. If force (F), work (W) and velocity (v) are taken as fundamental quantities.
What is the dimensional formula of time (T)?

Answer:

An electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor or space. It is measured as the net rate of flow of electric charge through a surface or into a control volume. ... In electric circuits the charge carriers are often electrons moving through a wire.

Answer:

The flow of moving electrons

Answer:

Energy Bill fluctuations are inevitable and depend on a variety of different factors. Two of the most important are the current weather your home is experiencing and the current price per Kilowatt Hour (which fluctuates more than you might think).

Answer:

below

Explanation:

Part A) This engine works  per cycle is 254.9 J.

Part B) The thermal efficiency is  23.42%

The thermal efficiency of any heat engine is represented in percentage of heat energy converted into work.

For isothermal expansion, work done is

W₁ =nRT₁ x ln(V₂/V₁)

W₁ = 0.1 x 8.314 x 800 x ln(5000/3000)

W₁ = 339.8 J =Q₁

For isochoric cooling ,

W₂ =0

Q₂ =nCvdT = 0.1 x 3R/2 x (T₂-T₁)

Q₂ = -748.3 J

For isothermal compression,

W₃ =nRT₂ ln (V₄/V₃)

W₃  = 0.1 x 8.314 x 200 x ln(3000/5000)

W₃ = -84.9J

For isochoric heating

W₄ =0

Q₄ =nCvdT = 0.1 x 3R/2 x (800-200)

Q₄  = -748.3 J

Total work done in all the process W = W₁ +W₂ +W₃ +W₄

W =254.9 J

Thus, the work done is 254.9 J

Thermal efficiency = Work done/Heat taken

η = W/ Q₁ +Q₄

η = [254.9 / 339.8 +748.3 ] x 100 %

η = 0.2342 x 100 %

η = 23.42%

Thus, the thermal efficiency is 23.42%

Learn more about thermal efficiency.

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Answer:

Loudness of the second sound is more than the first one.

Explanation:

There are two sounds, the second sound is identical to first but the loudness of second is more than the first one.

As the frequency is same so the itch is same for both the sounds.

As the loudness depends on the amplitude of the sound so the loudness of the second sound is more than the first sound.

Answer:

[tex] = \frac{30 \times {10}^{ - 3} }{1} \\ = 0.03 \: km \: per \: second[/tex]

Answer:

108 km/hr or 0.03 km/s

Explanation:

conversion factor for m/s to km/hr is 5/18

conversion factor for m/s to km/s is 1/1000

Explanation:

time=Distance/speed

t=32/8

t=4 seconds

Answer:

Boyle's law.

Explanation:

Robert Boyle was an Irish chemist and is famously referred to as the first modern chemist. He was born on the 25th of January, 1627 in Lismore, Ireland and died on the 31st, December 1691, London, United Kingdom.

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

[tex] PV = K[/tex]

[tex] P_{1}V_{1} = P_{2}V_{2} [/tex]

Where;

P1 is the original pressure.

P2 is the final pressure.

V1 is the original volume.

V2 is the final volume.

Hence, when you release some of the paint from a spray paint can by applying an amount of pressure and the can remains at the same temperature, the gas law which this represent is Boyle's law.

Answer:

[tex]N=675N[/tex]

Explanation:

From the question we are told that:

Force [tex]F=405N[/tex]

Generally the equation for Normal force in this case is is mathematically given by

 [tex]F=\mu_s N[/tex]

Where

Static Friction=[tex]\mu_s[/tex]

 [tex]\mu_s=0.6[/tex]

Therefore

 [tex]N=\frac{F}{\mu_s}[/tex]

 [tex]N=\frac{405}{0.6}[/tex]

 [tex]N=675N[/tex]

Explanation:

The initial kinetic energy [tex]KE_0[/tex] for both blocks is zero. Let [tex]m_1= m[/tex] and [tex]m_2 =3m[/tex]. So using the conservation law of linear momentum, we can write

[tex]0 = m_1v_1 - m_2v_2[/tex]

or

[tex]v_2 = \dfrac{m_1}{m_2}v_1 = \dfrac{m}{3m}v_1 = \dfrac{1}{3}v_1[/tex]

The final kinetic energies for the two masses are

[tex]KE_1 = \frac{1}{2}m_1v_1^2 = \frac{1}{2}mv_1^2[/tex]

[tex]KE_2 = \frac{1}{2}m_2v_2^2 = \frac{1}{2}(3m)(\frac{1}{3}v_1)^2 = \frac{1}{2}m(\frac{1}{3}v_1^2)[/tex]

Therefore, the ratio of their kinetic energies is

[tex]\dfrac{\Delta KE_2}{\Delta KE_1} = \dfrac{\frac{1}{2}(\frac{1}{3}v_1^2)}{\frac{1}{2}v_1^2} = \dfrac{1}{3}[/tex]

Answer:

Thus, the velocity at the time of strike is same as the velocity at the time of projection.

Explanation:

Let a projectile is projected vertically upwards with a speed of u and reaches to the maximum height H.

At maximum height , the speed is zero and then the projective comes back on the ground.

Use the third equation of motion

[tex]v^2 = u^2 + 2 g h \\\\0 = u^2 - 2 g H\\\\\u =\sqrt{2gH}[/tex]

Now let the velocity at the time of strike is v'.

Use third equation of motion, here initial velocity is zero.  

[tex]v'^2 = 0 + 2 g H \\\\v = \sqrt{2gH}[/tex]

Thus, the velocity at the time of strike is same as the velocity at the time of projection.

Answer: Hmmm im not sure but i'd go with 3.80x10^5

Explanation: Like i mentioned im not very good at physics...sorry if its wrong

Explanation:

ehb-pynw-ayo

joi n fast

by using v ^2 = u^2 + 2as we can find "a"

25 = 729 + 2 × a × 50

25 = 729 + 100a

a = - 7.04

so the answer is B

Answer:

A. liquid and solid

Explanation:

Answer:

a. 1 Newton = 100000 Dyne

b. a represents acceleration.

Explanation:

Newton is the standard unit (S.I) of measurement of force. Converting 1 Newton to dyne we have;

1 Newton = 10⁵ Dyne

1 Newton = 100000 Dyne

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

Force = mass * acceleration

[tex] F = ma[/tex]

Hence, we can deduce that a represents the acceleration of an object and it's measured in meters per seconds square.

Answer:

The system would be moving at [tex]17.5\; \rm m \cdot s^{-1}[/tex].

The kinetic energy of this system would be [tex]612500\; \rm J \![/tex] after the collision.

[tex]612500\; \rm J[/tex] (same amount) of kinetic energy would be lost.

Explanation:

The momentum of an object is the product of its mass [tex]m[/tex] and its velocity [tex]v[/tex]. That is: [tex]p = m \cdot v[/tex].

Assume that external forces (e.g., friction) have no effect on this system.  The total momentum of this system would stay the same before and after the collision.

Initial momentum of this system:

Hence, the momentum of this system would be [tex]70000\; \rm kg \cdot m \cdot s^{-1}[/tex] before the collision.

Under the assumptions, the collision would not change the momentum of this system. Hence, the momentum of this system would continue to be [tex]70000\; \rm kg \cdot m \cdot s^{-1}[/tex] after the collision.

However, with two identical trains stuck to each other, the mass of this system would be twice that of just one train: [tex]m = 2 \times 2000\; \rm kg[/tex].

Calculate the new velocity of this system:

[tex]\begin{aligned} v &= \frac{p}{m}\\ &= \frac{70000\; \rm kg \cdot m \cdot s^{-1}}{2 \times 2000\; \rm kg} = 17.5\; \rm m\cdot s^{-1}\end{aligned}[/tex].

Calculate the kinetic energy of this system before and after the collision.

Before the collision:

[tex]\begin{aligned}& \text{KE(before)} \\ =\; & \text{KE(moving train)} + \text{KE(stationary train)}\\ =\; & \frac{1}{2} \, m(\text{one train}) \cdot (v(\text{moving train}))^{2} + 0 \\ = \; &\frac{1}{2} \times 2000 \times (35\; \rm m\cdot s^{-1})^{2} \\ = \; & 1225000\; \rm J \end{aligned}[/tex].

After the collision:

[tex]\begin{aligned}& \text{KE(after)} \\ =\; & \frac{1}{2} \, m(\text{two trains}) \cdot v^{2} \\ = \; &\frac{1}{2} \times (2\times 2000\; \rm kg) \times (17.5\; \rm m\cdot s^{-1})^{2} \\ = \; & 612500\; \rm J \end{aligned}[/tex].

Change to the kinetic energy of this system:

[tex]1225000\; \rm J - 612500\; \rm J = 612500\; \rm J[/tex].

Answer:

Explanation:Force and acceleration are directly proportional. ... Mass and acceleration are inversely proportional. In this situation, acceleration changes in response to a change of mass, so mass is the independent variable and acceleration is the dependent variable.

Answer:

we need the block

Explanation:

1×2 =4 lest 74 =345

two electrons are created from a quanta of kight

Answer:

-30 N/C

Explanation:

Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m

Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V

Since E = -ΔV/Δx

substituting the values of the variables into the equation, we have

E = -ΔV/Δx

E = -0.30 V/0.01 m

E = -30 V/m

Since 1 V/m = 1 N/C.

E = -30 N/C

So, the average electric field is -30 N/C

Answer:

B

Explanation:

B

Answer:

B

Explanation:

The liquid evaporates in the solid is left in the dish..

Answer:

Approximately [tex]13\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] (assuming that the boiling point of water in this experiment is [tex]100\; \rm ^\circ C\![/tex].)

Explanation:

Latent heat of condensation/evaporation of water: [tex]2260\; \rm J \cdot g^{-1}[/tex].

Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to [tex]\rm J \cdot g^{-1}[/tex].

Specific heat of water: [tex]4.2\; \rm J \cdot g^{-1}\cdot \rm K^{-1}[/tex].

Specific heat of copper: [tex]0.39\; \rm J \cdot g^{-1}\cdot K^{-1}[/tex].

The temperature of this calorimeter and the [tex]250\; \rm g[/tex] of water that it initially contains increased from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex]. Calculate the amount of energy that would be absorbed:

[tex]\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^{-1}\cdot K^{-1} \times 50\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J \end{aligned}[/tex].

[tex]\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^{-1}\cdot K^{-1} \times 250\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J \end{aligned}[/tex].

Hence, it would take an extra [tex]585\; \rm J + 31500\; \rm J = 32085\; \rm J[/tex] of energy to increase the temperature of the calorimeter and the [tex]250\; \rm g[/tex] of water that it initially contains from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex].

Assume that it would take [tex]x[/tex] grams of steam at [tex]100\; \rm ^\circ C[/tex] ensure that the equilibrium temperature of the system is [tex]50\; \rm ^\circ C[/tex].

In other words, [tex]x\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] would need to release [tex]32085\; \rm J[/tex] as it condenses (releases latent heat) and cools down to [tex]50\; \rm ^\circ C[/tex].

Latent heat of condensation from [tex]x\; \rm g[/tex] of steam: [tex]2260\; {\rm J \cdot g^{-1}} \times (x\; {\rm g}) = (2260\, x)\; \rm J[/tex].

Energy released when that [tex]x\; {\rm g}[/tex] of water from the steam cools down from [tex]100\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex]:

[tex]\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^{-1}\cdot K^{-1}} \times (x\; \rm g) \times (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J \end{aligned}[/tex].

These two parts of energy should add up to [tex]32085\; \rm J[/tex]. That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex].

[tex](2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J[/tex].

Solve for [tex]x[/tex]:

[tex]x \approx 13[/tex].

Hence, it would take approximately [tex]13\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] for the equilibrium temperature of the system to be [tex]50\; \rm ^\circ C[/tex].

Answer:

- the power to the air is 850 MW

- mass flow rate of the air is 84577.11 kg/s

Explanation:

Given the data in the question;

Net power generated; [tex]W_{net[/tex] = 150 MW

Heat input; [tex]Q_k[/tex] = 1000 MW

Power to air = ?

For closed cycles

Power to air Q₀ = Heat input; [tex]Q_k[/tex] - Net power generated; [tex]W_{net[/tex]

we substitute

Power to air Q₀  = 1000 - 150

Q₀ = 850 MW

Therefore,  the power to the air is 850 MW

given that ΔT = 10 °C

mass flow rate of air required will be;

⇒ Q₀ / CpΔT

we know that specific heat of air at p=c ; Cp = 1.005 kJ/kg.K

we substitute

⇒ ( 850 × 10³ ) / [ 1.005 × 10 ]

⇒ ( 850 × 10³ ) / 10.05

⇒ 84577.11 kg/s

Therefore, mass flow rate of the air is 84577.11 kg/s

Answer:

(a) V = 3.75 x 10^5 V

(b) q = 5.2 x 10^-6 C

Explanation:

Diameter, d = 25 cm

radius, r = 12.5 cm = 0.125 m

Electric field, E = 3 x 10^6 V/m

(a) The maximum potential is given by

[tex]V = E \times r \\\\V = 3\times 10^6\times 0.125\\\\V = 3.75\times10^5 V[/tex]

(b) The charge is given by

[tex]V = \frac{k q}{r}\\\\3.75\times10^5=\frac{9\times10^9\times q}{0.125}\\\\q = 5.2\times 10^{-6} C[/tex]

Answer:

The answer should be clenching fists