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KengaRu [80]
2 years ago
13

Suppose an object is moving in a straight line at 50 mi/hr. According to Newton's first law of motion, the object will:

Physics
1 answer:
NeX [460]2 years ago
7 0

Answer:

The object will continue to move in straight line at 50 mi/hr except it is acted upon by an external force

Explanation:

Newton First Law of motion: A body will continue in its present state of  rest or, if it is in motion will continue to move with uniform speed in a straight line unless it is acted upon by a force. Newton first law of motion is also called the law of inertia.

Inertia: This is the tendency of a body to remain in its states of rest or uniform motion.

From the question, The object moving in a straight line at 50 mi/hr will continue to move in a straight line at 50 mi/hr, except it is acted upon by external force that will change the speed of the object.

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A car is traveling at 21.0 m/s. It slows to a stop at a constant rate over 5.00s. How far does the car travel during those 5.00
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Answer:

d = 105 m

Explanation:

Speed of a car, v = 21 m/s

We need to find the distance traveled by the dar during those 5 s before it stops. Let the distance is d. It can be calculated as :

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d = 21 m/s × 5 s

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So, it will cover 105 m before it stops.

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A heavy box weighing 1000 newtons sits on the floor. You lift upward on the box with a force of 450 newtons, but the box does no
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You push a bike pedal 10cm and the bike moves 100cm. Calculate the Speed ratio ( show all work ) ( use the formula and divide th
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2 years ago
A fully loaded cart with a mass of 2200 kg starts from the top of a 12-meter hill on a roller coaster.
Salsk061 [2.6K]

Answer:

A. potential energy is 258720 Joule

Explanation:

A.Gravitational potential energy is: PE = m × g × h

velocity =  15.33 m/s when the car reaches the bottom of the hill.

where, m = mass

            g = acceleration due to gravity

            h = height from the bottom of hill.

The potential energy is : m×g×h

                                     =(2200×9.8×12)

                                     =258720 Joule

B. at the bottom of the hill, the potential energy is converted into kinetic energy so PE at top = KE at bottom

                    kinetic energy= \frac{1}{2}(m*v^{2})

where v = velocity

          m= mass

therefore,               v=\sqrt\frac{2*K.E}{m} {}

                         or,  v=\sqrt{\frac{2*258720}{2200} }

                         or,   v=15.33 m/s

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3 years ago
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