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3 years ago

A galaxy containing substantial amounts of dark matter will.

Physics

2 answers:

pogonyaev3 years ago

5 0

Answer:

A galaxy containing substantial amounts of dark matter will spin faster

AveGali [126]3 years ago

4 0

Answer:

spin faster

Explanation:

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What is an ideal machine?

Svetlanka [38]

An ideal machine is one in which no part of the input energy to the machine gets wasted and the whole input energy is converted into the useful work. The <span>efficiency of such a machine is 100%.

i really hope this helps!</span>

5 0

3 years ago

4. Jimmy dropped a 10 kg bowling ball from a building that is 25 meters high.

KIM [24]

Answer:

The K.E of the bowling ball right before it hits the ground, K.E = 2450 J

Explanation:

Given data,

The mass of the bowling ball, m = 10 kg

The height of the building, h = 25 m

The total mechanical energy of the body is given by,

E = P.E + K.E

At height 'h' the P.E is maximum and the K.E is zero,

According to the law of conservation of energy, the K.E at the ground before hitting the ground is equal to the P.E at 'h'

Therefore, P.E at 'h'

P.E = mgh

= 10 x 9.8 x 25

= 2450 J

Hence, the K.E of the bowling ball right before it hits the ground, K.E = 2450 J

3 0

3 years ago

A 1 mW laser beam is incident onto a detector. Determine the fractional fluctuation in number of photons intercepted by the dete

Ugo [173]

Answer:

(a) $625 photons$

(b) $625\times 10^{6}photons$

Explanation:

Given that , the power of the laser beam is,

$P=1 mW\\P=1\times 10^{-3} W$

and time is given is,

$t=1\mu s\\t=10^{-6}s$

Now the energy formula for the laser beam is,

$E=P\times t$

Now,

$E=10^{-3}\times 10^{-6} \\E=10^{-9}J$

(a) The value of energy is given,

$E_{1}=10MeV\\E_{1}=10\times 10^{6}\times 1.6\times 10^{-19}J\\ E_{1}=16\times 10^{-13}J$

Now the no of photons's fraction fluctuation is,

$n=\frac{E}{E_{1} }\\ n=\frac{10^{-9} }{16\times 10^{-13} }\\ E_{1}=625 photons$

Therefore the no of photons is $625 photons$ .

(b)The value of energy is given,

$E_{z}=10eV\\E_{z}=16\times 10^{-19}J$

Now the no of photons's fraction fluctuation is,

$n=\frac{E}{E_{z} }\\ n=\frac{10^{-9} }{16\times 10^{-19} }\\n=625\times 10^{6}photons$

Therefore the no of photons is $625\times 10^{6}photons$ .

5 0

3 years ago

Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Sim

Nonamiya [84]

Answer:

The distance covered by puck A before collision is $z = 8.56 \ m$

Explanation:

From the question we are told that

The label on the two hockey pucks is A and B

The distance between the two hockey pucks is D 18.0 m

The speed of puck A is $v_A = 3.90 \ m/s$

The speed of puck B is $v_B = 4.30 \ m/s$

The distance covered by puck A is mathematically represented as

$z = v_A * t$

=> $t = \frac{z}{v_A}$

The distance covered by puck B is mathematically represented as

$18 - z = v_B * t$

=> $t = \frac{18 - z}{v_B}$

Since the time take before collision is the same

$\frac{18 - z}{V_B} = \frac{z}{v_A}$

substituting values

$\frac{18 -z }{4.3} = \frac{z}{3.90}$

=> $70.2 - 3.90 z = 4.3 z$

=> $z = 8.56 \ m$

8 0

3 years ago

The light from many stars can be seen from earth. but there is a time delay between the time the light is emitted from the star

mrs_skeptik [129]

Stars are located at a distance which are measured in terms of light years. Light year is an Astronomical unit used to measure distance between distant Celestial bodies.
1 light year = 9460730472580<span>800 metres
But no star is located at a distance of 1 light year. Some stars are located at millions of light years and light travels ~ 3 x 10</span>⁸ m/s. Thus light takes time to reach our atmosphere.

4 0

4 years ago