The layer of the sun that radiates most of the light that reaches earth is the

Answer:

(a) 5.73 * 10¹³ m/s²

(b) 4.99 * 10⁵ m/s

Explanation:

(a) We know that Electrical force is the product of electric charge and electric field:

F = qE

We also know that Force is the product of mass and acceleration:

F = ma

Therefore, equating both of them:

ma = qE

Acceleration, a, will be:

a = (qE) / m

Electric charge of a positron = 1.602 * 10^(-19) C

Mass of a positron = 9.11 * 10^(-31) kg

Acceleration will be:

a = (1.602 * 10^(-19)) * 9.11 * 10^(-31)) / 326

a = 5.73 * 10¹³ m/s²

(b) Acceleration is the time rate of change of velocity. It is given as:

a = v/t

Velocity, v, will then be:

v = a * t

Time, t = 8.7 * 10^(-9) s

Therefore, velocity will be:

v = 5.73 * 10^(13) * 8.7 * 10^(-9)

v = 49.85 * 10⁴ m/s = 4.99 * 10⁵ m/s

Answer:

Two major causes are outline bellow

1. The presence of air in the system

2. Clogged condenser

Explanation:

1. The presence of air in the system

One of the causes that have been established in relation to high compressor discharge pressure is the presence of air in the system. When this takes place, your best solution is to recharge the system.

2. Clogged condenser

Another is a clogged condenser in which case you will need to clean the condenser so that it will function properly. When you happen to spot that the discharge valve is closed and it is causing high discharge pressure on the compressor, you can solve that easily by opening the valve

The most likely cause of a refrigerant system's high-side pressure exceeding 375 psig is a blockage, overcharging, or a malfunctioning expansion valve. These issues prevent proper refrigerant flow and pressure regulation within the vapor compression system.

If a technician observes that the high-side pressure of a refrigerant system instantly exceeds 375 psig once the compressor is engaged, the most likely cause for this condition could be a blockage in the condenser or the liquid line, overcharging of the refrigerant, or a malfunctioning expansion valve that is not allowing the liquefied refrigerant to flow properly into the evaporator. The blockage could result in an excessively high pressure build-up because the refrigerant cannot circulate properly. Overcharging of the system adds too much refrigerant, which also leads to high pressure. Additionally, if the expansion valve is stuck closed or not functioning correctly, it would prevent the refrigerant from moving into the evaporator, causing the pressure to build up on the high side.

In a typical vapor compression system, as the electrically driven compressor introduces work (W), it raises the temperature and pressure of the refrigerant, forcing it into the condenser coils. Here, heat transfer occurs as the gas inside the coils is higher in temperature than the room, leading to condensation of the gas back into a liquid. The cooled pressurized liquid passes through the expansion, or pressure-reducing valve, and goes to the evaporator coils located outdoors where it is further cooled by expansion.

Answer:

N = 195 turns

Explanation:

The inductance of the inductor, L = 500 μH = 500 * 10⁻⁶H

The length of the tube, l = 12 cm = 0.12 m

The diameter of the tube, d = 4 cm = 0.04 m

Radius, r = 0.04/2 = 0.02 m

Area of the tube, A = πr² = 0.02²π = 0.0004π m²

[tex]\mu_{0} = 4\pi * 10^{-7}[/tex]

The inductance of a solenoid is given by:

[tex]L = \frac{\mu_{0}N^{2} A }{l}[/tex]

[tex]500 * 10^{-6} = \frac{4\pi *10^{-7} N^{2} *4\pi *10^{-4} }{0.12}\\500 * 10^{-6} = 0.00000001316N^{2} \\N^{2} = \frac{500 * 10^{-6}}{0.00000001316}\\N^{2} = 37995.44\\N = \sqrt{37995.44} \\N = 194.92 turns[/tex]

Answer:

The velocity of the 6.4 kg block immediately after the collision is 3.6 m/s in the original direction.

Explanation:

Given;

mass of first block, m₁ = 6.4 kg

initial speed of first block, u₁ =  5.4 m/s

mass of second block, m₂ = 12.8 kg

initial speed of second block, u₂ = 3.6 m/s

final speed of second block, v₂ = 4.5 m/s

To determine the final speed of 6.4 kg block immediately after the collision, we apply principle of conservation linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁  +  m₂u₂ = m₁v₁  + m₂v₂

where;

v₁ is the final speed of 6.4 kg block

v₂ is the final speed of 12.8 kg block

Substitute the given values of m₁, u₁, m₂,  u₂ , v₂ and calculate  v₁

6.4 x 5.4  +  12.8 x 3.6 = 6.4v₁  +  12.8 x 4.5

34.56 + 46.08  = 6.4v₁ + 57.6

80.64 = 6.4v₁  +  57.6

80.64 - 57.6 = 6.4v₁

23.04 = 6.4v₁

v₁ = 23.04 / 6.4

v₁  = 3.6 m/s

Therefore, the velocity of the 6.4 kg block immediately after the collision is 3.6 m/s in the original direction.

Answer:

(c) 10m/s

Explanation:

to find the speed of the waves you can use the following formula:

[tex]v=\frac{\lambda}{T}[/tex]

λ: wavelength of the wave

T: period

the wavelength is the distance between crests = 48m

the period is the time of a complete oscillation of the wave. In this case you have that the float takes 2.4 s to go from its highest to the lowest point. The period will be twice that time:

T = 2(2.4s)=4.8s

by replacing you obtain:

[tex]v=\frac{48m}{4.8s}=10\frac{m}{s}[/tex]

the answer is (c) 10m/s

The resistance means resist, means to resist the current, so we are here given that, three resistors are connected in series, so as we know that, if n resistors are connected in series combination, then the total resistance is the individual sum of all n Resistors, so if we apply the same formula here, we will be having :

[tex]{:\implies \quad \sf R_{Total}=3+4+5}[/tex]

[tex]{:\implies \quad \boxed{\bf{R_{Total}=12\Omega}}}[/tex]

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Answer:

Crust

Explanation:

When the convection currents flow in the mantle they also move the crust. This is best explained as that the crust gets a free ride with the convection currents in the mantle; just as a conveyor belt in a factory moves boxes.

Answer:

The ranking of the stars are 5 → 2 → 4 → 3 → 1

Explanation:

To solve the question, we note that an approaching star has a blue-shift, a regressing star has a red-shift and a crossing star will have no shift.

The magnitude of the blue-shift or red-shift is dependent on the speed of the stars.

Therefore we have the rank of the stars based on the Doppler shift that would be detected on earth from largest blue-shift to largest red-shift given as

5 → 2 → 4 → 3 → 1.

The Doppler shift affects the wavelengths of light emitted or reflected by objects in motion. Larger blueshifts indicate faster speeds towards us, while larger redshifts indicate faster speeds away from us. Stars with no shift do not have a detected Doppler shift.

The Doppler shift is a phenomenon that occurs when an object's motion causes a change in the wavelengths of light that it emits or reflects. In the case of stars, if a star is moving away from us, its light will be redshifted, meaning the wavelengths of the light will be longer. On the other hand, if a star is moving towards us, its light will be blueshifted, meaning the wavelengths of the light will be shorter.

To rank the stars based on the Doppler shift that we would detect on Earth, we need to understand that larger blueshifts indicate faster speeds towards us, and larger redshifts indicate faster speeds away from us. So, the star with the largest blueshift will have the highest ranking for blueshift, while the star with the largest redshift will have the highest ranking for redshift. Stars with no shift will have no Doppler shift detected.

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Answer:

Complete question

Luis is trying to push a box of new soccer balls across the floor.

If the box is not moving, which of the following must be true?

A. The box is exerting a larger force on Luis than he is exerting on the box

B. There is another force acting on the box that balances Luis's force.

C. Luis is applying a force that acts at a distance.

D. There is no force of friction acting on the box.

Explanation:

Using newton second law of motion

ΣF = ma

Now, for a body not to move when a force is acting on it means that the body is in equilibrium and acceleration a = 0

Therefore,

Fnet = 0

So, if he is applying a force F to push the box and the box is not moving then, there is an external force that is pushing the force back opposite the direction he his pushing and this force counterbalance is own force.

F—F' = 0

F' = F

So, F' is the counter balance force and it is equal to the force applied by Luis

Or it might be frictional force, because if the static friction is not overcome, then, the body will not leave it's state of rest. So if the fictional force is very high, then the box will not leave it rest position and we also know that frictional force opposes motion,

F—Fr=0

F = Fr

So using this explanation,.

The answer is B

B. There is another force acting on the box that balances Luis's force.

Answer:

Speed at which the bat have to fly is 90.895 m/s away from the human (listener) in positive direction.

Explanation:

Given:

Frequency of the bat that is source here, [tex]f_S[/tex] = 25.3 kHz = 25.3 * 10^3 Hz

Frequency of the listener (human), [tex]f_L[/tex] = 20 kHz = 20*10^3 Hz

We have to identify how fast the bat have to fly in order for a person to hear these chirps .

Let the velocity of bat that is source is "Vs" and "Vs" = "Vbat".

Doppler effects formulae :

Using the above formula and considering that the bat is moving away so that the human can listen the chirps also [tex]V_L=0[/tex] as listener is stationary.

⇒ [tex]f_L=(\frac{V+V_L}{V+V_S}) f_S[/tex]   ⇒ [tex]f_L=(\frac{V+0}{V+V_S}) f_S[/tex]

Re-arranging in terms of Vs.

⇒ [tex]V+V_S =\frac{V\times f_S}{f_L}[/tex]

⇒ [tex]V_S =\frac{V\times f_S}{f_L}-V[/tex]

⇒ [tex]V_S =\frac{343\times 25.3\times 10^3}{20\times 10^3}-343[/tex]

⇒ [tex]V_S=90.895[/tex] m/s

The speed at which the bat have to fly is 90.895 m/s away from the human (listener) in positive direction.

Answer:

The radius of strontium atom is

2.14 E-8cm

Explanation:

Go through the attached file for a comprehensive detailed explanation.

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

[tex]x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)[/tex]

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

[tex]\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}[/tex]

with this value you can compute the frequency:

a)

[tex]f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz[/tex]

b)

the mass of the block is given by the formula:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg[/tex]

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

[tex]\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s[/tex]

Finally, the amplitude is:

[tex]x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m[/tex]

Final answer:

Newton's law of universal gravitation states that all objects in the universe attract each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.

Explanation:

Newton's Law of Universal Gravitation:

Sir Isaac Newton's law of universal gravitation states that every object in the universe attracts every other object with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The gravitational force is universally attractive and only depends on mass and distance, following the equation F = G × (m1 × m2) / r², where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between the centers of the two masses.

a) 120 s

b) v = 0.052R [m/s]

Explanation:

a)

The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).

The graph of the problem is missing, find it in attachment.

To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.

The first point we take is t = 0, when the position of the book is x = 0.

Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.

Therefore, the period is

T = 120 s - 0 s = 120 s

b)

The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.

The perimeter of the wheel is:

[tex]L=2\pi R[/tex]

where R is the radius of the wheel.

The period of revolution is:

[tex]T=120 s[/tex]

Therefore, the tangential speed of the book is:

[tex]v=\frac{L}{T}=\frac{2\pi R}{120}=0.052R[/tex]

Natural gas burning on a stove is exothermic (δh < 0), isopropyl alcohol evaporating from the skin is endothermic (δh > 0), and water condensing from steam is exothermic (δh < 0).

(a) Natural gas burning on a stove: This is an exothermic process. The combustion of natural gas on a stove releases a huge amount of heat which cooks the food. The sign of δh (enthalpy change) in this process is negative (δh < 0) as heat is released.

(b) Isopropyl alcohol evaporating from the skin: This is an endothermic process. When isopropyl alcohol evaporates from our skin, it absorbs heat from our body which causes a cooling effect. Hence, the sign of δh is positive (δh > 0) as heat is absorbed.

(c) Water condensing from steam: This is an exothermic process. The condensation process releases heat and hence, the sign of δh is negative (δh < 0).

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Answer: The correct and is

D. Extrusion

Explanation:

What is extrusion?

Extrusion is a manufacturing process used to make basically pipes and hoses.

The pvc granules are melt into a liquid which is forced through a die, forming a long 'tube like' shape. The shape of the die determines the shape of the tube.

The extrusion is then cooled and forms a solid shape

Extrusion moulding is used to create products with a consistent cross-section.

Answer:

It is True

Explanation:.

A  commander assigns a zone reconnaissance mission when he seeks additional information on a zone before committing other forces in the zone. It is appropriate when the enemy situation is vague,  existing knowledge of the terrain is limited, or combat operations have altered the terrain. A zone  reconnaissance could include several route or area reconnaissance missions assigned to subordinate units.

Answer:

469.6KJ

Explanation:

Heat energy required can be calculated using the formula

H = mc∆t where

m is the mass of the water

c is the specific heat capacity of the water

∆t is the change in temperature of the water

Given m = 9.78kg

c = 4186j/kg-c.

∆t = 52.07°C - 40.82°C

∆t = 11.25°C

H = 9.78 × 4186 × 11.25

H = 460,564.65Joules

= 460.6KJ

Answer:

the amount of energy required to raise the temperature of the water is 460564.65 J

Explanation:

The energy required to raise the temperature of water can be calculated as follows;

Q = mcΔθ

where;

Q is the quantity of heat or energy required to raise the temperature of water

m is mass of water

c is specific heat capacity of water

Δθ is change in temperature = T₂ - T₁

Given;

m =  9.78kg

c = 4186j/kg-c

Δθ = T₂ - T₁ = 52.07°C - 40.82°C  = 11.25°C

Q = mcΔθ

Q = (9.78)(4186)(11.25)

Q = 460564.65 J

Therefore, the amount of energy required to raise the temperature of the water is 460564.65 J

If the same force acts upon another object whose mass is 13 kg the acceleration is [tex]4m/s^2[/tex].

Force is a fundamental concept in physics that describes the interaction between objects that can cause a change in their motion or shape. It's a vector quantity, meaning it has both magnitude and direction. In simpler terms, force is what can make things move, stop, change direction, or deform.

Given:

Masses, [tex]m_1 = 2\ kg[/tex] and [tex]m_2 = 13\ kg[/tex]

Acceleration, [tex]a = 26\ m/s^2[/tex]

The force is computed as:

[tex]F = m_1a\\F = 2\times26\\F = 52\ N[/tex]

The acceleration for mass 52 kilograms is:

[tex]F = m_2a\\a = F/m_2\\a = 52/ 13\\a = 4\ m/s^2[/tex]

Hence, if the same force acts upon another object whose mass is 13 kg the acceleration is [tex]4m/s^2[/tex].

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Applying Newton's second law of motion, the acceleration of an object is directly proportional to the force acting on it, but inversely proportional to its mass. The acceleration of a 13kg object, when acted upon by the same force that gives a 2kg object an acceleration of 26m/s^2, would be 4 m/s^2.

The question basically asks us to apply Newton's second law of motion, which states that the acceleration of a system is directly proportional to the net external force acting on the system, but inversely proportional to its mass. In equation form, this can be written as F=ma, where F is the force, m is the mass, and a is the acceleration. This is also referred to as the principle of constant acceleration, in which the force applied to an object will either accelerate or decelerate it at a constant rate, provided the mass of the object remains unchanged.

In this case, we know that a 2kg object accelerates at 26m/s^2 under the influence of a constant force. Therefore, the force (F) acting on it can be computed as F = m * a = 2kg * 26m/s^2 = 52N. Now, if the same force acts on another object with a mass of 13kg, the acceleration (a) of this object can be computed by rearranging the equation as a = F/m = 52N / 13kg = 4m/s^2. So, the acceleration of the second object, under the same force, would be 4 m/s^2.

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Answer:

Explanation:

Given that, .

Spring constant.

K = 6.1 N/m

Amplitude of oscillation

A = 10.3cm = 0.103m

Half wave, it speed is

V = 31.3cm/s = 0.313m/s

Mass of block

M =?

The is half of the distance between the equilibrium position and endpoint

Then,

x = A/2

Where,

A is the amplitude

x is the position of the block

x = A / 2 = 0.103/2

x = 0.0515m

The velocity if the block at any given position is given as

v = ω√(A²—x²)

Then,

ω = v / √(A²—x²)

Where

ω is angular frequency

v Is the velocity of the block

The force constant is given as

k = mω²

Where,

K is spring constant

ω is angular frequency

Substitute ω into k

Then,

k = m (v / √(A²—x²))²

k = mv² / (A²—x²)

Make m subject of formula since we want to find m

m = k(A²—x²) / v²

m = 6.1 (0.103²—0.0515²) / 0.313²

m = 6.1 × 7.96 × 10^-3 / 0.313²

m = 0.495 kg

The mass of the block is approximately 0.5kg

Answer:

The mass of the block is 496 g

Explanation:

Here we have by the principle of conservation of energy;

Energy in spring, E = 0.5·k·A²

Where:

k = Spring constant = 6.1 N/m

A = Amplitude of motion = 10.3 cm = 0.103 m

E = 0.5×6.1×0.103² = ‭0.03235745 J= 3.24 × 10⁻² J

At half way, we have

[tex]E = \frac{1}{2} k(\frac{A}{2} )^{2} + \frac{1}{2} mv^{2} = \frac{E}{4} + \frac{1}{2} mv^{2}[/tex]

Where:

m = Mass of the block

v = Velocity of block at the instant (Halfway between its equilibrium position and the endpoint)

Therefore,

[tex]\frac{3}{4}E = \frac{1}{2} mv^{2}[/tex] or

m = [tex]\frac{3}{2v^2}E[/tex] = [tex]\frac{3}{2\times 0.313^2} \times 3.24 \times 10^{-2}[/tex]= 0.496 kg = 496 g.

Answer:

Part(a): At [tex]\bf{t = 470~ms}[/tex] the center of mass will travel [tex]\bf{0.903~m}[/tex].

Part(b): The velocity of the first stone is [tex]\bf{4.606~m.s^{-1}}[/tex] and the velocity of the second stone is [tex]\bf{4.067~m.s^{-1}}[/tex].

Explanation:

Given:

The first stone is dropped at [tex]t_{1}=0~s[/tex].

The second stone is dropped at, [tex]t_{2}=55~ms=0.055~s[/tex]

The mass of the second stone is 3 times the mass of the first.

Both the stones are dropped from the same point.

Consider the mass of the first stone be [tex]m[/tex]. So the mass of the second stone is [tex]3m[/tex].

(a)

The formula to calculate the distance traveled by each stone is given by

[tex]y = \dfrac{1}{2}gt^{2}~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]

where [tex]g[/tex] is the acceleration due to gravity and [tex]t[/tex] is the time taken by each stone.

Substituting [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex], [tex]y_{1}[/tex] for [tex]y[/tex] and [tex]470~ms[/tex] or [tex]0.47~s[/tex] for first stone in equation (1), we have

[tex]y_{1}&=& \dfrac{1}{2}(9.8~m.s^{-2})(0.47~s)^{2}\\~~~~&=& 1.08~m[/tex]

where [tex]y_{1}[/tex] is the distance traveled by the first stone.

Substituting [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex], [tex]y_{2}[/tex] for [tex]y[/tex] and  [tex](470-55)~ms = 415~ms[/tex] or [tex]0.415~s[/tex] for second stone in equation (1), we have

[tex]y_{2}&=& \dfrac{1}{2}(9.8~m.s^{-2})(0.415~s)^{2}\\~~~~&=& 0.844~m[/tex]

The formula to calculate the distance traveled by the center of mass is given by

[tex]y_{c} &=& \dfrac{my_{1}+3my_{2}}{m+3m} \\~~~~&=& \dfrac{1.08m + 0.844(3m)}{4m}\\~~~~&=& 0.903~m[/tex]

(b)

The formula to calculate the velocity of each stone is given by

[tex]v=gt~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]

Substituting [tex]v_{1}[/tex] for [tex]v[/tex],  [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex] and [tex]0.47~s[/tex] for [tex]t[/tex] in equation (2), we have

[tex]v_{1} &=& (9.8~m.s^{-2})(0.47~s)\\~~~~&=& 4.606~m.s^{-1}[/tex]

where [tex]v_{1}[/tex] is the velocity of the first stone after [tex]470~ms[/tex].

Substituting [tex]v_{2}[/tex] for [tex]v[/tex],  [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex] and [tex]0.415~s[/tex] for [tex]t[/tex] in equation (2), we have

[tex]v_{2} &=& (9.8~m.s^{-2})(0.415~s)\\~~~~&=& 4.067~m.s^{-1}[/tex]

where [tex]v_{2}[/tex] is the velocity of the second stone after [tex]470~ms[/tex].

An increase in the level of nitrogen dioxide and a decrease in the level of ground-level ozone occurs.

Explanation:

Ozone gas is normally found in stratosphere, it protects us from solar radiation. Under certain circumstances, it can be formed on the ground level.

Conditions that have to be met in order for this to happen are existing of nitrogen oxides and volatile organic compounds and their reaction catalyzed by heat and light.

During dark, cold and cloudy days, due to the lack of heat and light from the Sun, formation of ozone will be decreased.

Also, this will enable building up of nitrogen dioxide, due to the same reason, leading to increased concentration.

Answer:

a. 60 V b. 6 A c. 360 W

Explanation:

a. Voltage in secondary

For an ideal transformer,

N₁/N₂ = I₂/I₁ = V₁/V₂ where N₁ = turns in primary = 50 turns, N₂ = turns in secondary = 250 turns, V₁ = voltage in primary = 12 V, V₂ = voltage in secondary = ? V

N₁/N₂ = V₁/V₂

V₂ = V₁N₂/N₁ = 250 × 12/50 = 60 V

b. Since V₂ = I₂R,

I₂ = V₂/R  R = 10 Ω

I₂ = 60 V/10 Ω

I₂ = 6 A

c. We first calculate the current in the primary from

N₁/N₂ = I₂/I₁  where I₁ = primary current

I₁ = N₂I₂/N₁ = 250 × 6 A/50 = 30 A

The power supplied to the primary is thus

P = I₁V₁ = 30 A × 12 V = 360 W

(a) The voltage supply at the secondary coil is of 60 V.

(b) The current flow through the 10- ohm device is 6 A.

(c)  The power supply to the primary coil is 360 W.

Given Data:

Number of turns of primary coil is, n1 = 50.

Number of turns of secondary coil is, n2 = 250.

The voltage supply to primary coil is, V1 = 12 V.

The resistance of device is, R = 10 ohm.

(a)

The first part of the given problem is based on the Transformer equation relating the number of turns of coils at primary and secondary and voltage supply at each coils.

Therefore,

n1 / n2 = V1 / V2

Here,

V2 is the voltage supply to the secondary coil.

Solving as,

50 / 250 = 12 / V2

V2 = 250 / 50 × 12

V2 = 60 V.

Thus, we can conclude that the voltage supply at the secondary coil is of 60 V.

(b)

Now in order to find the current flow through the 10 - ohm device, we can use the Ohm's law as,

V2 = I' × R

Solving as,

60 = I' × 10

I ' = 60 / 10

I' = 6 A

Thus, we can conclude that the current flow through the 10- ohm device is 6 A.

(c)

Now, we first calculate the current in the primary from

n1 / n2 = I' / I  

here,

I is the primary current

Solving as,

I = n2 × I' / n1

I = 250 × 6 /50

I = 30 A

The power supplied to the primary is thus

P1 = I × V1

P1 = 30  × 12  

P1 = 360 W

Thus, we can conclude that the power supply to the primary coil is 360 W.

Learn more about the transformer equation here:

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Answer:

Pseudoscience is not science - it involves beliefs and opinions.

Explanation:

i got it right on the test

Answer:

Pseudoscience is not science - it involves beliefs and opinions.

Explanation:

Answer:

The answer to your question is below

Explanation:

To explain what happens with the ball we must remember the Law of Conservation of Energy.

This law states that the energy can be neither created nor destroyed only converted from one form of energy to another.

Then,

At the top of the hill, the potential energy is maximum and the kinetic energy equals to zero.

When the ball starts to roll down the potential energy will be lower and the kinetic energy will have a low value.

At the middle of the hill, both energies have the same values.

At the end of the hill, the potential energy will be equal to zero and the kinetic energy will be maximum.

According to this law, energy cannot be generated or annihilated; it can only be transformed from one type of energy to another.

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Answer:electrical,kinetic,kinetic,electrical

Explanation:

Answer:

Electrical, kinetic, kinetic, electrical

Explanation:

Just did it

Answer:

D. Producing ligaments and tendons.

just to the quiz on edge

The skeletal system provides structure, assists in movement, protects internal organs, and produces both red and white blood cells. However, it does not produce ligaments and tendons, which are part of the muscular system and connective tissues. Thus, the correct option is D.

The skeletal system has multiple functions within the body. These include providing structure, facilitating movement, protecting internal organs, producing red blood cells, and producing white blood cells. These cells are crucial as they are responsible for oxygen transport around the body and defense against disease. However, producing ligaments and tendons is not a function of the skeletal system. Ligaments and tendons are part of the muscular system and connective tissues. Ligaments connect bones to each other, and tendons connect muscle to bone.

Therefore, the correct answer would be D. producing ligaments and tendons.

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Answer:

C) An atmosphere might moderate temperatures, making more or even, as on Earth

Explanation:

Atmosphere has moderating effect on a planet . It not only protects the planet from getting heated excessively during day time but it also protects it from getting too  much cooled during night . In day time , it filters excessive heat out and during night it traps the heat coming out . Both ways , it tries to keep the temperature of the planet evenly balanced in day and night.

The Answer Is C) An atmosphere might moderate temperatures, making more or even, as on Earth

Answer:

Change in momentum of the stone is 3.673 kg.m/s.

Explanation:

Given:

Mass of the ball on the horizontal the surface, m = 0.10 kg

Velocity of the ball with which it hits the stone, v = 20 m/s

According to the question it rebounds with 70% of the initial kinetic energy.

We have to find the change in momentum i.e Δp

Before that:

We have to calculate the rebound velocity with which the object rebounds.

Lets say that the rebound velocity be "v1" and KE remaining after the object rebounds be "KE1".

⇒ [tex]KE_1=0.7\times \frac{mv^2}{2}[/tex]    

⇒ [tex]KE_1=0.7\times \frac{0.10\times (20)^2}{2}[/tex]

⇒ [tex]KE_1=0.7\times \frac{0.10\times 400}{2}[/tex]

⇒ [tex]KE_1=14[/tex] Joules (J).

Rebound velocity "v1".

⇒ [tex]KE_1=\frac{m(v_1)^2}{2}[/tex]

⇒ [tex]v_1 = \sqrt{\frac{2KE_1}{m} }[/tex]

⇒ [tex]v_1 = \sqrt{\frac{2\times 14}{0.10} }[/tex]

⇒ [tex]v_1=16.73[/tex]

⇒ [tex]v_1=-16.73[/tex] m/s ...as it rebounds.

Change in momentum Δp.

⇒ [tex]\triangle p= m\triangle v[/tex]

⇒ [tex]\triangle p= 0.10\times (20-(-16.73)[/tex]

⇒ [tex]\triangle p= 0.10\times (20+16.73)[/tex]

⇒ [tex]\triangle p= 0.10\times (36.73)[/tex]

⇒ [tex]\triangle p = 3.673[/tex] Kg.m/s

The magnitude of the change in momentum of the stone is 3.673 kg.m/s.