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prisoha [69]
3 years ago
10

The speed of a bus increases uniformly from 15 ms per second to 60 ms per second in 20 seconds. calculate 1. the average speed 2

. the acceleration 3. the distance travelled during the entire period
Physics
2 answers:
cluponka [151]3 years ago
7 0

Answer:

Part a)

v_{avg} = 37.5 m/s

Part b)

a = 2.25 m/s^2

Part c)

d = 750 m

Explanation:

Part a)

Since we know that bus is moving uniformly from initial speed of 15 m/s to 60 m/s in 20 seconds

So we will have

v_{avg} = \frac{d}{t}

we know that

d = \frac{v_f + v_i}{2} t

so we have

v_{avg} = \frac{v_f + v_i}{2}

v_{avg} = \frac{15 + 60}{2}

v_{avg} = 37.5 m/s

Part b)

As we know that acceleration is rate of change in velocity

so it is given as

a = \frac{v_f - v_i}{t}

a = \frac{60 - 15}{20}

a = 2.25 m/s^2

Part c)

distance moved by the bus is given as

d = v_{avg} \times t

d = 37.5 \times 20

d = 750 m

joja [24]3 years ago
3 0
Vf= 60m/s
vs = 15m/s
t=20s
a = vf-vs/t
a = 45m/s / 20s
a = 2,25 m/s²
S = vs *t + 1/2 a*t²
S = 15 m/s*20s +1/2 * 2,25m/s² * 400 s²
S = 300m +  450m
S = 750m
v average = 750m/20s
v average = 37,5 m/s
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Answer:

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Explanation:

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The answer is "60.74^{\circ}".

Explanation:

Cavity and benzene should be extended in equal quantities.

\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to  (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\

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