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kkurt [141]
3 years ago
6

When a force is applied to an object for an amount of time, it is known as __________.

Physics
2 answers:
stiks02 [169]3 years ago
8 0

Answer:

Impulse

Explanation:

I took the test and got it correct.

Hope this helps! :)

NikAS [45]3 years ago
7 0

The correct answer to the question is : D) Impulse

EXPLANATION:

Before going to answer this question, first we have to understand impulse.

Impulse of a body is defined as change in momentum or the product of force with time.

Mathematically impulse = F × t = m ( v - u ).

Here, v is the final velocity

          u is the initial velocity

          F is the force acting on the body for time t.

Hence, the perfect answer of this question is impulse m i.e the force multiplied with time is known as impulse.


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A motorcycle moving at 18 m/s decelerates at a rate of 3.6 m/s². What is the car's
Gekata [30.6K]

Answer:

V = 3.6 m/s

Explanation:

Given:

V₀ = 18 m/s

a = - 3,6 m/s²  (The motorcycle is slowing down!)

t = 4 s

____________

V - ?

Motorcycle speed:

V = V₀ + a·t

V = 18 + (-3.6)·4 = 3.6 m/s

5 0
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An aircraft, traveling northward, lands on a runway with a speed of 77 m/s. Once it touches down, it slows to 6.3 m/s over 705 m
KengaRu [80]

Answer:

magnitude

a = 4.18 m/s^2

Direction

Opposite to the velocity of aircraft

Explanation:

As we know that the initial speed of the aircraft when it land on the ground is

v_i = 77 m/s

now the final speed of the aircraft when it covers 705 m on runway is given as

v_f = 6.3 m/s

so here we know that

v_f^2 - v_i^2 = 2 a d

so now we can say that

6.3^2 - 77^2 = 2ad

6.3^2 - 77^2 = 2(a)(705)

a = -4.18 m/s^2

5 0
3 years ago
Give at least two examples where people commonly use temperature to control reaction rates.
wariber [46]
The examples may be;
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2 years ago
Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was
True [87]

This problem is looking for the minimum value of μs that is necessary to achieve the record time. To solve this problem:


Assuming the front wheels are off the ground for the entire ¼ mile = 402.3 m, the acceleration a = µs·9.8 m/s².


For a constant acceleration, distance = 402.3


m = 1/2at^2 = 804.6 m / (4.43 s)^2 = a = µs·9.8 m/s^2



µs = 804.6 m / (4.43s)^2 / 9.8 m/s^2 = 4.18

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