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LiRa [457]
1 year ago
14

A car travels West at 35 mph for a time period of 2.5 hours and then travels West at 55 mph for an additional time of 4.0 hours.

How far will this car be from its starting point at the end of the journey?
Pls show work
Physics
1 answer:
OlgaM077 [116]1 year ago
7 0

The total distance travelled is 307.5 miles.

<h3>What is the distance covered?</h3>

We know that the distance covered can be obtained from the formula from speed since the motion all took place in the same direction.

Speed = distance/time

distance = speed * time

In the first case;

Distance =  35 mph  *  2.5 hours = 87.5 miles

In the second case;

Distance =  55 mph * 4.0 hours = 220 miles

Total distance = 87.5 miles + 220 miles = 307.5 miles

Hence the total distance travelled is 307.5 miles.

Learn more about total distance:brainly.com/question/20330990

#SPJ1

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What is the gravitational potential energy of a tiger that weighs 200 N standing on a rock that is 2 m above the ground?
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A block with a mass of 8.7 kg is dropped from rest from a height of 8.7 m, and remains at rest after hitting the ground. 1)If we
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On a horizontal frictionless floor, a worker of weight 0.900 kN pushes horizontally with a force of 0.200 kN on a box weighing 1
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Answer:

D) The worker will accelerate at 2.17  m/s²  and the box will accelerate at 1.08  m/s² , but in opposite directions.

Explanation:

Newton's third law

Newton's third law or principle of action and reaction states that when two interaction bodies appear equal forces and opposite directions. in each of them.

F₁₂= -F₂₁

F₁₂: Force of the box on the worker

F₂₁: Force of the worker on the box

Newton's second law

∑F = m*a

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Formula to calculate the mass (m)

m =  W/g

Where:

W : Weight (N)

g : acceleration due to gravity  (m/s²)

Data

W₁ =1.8 kN   : box weight

W₂ = 0.900 kN : worker weight

F₂₁ = 0.200 kN

F₁₂ = - 0.200 kN

g = 9.8 m/s²

Newton's second law for the box

∑F = m*a

F₂₁ = m₁*a₁    m₁=W₁/g

0.2 kN = (1.8kN)/(9.8 m/s² ) *a₁

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{1.8 kN}

a₁= 1.08 m/s² : acceleration of the box

Newton's second law for the worker

∑F = m*a

F₁₂ = m₂*a₂ , m₂=W₂/g

- 0.2 kN =( (0.9 kN) /(9.8 m/s² ) )*a₂

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{0.9 kN}

a₂=  -2.17 m/s² : acceleration of the worker

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3 years ago
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