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Which of the following is true about the mass of an object?

Butoxors [25]

I believe it is the first one

5 0

3 years ago

How is potential energy converted to thermal energy in a system?

Natali [406]

For example, an internal combustion engine converts the potential chemical energy in gasoline and oxygen into thermal energy which, by causing pressure and performing work on the pistons, is transformed into the mechanical energy that accelerates the vehicle (increasing its kinetic energy).

3 0

4 years ago

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A spherical conducting shell has charge Q. A point charge q is placed at the center of the cavity. The charge on the inner surfa

FinnZ [79.3K]

Answer:d

Explanation:

From Gauss law Electric field inside a surface is directly proportional to the charge enclosed in it.

Electric field inside a spherical shell is zero and hence there is no charge inside the spherical shell because q charge induces a -q charge on inside surface of spherical shell.

and to counter it there is q charge on the surface. So total charge outside the surface is Q+q

7 0

3 years ago

A rocket is fired with an initial VELOCITY OF 100m/s at an angle of 55° above the horizontal, It explodes On the mountain Side 1

GuDViN [60]

Answer

688.32m and 277.44m

Explanation :

⠀

$\large{\maltese{\textsf{\underline{To find :-}}}}$

The X and Y coordinates of the rocket relative of firing

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$\large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^{-1} \\ \sf angle ({\theta}_{1}) = 55.0{\degree} \\ \sf time (t) = 12s$

⠀

$\Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}}$

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The horizontal range of projectile at x.

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$\sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t}$

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$\large\textsf{\underline{Now substituting the required values}}$

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$\sf x = 300 \times \cos 55{\degree} \times 12 \\ \\ \sf x = 100 \times 0.5756 \times 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}}$

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The vertical position of projectile at y.

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$\sf \large y = v_{yi} \times t - (\frac{1}{2} \times g \times {t}^{2}) \\ \\ \sf \large y = v_i \times \cos \theta \times t - \frac{1}{2} g {t}^{2}$

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$\textsf{ \large {\underline{Now substituting the required values}} }$

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$\sf y = 100 \times \cos55{ \degree} \times 12 - \frac{1}{2} \times 9.80 \times {12}^{2} \\ \\ \sf y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\ \\ \sf y = 983.04 - 705.6 \\ \\ \underline{ \boxed{ \bold{y = 277.44m}}}$

⠀

<h3>Henceforth, the distance at horizon is 688.32m and at vertical is 277.44m.</h3>

8 0

2 years ago

A Doppler radar sends a pulse at

Rufina [12.5K]

Answer:

Explanation:

The problem is based on the concept of Doppler's effect of em wave .

Expression for apparent frequency can be given as follows

n = N x (V - v ) / ( V + v )

n is apparent frequency , N is real frequency , V is velocity of light and v is velocity of cloud.

n = 6 x 10⁹ ( 3 x 10⁸ - 8.52 ) / ( 3 x 10⁸ + 8.52 )

= 6 x 10⁹ ( 3 x 10⁸ ) ( 3 x 10⁸ + 8.52 )⁻¹

= 6 x 10⁹ ( 3 x 10⁸ ) ( 3 x 10⁸)⁻¹ ( 1 + 8.52/3 x 10⁸ )⁻¹

= 6 x 10⁹ ( 1 - 8.52/3 x 10⁸ )

= 6 x 10⁹ - 6 x 10⁹x 8.52/ (3 x 10⁸ )

= 6 x 10⁹ 1 - 170 .

So change in frequency = 170 approx.

7 0

3 years ago

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