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zimovet [89]
3 years ago
13

A body moving with a velocity of 20 m/s begins to accelerate at 3 m/s2. How far does the body move in 5 seconds?

Physics
2 answers:
sveta [45]3 years ago
5 0


Initial velocity = 20m/s
Accelerations = 3m/s^2
Time taken = 5s


Using second equation of motion
ut+(1/2)at^2


Where, s is the distance covered by the body
u is initial velocity
a is acceleration and t is time taken by that body

= (20)(5)+(1/2)(3)(5)^2
=100 + 37.5
= 137.5


So, the total distance covered by the body in next 5s would be 137.5m
Arlecino [84]3 years ago
5 0

The Correct answer to this question for Penn Foster Students is: 137.5 m

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You drag a suitcase of mass 8.2 kg with a force of f at an angle 41.9 ◦ with respect to the horizontal along a surface with kine
DedPeter [7]

Answer:

35.6 N

Explanation:

We can consider only the forces acting along the horizontal direction to solve the problem.

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where \mu=0.33, m=8.2 kg and g=9.8 m/s^2.

The two forces have opposite directions (because the frictional force is always opposite to the motion), and their resultant must be zero, because the suitcase is moving with constant velocity (which means acceleration equals zero, so according to Newton's second law: F=ma, the net force is zero). So we can write:

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8 0
3 years ago
Two motorcycles are traveling in opposite directions at the same speed when one of the cyclists blasts her horn, which has frequ
Galina-37 [17]

Answer:

6ms^-1

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Given that the frequency difference is

( 563- 544) = 19

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and there are 19 of this waves

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5 0
3 years ago
The graph below shows the velocity of a car over time.
jekas [21]

Answer:

D. Calculate the area under the graph.

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You can think of such a calculation as determining the area of a rectangle whose sides are velocity and time period. If you make the time period very very small, the rectangle will become a narrow "bar" - a bar with height determined by the average velocity during that corresponding short period of time. The area is, again, the distance made during that time. Now, you can cover the entire area under the curve using such narrow bars. Their areas adds up, approximately, to the total distance made over the entire span of motion. From this you can already see why the answer D is the correct one.

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4 years ago
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