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2 years ago

How long would it take a boat that started at rest to get to a velocity of 56 m/s if it was accelerating at 17 m/s/s?

Physics

1 answer:

noname [10]2 years ago

8 0

Answer:

3.29 s

Explanation:

Using v = u + at where u = initial velocity of boat = 0 m/s since it starts at rest, v = final velocity of boat = 56 m/s, a = acceleration of boat = 17 m/s² and t = time taken to accelerate to 56 m/s.

So, v -= u + at

t = (v - u)/a

substituting the values of the variables, we have

t = (56 m/s - 0 m/s)/17 m/s²

= 56 m/s ÷ 17 m/s²

= 3.29 s

It will take the boat 3.29 s to accelerate to 56 m/s at an acceleration of 17 m/s²

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Explanation:

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Usage of the equipment properly and also increasing the sample size.
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3 years ago

Read 2 more answers

A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in t

Margarita [4]

Answer: 996m/s

Explanation:

Formula for calculating velocity of wave in a stretched string is

V = √T/M where;

V is the velocity of wave

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Since the second wire is twice as far apart as the first, it will be L2 = 2L1

Let V1 and V2 be the speed of the shorter and longer wire respectively

V1 = √T/M1... 1

V2 = √T/M2... 2

Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1

The equations will now become

249 = √T/(m/L1) ... 3

V2 = √T/(m/2L1)... 4

From 3,

249² = TL1/m...5

From 4,

V2²= 2TL1/m... 6

Dividing equation 5 by 6 we have;

249²/V2² = TL1/m×m/2TL1

{249/V2}² = 1/2

249/V2 = (1/2)²

249/V2 = 1/4

V2 = 249×4

V2 = 996m/s

Therefore the speed of the wave on the longer wire is 996m/s

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2 years ago

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fredd [130]

Answer:

The correct option is;

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Similarly, when two like charges are brought closer together, the potential energy, or the energy available to push the two like charges apart increases charge as the as the charges are brought closer together

Therefore, the correct option is the potential energy between both like charges and like poles increases as they move closer together.

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