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leonid [27]
3 years ago
8

An object with velocity 141 ft/s has a kinetic energy of 1558.71 ft∙lbf, on a planet whose gravity is 31.5 ft/s2. What is its

mass in pounds
Physics
1 answer:
Sidana [21]3 years ago
6 0

Answer:

The mass of the object is 5.045 lbm.

Explanation:

Given;

kinetic energy of the object, K.E = 1558.71 ft.lbf

velocity of the object, V = 141 ft/s

The kinetic energy of the object is calculated as;

K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m  = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \  \times \ \ \ \ 1   \ lbf\ }

m  = \frac{(2 \ \times \ 1558.71  \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71  \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm

Therefore, the mass of the object is 5.045 lbm.

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Millikan is doing his oil drop experiment. He has a droplet with radius 1.6 µm suspended motionless in a uniform electric field
swat32

Answer:

The charge on the droplet is 3.106\times10^{-16}\ C.

Yes, quantization of charge is obeyed within experimental error.

Explanation:

Given that,

Radius = 1.6μm

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Density of oil = 0.085 g/cm³

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Put the value into the formula

q=\dfrac{\dfrac{4}{3}\times\pi\times(1.6\times10^{-6})^3\times85\times9.8}{46}

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We need to calculate the quantization of charge

Using formula of quantization

n = \dfrac{q}{e}

Put the value into the formula

n=\dfrac{3.106\times10^{-16}}{1.6\times10^{-19}}

n=1941.25

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Hence, The charge on the droplet is 3.106\times10^{-16}\ C.

Yes, quantization of charge is obeyed within experimental error.

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Answer with Explanation:

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We know that gravitational force=\frac{GmM}{r^2}

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F_m=F_M

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\frac{x}{D-x}=\sqrt{\frac{M}{m}}

Let R=\sqrt{\frac{M}{m}}

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x=\frac{DR}{1+R}

b.We have to find the ratio R of the mass of the mass of the planet to the mass of the moon when x=\frac{2}{3}D

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Hence, the ratio R of the mass of the planet to the mass of the moon=4:1

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