Please solve the Problem.

Answer:

It is the smarter option.

Explanation:

Renewable energy is the cheapest source of new power generation for more than two-thirds of the world and has no fuel costs. It can reduce the economic burden of energy bills by eliminating fuel charges — especially when coupled with energy-efficiency upgrades in our homes and businesses.

D. The input rating of the furnace is determined as 100,000 BTU.

Input power rating means the power, expressed in Watts or one of its multiples, for which the energy storage unit has been designed to operate at nominal conditions.

eff = (output rating / input rating) x 100%

80/100 = 80,000 BTU / input rating

0.8 = 80,000 BTU / input rating

input rating = 80,000 BTU/0.8

input rating = 100,000 BTU

Thus, the input rating of the furnace is determined as 100,000 BTU.

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The total head of the pipe is 430,400 Pa.

The total head of the pipe is the total pressure of the pipe and it is calculated as follows;

Pt = Pi + ¹/₂ρv² + ρgh

where;

Pt = ( 350,000) + ¹/₂(1000)(2)²  +  (1000 x 9.8 x 8)

Pt = 430,400 Pa

Thus, the total head or total pressure of the pipe depends on the density of water, speed of water and height above datum.

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Answer:

D. Object D

Explanation:

The hotter the temp the faster the molecules move

Answer:

Hi!!

It should be A, ratation

(sorry that I didnt spell that write)

Have a great day!

To solve this we must be knowing each and every concept related to energy and its different types. Therefore,  the correct option is option A among all the given options.

In physics, energy is the capacity to accomplish work. It can be potential, kinetic, thermal, electric, chemical, radioactive, or in other forms.

There is also heat and work—energy inside the process of being transferred by one body to the other. Energy is always classified according to its type once it has been transmitted. As a result, heat transported could become thermal energy, whereas labor done may emerge as mechanical energy. Radiation is a t type of energy that transfer energy that causes a sunburn  because particles of hot air move in space.

Therefore,  the correct option is option A among all the given options.

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The total distance traveled by the car at the given velocity and time is 900 m.

The given parameters:

The final time of motion of car before coming to rest is calculated as follows;

\(v_f = v_0 -at\\\\0 = 20 - 2t\\\\t = 10 \ s\)

The graph of the car's motion is in the image uploaded.

The total distance traveled by the car is calculated as follows;

\(total \ distance = A \ + B \ + C\\\\total \ distance = (\frac{1 }{2} \times 20 \times 20) \ + (20 \times 30) \ + (\frac{1}{2}\times 10 \times 20)\\\\total \ distance = 900 \ m\)

Thus, the total distance traveled by the car at the given velocity and time is 900 m.

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A golf ball is a unique ball made specifically for the game of golf.

A golf ball must adhere to specific velocity, distance, and symmetry requirements as well as have a mass of no more than 1.620 oz (45.9 g), a diameter of no less than 1.680 inches (42.7 mm), and no more than 1.620 oz (45.9 g) in total.  

Similar to golf clubs, golf balls must pass testing and approval by The R&A and the United States Golf Association before they can be used in competitions. 

Early in the 20th century, it was discovered that dotting the ball gave the player even more control over the trajectory, flight, and spin of the ball.

Therefore, A golf ball is a unique ball made specifically for the game of golf.

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c. Convection

Note that:

• Heat transfer involves the movement of heat energy across the molecules of a substance.

• When snow melts in the sun, there is a movement of heat energy from the air molecules to the snow. This movement of heat from the air also involves the movement of the air molecules

• Convection is the transfer of heat within a fluid, involving the actual movement of the molecules of the fluid

Therefore, from all the points highlighted above, we can conclude that snow melting in the sun is a good example of heat transfer by convection

Answer: B. DLC

B. Visual Updates

B. Competition and expectations

D. Every three months

D. Game expansion packs

C. Visual updates

A. Game expansion

C. Concept art books

A. Create DLC

A. Check licensing agreements

D. Separate her personal finances from the business finances

B. Soundtrack

C. Group dynamics

C. To open a business bank account

B. Onboarding activities

Explanation: :)

It takes same time for A and B to strike the plane.

Speed is measured as the ratio of distance to the time in which the distance was covered. Speed is a scalar quantity as it has only direction and no magnitude.

When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration directed towards the centre of the earth. The path of such a particle is called a projectile, and the motion is called projectile motion.

So both A and B strikes the plane at same time.

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Answer:

The speed of the ball when it reaches equilibrium position is 3.31 m/s

Explanation:

Given;

mass of the object, m = 0.25 kg

initial displacement of the object, h₁ = 0.56 m

spring constant, k = 105 N/m

displacement at equilibrium position, h₂ = 0

initial velocity of the object, v₁ = 0

velocity of the object at equilibrium position = v₂

The change in gravitational potential energy at the equilibrium position is given as;

ΔP.E = mg(h₂ - h₁)

The change in kinetic energy of the object at the equilibrium position is given as;

ΔK.E = ¹/₂m(v₂² - v₁²)  

Apply the principle of conservation of mechanical energy;

ΔK.E  +  ΔP.E = 0

¹/₂m(v₂² - v₁²)  +  mg(h₂ - h₁) = 0

¹/₂m(v₂² - 0)  +  mg(0 - h₁) = 0

¹/₂mv₂²  -  mgh₁  =  0

¹/₂mv₂²  = mgh

¹/₂v₂² = gh

v₂² = 2gh

v₂ = √2gh

v₂ = √(2 x 9.8 x 0.56)

v₂ = 3.31 m/s

Therefore, the speed of the ball when it reaches equilibrium position is 3.31 m/s

The force actually being used to push the mower along the ground is 191 N.

When the physics student exerts a force of 250 N along the handle of the push mower, it's important to consider the components of this force that contribute to the actual force used to push the mower along the ground.

To determine the force used to push the mower along the ground, we need to find the horizontal component of the applied force. The angle of 40° indicates that the applied force can be broken down into two components: the horizontal component and the vertical component. The vertical component of the force is perpendicular to the direction of motion and does not contribute to pushing the mower forward.

To find the horizontal component, we can use trigonometry. The horizontal component is given by the formula:

Horizontal component = Applied force * cos(angle)

Plugging in the values, we get:

Horizontal component = 250 N * cos(40°)

Calculating this value, we find that the horizontal component of the applied force is approximately 191 N.

Therefore, the force actually being used to push the mower along the ground is 191 N. This is the component of the applied force that contributes to the forward motion of the mower, while the remaining vertical component is directed perpendicular to the ground and does not assist in pushing the mower forward.

By exerting a force of 250 N along the handle at a 40° angle, the student effectively applies 191 N of force to push the mower along the ground, ensuring efficient use of their effort while considering the environmental impact.

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The charge on moving particles is calculated as q = (6.9 * 10⁻¹⁹ J) / (v * 325 N/C * 4 m)

Amount of work done in moving unit positive charge without acceleration from one point to another along any path between two points is called potential difference between any two points.

As ΔV = -ΔPE/q

ΔV is potential difference, ΔPE is change in electrical potential energy, and q is charge on the moving particle.

F = qvB

F is magnetic force, v is velocity of the charged particle, and B is magnetic field strength.

W = Fd = qvBd

qvbD = 6.9 * 10⁻¹⁹  J

q = ΔPE / (vBd)

ΔPE is change in electrical potential energy, v is velocity of charged particle, B is magnetic field strength, and d is distance moved in the direction of magnetic field.

q = (6.9 * 10⁻¹⁹ J) / (v * 325 N/C * 4 m)

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Assuming constant acceleration, if v is the final velocity of the ball, then

v ² - (16 m/s)² = 2 (-2 m/s²) (4 m)

v ² = 240 m²/s²

v ≈ 15.5 m/s

We can see here that the diagram that reveals the absorption of photons and production of electrons (photoelectric effect) used in solar PV systems is seen below.

A photon is a fundamental particle or quantum of light.

It is the smallest indivisible unit of electromagnetic radiation, which also includes radio waves, microwaves, X-rays, visible light, ultraviolet light, and infrared radiation. In scientific calculations, photons are represented by the symbol "c" and move at the speed of light in a vacuum while carrying energy.

Wave-particle duality is the idea that photons are both particles and have wave-like characteristics. This means that they can exhibit wave-like properties like interference and diffraction as well as discrete particle behavior, each with a distinct energy and momentum.

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Answer:

Explanation:

The electric flux through a surface is given by the dot product of the electric field and the area vector of the surface:

Φ = E · A

where Φ is the electric flux, E is the electric field, and A is the area vector of the surface.

(a) If the surface lies in the yz plane, its area vector is in the x direction. Therefore, the area vector can be written as A = Ax i, where Ax is the magnitude of the area. The electric field is given as E = ai + bj. Therefore, the flux through the surface is:

Φ = E · A = (ai + bj) · (Ax i) = aAx

(b) If the surface lies in the xz plane, its area vector is in the y direction. Therefore, the area vector can be written as A = Ay j, where Ay is the magnitude of the area. The electric field is given as E = ai + bj. Therefore, the flux through the surface is:

Φ = E · A = (ai + bj) · (Ay j) = bAy

(c) If the surface lies in the xy plane, its area vector is in the z direction. Therefore, the area vector can be written as A = Az k, where Az is the magnitude of the area. The electric field is given as E = ai + bj. Therefore, the flux through the surface is:

Φ = E · A = (ai + bj) · (Az k) = 0

since the dot product of perpendicular vectors is zero.

Answer:

yes its okay i think so it is the correct answer

Answer:

occipital parietal

Explanation:

Hope this helps, plus I took the test

Answer:

Explanation:

The kinetic energy will convert to heat energy (provided the car has friction brakes and not regenerative brakes as might be found on an electric or hybrid) Also assuming level road.

E = ½mv² = ½(1000)30² = 450,000 J

The final speed of cart 2 is -1.0 m/s, and the final speed of cart 3 is 0.5 m/s.

To find the final speeds of cart 2 and cart 3 after the collision, we can use the principles of conservation of momentum and kinetic energy.

Given:

Mass of cart 1 (approaching cart): m = 0.20 kg

Initial velocity of cart 1: v0 = 1.5 m/s

Mass of cart 2: 2m

Mass of cart 3: 3m

Let's denote the final velocities of cart 2 and cart 3 as v2 and v3, respectively.

According to the conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.

Initial momentum = Final momentum

(m * v0) + (2m * 0) + (3m * 0) = m * v2 + 2m * v3 + 3m * v3

Simplifying the equation, we have:

m * v0 = m * v2 + 5m * v3    ...(1)

Since the collision is elastic, the total kinetic energy before the collision should be equal to the total kinetic energy after the collision.

Initial kinetic energy = Final kinetic energy

(1/2) * m *\(v0^2 = (1/2) * m * v2^2 + (1/2) * 2m * v3^2 + (1/2) * 3m * v3^2\)

Simplifying the equation, we have:

(1/2) * m *\(v0^2 = (1/2) * m * v2^2 + m * v3^2 + (3/2) * m * v3^2\)

m * \(v0^2 = m * v2^2 + 2m * v3^2 + 3m * v3^2\)

\(v0^2 = v2^2 + 2v3^2 + 3v3^2\)    ...(2)

Now, we have two equations (equation 1 and equation 2) with two unknowns (v2 and v3). We can solve these equations simultaneously to find the values of v2 and v3.

From equation 1, we can rewrite it as:

v2 = v0 - 5v3

Substituting this expression into equation 2, we get:

\(v0^2 = (v0 - 5v3)^2 + 2v3^2 + 3v3^2\)

Expanding and simplifying the equation, we have:

\(v0^2 = v0^2 - 10v0v3 + 25v3^2 + 2v3^2 + 3v3^2\)

\(0 = -10v0v3 + 30v3^2\)

Rearranging the equation, we get:

10v0v3 = 30\(v3^2\)

v0 = 3v3

Solving for v3, we find:

v3 = v0/3 = (1.5 m/s) / 3 = 0.5 m/s

Substituting this value of v3 back into the expression for v2, we have:

v2 = v0 - 5v3 = 1.5 m/s - 5 * 0.5 m/s = 1.5 m/s - 2.5 m/s = -1.0 m/s

Therefore, the final speed of cart 2 is -1.0 m/s (indicating it moves in the opposite direction with respect to the initial velocity), and the final speed of cart 3 is 0.5 m/s.

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1 Light

1 switch

2 cells

1 Voltmeter

The circuit is drawn in black, the arrows near to the wire give the direction for current and electrons. Remember that the electrons go in the opposite direction of electrons.

The rest of the arrows in some parts of the drawing are symbols for cells and lights.

Taking into account the definition of kinetic energy, the velocity of the object with a kinetic energy of 250 J and a mass of 32 kg is 3.95 m/s².

Kinetic energy is the energy possessed by a body or system due to its movement.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a certain mass and in a position of rest, until it reaches a certain speed. Once the final speed is reached, the amount of kinetic energy accumulated will remain constant, that is, it will not vary, unless another force acts on the body again.

Kinetic energy is represented by the following expression:

Ec = 1/2×m×v²

Where:

In this case, you know:

Replacing in the definition of kinetic energy:

250 J = 1/2× 32 kg×v²

Solving:

250 J÷ (1/2× 32 kg) = v²

15.625 J÷kg = v²

√15.625 J÷kg = v

3.95 m/s² = v

Finally, the velocity of the object is 3.95 m/s².

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Answer:

Explanation:

Force of friction

F = μ mg

μ is coefficient of friction , m is mass and g is acceleration due to gravity .

If f be the force applied to pull the sled , the horizontal component of force should be equal to frictional  force

The vertical component of applied force will reduce the normal force or reaction force from the ground

Reaction force R = mg - f sin28.3

frictional force = μ R where μ is coefficient of friction

frictional force = μ x (mg - f sin28.3 )

This force should be equal to horizontal component of f

μ x (mg - f sin28.3 ) = f cos 28.3

μ x mg = f μsin28.3 + f cos 28.3

f = μ x mg / (μsin28.3 + cos 28.3 )

a )

work done by pulling force  = force x displacement

f cos28.3 x d

μ x mg d cos28.3  / (μsin28.3 + cos 28.3 )

b ) Putting the given values

= .155 x 53.1 x 9.8 x 25.3 cos28.3 / ( .155 x sin28.3 + cos 28.3 )

= 1796.76 / (.073 + .88 )

= 1885.37  J

c )

Work done by frictional force

= frictional force x displacement

=  -  μ x (mg - f sin28.3 ) x d  

= -  μ x mgd + f μsin28.3  x d

= -  μ x mgd + μsin28.3  x d x μ x mg / (μsin28.3 + cos 28.3 )

d )

Putting the values in the equation above

- .155 x 53.1 x 9.8 x 25.3 +

.155 x .474 x 25.3 x .155 x 53.1 x 9.8 /( .155 x .474 + .88)

= -2040.67 + 149.92 / .95347

= -2040.67 + 157.23

= -1883.44 J .

Answer:

the final speed of object A changed by a factor of  \(\frac{1}{\sqrt{3} }\) = 0.58

the final speed of object B changed by a factor of \(\sqrt{\frac{5}{3} }\) = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = \(m_A\)

then, the mass of object B = \(m_B = \frac{m_A}{4}\)

work done on object A = -18 J

work done on object B = -18 J

let \(v_i\) be the initial speed

let \(v_f\) be the final speed

For object A;

\(K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2 = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ - v_i^2 )\ =- 18\\\\v_f^2 \ - v_i^2 = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ - v_i^2 = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\\)

\(v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\\)

Thus, the final speed of object A changed by a factor of  \(\frac{1}{\sqrt{3} }\) = 0.58

To obtain the change in the final speed of object B, apply the following equations.

\(K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B = \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\\)

\(\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i\)

Thus, the final speed of object B changed by a factor of \(\sqrt{\frac{5}{3} }\) = 1.29

Velocity is the rate of change of position with respect to time. The velocity of objects A and B is 0.577 and  1.29 times the initial velocities.

Velocity is the rate of change of position with respect to time.

Given to us

The kinetic energy of object A before work is applied,  \(KE_A = 27\rm\ J\)

Mass of object A = \(m_a\)

Mass of object B = \(m_b = \dfrac{m_a}{4}\)

Work done on both the objects, w = -18 J

We know that the kinetic energy of object A is 27 j, therefore,

\(KE_A = 27 J\\\\\dfrac{1}{2}m_av_i^2 = 27\\\\m_av_i^2 = 54\\\\m_a = \dfrac{54}{v_i^2}\)

We know that the work done on object A is -18 J, therefore,  the difference in the Kinetic energy of the object will be work,

\(KE_f - KE_i = -18\\\\\dfrac{1}{2}m_Av_f^2 -\dfrac{1}{2}m_Av_i^2 = 18\\\\v_f^2 -v_i^2 = \dfrac{-18 \times 2}{m_A}\\\\v_f^2 -v_i^2 = \dfrac{-36}{m_A}\)

Substitute the value of \(m_a\),

\(v_f^2 -v_i^2 = \dfrac{-36}{m_A}\\\\v_f^2 -v_i^2 = \dfrac{-36}{\dfrac{54}{v_i^2}}\\\\v_f^2 -v_i^2 = \dfrac{-36 v_i^2}{54}\\\\v_f^2 = \dfrac{1v_i^2}{3}+v_i^2\\\\v_f^2 = \dfrac{-36 v_i^2}{54}\\\\v_f = v_i\sqrt{\dfrac{1}{3}}\\\\v_f = 0.577v_i\)

Hence, the velocity of object A is 0.577 times the initial velocity.

Kinetic Energy of the object B,

\(KE_B = \dfrac{1}{2}m_B v_i^2\\\\KE_B = \dfrac{1}{2}\dfrac{m_A}{4} v_i^2\\\\KE_B =\dfrac{m_A}{8} v_i^2\\\\KE_B =\dfrac{m_A}{8} v_i^2\\\\KE_B =\dfrac{54}{8} \\\)

We know that the work done on object B is -18 J, therefore,  the difference in the Kinetic energy of the object will be work,

\(KE_f - KE_i = -18\\\\\dfrac{1}{2}m_Bv_f^2 -\dfrac{1}{2}m_Bv_i^2 = 18\\\\v_f^2 -v_i^2 = \dfrac{-18 \times 2}{m_B}\\\\v_f^2 -v_i^2 = \dfrac{-36}{m_B}\\\\v_f^2 -v_i^2 = \dfrac{-36}{\dfrac{m_A}{4}}\\\\v_f^2 -v_i^2 = \dfrac{-144}{m_A}\\\\}}\\\\v_f^2 -v_i^2 = \dfrac{-144}{\dfrac{54}{v_i^2}}\)

\(v_f^2 -v_i^2 = \dfrac{-144}{\dfrac{54}{v_i^2}}\\\\v_f^2 -v_i^2 = \dfrac{-144v_i^2}{54}\\\\v_f^2 = \dfrac{-144v_i^2}{54} + v_i^2\\\\v_f^2 = \dfrac{-144v_i^2}{54} + v_i^2\\\\v_f^2 = \dfrac{-5v_i^2}{3}\\\\ | v_f |= \sqrt{\dfrac{5v_i^2}{3}}\\\\v_f = \sqrt{\dfrac{5v_i^2}{3}}\\v_f = v_i\sqrt{\dfrac{5}{3}}\\\\v_f = 1.29 v_i\)

hence, the velocity of object B is 1.29 times the initial velocity.

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Answer:

In an ionic bond the valence electrons are transferred from the metal

Explanation:

Answer:

\(3.1\times 10^{5}m/s\)

Explanation:

The computation of the speed does the proton gain is shown below:

The potential difference is the difference that reflects the work done as per the unit charged

So, the work done should be

= Potential difference × Charge

Given that

Charge on a proton is

= 1.6 × 10^-19 C

Potential difference = 500 V

\(v= \sqrt{\frac{2.q.\Delta V}{m_{p}}} \\\\\\= \sqrt{\frac{2\times 1.6\times 10^{-19}\times 5\times 10^{2}}{1.67\times 10^{-27}}}\)

\(v= \sqrt{9.58\times 10^{10}}m/s \\\\= 3.095\times 10^{5}m/s\\\\\approx 3.1\times 10^{5}m/s\)

Simply we applied the above formulas

The induced e.m.f is -0.02sin(10^3t) (V) which is option D

Induced voltage refers to the electrical voltage that is generated in a conductor or coil due to a changing magnetic field.

This phenomenon is known as electromagnetic induction and is the basis for many electrical devices such as generators and transformers.

Induced voltage can be calculated using Faraday's law, which states that the induced electromotive force (EMF) is equal to the rate of change of magnetic flux.

The induced voltage can be either positive or negative depending on the direction of the changing magnetic field and the orientation of the conductor or coil.

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