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Dimas [21]
3 years ago
10

Assume the equation x 5 At3 1 Bt describes the motion of a particular object, with x having the dimension of length and t having

the dimension of time. Determine the dimen- sions of the constants A and B. (b) Determine the dimen- sions of the derivative dx/dt 5 3At2 1 B.
Physics
1 answer:
igomit [66]3 years ago
6 0

Answer:

(a) A = m/s^3, B = m/s.

(b) dx/dt = m/s.

Explanation:

(a)

x = At^3 + Bt\\m = As^3 + Bs\\m = (\frac{m}{s^3})s^3 + (\frac{m}{s})s

Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.

(b) \frac{dx}{dt} = 3At^2 + B = 3(\frac{m}{s^3})s^2 + \frac{m}{s} = m/s

This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.

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Answer:

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Explanation:

All matter is attracted and pulled downward toward the ground.

Most matter is attracted and pulled downward toward the ground. This option is incorrect, because all matter is attracted and pulled downward the ground.

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Only surface matter is incorrect, all matter is attracted and pulled downward.

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3 years ago
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Radius of the path of electron which is moving perpendicular to magnetic field is defined as,

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Given that, the velocity of electron is, v=4000m/s.

And the magnetic field is B=1.5T.

And the mass of electron is, m=9.11\times10^{-31}kg

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Therefore, the path radius of a moving electron is 15.18\times 10^{-9} m

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5 UNDER 5

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