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3 years ago

A body moving at .500c with respect to an observer

Physics

1 answer:

lakkis [162]3 years ago

6 0

Answer:

0.8c and -0.14c

Explanation:

The first fragment will have a speed of +0.5c respect of a frame of reference moving at +0.5c

Lest name v the velocity of the frame of reference, and u' the velocity of the object respect of this moving frame of reference.

The Lorentz transform for velocity is:

u = (u' + v) / (1 + (u' * v) / c^2)

u = (0.5c + 0.5c) / (1 + (0.5c * 0.5c) / c^2) = 0.8c

The other fragment has a velocity of u' = -0.6c respect of the moving frame of reference.

u = (-0.6v + 0.5c) / (1 + (0.5c * 0.5c) / c^2) = -0.14c

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A steel ball with mass m is suspended from the ceiling at the bottom end of a light, 17.0-m-long rope. The ball swings back and

iogann1982 [59]

Answer:

1. 18.25 m/s

2. 0 m/s

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1.So the centripetal acceleration of the ball at this lowest point must be, taking gravity into account

$a_c = \frac{T - mg}{m} = \frac{3mg - mg}{m} = 2g$

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3 years ago

The wind is blowing horizontally at 30 m / s in a storm at P o , 20°C toward a wall, where it comes to a stop (stagnation) and l

aleksklad [387]

Answer:

To=20.44 °C

Explanation:

Given that

Velocity , v= 30 m/s

Temperature , T= 20°C

We know that specific heat capacity for air ,Cp=1.005 kJ/kg.K

By using energy conservation ,the stagnation temperature is given as

$T_0=T+\dfrac{v^2}{2C_p}$

Now by putting the values in the above equation we get

$T_0=(20+273)+\dfrac{30^2}{2\times 1.005\times 1000}\ K$

To= 293.44 K

To= 293.44 - 273 °C

To=20.44 °C

Therefore the stagnation temperature will be 20.44 °C.

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3 years ago