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Tcecarenko [31]
3 years ago
10

a bicycle accelerates at 1 m/s² from an initial velocity of 4 m/s² for 10s. Find the distance moved by it during this interval o

f time?​
Physics
1 answer:
BaLLatris [955]3 years ago
6 0

Answer:

90 m

Explanation:

Acceleration, a=\frac {v-u}{t} where v and u are final and initial velocities respectively, t is the time taken

Substituting 1 m/s^{2} for a,  4 m/s for u and 10 s for t then

1*10=v-4

v=14 m/s

From kinematic equations

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

Making s the subject then

s=\frac {v^{2}-u^{2}}{2a}=\frac {14^{2}-4^{2}}{2\times 1}=90 m

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(b) A car of mass 3000 kg travels at a constant velocity of 5.0 m/s.
Tamiku [17]

Answer:

16.7 s

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T= <u>Vf - Vo</u>          a= <u>F</u>

         a                    m

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4 0
3 years ago
the electrical resistance of copper wire varies directly with its length and inversely with the square of the diameter of the wi
leonid [27]

Answer:24.47

Explanation:

Given

L_1=30 m

d_1=3 mm

R_1=25 \Omega

L_2=40 m

d_2=3.5 mm

we know Resistance R=\frac{\rho L}{A}

Where R=resistance

\rho =resistivity

L=Length

A=area of cross-section

A=\frac{\pi d^2}{4}

Thus R\propto \frac{L}{d^2}

therefore

R_1\propto \frac{L_1}{d_1^2}--------1

R_2\propto \frac{L_2}{d_2^2}--------2

divide 1 and 2 we get

\frac{R_1}{R_2}=\frac{L_1}{L_2}\times \frac{d_2^2}{d_1^2}

\frac{R_1}{R_2}=\frac{30}{40}\times \frac{3.5^2}{3^2}

R_2=25\times 0.734\times 1.33

R_2=24.47 \Omega

8 0
3 years ago
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