what causes the moon to change its appearance over a month?​

Answer:

Produces 60-cycle AC electricity; it is usually an off-the-shelf induction generator. High-speed shaft: Drives the generator. Low-speed shaft: Turns the low-speed shaft at about 30-60 rpm. Nacelle: Sits atop the tower and contains the gear box, low- and high-speed shafts, generator, controller, and brake.

Explanation:

Explanation:

1200 is your answer for this question

Explanation:

Using the ideal gas equation, which I presume you are since you didn't specify using any other EOS, we have PV=nRT. Solving for what changes, i.e. pressure(P) and temperature(T), we have P/T=nR/V. Now, we can set up a relationship between the two pressures and temperatures and solve for what's necessary.

So, we have:

P1/T1=P2/T2

Solving for P2, we have:

P2=(P1*T2)/T1

NOTE: We MUST convert our temperatures to kelvin, otherwise you will end up with a NEGATIVE AND INCORRECT pressure!

Plugging in our values of P1=3.00x10^5 N/m^2, T1 of 308.15K, and T2 of 235.15K. Now we are free to evaluate:

P2=[(3.00x10*5 N/m^2)(235.15K)]/[308.15K]

P2=228930.7156 N/m^2

Or, to the appropriate amount of significant figures: 2.29x10^5 N/m^2

Which makes sense intuitively, as things tend to deflate slightly when the temperature drops!

Hope this helps!

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Answer:

lungs

Explanation:

lungs bring oxygen

Answer:

lungs

Explanation:

i took the quiz

Answer:

v = √{2(GmE(1/RE - 1/(RE + h)]}

Explanation:

From the law of conservation of energy, the total kinetic + potential energy at h = total kinetic  + potential energy at the surface of the earth(h = 0)

So, K + U = K' + U'

where K = kinetic energy of hammer at h = 0

U = gravitational potential energy of hammer at h = -GmEm/(RE + h)

K' = kinetic energy of hammer at the earth's surface (h = 0) = 1/2mv²

U = gravitational potential energy of hammer at earth's surface (h = 0) = -GmEm/RE

So, K + U = K' + U'

0 + [-GmEm/(RE + h)] = 1/2mv² +  [-GmEm/RE]

0 - GmEm/(RE + h) = 1/2mv² - GmEm/RE

collecting like terms,we have

1/2mv² = GmEm/RE - GmEm/(RE + h)

Factorizing GmE and multiplying both sides by 2, we have

v² = 2(GmE(1/RE - 1/(RE + h)]

taking square-root of both sides, we have

v = √{2(GmE(1/RE - 1/(RE + h)]}

The expression for the speed of the hammer will be:

[tex]V=\sqrt{2(Gm_e(\dfrac{1}{R_e}-\dfrac{1}{R_e+h}) }[/tex]

The law of conservation says that the total energy at the height h and the total energy at the surface of the earth will remain constant.

From the law of conservation of energy

[tex]KE+PE=KE'+PE'[/tex]

where

KE = kinetic energy of hammer at h = 0

PE = gravitational potential energy of hammer at h height

[tex]PE=\dfrac{-Gm_em}{R_e+h}[/tex]

KE' = kinetic energy of hammer at the earth's surface (h = 0)

[tex]KE'=\dfrac{1}{2} mv^2[/tex]  

PE' = gravitational potential energy of hammer at earth's surface (h = 0)

[tex]PE'=\dfrac{Gm_em}{R_e}[/tex]

Now by putting the values

[tex]KE+PE=KE'+PE'[/tex]

[tex]0+ \dfrac{-Gm_em}{R_e+h}[/tex]   [tex]=\dfrac{1}{2} mv^2[/tex] [tex]-\dfrac{Gm_em}{R_e}[/tex]

[tex]\dfrac{1}{2} mv^2=\dfrac{Gm_em}{R_e} -\dfrac{Gm_em}{R_e+h}[/tex]

[tex]v^2=2Gm_e(\dfrac{1}{R_e} -\dfrac{1}{R_e+h} )[/tex]

[tex]v=\sqrt{2Gm_e(\dfrac{1}{R_e} -\dfrac{1}{R_e+h} )}[/tex]

Thus the expression for the speed of the hammer will be:

[tex]V=\sqrt{2(Gm_e(\dfrac{1}{R_e}-\dfrac{1}{R_e+h}) }[/tex]

To know more about the speed follow

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Answer:

(a) 6.283 Wb (b) 69.11 Wb (c) I = 0.628 A

Explanation:

Given that,

The diameter of the loop, d = 40 cm

Radius, r = 20 cm

Initial magnetic field, B = 5 mT

Final magnetic field, B' = 55 mT

Initial magnetic flux,

[tex]\phi_i=BA\\\\=5\times 10^{-3}\times \pi \times 20^2\\\\=6.283\ Wb[/tex]

Final magnetic flux,

[tex]\phi_f=B'A\\\\=55\times 10^{-3}\times \pi \times 20^2\\\\=69.11\ Wb[/tex]

Due to change in magnetic field an emf will be generated in the loop. It is given by :

[tex]\epsilon=-\dfrac{d\phi}{dt}\\\\=\phi_f-\phi_i\\\\=69.11-6.283\\\\=62.827\ V[/tex]

Let I be the current in the loop. We can find it using Ohm's law such that,

[tex]\epsilon=IR\\\\I=\dfrac{\epsilon}{R}\\\\I=\dfrac{62.827}{100}\\\\=0.628\ A[/tex]

Hence, this is the required solution.

Answer:

19 protons, 20 neutrons and 18 electrons.

Explanation:

The atomic number gives the number of protons 19

p

=

19

The atomic mass is the sum of the protons and neutrons

p

+

n

=

39

p

=

19

put p into the equation and solve for n the neutrons.

19

+

n

=

39

Subtract 19 from both sides

19

−

19

+

n

=

39

−

19

n

=

20

The number of electrons equals the number of protons in a neutral atom. The positive charge equals the negative charge. The negative charge is the number of electrons. This ion has a charge of +1. So solve for the negative charge.

−

19

+

1

=

−

18

The negative charge is -18 so

e

=

18

Answer:

Answer D. Inertial mass constant, weight decreases.

Explanation:

Khan Academy

Answer:

4m/s

Explanation:

Speed=distance/time

So the speed will be 16m/4s which gives 4m/s

Answer:

Could I get brainest answer

I think it is B and e

Answer:

The answer is D hope it helps

Explanation:

Answer:

Yes, the value will be the same.

Explanation:

Yes, or at least to some degree, that value of K will remain the same. You're looking for a difference in absorbance, and the difference should be visible at all wavelengths, not only at the limit. That being said, resolution varies, and if we don't read the value to the maximum, we can get a less accurate reading.

Answer:

(A) The total resistance of the circuit is 25 Ω

(B) The current through each resistor is 4.4 A

(C) For 10Ω: Potential drop = 44 V

     For 15Ω: Potential drop = 66 V

Explanation:

Given;

potential difference, V = 110V

resistors in series, = 10Ω and a 15Ω

(A) The total resistance of the circuit is calculated as follows;

Rt = 10Ω + 15Ω = 25Ω

(B) The current through each resistor;

Same current will flow through the two resistors since they are in series.

I = V/Rt

I = 110 / 25

I = 4.4 A

(C) The voltage drop across each resistor;

For 10Ω: Potential drop = IR₁ = 4.4 x 10 = 44 V

For 15Ω: Potential drop = IR₂ = 4.4 x 15 = 66 V

Answer:

P = 3.92 10¹⁰ W

Explanation:

The power is data by the expression

         P = W / t

the work of a force is

         W = F. y

the bold ones represent vectors. In this case the displacement is vertical upwards and the vertical forces upwards, therefore the angle is zero and the cos 0 = 1

          W = F y

we substitute

          P = F y / t

          P = F v

as the body rises at constant speed the acceleration is zero and from the equilibrium condition

            F -W = 0

            F = mg

we substitute

           P = m g v

let's calculate

           P 1.00 10⁹ 9.8  4

           P = 3.92 10¹⁰ W

Answer:

I am calculating the total area of a solar panel for a particular load demand by ... designing according to energy demand for a day then how will it affect total solar area? ... Total Power Output=Total Area x Solar Irradiance x Conversion Efficiency ... would need is a 1 m2 solar panel to produce 1000 Watts of electrical energy.

Explanation:

Answer:

Explanation:

The light waves in the reflected sunlight are horizontally polarized, which illustrates that they oscillate at a [tex]90^o[/tex] angle related to a vertical line.

Depending on the condition of the height of the light, the glare can be almost entirely horizontally polarized. Furthermore, all reflections from over-water surfaces are partially polarized. The water becomes more translucent when using polarized sunglasses.

If polarized sunglasses are to be efficient against solar glare, the transmission axis should be positioned at an angle of [tex]\theta = 45^{o}[/tex]

Answer:

The balloon prohibits the flow of air through the air capacitor.

Explanation:

Just like an electric capacitor has an insulator between the plates, the air capacitor has a balloon between the chambers.

The balloon analogy is frequently used in electrical capacitors to assist visualise the notion of the capacitor's behaviour.

The balloon illustrates the capacitor's physical structure, specifically the two conducting plates and the dielectric material between them.

The capacitor's stored energy is equivalent to the air pressure within the balloon. When the voltage is withdrawn, the stored charge is released, and the capacitor returns to its uncharged condition, exactly as the balloon deflates when the air exits.

The balloon analogy aids in understanding how a capacitor stores electrical energy in its electric field and how that energy may be released when linked to a circuit.

Thus, it depicts the relationship between a capacitor's voltage, charge, and capacitance.

For more details regarding capacitance, visit:

https://brainly.com/question/31871398

#SPJ2

Answer:

ffff

Explanation:

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Answer:

B

Explanation:

the answer as said is picture 2

Answer:

Here we only need to look at the vertical problem, so first, let's look at the forces acting vertically on the wallet.

When the wallet starts to fall, the only force acting on it will be the gravitational force (where we are ignoring the effects of air friction).

Then the acceleration of the wallet will be equal to the gravitational acceleration, g = 9.8m/s^2

Then we can write:

a(t) = (-9.8m/s^2)

Where the negative sign is because this acceleration is downwards.

To get the vertical velocity equation of the wallet we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the constant of integration, and in this case is the initial velocity of the wallet, which we know is equal to 6m/s, then the velocity equation is:

v(t) = (-9.8 m/s^2)*t + 6m/s

To get the position equation we need to integrate over time again, we get:

p(t) = (1/2)*(-9.8 m/s^2)*t^2 + (6m/s)*t + p0

Where p0 is the initial vertical position, in this case, is the height at which the wallet is dropped, which is also the altitude of your jet pack when the wallet falls.

Now we want to know two things:

Determine the speed of your wallet when it hits the ground at t = 8s

Here we just need to evaluate the velocity equation in t = 8s.

v(8s) =  (-9.8 m/s^2)*8s + 6m/s = -72.4 m/s

We also want to determine the altitude of the jet pack (when the wallet drops).

To find this, we can use the fact that the wall hits the ground at t = 8s.

The wallet hits the ground when it's vertical position is equal to zero, then:

p(8s) = 0m = (1/2)*(-9.8 m/s^2)*(8s)^2 + (6m/s)*8s + p0

Now we can solve this for p0.

0m = (1/2)*(-9.8 m/s^2)*(8s)^2 + (6m/s)*8s + p0

(1/2)*(9.8 m/s^2)*(8s)^2 -  (6m/s)*8s = p0

265.6m = p0

This means that the altitude of the jet pack when the wallet drops is 265.6m

Answer:

E = 3.6 x 10⁷ N/C

Explanation:

The electric field strength due to a point charge is given by the following formula:

[tex]E = \frac{kq}{r^2}\\\\[/tex]

where,

E = Electrical Field Strenght = ?

k = Colomb's Constant = 9 x 10⁹ Nm²/C²

q = magnitude of charge = 10 μC = 1 x 10⁻⁵ C

r = distance = 5 cm = 0.05 m

Therefore,

[tex]E = \frac{(9\ x\ 10^9\ Nm^2/C^2)(1\ x\ 10^{-5}\ C)}{(0.05\ m)^2}[/tex]

E = 3.6 x 10⁷ N/C

Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.

initial instant. before the crash

         p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

         p_f = (m + M) v

the moment is preserved

          p₀ = p_f

          m v₀ = (m + M) v

          v = [tex]\frac{m}{m+M} \ v_o[/tex]

for when n balls have collided

          v = [tex]\frac{m}{n \ m + M}[/tex]  v₀

Now let's analyze the case of the bouncing ball (elastic)

initial instant

        p₀ = m v₀

final moment

        p_f = m v_{1f} + M v_{2f}

        p₀ = p_f

        m v₀ = m v_{1f} + M v_{2f}

       m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

        K₀ = K_f

        ½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

        v₁ = v_{1f}            v₂ = v_{2f}

        m (v₀² - v₁²) = M v₂²

let's use the identity

         (a² - b²) = (a + b) (a-b)

we write our equations

         m (v₀ - v₁) = M v₂                       (1)

         m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

         v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

          m (v₀ - v₁) = M (v₀ + v₁)

          v₀ (m -M) = (m + M) v₁

          v₁ = [tex]\frac{m-M}{m + M}[/tex]   v₀

we substitute in equation 1 to find v₂

            [tex]\frac{M}{m}[/tex]  v₂ = v₀ -  [tex]\frac{m-M}{m+M}[/tex]   v₀

            v₂ = [tex]\frac{m}{M} ( 1 - \frac{m-M}{m+M} ) \ v_o[/tex]

            v₂ = [tex]\frac{m}{M} ( \frac{2M}{m+M} ) \ \ v_o[/tex]

            v₂ = [tex]\frac{2m}{m +M} \ v_o[/tex]  

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.

Answer:  

A. α = - 1.047 rad/s²  

B. θ = 14.1 rad  

C. θ = 2.24 rev  

Explanation:  

A.  

We can use the first equation of motion to find the acceleration:

[tex]\omega_f = \omega_i + \alpha t[/tex]  

where,  

ωf = final angular speed = 0 rad/s  

ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s  

t = time = 3 s  

α = angular acceleration = ?  

Therefore,

[tex]0\ rad/s = 3.14\ rad/s + \alpha(3\ s)[/tex]  

α = - 1.047 rad/s²

B.  

We can use the second equation of motion to find the angular distance:

[tex]\theta = \omega_it +\frac{1}{2}\alpha t^2\\\theta = (3.14\ rad/s)(3\ s) + \frac{1}{2}(1.04\ rad/s^2)(3)^2[/tex]  

θ = 14.1 rad

C.  

θ = (14.1 rad)(1 rev/2π rad)  

θ = 2.24 rev

[tex]\huge \fbox \pink {A}\huge \fbox \green {n}\huge \fbox \blue {s}\huge \fbox \red {w}\huge \fbox \purple {e}\huge \fbox \orange {r}[/tex]

A black hole is a tremendous amount of matter crammed into a very small — in fact, zero — amount of space. The result is a powerful gravitational pull, from which not even light can escape — and, therefore, we have no information or insight as to what life is like inside. A black hole is not empty, It's actually a lot of matter condensed into a single point. This point is known as a singularity.