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3 years ago

A racecar drives at a constant speed down a straight track. The car is in _?_

Physics

1 answer:

vagabundo [1.1K]3 years ago

7 0

Answer:

the answer is

the car is in motion

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<em>Hence, option B is the correct answer.</em>

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3 years ago

Read 2 more answers

A horizontal spring with spring constant of 9.80 N/m is attached to a block with a mass of 1.20 kg that sits on a frictionless s

Andreas93 [3]

Answer:

Explanation:

Given

at certain point displacement and velocity is 0.345 m and 0.54 m/s

Therefore Potential and Kinetic Energy associated is

$U=\frac{1}{2}kx^2$

$U=\frac{1}{2}\times 9.8\times (0.345)^2$

$U=0.583 J$

Kinetic Energy $K=\frac{1}{2}mv^2$

$k=\frac{1}{2}\times 1.2\times (0.54)^2=0.174 J$

Total Energy $T=U+K=0.583+0.174=0.757 J$

Total Energy is conserved hence at maximum displacement all energy will be Potential energy

$T=\frac{1}{2}kA^2$

where A=maximum displacement

$0.757=\frac{1}{2}\times 9.8\times A^2$

$0.1544=A^2$

$A=0.393 m$

Maximum speed occurs at equilibrium Position where Potential Energy is zero

thus $T=\frac{1}{2}mv^2$

$0.757=\frac{1}{2}\times 1.2\times v_{max}^2$

$v_{max}=1.123 m/s$

(c)When block is at 0.2 m from Equilibrium speed then its Potential Energy is

$U=\frac{1}{2}kx^2=\frac{1}{2}\times 9.8\times (0.2)^2=0.196 J$

$T=U+K$

$K=0.757-0.196=0.561 J$

$K=\frac{1}{2}mv^2$

$0.561=\frac{1}{2}\times 1.2\times v^2$

$v=0.966 m/s$

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4 years ago