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Verdich [7]
2 years ago
15

2. A car accelerates uniformly from +10.0 m/s to +50.0 m/s over a distance of 225 m. How long did it take to go that distance? S

how all your work, including the equation used, given and unknown quantities, and any algebra required. Make sure your answer has the correct number of significant figures.
Physics
1 answer:
artcher [175]2 years ago
7 0
Let's call the constant acceleration a.
At a time t, its speed will thus be v(t)=a*t+v0 where v0 is its initial speed, here 10 m/s. Hence v(t)=a*t+10.

From there we can deduce the position P(t)=a*t^2/2+10t+p0 where p0 is the initial position, here 0.

Hence P(t)=a*t^2/2+10t

Let's call T the time at which it's at 50 m/s, we know that P(T)=225m and that v(T)=50 m/s hence a*T+10=50 thus a=40/T and P(T)=(40/2+10)T=30T

Hence T=225/30=7.5

It took 7.5 seconds


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a stationary object stays still
a moving object continues to move at the same speed and in the same direction
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(b) Suppose at a certain instant the kinetic energy is twice the elastic potential energy. Write an equation describing this sit
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1/2mv² = ke²

Explanation:

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Its elastic potential energy will be the work done on the spring when stretched which is equal to 1/2kx²

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8 0
3 years ago
Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1200 km/s , measured relative to the
Radda [10]

Answer:

The maximum electrical force is 2.512\times10^{-2}\ N.

Explanation:

Given that,

Speed of cyclotron = 1200 km/s

Initially the two protons are having kinetic energy given by

\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2

When they come to the closest distance the total kinetic energy is converts into potential energy given by

Using conservation of energy

mv^2=\dfrac{kq^2}{r}

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r=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{1.67\times10^{-27}\times(1200\times10^{3})^2}

r=9.57\times10^{-14}\ m

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Using formula of force

F=\dfrac{kq^2}{r^2}

F=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{(9.57\times10^{-14})^2}

F=2.512\times10^{-2}\ N

Hence, The maximum electrical force is 2.512\times10^{-2}\ N.

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