Q3. The following potential differences were measured in a circuit: A is +4 volts with respect to (wrt) B B is +3 volts wrt C C is +2 volts wrt D D is +1 volts wrt to ground What is the potential difference of: i) ii) iii) iv) A wrt ground B wrt ground A wrt C B wrt D​

Answer: The scenario violates the second law of thermodynamics.

Explanation: The second law states that heat cannot be converted into work without some loss of usable energy, and that the amount of usable energy in a closed system will always decrease over time. Therefore, the machine described in the scenario cannot exist because it would violate the second law by converting all of the heat into electricity without any loss of usable energy.

Answer:

Inelastic

Explanation:

If a football player running down the field is tackled by another football player who holds onto the first football player so that they both fly out of bounds together, that collision is inelastic.

An inelastic collision can be defined as a type of collision in which the total kinetic energy possessed by the colliding bodies doesn't remain the same as a result of internal friction.

This ultimately implies that, in an inelastic collision the maximum amount of kinetic energy possessed by the colliding bodies isn't conserved (kinetic energy is lost) after the collision, this is due to the action of an internal friction.

Mathematically, inelastic collision is given by the formula;

\( V=\frac{(M_{1}V_{1}+M_{2}V_{2})}{(M_{1}+M_{2})}\)

Where;

Answer: Initial Velocity

Explanation:

The formula is:

Δx = Vit + 1/2at^2

Therefore the initial velocity is zero

Objects 1 and 2 attract each other with an electrostatic force of 36.0 units. If the distance separating Objects 1 and 2 is tripled, then the new electrostatic force will be four units.

Coulomb's law can be expressed as:

F = k × (q1 × q2) / r²

In which:

F = electrostatic force

k = electrostatic constant (k = 9 × 10⁹ N·m²/C²)

q1 and q2 = the charges of the objects

r =  distance between the objects

Let's consider that the initial electrostatic force in between objects 1 and 2 is 36.0 units.

F1 = 36.0 units

Next, if the distance is considered between the objects is tripled, the new distance (r') changes into three times the initial distance (r):

r' = 3 ×  r

To determine the new electrostatic force (F'), replacement r' into Coulomb's law:

F' = k  × (q1  × q2) / (r')²

Place r' = 3r:

F' = k × (q1 × q2) / (3r)²

= k × (q1 × q2) / 9r²

The new force will be one-ninth (1/9) of the initial force since the electrostatic force (F') is directly proportional to (q1 q2) and inversely proportional to r2.

F' = (1/9) ×  F1

= (1/9) × 36.0

= 4.0 units

Thus, objects 1 and 2 attract each other with an electrostatic force of 36.0 units. If the distance separating Objects 1 and 2 is tripled, then the new electrostatic force will be 4 units.

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Answer:

area of point has more pressure.

Explanation:

formula's : pressure = force ÷ area

1st pressure,

200 ÷ 0.2 = 1000 Pa

2nd pressure,

200 ÷ 0.004 = 50000 Pa

Answer:

longitudinal wave

i guess

sry if its wrong

A. The output voltage, Vout, as a function of time, t, in a series RL circuit can be determined using the equation: Vout(t) = Vmax * (1 - e^(-t/τ)), where τ = L/R.

B. When the time interval T is very short (T → 0) and the maximum voltage Vmax is quite high (VmaxT ≈ Φimp), we can approximate the output voltage Vout using the equation: Vout(t) ≈ Φimp * e^(-t/τ), where τ = L/R.

A. To determine the output voltage Vout as a function of time t in a series RL circuit, we use the following equation:

Vout(t) = Vmax * (1 - e^(-t/τ))

Here, Vmax is the maximum input voltage, τ = L/R is the time constant of the circuit (where L is the inductance and R is the resistance).

B. When the time interval T is very short (T → 0) and the maximum voltage Vmax is quite high (VmaxT ≈ Φimp), we can make the following approximation:

Vout(t) ≈ Vmax * e^(-t/τ)

In this case, we substitute VmaxT with Φimp, which is the total magnetic flux in the circuit.

Rearranging the equation, we get:

Vout(t) ≈ Φimp * e^(-t/τ)

This approximation is valid when the time interval T is very small compared to the time constant τ of the circuit and when the maximum voltage is sufficiently high.

The time constant τ is determined by the values of inductance (L) and resistance (R) in the circuit. It represents the characteristic time scale over which the current and voltage in the circuit change in response to a voltage or current input.

Therefore, in the given scenario, when T is very small and Vmax is high, we can approximate the output voltage Vout(t) in the series RL circuit by the equation: Vout(t) ≈ Φimp * e^(-t/τ), where τ = L/R.

Note: The symbol Φimp in the equation represents the total magnetic flux in the circuit.

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Answer:

To find the net charge on the sphere using electric flux, we can use the formula:  

Φ = Q/ε0

Where Φ is the electric flux, Q is the charge, and ε0 is the permittivity of free space.  

Given that the electric field 20 cm from the center of the sphere is N/C and points radially inwards, we can use the formula for electric field due to a charged sphere to find the charge on the sphere:

E = kQ/r^2  

Where E is the electric field, k is Coulomb's constant, Q is the charge, and r is the distance from the center of the sphere.  

Substituting the given values, we get:  

20 = (1/4πε0)(Q)/(0.2)^2  

Solving for Q, we get:  

Q = (20)(0.2)^2(4πε0)  

Q = 0.64πε0 C  

Now, substituting this value of Q in the formula for electric flux, we get:  Φ = Q/ε0 = (0.64πε0)/(ε0) = 0.64π C  

Therefore, the net charge on the sphere is 0.64π C.

Answer:

solar energy stored by plants

The part of a road vehicle which must be tested to ensure that there is sufficient friction to stop the vehicle in an emergency is the tyre.

This is referred to as a force that resists the motion of one object against another when they roll or slide against each other.

When dealing with braking, the main factor is to have sufficient friction between the road surface and tyre to bring the vehicle to a standstill. If the tyres are wornout there won't be enough friction to make the vehicle stop during emergencies which is therefore the reason why it was chosen as the correct choice.

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Answer:

the attraction between particles

(a) The velocity distribution is, u = (U/b)y.

(b) The flow rate per unit width is \($\frac{1}{2}Ub$\).

We can start by writing down the Navier-Stokes equations for an incompressible fluid:

\(\rho (\dfrac{\partial{u}}{\partial{t}} + \vec{u}.\nabla{u}}) = -\nabla{p} + \mu\nabla^2{u} +f\)

where \($\rho$\) is the fluid density, \($\mathbf{u}$\) is the velocity vector, p is the pressure, \($\mu$\) is the dynamic viscosity, and \($\mathbf{f}$\) is any external forces. Since the flow is steady and there are no external forces, we can simplify the equation to:

\(\mu \dfrac{\partial^2{u}}{\partial{y}^2} = 0\)

where u is the velocity in the x direction and y is the distance from the bottom plate. Integrating this equation twice, we get:

u = (U/b)y

where U is the velocity of the top plate and we have used the no-slip boundary condition at the bottom plate (u=0 at y=0) and the top plate (u=U at y=b).

(b) The flow rate per unit width is given by:

\(Q = \int_0^bu(y)dy\)

Substituting the velocity distribution from part (a), we get:

\(Q = \int_0^b\dfrac{U}{b}(y)dy\\=\dfrac{1}{2}\dfrac{U}{b} b^2\)

Therefore, the flow rate per unit width is \($\frac{1}{2}Ub$\).

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Answer:

In 1924 Louis de Broglie introduced the idea that particles, such as electrons, could be described not only as particles but also as waves. This was substantiated by the way streams of electrons were reflected against crystals and spread through thin metal foils.

Explanation:

I know I probably didn't answer your question, I just used all of my knowledge that I learned about Louie De Broglie. Hope it helps!

Answer:

The acceleration of the crate is 7.5 m/s²

Explanation:

Given;

force exerted on the crate, F₁ = 500 N

mass of the crate, m = 40 kg

frictional force between the floor and the crate, F₂ = 200 N

The net horizontal force is given by;

∑Fₓ = F₁ - F₂

∑Fₓ = 500 N - 200 N

∑Fₓ = 300 N

Apply Newton's second law of motion to determine the acceleration of the crate;

∑Fₓ = ma

a = ∑Fₓ / m

a = 300 N / 40 kg

a = 7.5 m/s²

Therefore, the acceleration of the crate is 7.5 m/s²

If the same motor is to be driven from a 440-V, 60 Hz power supply, The slip at pullout torque is 0.0455.

The slip at pullout torque, pullout torque, and rotor speed at the pullout torque of the motor are given by Slip

\(s = Pmech/ (Xm\times 1.5\times V^2/1000)\)

Pullout torque is given by \((0.5 \times V^2 / X2) \times (R2 / (R1^2 + (X1 + X2)^2))\)

Rotor speed at pullout torque is given by Ns = (120f/p)(1 - s)

The given parameters of the motor are R1 = 0.075 Ohms, R2 = 0.065 Ohms, Xm = 7.2 Ohms, X1 = 0.17 Ohms, X2 = 0.17 Ohms, Pmech = 1.0 KW W, Pstray = 150 W, Pcore = 1.1 KW.

The motor is rated at 75 KW with a power factor of 0.85.Assuming the motor is running at unity power factor.

The output power of the motor is Pout = 75 KW and the input power to the motor is Pin = 75 KW / 0.85 = 88.24 KW

The input current to the motor is \(Iin = 88.24 KW / (3 \times 440 V \times sqrt(3)) = 89.5 A\)

Therefore, the torque developed by the motor is \(T = 1000 \times Pout / (2 \times pi \times N)\)

The slip at the pullout torque is given by \(s = sqrt((Pcore + Pstray) / Pout) = sqrt((1.1 KW + 150 W) / 75 KW) = 0.0455.\)

The pullout torque is given by \((0.5 \times 440^2 / 0.17) \times (0.065 / (0.075^2 + (0.17 + 0.17)^2)) = 411.7 Nm.\)

The rotor speed at pullout torque is \(Ns = (120 \times 50 / 2) \times (1 - 0.0455) = 2846 rpm.\)

The pullout torque of the motor if driven from a 440 V, 60 Hz power supply is given by

\((0.5 \times 440^2 / 0.17) \times (0.065 / (0.075^2 + (0.17 + 0.17)^2)) \times (60 / 50) = 548.4 Nm.\)

The slip at pullout torque is given by \(s = sqrt((1.1 KW + 150 W) / 75 KW) = 0.0455.\)

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Answer:

Here u go

Explanation:

the answer is a. bc carpet has lot and tile has more than gravel

Answer:The ampere is unique in that it uses the base unit of time (t) in its definition (second). All other electrical and magnetic units (such as voltage, power, and magnetic flux) use various combinations of base units in their definitions and are called derived units.

Explanation:

1) The force acting on the object during the time interval 0-3 seconds is 1/3 N.

2) The force acting on the object during the time interval 6-10 seconds is -0.5 N.

3) There is no time interval in which no force acts on the object.

(i) Force acting on the object in time interval 0-3 seconds. Force acting on the object is equal to the product of its mass and acceleration, i.e.,F = ma.

In the given velocity-time graph, the acceleration of the object can be determined by determining the slope of the velocity-time graph from 0 to 3 seconds.

Slope = (change in velocity) / (change in time)= (20-0) / (3-0) = 20/3 m/s^2

Acceleration, a = slope= 20/3 m/s^2

Mass of the object, m = 50 g = 0.05 kg

∴ Force acting on the object, F = ma= 0.05 × 20/3= 1/3 N.

Therefore, the force acting on the object during the time interval 0-3 seconds is 1/3 N.

(ii) Force acting on the object in time interval 6-10 seconds. Similar to the first question, the force acting on the object in time interval 6-10 seconds can be determined by determining the acceleration of the object during this time interval.

The slope of the velocity-time graph from 6 seconds to 10 seconds can be determined as follows:

Slope = (change in velocity) / (change in time)= (-20-20) / (10-6) = -40/4= -10 m/s^2 (negative sign indicates that the object is decelerating)

Mass of the object, m = 50 g = 0.05 kg

∴ Force acting on the object, F = ma= 0.05 × (-10)= -0.5 N.

Therefore, the force acting on the object during the time interval 6-10 seconds is -0.5 N.

(iii) Time interval in which no force acts on the object. There is no time interval in which no force acts on the object. This is because, as per Newton's first law of motion, an object will continue to remain in a state of rest or uniform motion along a straight line unless acted upon by an external unbalanced force.In other words, if the object is moving with a constant velocity, there must be a force acting on the object to maintain its motion.

Therefore, there is no time interval in which no force acts on the object.

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Answer:

Suppose the  object were stationary and emitting waves that had a distance of 1 m between crests - the receiver would receive waves that had a distance of 1  between crests

Suppose the object were moving towards the receiver, then there would no longer be 1 m between the crests as measured in the laboratory frame because of movement of  the object.

Then the receiver would receive waves that were less than 1 m apart and would report a higher frequency than if the object were stationary,

The speed of sound of a wave having frequency 200 Hz in air is 340 m/s.

The following equation can be used to compute wavelength: wavelength = wave velocity/frequency. In most cases, wavelength is measured in meters. Greek letter lambda is used to represent wavelength, so = v/f.

The wavelength, which will also apply to troughs, is the separation between two wave crests. The frequency is measured in cycles per second (Hz), which is the quantity of vibrations that cross a specific area in a second (Hertz).

Wavelength = (speed) / (frequency) =

(3 x 10⁸ m/s) / (200HZ) =  340m/s

The speed of radio waves is 3*10^8 m/s, the same as the speed of light.

The speed of sound of a wave having frequency 200 Hz in air is 340 `m//s`.

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The correct answer is B. No, because she is not working on the atomic structures of a solid

Explanation:

Solid-state physics is a sub-discipline of physics that focuses on studying solids, this includes analyzing solids structures, features, and other phenomena that occur in substances in this state of the matter. This means a solid-state physics will not study or gases.

In this context, the fact the scientist is trying to create an electromagnetic field by using solutions and the flow of these show the scientists is not working with solids but with liquids or gases as solids do not flow. Also, her focus is not solids, and therefore she is not a solid-state physicist. Thus, it can be concluded she is not a solid-state physicists because she is not working on the structures of solids.

Answer:

the energy of the spring at the start is 400 J.

Explanation:

Given;

mass of the box, m = 8.0 kg

final speed of the box, v = 10 m/s

Apply the principle of conservation of energy to determine the energy of the spring at the start;

Final Kinetic energy of the box = initial elastic potential energy of the spring

K.E = Ux

¹/₂mv² = Ux

¹/₂ x 8 x 10² = Ux

400 J = Ux

Therefore, the energy of the spring at the start is 400 J.

Answer:

C

Explanation:

C