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Leviafan [203]
3 years ago
12

A cartoon shows two friends watching an unoccupied car in free fall after it has rolled off a diff. One friend says to the other

, "It goes from zero to sixty mphin about three seconds." Is this statement correct? (Note that 60 mph converts to 26.8 m/s.)
Physics
1 answer:
olga55 [171]3 years ago
8 0

Answer:

The statement is not correct.

Explanation:

To know if the statement is correct, we shall determine the velocity of the car after 3 s. This is illustrated below.

Data obtained from the question include:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 3 s

Final velocity (v) =?

v = u + gt

v = 0 + (9.8 × 3)

v = 0 + 29.4

v = 29.4 m/s

Thus, the velocity of the car after 3 s is 29.4 m/s.

Hence, the statement made by the friend is not correct as the car has a falling velocity of 29.4 m/s after 3 s.

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In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst
Archy [21]

Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

E = \frac{1}{2}mv^{2}

So, v = \sqrt{\frac{2E}{m}}   .... (1)

Now,

r = \frac{mv}{Bq}

Substitute the value of v from equation (1), we get

r = \frac{\sqrt{2mE}}{Bq}

Let the radius of the alpha particle is r2.

For proton

So, r_{1} = \frac{\sqrt{2m_{1}E}}{Bq_{1}}    ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, r_{2} = \frac{\sqrt{2m_{2}E}}{Bq_{2}}    ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

\frac{r_{1}}{r_{2}}=\frac{q_{2}}{q_{1}}\times \sqrt{\frac{m_{1}}{m_{2}}}

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

\frac{r_{1}}{r_{2}}=\frac{2q}}{q}}\times \sqrt{\frac{m}}{4m}}=1

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

8 0
3 years ago
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atroni [7]

Answer:

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Explanation:

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Answer

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√h/1/2a

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√8/4.905

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