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2 years ago

Find the speed of a wave (in meters/second) whose wavelength is 4 meters and

Physics

1 answer:

Gemiola [76]2 years ago

8 0

Wave speed = (wavelength) x (frequency)

Wave speed = (4 m) x (3.5 /s)

<em>Wave speed = 14.0 m/s</em>

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The reason an astronaut in an earth satellite feels weightless is thatThe reason an astronaut in an earth satellite feels weight

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The reason an astronaut in an earth satellite feels weightless is that the astronaut is falling.

Option a

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The other options except Option is not applicable since the gravitational force is a long range force, in which the satellite revolves very close to the surface of the Earth where the gravity is felt.The zero weight experienced by the astronaut in a satellite is due to the earth pulling along with satellite. Due to gravitational force of the Earth,the astronaut falls freely .

But why not the satellite comes down due to gravity when its launched in space. The fact is that the satellite is launched with velocity of tangent direction and it is very high. The centripetal force balances the gravity.

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3 years ago

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2 years ago

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

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2 years ago

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The fast sports car does more damage then the slow semi truck

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