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11Alexandr11 [23.1K]
2 years ago
10

Please help me out !!

Physics
1 answer:
blagie [28]2 years ago
4 0
I think the answer would be letter a
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You are driving a 2400.0-kg car at a constant speed of 14.0 m/s along a wet, but straight, level road. As you approach an inters
olya-2409 [2.1K]

Answer:0.43

Explanation:

Given

mass of car m=2400 kg

Speed of car u=14 m/s

Distance traveled before coming to halt s=23.2 m

Let \muthe coefficient of friction

Maximum deceleration road can provide during motion is

a=\mu g

using v^2-u^2=2 as

0-14^2=2\cdot (-\mu g)\cdot 23.2

\mu =\frac{196}{454.72}

\mu =0.431

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3 years ago
Would earth orbiting the sun be known as
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C. Newton’s Third Law of Motion.

Because...
Newtons third law implies conversation of momentum it can also be seen as following from the second law: when one object pushes a second object at some point of contact using an applied force, there must be an equation of opposite force from the second object that cancels the applied force. Otherwise, there would be a nonzero net force on a massless point which, by the second law, would accelerate the point of contact by an infinite amount.
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3 years ago
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Populations of organisms that reproduce asxually tend to show less variation from one individual to another.Why might this be?
algol13

Answer:

This might be because since they reproduce without mating, the genes of their offspring have are not mixed with anyone elses, making them have very little variation in contrast with those that reproduce our way.

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A student conducts an experiment to determine how the temperature of water affects the time for sugar to dissolve. In each trial
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Answer:

DGFGDFDDF

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4 0
3 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.80×106 N , one at an angle 14.0 ∘ west of north,
Leviafan [203]

Answer:

W = 1,049 10⁹ J

Explanation:

Work is defined by the relation

         W = F. d = F d cos θ

where tea is the angle between the forces and the displacement.

The total work is the sum of the work of each tug.

Tug 1

       W₁ = F d cos θ₁

 

the angle measured from the positive side of the x-axis is

       θ₁ = 14 + 90 = 104º

           

tugboat 2

             W₂ = F d cos θ₂

             θ₂ = 14

we substitute

             W = F d cos θ₁ + F d cos θ₂

             W = F d (cos θ₁ + cos θ₂)

               

let's calculate

             W = 1.80 10⁶  800 (cos 104 + cos 14)

             W = 1,049 10⁹ J

5 0
3 years ago
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