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4 years ago

Doorknob larger radius than shaft, why?

Physics

1 answer:

Darina [25.2K]4 years ago

5 0

I believe because to open the door the knob has to turn a greater distance than the shaft is capable of turning

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At a certain instant, the earth, the moon, and a station- ary 1250-kg spacecraft lie at the vertices of an equilateral triangle

Murrr4er [49]

Answer:

a). F = 3.376 N, θ = 59.18°

b). W = 1.3x $10^{9}$ J

Explanation:

We know

Gravitational constant, G = 6.673 x $10^{-11}$ N- $m^{2}$ / $kg^{-2}$

Mass of the earth, M = 5.97 x $10^{24}$ kg

mass of the moon, m = 7.35 x $10^{22}$ kg

Mass of the satellite, $m_{s}$ = 1250 kg

Distance between the objects, r = 3.84 x $10^{5}$ km

= 3.84 x $10^{8}$ m

Now

The force on the satellite due to moon

$F_{m}= \frac{G\times m\times m_{s}}{r^{2}}$

$F_{m}= \frac{6.673\times 10^{-11}\times 7.35\times 10^{22}\times 1250}{(3.84\times 10^{8})^{2}}$

$F_{m}$ = 0.0415 N ( in the positive x direction )

The force on the space craft due to the earth

$F_{m}= \frac{G\times M\times m_{s}}{r^{2}}$

$F_{m}= \frac{6.673\times 10^{-11}\times 5.97\times 10^{24}\times 1250}{(3.84\times 10^{8})^{2}}$

$F_{m}$ = 3.377 N ( at 60° to x axis )

Now component of force of earth along x axis

$F_{e_{x}} = F_{e}\times cos 60$

= 3.377 x 0.5

= 1.6885 N

Now component of force of earth along y axis

$F_{e_{y}} = F_{e}\times sin 60$

= 3.377 x 0.86

= 2.90 N

∴ Net force on the space craft due to earth and moon along x axis

$F_{x}$ = $F_{e}$ cos 60+ $F_{m}$

= 1.3885+0.0415

= 1.73 N

Net force on the space craft due to earth and moon along y axis

$F_{x}$ = $F_{e_{y}}$

= 2.90 N

Therefore, total force F = $\sqrt{(F_{x}^{2})+(F_{y}^{2})}$

F = $\sqrt{(1.73^{2})+(2.90^{2})}$

F = 3.376 N

∴ Magnitude of the net gravitational force on the space craft is 3.376 N

Direction of net force on the space craft is given by

$\Theta = \arctan \left (\frac{F_{y}}{F_{x}}\right )$

$\Theta = \arctan \left (\frac{2.90}{1.73}\right )$

$\Theta = 59.18$ °

Therefore this direction is 59.18° from the line joining earth and the space craft.

b).

∴ Gravitational potential energy of the space craft is given by

$E = \frac{G.M.m_{s}}{r}+\frac{G.m.m_{s}}{r}$

$E = \frac{G\times m_{s}\left ( M+m \right )}{r}$

$E = \frac{6.673\times 10^{-11}\times 1250\left ( 5.97\times 10^{24}+7.35\times 10^{22} \right )}{3.84\times 10^{8}}$

E = 1312769385 J

E = 1.3 x $10^{9}$ J

Therefore minimum work done is 1.3x $10^{9}$ J

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Large group of asteroids that orbit the sun between Mars and jupiter

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That would be the Main Asteroid Belt

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3 years ago

The electron structures of atoms are not involved in the emission of:

nekit [7.7K]

The correct one is gamma rays. Lets go over them one by one.

Gamma rays are rays that arise from gamma decay, a type of radioactive decay. Often, after another decay, the nucleus is still unstable and it gives off energy in the form of gamma rays to stabilize itself. Hence, gamma rays have nothing to do with the electron structure, only with the nucleus of the atom.

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