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3 years ago

The acronym laser stands for light amplification by ____ emission of radiation

Physics

1 answer:

AleksAgata [21]3 years ago

6 0

light amplification by stimulated emission of radiation

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100Gw x 35ms (note: a watt is a joule per second)

Leviafan [203]

'G' = "giga..." = billion = 10⁹

'ms' = "millisecond" = 0.001 second

100 Gw = 10¹¹ watts

35 ms = 0.035 second

100 Gw x 35 ms = 10¹¹ watts x 0.035 second = 3.5 x 10⁹ J

= 3.5 G-joules

7 0

3 years ago

Describe how the amount of voltage applied to a circuit affects current flow.

posledela

If you’re on Usa Test Prepnor something, the answer will be C

8 0

4 years ago

Read 2 more answers

A pendulum consists of a 1.5 kg stone swinging on a 4.3 m string of negligible mass. The stone has a speed of 8.4 m/s when it pa

Ad libitum [116K]

Answer:

(a) v = 1.54m/s, (b) Θ = 80.6° and (c) E = 52.92J

Explanation:

(a) At the lowest point, the total energy is giving by:

$E_{T}_{0} = E_{K}_{0} + E_{P}_{0} = \frac {1}{2} mv_{0}^{2} + mgy_{0}$

where, $E_{K}$ : kinetic energy, $E_{P}$ : potential energy, m: pendulum's mass, v: speed of the pendulum, g:gravity and y: heigh of the pendulum.

$E_{T}_{0} = \frac {1}{2} mv_{0}^{2} + mg(0) = \frac {1}{2} (1.5)(8.4)^{2} = 52.92 J$ (1)

When the string is at 64° from the vertical, the total energy is:

$E_{T}_{f} = \frac {1}{2} mv_{f}^{2} + mgy_{f}$ (2)

At this point, $y_{f}$ is:

$y_{f} = L - Lcos(\Theta)$ (3)

where L: longitude of the pendulum

$y_{f} = 4.3 m (1 -cos(64)) = 2.42 m$

By conservation of energy, we can calculate the speed of the string, at 64 ° to the vertical. Equaling equations (1) and (2):

$E_{T}_{0} = E_{T}_{f} = \frac {1}{2} mv_{f}^{2} + mgy_{f}$

$v_{f} = \sqrt \frac{2(E_{T}_{0} - mgy_{f})}{mg}}$

$v_{f} = \sqrt \frac{2(52.92 - 1.5 \cdot 9.8 \cdot 2.42)}{1.5 \cdot 9.8}} = 1.54 \frac{m}{s}$

(b) Smilarly, by conservation of energy we can find the greatest angle, assuming that vf = 0 at the greatest angle reached:

$\frac {1}{2} mv_{0}^{2} + mgy_{0} = \frac {1}{2} mv_{f}^{2} + mgy_{f}$

$y_{f} = \frac {E_{0}}{mg} = \frac {52.92}{(1.5)(9.8)} = 3.6 m$

Using the heigh calculated in equation (3) we can find the angle:

$\Theta = Arccos (\frac {L - y_{f}}{L}) = Arccos (\frac {4.3 - 3.6}{4.3}) = 80.6^\circ$

(c) The total mechanical energy at the lowest point is giving by:

$E_{T} = E_{K} + E_{P} = \frac {1}{2} mv_{0}^{2} + 0 = \frac {1}{2} (1.5)(8.4)^{2} = 52.92 J$

Have a nice day!

8 0

3 years ago

A charged ball of mass 10g and charge 2 C is placed at rest in a field of field strength 5NC as shown.

Natali [406]

Answer:

1000 m/s

Explanation:

Given that

Mass of the ball, m = 10 g = 0.01 kg

Charge of the ball, q = 2 C

Field strength of the field, E = 5 N/C

Acceleration of the ball, a = ? m/s²

We know the the equation holds,

F = qE, on substituting, we have

F = 2 C * 5 N/C

F = 10 N

Also, we know that

F = ma, and as such,

a = F / m, on substituting we have

a = 10 N / 0.01 kg

a = 1000 m/s²

Therefore, the acceleration of the ball is 1000 m/s²

8 0

3 years ago

What is the si unit of force

Tcecarenko [31]

Explanation:

SI unit of force is Newton

3 0

3 years ago

Read 2 more answers