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mixer [17]
3 years ago
5

A cheetah can run at approximately 100 km/hr and a gazelle at 80.0 km/hr. If both animals are running at fullspeed, with the gaz

elle 70.0 m ahead, how long before the che
Physics
1 answer:
brilliants [131]3 years ago
6 0

Answer:

12.6 seconds

Explanation:

With respect to the gazelle, the cheetah has a relative speed of 100 - 80 = 20 km/hr.

We convert from km/hr to m/s.

\dfrac{20000\text{ m}}{3600\text{ s}} = \dfrac{50}{9}\text{ m/s}

At this relative speed, the cheetah will overtake the gazelle in time <em>t</em> given by

v= \dfrac{d}{t}

<em>d</em> is the distance and <em>v</em> is the relative speed.

t = \dfrac{d}{v} = \dfrac{70.0}{50/9} = 12.6\text{ s}

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Rudik [331]

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3 years ago
What does the area under the velocity-time graph represent.
V125BC [204]

Answer:

The distance traveled!

Explanation:

This is a velocity time graph of an object moving in a straight line due North.

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3 years ago
The speedometer in your car reads 55mi/h this represents ___ of the car
Readme [11.4K]

instantaneous speed of the car

8 0
3 years ago
A bicycle is ridden along a horizontal road with a driving force of 400, n,400n. its speed is constant at 12, m, slash, s,12m/s.
mr_godi [17]

The magnitude of the sum of the frictional forces acting on the bike and its rider is 400N.

<h3>What is friction force?</h3>

The friction force is the opposing force which acts on the object which is in relative motion.

The driving force is equal and opposite to the friction force acting between road and bicycle.

Friction force = 400N

The friction force between rider and bike is zero.

So the magnitude of sum of friction force = 400N +0 = 400N

Thus, the magnitude of the sum of the frictional forces acting on the bike and its rider.

Learn more about friction force.

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4 0
2 years ago
A 70 ft rope hangs from a helicopter above this room. The rope has a mass per unit length of 2 lb/ft. In order to be rescued fro
Mrac [35]

Answer:

The work done to get you safely away from the test is  2.47 X 10⁴ J.

Explanation:

Given;

length of the rope, L = 70 ft

mass per unit length of the rope, μ = 2 lb/ft

your mass, W = 120 lbs

mass of the 70 ft rope  = 2 lb/ft x 70 ft

                                         = 140 lbs.

Total mass to be pulled to the helicopter, M = 120 lbs  + 140 lbs  

                                                                       = 260 lbs

The work done is calculated from work-energy theorem as follows;

W = Mgh

where;

g is acceleration due gravity = 32.17 ft/s²

h is height the total mass is raised = length of the rope = 70 ft

W = 260 Lb x 32.17 ft/s²  x 70 ft

W = 585494 lb.ft²/s²

1 lb.ft²/s² = 0.0421 J

W = 585494 lb.ft²/s²  = 2.47 X 10⁴ J.

Therefore, the work done to get you safely away from the test is  2.47 X 10⁴ J.

4 0
3 years ago
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