Can anyone help please ​

Answer:

ffff

Explanation:

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Answer:

D

Explanation:

It’s D

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:

Answer:

9. 4.8 Ω

10. 0.92 Ω

Explanation:

9. Determination of the equivalent resistance.

Resistor 1 (R₁) = 10 Ω

Resistor 2 (R₂) = 20 Ω

Resistor 3 (R₃) = 30 Ω

Resistor 4 (R₄) = 40 Ω

Equivalent Resistance (R) =?

The equivalent resistance can be obtained as follow:

1/R = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄

1/R = 1/10 + 1/20 + 1/30 + 1/40

Find the least common multiple (lcm) of 10, 20, 30 and 40. The result is 120. Divide 120 by each of the denominator and multiply the result obtained by the numerator as shown below:

1/R = (12 + 6 + 4 + 3) / 120

1/R = 25 / 120

Invert

R = 120 / 25

R = 4.8 Ω

Thus, the equivalent resistance is 4.8 Ω

10. Determination of the equivalent resistance.

Resistor 1 (R₁) = 2 Ω

Resistor 2 (R₂) = 3 Ω

Resistor 3 (R₃) = 4 Ω

Equivalent Resistance (R) =?

The equivalent resistance can be obtained as follow:

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/2 + 1/3 + 1/4

Find the least common multiple (lcm) of 2, 3 and 4. The result is 12. Divide 12 by each of the denominator and multiply the result obtained by the numerator as shown below:

1/R = (6 + 4 + 3) / 12

1/R = 13 / 12

Invert

R = 12 / 13

R = 0.92 Ω

Thus, the equivalent resistance is 0.92 Ω

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:

D

Explanation:

D is high pitch c is low pitch

Answer:

D creo que esa es la nota mas alta, la segumda

Answer:

10kg

Explanation:

Weight is "how much does gravity drag this down".

Mass is "how much matter is there here".

The relation is:

[tex]F_g = mg[/tex]

where [tex]F_g[/tex] is the weight, [tex]m[/tex] is the mass and [tex]g[/tex] is the gravitational acceleration (roughly equal to 10N/kg on Earth).

From the task we know that:

[tex]F_g = 100N\\g = 10\frac{N}{kg}[/tex]

So let's input it into the relation:

[tex]100N = m\cdot 10\frac{N}{kg}\\10N = m \cdot 1\frac{N}{kg}\\10N \cdot \frac{kg}{N} = m\\~\\m = 10kg[/tex]

Answer:

Could I get brainest answer

I think it is B and e

Answer:

The answer is D hope it helps

Explanation:

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:

A.  Kepler's second law says that the radius vector sweeps equal areas in             equal times.

Answer:

lungs

Explanation:

lungs bring oxygen

Answer:

lungs

Explanation:

i took the quiz

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

Answer:

13.04

Explanation:

credit to the comment above

A lens with f= 20.0 cm creates a virtual image at -37.5 cm, then the object distance is approximately 0.0767 cm.

To determine the object distance in this scenario, we can use the lens formula:

[tex]\rm \frac{1}{f} =\frac{1}{v} -\frac{1}{u}[/tex]

Here, we have:

f = 20.0 cm (positive for a converging lens)

v = -37.5 cm (negative because it is a virtual image formed in front of the lens)

Substituting the values into the lens formula:

1÷20.0 = 1 ÷ (-37.5) - 1 ÷ u

Simplifying the equation:

1 ÷ 20.0 + 1÷ 37.5 = 1 ÷ u

Multiplying through by the common denominator:

(37.5 + 20.0) ÷ (20.0 * 37.5) = 1 ÷ u

57.5 ÷ 750 = 1 ÷ u

Dividing both sides by 57.5:

1 ÷ u = 750 ÷ 57.5

u = 57.5 ÷ 750

u ≈ 0.0767 cm

Thus, the object distance is approximately 0.0767 cm.

For more details regarding lens visit:

https://brainly.com/question/29834071

#SPJ5

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:

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:

4m/s

Explanation:

Speed=distance/time

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

Answer:

The reason we see different phases of the Moon here on Earth is that we only see the parts of the Moon that are being lit up by the Sun. When the Moon is between Earth and the Sun, the lit side is hidden from us. As it moves around Earth, more and more of the lit side comes into view. Then it begins to disappear again, creating the different phases we see.

Explanation:

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.

This question is incomplete, the complete question is;

The entropy of an ?-ideal gas changes in the following way as a function of temperature and volume:

ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])

For an adiabatic process, the change in T is determined by the change in V. In this problem you will compute the contributions to S from the V and T terms separately, then add them up to find the total entropy change for an adiabatic process.

Argon gas, initially at pressure 100 kPa and temperature 300 K, is allowed to expand adiabatically from 0.01 m³ to 0.026 m³ while doing work on a piston.

1) What is the change in entropy due to the volume change alone, ignoring any effects of changing internal energy? ΔS = ? J/K

2) For this adiabatic expansion, what is the final temperature?  T[tex]_f[/tex] =  ? K

Answer:

1) the change in entropy due to the volume change alone, ignoring any effects of changing internal energy is 3.185 J/K.

2) the final temperature is 158.66 K

Explanation:

Given the data in the question;

ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])

P[tex]_i[/tex] = 100 kPa = 100000 Pa

V[tex]_i[/tex] = 0.01 m³

V[tex]_f[/tex] = 0.026 m³

T[tex]_i[/tex] = 300 K

1)  the change in entropy due to the volume change alone

from the question; ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])

so change in entropy due to the volume change alone is;

ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex])

we know that, from ideal gas law; PV = nRT

so, nR = P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex]  ---- let this be equation 1

∴ ΔS = P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] × ln(V[tex]_f[/tex]/V[tex]_i[/tex])

we substitute

ΔS = [( 100000 Pa ×  0.01 m³) / 300 K ] × ln(0.026m³ / 0.01m³ )

ΔS = 3.185 J/K  

Therefore, the change in entropy due to the volume change alone, ignoring any effects of changing internal energy is 3.185 J/K.

2)  Final temperature

we know that, in an adiabatic expansion;

[tex]PV^Y[/tex] = K

where Y = 5/3

so

[tex]P_i[/tex][tex]V_i^{(5/3)[/tex] = [tex]P_f[/tex][tex]V_f^{(5/3)[/tex]

[tex]P_f[/tex] = [tex]P_i[/tex][tex]( \frac{V_i}{V_f})^{(5/3)[/tex]

we substitute

[tex]P_f[/tex] = ( 100000 Pa) [tex]( \frac{0.01 m^3}{0.026 m^3})^{(5/3)[/tex]

[tex]P_f[/tex] = 20341.255 Pa

Also from ideal gas law;

PV = nRT

T = PV / nR

so

T[tex]_f[/tex] = P[tex]_f[/tex]V[tex]_f[/tex] / nR

but from equation 1; nR = PV/T

so

T[tex]_f[/tex] = (P[tex]_f[/tex]V[tex]_f[/tex]) / (P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] )

T[tex]_f[/tex] = ( P[tex]_f[/tex]V[tex]_f[/tex]T[tex]_i[/tex] / P[tex]_i[/tex]V[tex]_i[/tex] )

we substitute

T[tex]_f[/tex] = ( 20341.255 Pa × 0.026 m³ ×  300 K) / 100000 Pa × 0.01 m³ )

T[tex]_f[/tex] = 158.66 K

Therefore, the final temperature is 158.66 K

Answer:

After a little online search, I've found the graph of this question, the graph can be seen below.

a) The displacement is defined as the distance between the final position and the initial position.

In the graph, the vertical axis represents the distance. We also can see that both of the lines start in position 0, so at any given time, the displacement of the objects is given by the vertical position in the graph.

Thus, at t = 2 seconds, both lines have the same y-value, this means that the displacements have the same magnitude.

b) The velocity is related to the slope of the curve,

We can clearly see that the slope of graph A and the slope of gaph B are different at t = 2 seconds (graph A is steeper) then we can conclude that the velocities do not have the same magnitude.

c) By Newton's second law, we know that F = m*a

Force equals mass times acceleration.

Acceleration is the rate of change of velocity. When the position graph is a straight line, in any point of the line the slope will be the same, thus the object has always the same velocity, thus the object is not accelerated.

If we do not have a straight line (like in graph A) then the velocity is changing, then we have acceleration, then we have a force.

Then object A has a greater net force (because object B has a net force equal to 0)

d) It is already explained in point c.

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:

The current can be adjusted to control the strength of the

magnetic field.

Answer:

Volume of metal piece = 0.0069 m³ (Approx.)

Explanation:

Given:

Weight of metal in air = 300 N  

Weight of metal in water = 232.5 N

Find:

Volume of metal piece

Specific gravity of metal

Computation:

We know that;

Density of water = 1,000 kg/m³

Buoyant force applied on metal piece = Weight of metal in air - Weight of metal in water

Buoyant force applied on metal piece = 300 N - 232.5 N

Buoyant force applied on metal piece = 67.5 N

Buoyant force = Volume of metal x Density of water x Gravitational force

67.5 = Volume of metal x 1,000 x 9.8

Volume of metal piece = 0.0069 m³ (Approx.)

Answer:

The total energy must be .7 J + .3 J = 1 J     for a particle at the endpoint or midpoint of motion.

Also,    omega = (k / m)^1/2

f = omega / (2 * pi)

omega^2 = 4 pi^2 * f^2 = k / m    

k = 4 * pi^2 * f^2 * m = 40.05

Max KE or PE = 1/2 k A^2

A^2 = 2 * E / k = 2 * 1 / k = .0499

A = .223 meters