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3 years ago

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Physics

1 answer:

SpyIntel [72]3 years ago

5 0

Answer:

Determine how to separate gasoline from other substances in petroleum

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Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect a

rjkz [21]

Answer:

The speed will be "18km/s". A further explanation is given below.

Explanation:

According to the question, the values are:

Wavelength,

$\lambda = 656.46 \ nm$

$\Delta \lambda = 0.04$

$c=3\times 10^8$

As we know,

⇒ $\frac{\Delta \lambda}{\lambda} =\frac{v}{c}$

On substituting the values, we get

⇒ $\frac{656.46}{0.04} =\frac{v}{3\times 10^8}$

⇒ $v=\frac{656.46}{0.04} (3\times 10^8)$

⇒ $=16411.5\times 3\times 10^8$

⇒ $=18280 \ m/s$

or,

⇒ $=18 \ km/s$

8 0

3 years ago

A 1500-W electric heater is plugged into the outlet of a 120-V circuit that has a 20-A circuit breaker. You plug an electric hai

Korolek [52]

Answer:

b) 900 W

Explanation:

The breaker trips when the current is equal to 20 A. The power (P) is the ddp (V) multiplied by the current. So, for the electric heater, the current is:

P = V*i

1500 = 120*i

i = 12.5 A

So, to become in 20 A, it's needed 7.5 A, which must come from the hairdryer. Its power must be:

P = 120*7.5

P = 900 W

8 0

3 years ago

What's an easy way to create an interference pattern of waves?

igor_vitrenko [27]

Answer:

Explanation:

5 0

3 years ago

A 6V radio with a current of 2A is turned on for 5 minutes. Calculate the energy transferred in joules

iogann1982 [59]

Answer:

R=V/I=6/2=3ohm

time =5minutes =5*60=300seconds

I=2A

Heat =I^2Rt=(2)^2*3*300=4*900=3600J

7 0

2 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

3 years ago