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leonid [27]
2 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]2 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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Two particles each have the same mass but particle #1 has four times the charge of particle #2. Particle #1 is accelerated from
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Answer:

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Explanation:

Given:

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Solution:

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Where, r is the distance from one end to another.

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- Equate the two expressions:

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- The speed of the particle in an electric field is given by third kinetic equation of motion.

                                      v_f^2 - v_i^2 = 2*a*r

Where, v_f is the final velocity,

            v_i is the initial velocity = 0

                                      v_f^2 - 0 = 2*a*r

Substitute the expression for acceleration in equation of motion:

                                       v_f^2 = 2*(Q*V / m*r)*r

                                       v_f^2 = 2*Q*V / m

                                       v_f = sqrt (2*Q*V / m)

- The velocity of first particle is v:

                                       v = sqrt (20*Q / m)

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                                       v_2 = 2*sqrt (20*Q / m)

                                       v_2 = 2*v  

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Hope this helps!

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Question in pic.. thanks!
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That's false. Displacement would be (r2 - r1) .

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