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Llana [10]
3 years ago
7

What is the force on an object that goes from 35 m/s to 85 m/s in 20 seconds and has a mass of 148 kg

Physics
1 answer:
Sever21 [200]3 years ago
3 0
F=ma
F = 148×(85-35)÷20
F = 148×(50÷20)
F = 148×2.5
F = 370N
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Answer:

0\:\mathrm{ m/s^2}

Explanation:

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Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.

We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).

We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.

Substituting values in our formula, we have:

\displaystyle a=\frac{25-25}{5}=\frac{0}{5}=\boxed{0\:\mathrm{m/s^2}}

Alternative:

Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!

8 0
2 years ago
A bowling ball is dropped from rest. what is the ball’s velocity after falling for 4 seconds?
muminat

Answer:

it would be 39.2 m/s

Explanation:

After one second, you're falling 9.8 m/s. After two seconds, you're falling 19.6 m/s, and so on.

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Now, we know that L_0 = 29.0m and v = 0.95c, and putting these into (1) we get:

L = 29\sqrt{1-\dfrac{(0.95c)^2}{c^2} }

L = 29\sqrt{1-0.95^2 }

\boxed{L = 9.055m}

Thus, an observer moving at 0.95c observes the space-probe to be 9.055m long.

3 0
3 years ago
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nika2105 [10]

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6 0
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Read 2 more answers
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P = \frac{F}{A}

Replacing F with N'

P = \frac{N'}{A}

P = \frac{\frac{mgcos\theta}{2}}{A}

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m = 58kg

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Replacing we have that

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