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pishuonlain [190]
4 years ago
7

What should you keep in mind when selecting a ball hitch and coupler for towing your boat?

Physics
1 answer:
mrs_skeptik [129]4 years ago
7 0
<span>There is only one thing that you must keep in mind, and that is that the ball hitch and coupler for towing have to be the same size. If not, then you can create a problem which can in turn result with damage to you and your boat and you would have to spend a lot of money to fix things or get a new boat.</span>
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A 62.0 kg sprinter starts a race with an acceleration of 1.44 m/s2. If the sprinter accelerates at that rate for 30 m, and then
Dominik [7]

Answer:

The time taken for the race is 17.20 s.

Explanation:

It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.

Calculate the final speed of the sprinter.

The expression for the equation of the motion is as follows;

v^{2}-u^{2}= 2as

Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.

Put u= 0, s=30 m and a= 1.44 ms^{-2}.

v^{2}-(0)^{2}= 2(1.44)(30)

v= 9.3 m^{-1}

Calculate time taken to cover 30 m distance.

The expression for the equation of motion is as follows;

s=ut+\frac{(1)}{2}at^2

Put u= 0, s=30 m and a= 1.44 ms^{-2}.

30=(0)t+\frac{(1)}{2}(1.44)t^2

t=6.45 s

Calculate the time taken to complete his race.

T= t+t'

Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.

T=6.45+\frac{s'}{v}

Put s= 30 m, v= 9.3 m^{-1} and s'= 100 m.

T=6.45+\frac{(100)}{9.3}

T= 17.20 s

Therefore, the time taken for the race is 17.20 s.

6 0
4 years ago
The power in an electrical circuit is given by the equation P= RR, where /is
Kobotan [32]

Answer:

The power in an electrical circuit is given by the equation P

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Explanation:

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Explanation:

The newton (symbol: N) is the International System of Units (SI) derived unit of force.

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<em>friction transforms KE into thermal energy (a)</em>

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