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m_a_m_a [10]
3 years ago
14

A driver of a car traveling at -15m/s applies the brakes, causing a uniform acceleration of +2.0m/s2. If the brakes are applied

for 2.5s, what is the velocity of the car at the end of the braking period? How far has the car moved during the braking period?
Physics
1 answer:
klemol [59]3 years ago
3 0

Given :

Initial velocity, u = -15 m/s.

Acceleration , a = 2 m/s².

Time taken to applied brake, t = 2.5 s.

To Find :

The velocity of the car at the end of the braking period.

How far has the car moved during the braking period.

Solution :

By equation :

v = u+at\\\\v=-15 + 2\times 2.5\\\\v=-10 \ m/s

Now, distance covered by car is :

s=ut+\dfrac{at^2}{2}\\\\s=(-15)(2.5)+\dfrac{2(2.5)^2}{2}\\\\s=-31.25\ m

Hence, this is the required solution.

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Answer:

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Circular Motion A 650-kg car moving at 8.5 m/s takes a turn around a circle with a radius of 48.0 m. Determine the acceleration
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Answer:

Explanation:

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a=\frac{v^2}{r} and filling in

a=\frac{(8.5)^2}{48.0} and we need 2 significant digits in our answer. That means that

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8 0
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Which scientist was involved in the War of Currents?<br> Help
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6 0
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A 30-km, 34.5-kV, 60-Hz, three-phase line has a positive-sequence series impedance z 5 0.19 1 j0.34 V/km. The load at the receiv
zmey [24]

Answer:

(a) With a short line, the A,B,C,D parameters are:

    A = 1pu    B = 1.685∠60.8°Ω    C = 0 S    D = 1 pu

(b) The sending-end voltage for 0.9 lagging power factor is 35.96 KV_{LL}

(c) The sending-end voltage for 0.9 leading power factor is 33.40 KV_{LL}

Explanation:

(a)

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Compare the line equations with the A,B,C,D parameter equations.

(b)

Determine the receiving-end current for 0.9 lagging power factor.

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Determine the sending end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

(c)

Determine the receiving-end current for 0.9 leading power factor.

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