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Lunna [17]
3 years ago
7

A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. H

ow far does the puck move from rest in 2.25 s?

Physics
1 answer:
lukranit [14]3 years ago
4 0

Answer:

d = 6.32 m

Explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2

Let d is the distance moved in 2.25 s. Using second equation of motion,

d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m

So, it will move 6.32 m from rest in 2.25 seconds.

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Rama09 [41]

Answer:

21 psi

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3 years ago
Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 53 km/h. After six hour
Serhud [2]
For the answer to the question above, first find out the gradient. 

<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>

<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>

<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>

<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
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<span>y = 3x - 9 + 53 </span>
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<span>y is the velocity of the car, x is the time.
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The electronic configuration will be 1s^2

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