(0) + V_x,i*t

                                 150 = 0 + V_x,i*3.349

                                  V_x,i = 150 / 3.349 = 44.8 m/s

- Use the first equation of motion in y direction:

                                 V_y,f = V_y,i + g*t

                                 V_y,f = 0 + 9.81*3.349

                                 V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

                                 V^2 = V_y,f^2 + V_x,i^2

                                 V = sqrt (32.85^2 + 44.8^2)

                                 V = 55.55 m/s

5 0
3 years ago
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Akimi4 [234]
D.) White Dwarf

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Hope this helps!
3 0
3 years ago
Read 2 more answers
How does changes in distance affect the gravitational pull between two objects? Describe and give one example.
maxonik [38]
The formula is

F_grav = G * m1 * m2 / r^2

G m1 and m2 are going to stay the same once chosen no matter what the distance is. The only thing that will change is the distance.

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As the distance decreases, the gravitational force will Increase.

The relationship is inverse. The moon travelling around the earth is one example. The earth travelling around the sun is another.
8 0
3 years ago
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