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deff fn [24]
3 years ago
14

Which of these equations is dimensionally correct?

Physics
1 answer:
8_murik_8 [283]3 years ago
3 0

Answer: first and third.

Explanation:

An equation is dimensionally correct if the units are the same in both sides of the equation.

first, let's define the units used:

{m} = kg

{v} = m/s

{F} = kg*m/s^2

{x} = m

{t} = s

{a} = m/s^2

Now, let's analyze each option:

1) m*v/t = F

in the left side the units are:

{m}*{v}/{t} =  kg*(m/s)*(1/s) = kg*m/s^2

And as is written above, these are the units of F, so this is correct.

2) x*v^2 = F*(x^3/x^2)

This is more trivial, in the right side we can see an F, that has mass units (kg)  and in the left side we have x and v, and we know that none of these have mass units, so this expression is not correct.

3) xt= vt^2+at^3

the units in the right side are:

{x}*{t] = m*s

in the right side are:

{v}*{t}^2 + {a}*{t}^2 = (m/s)*s^2 + (m/s^2)*s^3 = m*s + m*s

So in both sides of the equation we have the same units, then this equation is dimensionally correct.

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<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>  

The ball rotates 6.78 revolutions.

     

Explanation:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>        

At the bottom the ball has the following angular speed:

\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m

To find the revolutions we need the time, which can be found using the following equation:                

v_{f} = v_{0} + at  

t = \frac{v_{f} - v_{0}}{a} (1)

So first, we need to find the acceleration:

v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}    (2)  

By entering equation (2) into (1) we have:

t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}

Since it starts from rest (v₀ = 0):  

t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s

Finally, we can find the revolutions:  

\theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!                                                                                                                                                                                          

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