Multi-part question If a galaxy moving away from the Earth has a speed of 1000 km/s and emits 656 nm light characteristic of hydrogen (the most common element in the universe). What wavelength would we observe on the Earth

[tex]\boxed{\sf E=QV^2}[/tex]

[tex]\\ \sf\longmapsto Q=\dfrac{E}{V^2}[/tex]

[tex]\\ \sf\longmapsto Q=\dfrac{3.92\times 10^{-4}}{24^2}[/tex]

[tex]\\ \sf\longmapsto Q=\dfrac{3.92\times 10^{-4}}{576}[/tex]

[tex]\\ \sf\longmapsto Q=0.006\times 10^{-4}C[/tex]

[tex]\\ \sf\longmapsto Q=6\times 10^{-1}C[/tex]

[tex]\\ \sf\longmapsto Q=0.6C[/tex]

Now

[tex]\boxed{\sf Q=CV}[/tex]

[tex]\\ \sf\longmapsto C=\dfrac{Q}{V}[/tex]

[tex]\\ \sf\longmapsto C=\dfrac{0.6}{24}[/tex]

[tex]\\ \sf\longmapsto C=0.025F[/tex]

Note:-

Answer:

The acceleration due to gravity at Pluto is 0.0597 m/s^2.

Explanation:

Length, L = 1 m

10 oscillations in 257 seconds

Time period, T = 257/10 = 25.7 s

Let the acceleration due to gravity is g.

Use the formula of time period of simple pendulum

[tex]T = 2\pi\sqrt{\frac{L}{g}}\\\\25.7 = 2 \times 31.4\sqrt{\frac{1}{g}}\\\\g = 0.0597 m/s^2[/tex]

Answer:

38500

Explanation:

I looked it up so it may be wrong

Answer:

0.48N

Explanation:

according to the second law of motion

force=mass×acceleration

the mass in this question is 0.025,the acceleration 19

therefore f=0.025×19

=0.48N

I hope this helps

Explanation:

initial velocity(u)=30km/hr = 30*1000/60*60=8.33 m/s

final velocity(v) =60km/hr = 60*1000/60*60 =16.67 m/s

time taken(t) = 1 minutes

= 60 seconds

Now,

Average velocity = u+v/2

= 8.33 m/s + 16.67m/s÷2

=12.5 m/s

Answer: Remove your foot from the gas pedal and slow down

Explanation:

Hydroplaning also refers to aquaplaning and this refers to the scenario whereby there's a layer of water that builds between the tyres of a car and the surface of the road which then brings about a loss of traction which eventually leads to a situation whereby the car doesn't respond to control inputs.

This lack of traction with the pavement takes away your control of the vehicle and to regain control of the vehicle, one should remove your foot from the gas pedal and slow down.

Answer:

Explanation:

It was Newton's genius that saw that it was true. He also realized that if it was true, the motion the earth experience would be unmeasurably small, especially in his time. Their instruments were just not good enough: crude and too approximate would be better descriptions.

Answer:

I dont know answer Sorry For that thank u

The approximate temperature of the star as determined is D) 5500 K.

The Wien's displacement law relates the maximum wavelength of a body to its absolute temperature. Wien's displacement law states that:

λ = [tex]\frac{b}{T}[/tex]

where λ is the maximum wavelength of the body, b is the constant of proportionality and T is the absolute temperature.

Thus from the given question, λ = 525 nm (525 x [tex]10^{-9}[/tex]), and b = 2.9 x [tex]10^{-3}[/tex] mK.

So that,

525 x [tex]10^{-9}[/tex] = [tex]\frac{2.9*10^{-3} }{T}[/tex]

Make T the subject of the formula to have;

T = [tex]\frac{2.9*10^{-3} }{525*10^{-9} }[/tex]

  = 5523.81

T = 5523.81 K

T ≅ 5500.00 K

The approximate temperature of the star in Kelvin is 5500 K.

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The relative velocity between Car A and Car B is 0 m/s

Since Car A is heading East at 25 m/s, its velocity is u = + 25 m/s. (since it is moving in the positive x- direction)

Also, car B is also heading East at 25 m/s, its velocity is v = + 25 m/s. (since it is moving in the positive x- direction)

So, the relative velocity between car A and Car B is V = velocity of car A - velocity of car B.

So, V = u - v

V = + 25 m/s - (+ 25 m/s)

V = + 25 m/s - 25 m/s

V = 0 m/s

So, the relative velocity between Car A and Car B is 0 m/s

The passenger in Car A would describe Car B's motion as stationary.

Since Car A is heading East at 25 m/s and car B is also heading East at 25 m/s, both cars are moving at a velocity of + 25 m/s.

Since their relative velocity is 0 m/s, it would appear to a passenger in Car A that car B is stationary according to motion from car A's reference frame.

So, the passenger in Car A would describe Car B's motion as stationary.

Answer:

Explanation:

Givens

t = 10 seconds

vi = 0

vf = 3600 mph

a = ?

d = 10000 feet

Formula

a = (vf - vi)/t

Solution

vf = 3600 mph * 1 hr / 3600 seconds * 5280 feet / 1 mile = 5280 ft / sec

a = (5280 - 0)/10

a = 528 ft/sec^2

Answer:

The emf of a cell is the sum of the electric potential differences (PDs) produced by a separation of charges (electrons or ions) that can occur at each phase boundary (or interface) in the cell. The magnitude of each PD depends on the chemical nature of the two contacting phases.

Explanation:

pls mark brainliest

Answer:

could and largejsjisj and we look like they can get to

Explanation:

The acceleration g varies by about 1/2 of 1 percent with position on Earth's surface, from about 9.78 metres per second per second at the Equator to approximately 9.83 metres per second per second at the poles.

Answer:

There is a 1 in roughly 5.2 to the 61st power chance that if you were to slap a table, all of the molecules in your hand and the table would miss each other and your hand would go through it.

Explanation:

I'm not sure but I hope it helps

Answer:

E_O = 96J

Explanation:

E_O = E_I*(%e/100%)

E_O = 120J*(80%/100%)

E_O = 96J

Answer:

a. 5 batteries b. 1050 mAh

Explanation:

Here is the complete question

A student project is required to be portable and hand held. It requires 6 V DC power at a current of 150 mA. The batteries for the power supply must last for a minimum of 7 hours of continuous operation. NiMH rechargeable batteries in AA size are to be used. A) How many batteries are needed? B) What mAh capacity should the batteries have?

Solution

A) How many batteries are needed?

Since the nominal voltage for a single NiMH battery is 1.2 V per battery and we require 6V DC power, we combine the batteries in series to obtain a total voltage of 6 V. The number of batteries required, n = total voltage/voltage per cell = 6V/1.2V per battery = 5 batteries

So, the number of batteries needed is 5.

B) What mAh capacity should the batteries have?

Since the batteries are in series, they would each deliver a current of 150 mA. Since we require a current of 150 mA for 7 hours, the number of milliampere-hours capacity mAh of batteries required is Q = It where I = current = 150 mA and t = time = 7 hours.

So, Q = It = 150 mA × 7 h = 1050 mAh.

So, the batteries should have a mAh of 1050 mAh

Answer:

Explanation:

        16093.4

Our sun is 27 million times larger than our moon. Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth. If tidal forces were based solely on comparative masses, the sun should have a tide-generating force that is 27 million times greater than that of the moon. However, the sun is 390 times further from the Earth than is the moon. Thus, its tide-generating force is reduced by 3903, or about 59 million times less than the moon. Because of these conditions, the sun’s tide-generating force is about half that of the moon.

Answer:

Remember the relation:

Speed*Time = Distance.

We can estimate that the speed at which an earthquake "moves", in the surface, is:

S = 6km/s  (this is a low estimation actually)

Then if the wave is detected 12.5 minutes after the earthquake, we know that it traveled for 12.5 minutes before reaching the detector.

So we know the speed of the wave and the time it took to reach the detector, then we can use the equation:

Speed*Time = Distance.

to find the distance.

First, we should write the time in seconds

we know that:

1 min = 60 s

then:

12.5 min = 12.5*(60 s) = 750 s

Then, the wave traveled with a speed of 6 km/s for 750 seconds until it reached the detector, then the distance that it traveled is:

(6km/s)*750s = 4500 km

The distance between the detector and the site of the quake is around 4500 km.

Answer:

The velocity at that time would be [tex](-67.975\; \rm s)[/tex]. The velocity of this particle is continuously changing.

Explanation:

Differentiate the expression for position [tex]x[/tex] with respect to time [tex]t[/tex] to find an expression for velocity.

[tex]\begin{aligned}v(t) &= \frac{d}{dt}[x(t)] \\ &= \frac{d}{dt} \left[ 7.4 + 9.2\, t - 2.1\, t^{3}\right]\\ &= 9.2 - 6.3\, t^{2}\end{aligned}[/tex].

Hence, at [tex]t = 3.5\; \rm s[/tex], velocity would be [tex]v(3.5) = 9.2 - 6.3 \times (3.5)^{2} = -67.975\; \rm m[/tex].

Since velocity [tex]v(t)[/tex] changes with time [tex]t[/tex], the velocity of this particle is continuously changing.

Explanation:

The three importance of measurement are as follows :-

Hope it is helpful to you

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Sabemos todo material tiene asociado un coeficiente de difracción n.

De tal forma que la velocidad de una onda electromagnetica que viaja a travez de dicho material, será:

v = c/n

donde c es la velocidad de la luz.

Con la información dada, podremos concluir que el coeficiente de difracción del medio Q es dos veces el del medio P.

Ahora veamos como llegamos a esto:

Sabemos que en el medio P, la onda recorre una distancia 8*D en un tiempo 4*T

entonces la velocidad de la onda en el medio P es:

[tex]V_p = \frac{8*D}{4*T} = 2*\frac{D}{T}[/tex]

Mientras que en el medio Q, recorre una distancia 4*D en un tiempo 4*T, por lo que la velocidad en el medio Q será:

[tex]V_Q = \frac{4*D}{4*T} = \frac{D}{T}[/tex]

Podemos ver que:

[tex]V_P = 2*V_Q[/tex]

Reescribiendo las velocidades como el cociente entre la velocidad de la luz y el correspondiente coeficiente de difracción obtenemos:

[tex]\frac{c}{n_P} = 2*\frac{c}{n_Q} \\\\n_Q = 2*n_P[/tex]

Es decir, podemos concluir que el coeficiente de difracción del medio Q es dos veces el del medio P.

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Answer:The answer is A

Explanation:just did it on a test

Answer:

I THINK IT'S D....

HOPE SO

The answer are :

A. Distance = 15.71 m , The displacement = 10 m

B. distance traveled will increase, displacement will decrease

C. The distance = 31.42 m , Displacement = 0

Distance is a scalar quantity, while displacement is a vector quantity. Displacement is the distance travelled in a specific direction.

Given that a child rides a pony on a circular track with a radius of 5 m, after the child has gone halfway around the track

(A)

The distance traveled will be = 2[tex]\pi[/tex]r / 2

The distance = πr

The distance = 22/7 x 5

The distance = 15.71 m

The displacement = 2r

The displacement = 2 x 5

The displacement = 10 m

(B) The distance traveled will increase as the child completes one circle of the track but the displacement will decrease because the displacement in one cycle of a circular motion is zero since it is a vector quantity.

(C) The distance after a complete circuit of the track = 2πr

The distance = 2 × 22/7 × 5

The distance = 31.42 m

The displacement after a complete circuit of the track will be zero.

That is;

Displacement = 0

Therefore, the distance traveled and the displacement after the child has gone halfway around the track are 15.7 m and 10 m respectively. While the distance and displacement after a complete circle of the track are 31.42 m and 0 respectively.

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The height of the window above the ground is 11.025m

The first step is to write out the parameters given in the question

U(initial velocity)= 0

Time= 1.5 seconds

Acceleration due to gravity= 9.8

Therefore the height of the window can be calculated as follows

S= ut + 1/2gt²

= 0(1.5) + 1/2(9.8)(1.5²)

= 0 + 1/2(9.8)(2.25)

=  1/2(22.05)

= 0.5×22.05

= 11.025

Hence the height of the window above the ground is 11.025m

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