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krok68 [10]
3 years ago
11

A bike first accelerates from 0.0m/s to 4.5m/s in 4.5 s, the continues at this constant speed for another 6.0 s. What is the tot

al distance traveled by the bike?
Physics
1 answer:
Alex_Xolod [135]3 years ago
3 0

Answer:

37.125 m

Explanation:

Using the equation of motion

s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration

<u>Distance during acceleration</u>

Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.

Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be

a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}

Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion

s=0*4.5+ 0.5*1*4.5^{2}=0+10.125 =10.125 m

<u>Distance at a constant speed</u>

At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time

Distance=4.5 m/s*6 s=27 m

<u>Total distance</u>

Total=27+10.125=37.125 m

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Answer:

The speed of the electron is 1.371 x 10⁶ m/s.

Explanation:

Given;

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The energy of the incident light is given by;

E = hf

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f = c / λ

E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J

Photo electric effect equation is given by;

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8 0
3 years ago
I would like help with this physics problem
Darina [25.2K]

(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.

Recall that

v_f=v_i+at

We have v_f=-v_i, so that

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(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.

(c) The ball's average velocity is 0. Average velocity is given by \dfrac{v_i+v_f}2, and we know that v_f=-v_i.

(d) The position of the ball x_f at time t is given by

x_f=x_i+v_it+\dfrac12at^2

Take the starting position to be the origin, x_i=0. Then after 6 seconds,

x_f=\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2=42\,\mathrm m

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We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

x_f-x_i=\dfrac{v_i+v_f}2t\implies42\,\mathrm m=\dfrac{8\,\frac{\mathrm m}{\mathrm s}+v_f}2(6\,\mathrm s)\implies v_f=6\,\dfrac{\mathrm m}{\mathrm s}

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