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kumpel [21]
4 years ago
9

A 2498 kg car is moving at 17.1 m/s slams on its brakes and slows to 2.6 m/s. What is the magnitude (absolute value) of the impu

lse the car experiences during this time?
Physics
1 answer:
bija089 [108]4 years ago
8 0

Answer:

<em>J=36221 Kg.m/s</em>

Explanation:

<u>Impulse-Momentum Theorem</u>

These two magnitudes are related in the following way. Suppose an object is moving at a certain speed v_1 and changes it to v_2. The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by

p=mv

The initial and final momentums are, respectively

p_1=mv_1,\ p_2=mv_2

The change of momentum is

\Delta p=p_2-p_1=m(v_2-v_1)

It is numerically equal to the Impulse J

J=\Delta p

J=m(v_2-v_1)

We are given

m=2498\ kg,\ v_1=17.1\ m/s,\ v_2=2.6\ m/s

The impulse the car experiences during that time is

J=2498(2.6-17.1)=2498(-14.5)

J=-36221 Kg.m/s

The magnitude of J is

J=36221 Kg.m/s

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Ksju [112]

Answer:

No. of laps of Hannah are 7 (approx).

Solution:

According to the question:

The total distance to be covered, D = 5000 m

The distance for each lap, x = 400 m

Time taken by Kara, t_{K} = 17.9 min = 17.9\times 60 = 1074 s

Time taken by Hannah, t_{H} = 15.3 min = 15.3\times 60 = 918 s

Now, the speed of Kara and Hannah can be calculated respectively as:

v_{K} = \frac{D}{t_{K}} = \frac{5000}{1074} = 4.65 m/s

v_{H} = \frac{D}{t_{H}} = \frac{5000}{918} = 5.45 m/s

Time taken in each lap is given by:

(v_{H} - v_{K})t = x

(5.45 - 4.65)\times t = 400

t = \frac{400}{0.8}

t = 500 s

So, Distance covered by Hannah in 't' sec is given by:

d_{H} = v_{H}\times t

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No. of laps taken by Hannah when she passes Kara:

n_{H} = \frac{d_{H}}{x}

n_{H} = \frac{2725}{400} = 6.8 ≈ 7 laps

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3 years ago
How can a large force result in a relatively small power?
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Answer:

hey mate here is your answer

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