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Charra [1.4K]
3 years ago
14

A car moving at a speed of 10 m/s enters a Q. highway and accelerates at 4 m/s2. How fast will the car be moving after it accele

rated for 60m?
answer choices
580 m/s
4.8 m/s
24 m/s
480 m/s m​
Physics
1 answer:
geniusboy [140]3 years ago
5 0
  • Initial velocity=u=10m/s
  • Final velocity=v
  • Acceleration=a=4m/s^2
  • Distance=s=60m

According to third equation of kinematics

\\ \rm\rightarrowtail v^2=u^2+2as

\\ \rm\rightarrowtail v^2=10^2+2(4)(60)

\\ \rm\rightarrowtail v^2=100+480

\\ \rm\rightarrowtail v^2=580

\\ \rm\rightarrowtail v\approx 24m/s

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ANSWER

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Given:

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\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}

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\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

T_1\cos 30\degree-m\cdot g=0

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T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N

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T_2-T_1\sin 30\degree=0

Solve for T₂,

T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

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