In a cyclic process, a gas performs 123 J of work on its surroundings per cycle. What amount of heat, if any, transfers into or out of the gas per cycle?

123 J transfers out of the gas

123 J transfers into the gas

246 J transfers into the gas

0 J (no heat transfers)

Answer:

The maximum height of the ball after the bounce is 1.2 m

Explanation:

Potential Energy = mass * Height * acceleration of gravity

PE=mgh

= 0.2 x 9.8 x 1.5

P.E = 2.94 J

During bounce of ball 0.60 J of energy is lost. So

  2.94 - 0.6 = 2.34 J

now  new energy is 2.34

New P.E = mgh

2.34 = 0.2 x 9.8 x h

h = 2.34 / 0.2 x 9.8

h =  1.2 m

The maximum height of the ball after the bounce is 1.2 m

Explanation:

Given:

Mass = 0.2 kg

Height = 1.5 m

Potential energy of the drop, PE = m × g × h

= 0.2 x 9.81 x 1.5

= 2.94 J

After the drop, 0.6 J of energy is dissipated, so amount of energy left

= 2.94 - 0.6

= 2.34 J

This energy is eqal to thenew potential energy which is:

m × g × h = 2.34

0.2 × 9.81 × h = 2.34

= 2.34/1.962

= 1.19 m.

Answer: A.True

Explanation: simply put the magnitude of the speed is a function of Distance while the magnitude of velocity is a function of Displacement. Displacement is the average Distance moved in different directions which will be smaller in magnitude compared to the Total Distance used the calculating the magnitude of speed.

Answer:

The answer for this one is that wood is the oldest mean of energy and it is also cheap that is why people in less developed coubtries use wood as a primary energy source.

Explanation:

The Developing countries depends on the wood and other forest goods for their everyday cooking and heating needs, prompting the public to assist with tropical deforestation and insecurity through the usage of these tools.

Both ideas— that forest-derived energy is mainly used in the developed nation and its value in the energy portfolios of advanced economies is negligible— fail to catch

Final answer:

People in less developed countries rely on wood and biomass for energy because it's inexpensive and accessible; however, this reliance can lead to deforestation and pollution. In contrast, developed nations are seeing an increased biomass use due to rising fossil fuel costs, though they have a greater mix of energy sources.

Explanation:

People in less developed countries often use wood as a primary energy source for multiple reasons. Wood and other forms of biomass, such as animal dung, are inexpensive, relatively efficient, and readily available. These sources of energy are crucial for domestic uses including heating, sanitation, and cooking. In areas where electricity and modern fuels are not accessible or are too costly, biomass is a vital resource. Additionally, due to rapid population growth and poverty, some regions have a higher demand for firewood as they lack alternative energy sources. This reliance often leads to environmental issues like deforestation, as the use of wood outpaces the replenishment of forests.

Furthermore, in developed nations, the consumption of energy includes a mix of sources with a declining reliance on biomass due to access to electricity and other modern energy forms. However, as fossil fuel prices increase and availability declines, biomass use is also growing in these nations. Despite being renewable, the environmental impact of biomass as an energy source includes deforestation and the release of harmful pollutants such as carbon monoxide and particulate matter from burning wood.

Answer:

A

Explanation:

- The complete question is:

" A crate is lifted vertically 1.5 m and then held at rest. The crate has weight 100 N (i.e., it is  supported by an upward force of 100 N).Now suppose the crate is lifted so rapidly that air resistance was significant during the raising. How much work was done by the lifting force as the box was raised 1.5 m?"

Options:

1. More than 150 J

2. A bit less than 150 J because the air partially supported the crate.

3. No work was done.

4. Still 150 J

5. None of these

Solution:

- As the box is raised the work is done against gravity and air resistance opposes the motion of box. Hence, both the force of gravity ( Weight and air resistance act downward).

- The amount of work done is the sum of both work done against gravity and against air resistance.

-                         W = W_g + W_r

                         W > W_g

                         W > 100*1.5

                         W > 150 J

- Hence, the work done is greater than 150 J i.e work is also done against ai r resistance.  

Answer:

Intensity of beam 18 feet below the surface is about 0.02%

Explanation:

Using Lambert's law

Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium

then dI / I = kdt

taking log,

ln(I) = kt + ln C

I = Ce^kt

t=0=>I=I(0)=>C=I(0)

I = I(0)e^kt

t=3 & I=0.25I(0)=>0.25=e^3k

k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

Intensity of beam 18 feet below the surface is about 0.2%

Answer:

B

Explanation:

Because they have a full valence shell they don't need to make bonds to stabilize because they are already stable. These will be your Noble Gasses (ie Neon)

Answer:

2.65 m

Explanation:

From work-kinetic energy principle,

workdone by friction + workdone by gravity on sled = kinetic energy change of sled

Let h be the  vertical height moved and d the distance moved along the incline. h = dsinθ   where θ is the angle of the incline = 20°.

The workdone by gravity on the sled is mghcos180 = -mgh = -mgdsinθ

The frictional force = -μmgcosθ where μ = 0.20 and the work done by friction = -μmgcosθd

The kinetic energy change = 1/2m(v₂² - v₁²) where v₁ = initial speed = 2.0 m/s and v₂ = final speed = 0 m/s (since the sled stops)

So, -mgdsinθ - μmgcosθd = 1/2m(v₂² - v₁²)

-gd(sinθ - μcosθ) = 1/2(v₂² - v₁²)

d = (v₂² - v₁²)/-2g(sinθ - μcosθ)

substituting the values for the variables,

d = (0 - 2²) /[- 2 × 9.8(sin20 - 0.2cos20)]

d = -4 /[- 2 × 9.8(sin20 - 0.2cos20)]

d = -4/ - 1.51 = 2.65 m

It has moved up the incline a distance of 2.65 m.

Answer:

Acceleration, a= 1.65m/s^2

Acceleration is downward.

Explanation:

y: N- m1gcosalpha=0

x: Ff- FT- m1gsinalpha= m1× alpha

Ff= uN= um1gcosalpha

FT=m2g

Acceleration ,a = g(ucosalpha- (m2/m1)-sin alpha)

a= (m2/m1) + sinalpha= 1.976

Cos alpha= 0.766

U2= 2.8

a= g(2.8×0.766)- 1.976)

a= g× 0.1688

a= 9.8× 0.1688

a=1.65m/s^2

Answer:

Maximum height =1031km

Explanation:

Given:

Velocity,Vo= 1.445km/s= 1445m/s

Mass of moon= 7.35×10^22kg

Radius of moon= 1737km= 1737000m

Using conservation energy

Ui + Ki= Uf + Kf

--'>(GMm/R) + 1/2 (m ^2)

-GMm/(R + h) - 0

Vo^2= 2Gm(1/R - 1/R+h)

1445^2= 2× (6.67x10^-11)×(7.35×10^22)[(1/1737000)- (1/1737000 + h)]

h= 1031km

Answer:

104.4km

Explanation:

Projectile motion occurs when object is launched into air and allowed to fall freely under the influence of gravity.

Maximum height reached by the object is expressed as;

H = u²sin²(theta)/2g where;

u is the initial velocity = 1.445km/s

u = 1445m/s since 1000m is equivalent to 1km

theta is the angle of launch

g is the acceleration due to gravity = 10m/s²

theta = 90° (since object is launched vertically)

Substituting the values in the formula we have;

H = 1445²(sin90°)²/2(10)

H = 1445²/20

H = 104,401.25m

H = 104.4km

The the maximum height reached by the capsule is 104.4km

Answer:

x2=0.732m

Explanation:

We can calculate the spring constant using the equilibrium equation of the block m1. Since the spring is in equilibrium, we can say that the acceleration of the block is equal to zero. So, its equilibrium equation is:

[tex]m_1g-k\Delta x_1=0\\\\\implies k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(7.5kg)(9.8m/s^{2})}{0.84m-0.69m}=490N/m[/tex]

Then using the equilibrium equation of the block m2, we have:

[tex]m_2g-k\Delta x_2=0\\\\\\implies x_2=x_0+\frac{m_2g}{k} \\x_2=0.69m+\frac{(2.1kg)(9.8m/s^{2})}{490N/m}= 0.732m[/tex]

In words, the lenght x2 of the spring when the m2 block is hung from it, is 0.732m.

Answer:

a) x_max = 0.20794 m

b)  v_max = 3.8436 m/s

c) P = 0.05883 W

Explanation:

Given:

- The stiffness k = 205 N / m

- The mass m = 0.6 kg

- initial compression of the spring xi = 13 cm

- initial speed of the mass vi = 3 m/s

Find:

(a) What is the maximum stretch during the motion? m

(b) What is the maximum speed during the motion? m/s

(c) Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system. What is the average power input in watts required to maintain a steady oscillation?

Solution:

- Conservation of energy principle can be applied that the total energy U of the system remains constant. So the Total energy is:

                          U = K.E + P.E

                          U = 0.5*m*v^2 + 0.5*k*x^2

- We will take initial point with given values and maximum compression x_max when v = 0.

                          0.5*m*vi^2 + 0.5*k*xi^2 = 0.5*k*x_max^2

                          (m/k)*vi^2 + xi^2 = x_max^2

                          x_max = sqrt ( (m/k)*vi^2 + xi^2 ) = sqrt ( (.6/205)*3^2 + .13^2  

                          x_max = 0.20794 m

- The angular speed w of the harmonic oscillation is given by:

                          w = sqrt ( k / m )

                          w = sqrt ( 205 / 0.6 )

                          w = 18.48422 rad/s

- The maximum velocity v_max is given by:

                          v_max = - w*x_max

                          v_max = - (18.48422)*(0.20794)

                          v_max = 3.8436 m/s

- The amount of power required to stabilize each oscillation is given by:

                         P = E_cycle / T

Where, E = Energy per cycle  = 0.02 J

             T = Time period of oscillation

                         T = 2π/w

                         P = E_cycle*w / 2π

                         P = (0.02*18.48422) / 2π

                         P = 0.05883 W

The maximum stretch and maximum speed of the spring can be obtained from the conservation of energy. The average power required to maintain a steady oscillation can be calculated using the energy dissipation and the period of oscillation.

This problem involves

conservation of energy

(kinetic and potential) which is given by the equation E = K + U where K is the kinetic energy = 0.5*m*v^2, m is the mass and v is the speed. U is the potential energy = 0.5*k*x^2 where k is the spring stiffness and x is the spring displacement. The maximum stretch of the spring occurs when all the kinetic energy has been transferred into potential energy (maximum potential energy). At this maximum stretch, v = 0, thus the total energy E becomes 0.5*k*x_max^2, from which we can calculate x_max. The maximum speed occurs when the spring is at its equilibrium position, at which point all the potential energy has been transferred into kinetic energy (maximum kinetic energy). At this point, x = 0, and the total energy E becomes 0.5*m*v_max^2, from which we can calculate v_max. The average power P required to maintain a steady oscillation with energy dissipation is P = energy dissipation / period of oscillation. The period T of the oscillation of a spring-mass system is given by T = 2*pi*sqrt(m/k).

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Answer:

A solid moment of inertia is  [tex]I = \frac{mr^2}{2}[/tex].

Here, both the solid cylinder and the cylindrical shell have the same mass, the same radius, and turn on a horizontal, friction-less axle.

The solid cylinder has less inertia than the cylindrical shell, and it requires less torque to rotate, meaning that the solid cylinder weight block falls faster than the cylindrical shell itself.

Fill in the blanks, in order.

Gravitational potential energy, Translation kinetic energy, Kinetic energy;

Hallow, Solid, Hallow, Solid;

Hallow, Transitional kinetic energy, Hallow

Answer:

The answer to the queations are;

A) The block attached to the solid cylinder would hit the ground first.

B) By the time the blocks reach the ground, they have transformed identical amounts of _gravitational potential_______energy into_____rotational kinetic________ energy of the cylinders. energy of the blocks and _______translational kinetic_____But the moment of inertia of a ____hollow______cylinder is higher than that of a ____solid_______ cylinder of the same mass, so more of the energy of the system is in the form of rotational kinetic energy for the ______hollow_____________cylinder than for the ___solid_______ one. This leaves less energy in the form of translational kinetic energy for the ____hollow________cylinder. But it is the ____translational kinetic________ energy that determines the speed of the block. So the block moves more slowly for the system with the ______hollow______cylinder, and so its block reaches the ground last.

Explanation:

To solve the question, we note that

The total energy of motion of the moving cylinders is equal to

K[tex]_{TOT[/tex] = 1/2·m·v² + 1/2·I·ω²

Where

m = Mass

v = Velocity

ω = Angular velocity

I = moment of inertia where I for hollow cylinder = MR² and

I for solid cylinder = 1/2·MR².

Therefore we have

K[tex]_{TOT[/tex] for solid cylinder = 1/2·m·v² + 1/2·I·ω² = 1/2·m·v² + 1/2·1/2·MR²·ω²

= 1/2·m·v² + 1/4·MR²·v²/r² = 1/2·m·v² + 1/4·M·v² = 3/4·m·v²

For the hollow cylinder, we have

K[tex]_{TOT[/tex] = 1/2·m·v² + 1/2·MR²·ω² = 1/2·m·v² + 1/2·MR²·v²/r² = 1/2·m·v² + 1/2·m·v²

= m·v²

From conservation of energy the initial potential energy is transformed into potential energy as follows

PE = m·g·h

Where:

m = Mas

g = Gravitational acceleration

h = height

Therefore

For the solid cylinder 3/4·m·v² = m·g·h and  v² = [tex]\frac{3}{4} \frac{m*g*h}{m}[/tex] and v  = [tex]\sqrt{\frac{4}{3} gh}[/tex]

For the hollow cylinder m·v² = m·g·h and  v² = [tex]\frac{m*g*h}{m}[/tex] and v  = [tex]\sqrt{gh}[/tex]

This shows that the solid cylinder has a higher downward velocity and the block attached to the solid cylinder would hit the ground first

B) By the time the blocks reach the ground, they have transformed identical amounts of _gravitational potential_______energy into_____rotational kinetic________ energy of the cylinders. energy of the blocks and _______translational kinetic_____But the moment of inertia of a ____hollow______cylinder is higher than that of a ____solid_______ cylinder of the same mass, so more of the energy of the system is in the form of rotational kinetic energy for the ______hollow_____________cylinder than for the ___solid_______ one. This leaves less energy in the form of translational kinetic energy for the ____hollow________cylinder. But it is the ____translational kinetic________ energy that determines the speed of the block. So the block moves more slowly for the system with the ______hollow______cylinder, and so its block reaches the ground last.

Answer:

Using a good pair of binoculars, you observe a section of the sky where there are stars of many different apparent brightnesses. You find one star that appears especially dim. This star looks dim because it is farther away or it has a small radius.

Explanation:

Apparent magnitude in astronomy is the apparent brightness of a star that is seen from the Earth, that brightness can variate according to the distance at which the star is from the Earth or due to its radius.

That can be demonstrate with the next equation:

[tex]F = \frac{L}{4\pi r^2}[/tex] (1)

Where F is the radiant flux received from the star, L its intrinsic luminosity and r is the distance.

For example, an observer sees two motorbikes approaching it with its lights on but one of the motorbikes is farther, so the light of this one appears dimmer, even when the two lights emit the same amount of energy per second.  

That is because the radiant flux decreases with the square distance, as can be seen in equation 1.

In the other hand, a bigger radius means that the gravity in the surface of the star will be lower, allowing that light can escape more easily:

[tex]g = \frac{GM}{R^2}[/tex] (2)

Where g is the surface gravity in the star, G is the gravitational constant, M is the mass of the star and R is the radius of the star.

The potential difference across the capacitor is 7.5 V both before and after the Mylar is inserted. To calculate charge and electric field values, additional information about the system, such as the dielectric constant of Mylar or capacitance, is required. Without this information, only the potential difference can be definitively determined.

The subject concerns Physics, specifically the study of capacitors in the context of their behavior before and after a dielectric material is inserted. We are given the dimensions of the capacitor plates and the potential difference applied by a battery, and we need to deduce several properties of the capacitor both before and after the insertion of the Mylar dielectric sheet. The capacitance of the capacitor is not provided directly, but can be inferred using the provided dimensions and the permittivity of free space or air, as required.

The potential difference across a capacitor is determined by the voltage applied by the battery. Thus, the capacitor's potential difference before the Mylar is inserted is 7.5 V.

To find the electric field in the capacitor before the Mylar is inserted, we can use the formula E = V/d, where V is the potential difference and d is the separation between the plates. Thus, E = 7.5 V / 0.1 mm = 75,000 V/m.

The charge on a capacitor is given by Q = CV, where C is the capacitance and V is the potential difference. However, without the capacitance, we cannot calculate the charge directly. More information, like the dielectric constant of the Mylar or the capacitance of the capacitor with air between the plates, would be needed.

Once the Mylar is inserted, if we assume the dielectric is not discharged, the potential difference across the capacitor remains at 7.5 V, since the battery is still connected.

If the Mylar has a dielectric constant, we can calculate the new electric field using the formula E' = E / K, where K is the dielectric constant of the Mylar. But again, without K, we cannot calculate E' directly.

The charge on the capacitor after the Mylar is inserted will be the same as before, provided the capacitor is still connected to the battery, since potential difference and charge are related by Q = CV, and V remains unchanged at 7.5 V.

A) Potential difference before the Mylar is inserted: [tex]\( 7.5 \, \text{V} \).[/tex] B) Electric field before the Mylar is inserted: [tex]\( 7.5 \times 10^4 \, \text{V/m} \)[/tex]. C) Charge before the Mylar is inserted: [tex]\( 1.66 \times 10^{-11} \, \text{C} \)[/tex]. D) Potential difference after the Mylar is inserted: [tex]\( 7.5 \, \text{V} \)[/tex]. E) Electric field after the Mylar is inserted: [tex]\( 7.5 \times 10^4 \, \text{V/m} \).[/tex] F) Charge after the Mylar is inserted: [tex]\( 5.14 \times 10^{-11} \, \text{C} \).[/tex]

To solve this problem, we need to use the concepts of capacitance, electric field, and the effect of a dielectric on a capacitor.

- Electrode area [tex]\( A = 5.0 \, \text{mm} \times 5.0 \, \text{mm} = 25.0 \, \text{mm}^2 = 25.0 \times 10^{-6} \, \text{m}^2 \)[/tex]

- Separation distance [tex]\( d = 0.10 \, \text{mm} = 0.10 \times 10^{-3} \, \text{m} \)[/tex]

- Voltage [tex]\( V = 7.5 \, \text{V} \)[/tex]

- Dielectric constant of Mylar [tex]\( \kappa \approx 3.1 \)[/tex]

A) Potential difference before the Mylar is inserted

The potential difference across the capacitor before the Mylar is inserted is simply the voltage of the battery since the battery is connected directly to the capacitor.

[tex]\[ V_{\text{before}} = 7.5 \, \text{V} \][/tex]

B) Electric field before the Mylar is inserted

The electric field [tex]\( E \)[/tex] in a parallel-plate capacitor is given by:

[tex]\[ E = \frac{V}{d} \][/tex]

Substituting the given values:

[tex]\[ E_{\text{before}} = \frac{7.5 \, \text{V}}{0.10 \times 10^{-3} \, \text{m}} \][/tex]

[tex]\[ E_{\text{before}} = 7.5 \times 10^{4} \, \text{V/m} \][/tex]

C) Charge on the capacitor before the Mylar is inserted

The capacitance [tex]\( C \)[/tex] of a parallel-plate capacitor without a dielectric is given by:

[tex]\[ C = \frac{\epsilon_0 A}{d} \][/tex]

where [tex]\( \epsilon_0 \)[/tex] is the permittivity of free space [tex](\( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m} \)).[/tex]

[tex]\[ C_{\text{before}} = \frac{(8.85 \times 10^{-12} \, \text{F/m})(25.0 \times 10^{-6} \, \text{m}^2)}{0.10 \times 10^{-3} \, \text{m}} \][/tex]

[tex]\[ C_{\text{before}} \approx 2.21 \times 10^{-12} \, \text{F} \][/tex]

The charge [tex]\( Q \)[/tex] on the capacitor is given by:

[tex]\[ Q = CV \][/tex]

[tex]\[ Q_{\text{before}} = (2.21 \times 10^{-12} \, \text{F})(7.5 \, \text{V}) \][/tex]

[tex]\[ Q_{\text{before}} \approx 1.66 \times 10^{-11} \, \text{C} \][/tex]

D) Potential difference after the Mylar is inserted

When the Mylar is inserted, the potential difference across the capacitor remains the same because the battery is still connected.

[tex]\[ V_{\text{after}} = 7.5 \, \text{V} \][/tex]

E) Electric field after the Mylar is inserted

The electric field [tex]\( E \)[/tex] in a capacitor with a dielectric is given by the same formula:

[tex]\[ E = \frac{V}{d} \][/tex]

Since the voltage and the separation remain the same, the electric field remains the same:

[tex]\[ E_{\text{after}} = \frac{7.5 \, \text{V}}{0.10 \times 10^{-3} \, \text{m}} \][/tex]

[tex]\[ E_{\text{after}} = 7.5 \times 10^{4} \, \text{V/m} \][/tex]

F) Charge on the capacitor after the Mylar is inserted

The capacitance with the dielectric [tex]\( C' \)[/tex] is given by:

[tex]\[ C' = \kappa \cdot C \][/tex]

[tex]\[ C_{\text{after}} = 3.1 \times 2.21 \times 10^{-12} \, \text{F} \][/tex]

[tex]\[ C_{\text{after}} \approx 6.85 \times 10^{-12} \, \text{F} \][/tex]

The charge [tex]\( Q \)[/tex] on the capacitor is given by:

[tex]\[ Q = CV \][/tex]

[tex]\[ Q_{\text{after}} = (6.85 \times 10^{-12} \, \text{F})(7.5 \, \text{V}) \][/tex]

[tex]\[ Q_{\text{after}} \approx 5.14 \times 10^{-11} \, \text{C} \][/tex]

The required flow work of the compressor is calculated using the flow work equation and the ideal gas law. First, we determine the final volume using the ideal gas law. Then we substitute these figures into the flow work equation, giving the result as 8.08 kJ/kg.

The flow work required by the compressor is calculated using the equation

flow work = pressure * volume.

Given the initial pressure P1 = 120 kPa, volume V1 = 6 L and final pressure P2 = 1000 kPa,

we can substitute these values into the equation.

However, the volume at the end of compression is not given. To find this, we need to use the ideal gas law, P1V1/T1=P2V2/T2,

where T1 is the initial temperature and T2 is the final temperature.

Convert the temperatures to kelvins (T1 = 22 + 273 = 295 K, T2 = 400 + 273 = 673 K) and volume to m3 (V1 = 6 / 1000). Solving for V2 gives V2 = P1V1T2 / P2T1 = 0.00147 m3.

Now, substituting again into the flow work equation gives

flow work = (1000 kPa)(0.00147 m3) = 1.47 kJ.

This is the energy per unit volume, to find it per unit mass, divide it  by the specific volume

v2 = V2 / m = R*T2 / P2 = 0.182 m3/kg,

Therefore, the required flow work = 1.47 kJ / 0.182 kg = 8.08 kJ/kg.

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The flow work required by the compressor is approximately 622.476 kJ/kg.

The flow work required by the compressor is given by the equation:

[tex]\[ w_{flow} = \int_{1}^{2} v \, dp \][/tex]

where [tex]\( v \)[/tex] is the specific volume of the air and [tex]\( dp \)[/tex] is the differential pressure change. For an ideal gas, the specific volume can be calculated using the ideal gas law:

[tex]\[ v = \frac{RT}{p} \][/tex]

where [tex]\( R \)[/tex] is the gas constant, [tex]\( T \)[/tex] is the absolute temperature in Kelvin, and [tex]\( p \)[/tex] is the pressure.

Given:

- Initial pressure [tex]\( p_1 = 120 \)[/tex] kPa

- Final pressure [tex]\( p_2 = 1000 \)[/tex] kPa

- Initial temperature [tex]\( T_1 = 22 + 273.15 = 295.15 \)[/tex] K (converting from Celsius to Kelvin)

- Final temperature [tex]\( T_2 = 400 + 273.15 = 673.15 \)[/tex] K

- Gas constant [tex]\( R = 0.287 \) kPa/m^3/kg/K[/tex]

Since the process is adiabatic and the air is being compressed, we can assume that the specific volume at the initial state [tex]\( v_1 \)[/tex] can be calculated using the initial conditions:

[tex]\[ v_1 = \frac{RT_1}{p_1} \][/tex]

Substituting the values:

[tex]\[ v_1 = \frac{0.287 \times 295.15}{120} \][/tex]

[tex]\[ v_1 = \frac{84.80}{120} \][/tex]

[tex]\[ v_1 = 0.7067 \text{ m}^3/\text{kg} \][/tex]

The flow work can be approximated if we assume the process to be isothermal at the inlet temperature (which is a common simplification for flow work calculation):

[tex]\[ w_{flow} = v_1(p_2 - p_1) \][/tex]

Substituting the values:

[tex]\[ w_{flow} = 0.7067(1000 - 120) \][/tex]

[tex]\[ w_{flow} = 0.7067 \times 880 \][/tex]

[tex]\[ w_{flow} = 622.476 \text{ kJ/kg} \][/tex]

Answer:

Average power output of insect is 2.42W

Explanation:

Workdone by constant force during displacement is given by:

W= F× d cos theta

Where theta is angle between F and d.

Power output due to the force over the interval time is given by:

P= Workdone/change in time

Ginen:

Mass of insect,m= 7.0g= 7/1000 = 0.07kg

Downward force applied by insect,F= 2mg

Distance moved by the wing each stroke=1.5cm=1.5/100= 0.015m

W= F× d cos theta

Where theta=0° since force is in the same direction as the displacement.

W= 2mg×d

W= 2× 0.07 × 9.8 × 0.015

W= 0.02058J

Power output = W/ change in time

Since wings make 117strokes each second time interval is 1/117 = 8.5×10^-3seconds

Power= 0.02058/(8.5×10^-3)

Power= 2.42W

Answer:

0.25m

Explanation:

Using the expression that relates the angular velocity(w) and the linear velocity (v).

v= wr where;

w is the angular speed= 128rad/s

v is the linear speed = 32m/s

r is the radius

r = v/w

r = 32/128

r = 0.25m

The radius of the blade is 0.25m

Answer:

[tex]8000\ N[/tex]

Explanation:

The question is incomplete.

The complete question would be

B) Suppose the magnitude of the gravitational force between two spherical objects is 2000 N when they are 100 km apart. What is the gravitational force [tex]F_g[/tex] between the two objects described in Part B if the distance between them is only 50 km.

Given gravitational force between two object [tex]F_g=2000\ N[/tex] when objects are placed [tex]100\ km[/tex] apart.

We need to determine gravitational force when they are kept [tex]50\ km[/tex] apart.

As we know the gravitational force [tex]F_g[/tex] is inversely proportional to the square of distance between objects [tex](d)[/tex].

[tex]F_g=\frac{K}{d^2}[/tex]

Where [tex]K=Gm_1m_2[/tex] that will be constant. Because the mass of the object remain same in both cases. And [tex]G[/tex] is already gravitational constant.

Given,

[tex]2000=\frac{K}{100^2}\\ So,\ K=2000\times 100^2[/tex]

Let [tex]F'_g[/tex] is the force between objects when they were kept [tex]50\ km[/tex] apart.

[tex]F'_g=\frac{K}{50^2} \\\\F'_g=\frac{2000\times 100^2}{50^2}\\ \\F'_g=2000\times 4=8000\ N[/tex]

So, [tex]8000\ N[/tex] is the gravitational force when two objects were kept [tex]50\ km[/tex] apart.

The gravitational force between two objects can be calculated using Newton's law of gravitation formula, which requires knowledge of both objects' masses and the distance between them. Without exact masses of the objects in the question, it's not possible to provide a numeric answer, but generally, an increased distance leads to a decreased gravitational force.

Calculating Gravitational Force Between Two Objects

To calculate the gravitational force (Fg) between two objects using Newton's law of gravitation, you need to know the masses of the two objects and the distance between them. The formula to use is F = Gm1m2 / r2, where F is the gravitational force, G is the gravitational constant (6.673x10-11 N·m²/kg²), m1 and m2 are the masses of the objects, and r is the distance between their centers. If the masses of the objects are not provided in the question, you cannot calculate the force accurately. However, assuming identical conditions to the provided reference, where two objects have a mass of 50 kg each and are 0.50 meters apart, and you change the distance to 50 km (50,000 meters), the gravitational force will be significantly smaller due to the increase in distance according to the inverse square law of gravitation.

If you had the exact masses, the gravitational force at the new distance of 50 km could be calculated directly using the formula by substituting the known values for mass and distance. The result would show how much less the gravitational force is at the greater distance compared to the initial 0.50 meters. This calculation demonstrates the concept that as distance increases, the gravitational force between two masses decreases rapidly.

Answer: 1.9394m/s

Explanation: m1 (mass of 3 billiard ball) = 0.16kg

U1(initial velocity of the 3 billiard ball) = 4.0m/s

M2 (mass of the cue ball) = 0.17kg

U1 (initial velocity of the cue ball) = 0m/s

Final velocities of both the cue ball and 3 billiard ball after collision = ?

According to the principle of conservation of linear momentum, total momenta before collision = total momenta after collision:

M1U1 + M2U2 = (M1+M2)V

0.16*4.0 + 0.17* 0 = (0.16 + 0.17)V

0.64 + 0 = 0.33V

V = 0.64/0.33 = 1.9394m/s

Answer: True

Explanation:

Electromagnetic wave is defined as the wave which is associated with both electrical and magnetic component associated with them. They can travel in vacuum as well and travel with the speed of light.

Electromagnetic spectrum consists of electromagnetic waves called as radio waves, microwaves, infrared, Visible, ultraviolet , X rays and gamma rays in order of increasing frequency and decreasing wavelength.

Thus the statement that All light waves move through a vacuum with a constant speed is true.

Answer:

The electrostatic potential within the conductor will be negative and constant throughout the conductor.

Explanation:

Whenever a conductor is charged the total charge is spread throughout the entire surface uniformly. Inside the conductor the net electric field is always zero. Also we know if [tex]\overrightarrow{E}[/tex] e the electric field and V is the potential, then

[tex]\overrightarrow{E}= - \overrightarrow{\nabla}V[/tex]

Now as in this case electric field within the conductor is zero, so potential (V) has to be constant.

Final answer:

The electric potential within a charged conductor is constant and uniform, though not necessarily zero, while outside the conductor, the potential is nonzero. This is because the electric field inside a conductor in electrostatic equilibrium is zero, leading to a uniform potential throughout the conductor's inside and surface.

Explanation:

When considering an arbitrarily shaped piece of conductor with a net negative charge in space, we can assert that the electric potential within the conductor is constant throughout its volume. By the principle that in electrostatic equilibrium, the electric field inside a conductor is zero, the potential has the same value in all points inside the conductor. However, this potential value is not necessarily zero; it is referenced to a point far away from the conductor, where the potential is assumed to be zero. Outside of the conductor, the potential is nonzero and governed by the distribution of charge on the conductor's surface and the distance from it.

Considering a perfect conductor, when a test charge is moved from one point to another within or on the surface of the conductor, no work is done, because the electric field is zero. Hence, the electric potential difference between any two points is zero, and all points in and on the conductor share the same potential value with respect to infinity or ground.

Despite the conductor’s potential not being zero in general, its uniformity throughout highlights a fundamental concept in electrostatics: that upon reaching equilibrium, charge distributes on the conductor's surface to ensure the electric field within the conductor is nonexistent, thus establishing a uniform potential.

Explanation:

Below is an attachment containing the solution.

Answer:

Jupiter

Explanation:

This is because it has the largest mass. Jupiter  is massive and has the highest mass in the solar system.

Gravitational pull is dependent on the mass of body based on the newton's law of gravitation.

I hope this was helpful, please mark as brainliest

Answer:i had this on usatest prep it is jupiter

Explanation:

The unstretched length of the spring is 0.15 m

Explanation:

Given-

Length of the stretched spring, [tex]x_{f}[/tex] = 20 cm = 0.2 m

K = 200 N/m

U = 0.25 J

unstretched length of the spring, x₀ = ?

We know,

U = [tex]\frac{1}{2} (k)[/tex] (Δx)²

Δx = [tex]x_{f}[/tex] - x₀

Δx = 0.2 - x₀

0.25 = 100 (0.2 - x₀)²

0.0025 = (0.2 - x₀)²

0.05 = 0.2 - x₀

x₀ = 0.15 m

Therefore, the unstretched length of the spring is 0.15 m

Answer:

Explained below:

Explanation:

Range of Motion is the measurement of action throughout a specific joint or body part. If your goniometer breaks, no need to worry because an inclinometer is also helpful to measure the range of motion in a joint by measuring the joint angles at length and flexion and in order to verify that there is improvement being made on rising the range of motion in a joint, the physical therapist measures the joint angle before the treatment and keep on doing to do so over time.

It is possible to measure a range of motion by using a protractor and a ruler.

Motion or movement is a measure to indicate the distance that travels a particular object and/or material.

In conclusion, it is possible to measure a range of motion by using a protractor and a ruler.

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Answer:

The frictional force will also act on the box. So, we can say that the 10 N force is an unbalanced force.

Explanation:

It is given that, a 10-Newton force is applied to a 2-kg block. The block slides across the floor at a constant speed of 5 m/s.

The block is sliding with a constant speed. This shows the acceleration of the block is zero.

When an object slides that means the force acting on it is unbalanced i.e. the object will move in a particular direction.

A)  The force of friction was 10-newtons.

The force of friction was 10-newtons. If the block is moving at constant speed, all the forces on the block must be balanced.

Answer:

3seconds

Explanation:

Time of flight of an object is the time taken by the object to spend in the air before landing. Mathematically it is represented as;

T = 2u/g

Where g is the acceleration due to gravity = 10m/s²

If the vertical velocity component (u) is given as 15m/s, the time of flight will be;

T = 2(15)/10

T = 30/10

T = 3seconds.

Therefore the total time by the ball in the air is 3seconds

Explanation:

Below is an attachment containing the solution.

Explanation:

Below is an attachment containing the solution.

The rider is rising at a speed of approximately 15.71 m/min when the rider is 9 m above ground level on a Ferris wheel with a radius of 5 m rotating at a rate of one revolution every 2 minutes.

To find the speed at which the rider is rising, we can use the concept of angular velocity. The angular velocity of the Ferris wheel can be calculated by taking the circumference of the circle formed by the rider's position and dividing it by the time it takes to complete one revolution. In this case, the circumference is equal to 2π multiplied by the radius of the Ferris wheel. The time it takes to complete one revolution is given as 2 minutes.

The formula for angular velocity is ω = θ/t, where ω is the angular velocity, θ is the angle swept, and t is the time taken. Since one revolution is equal to 360 degrees or 2π radians, the angular velocity can be calculated as:

ω = (2π rad)/(2 min) = π rad/min

Now, to find how fast the rider is rising, we can use the relationship between linear velocity (v) and angular velocity (ω) given by v = rω, where r is the radius of the Ferris wheel. In this case, the radius is given as 5 m. Plugging in the values, the linear velocity is:

v = (5 m)(π rad/min) = 5π m/min ≈ 15.71 m/min

Therefore, when the rider is 9 m above ground level, the rider is rising at a speed of approximately 15.71 m/min.

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Answer:

The magnetic force on a free moving charge depends on the velocity of the charge and the magnetic field, direction of the force is given by the right hand rule. While gravitational depends on the mass and distance of the moving particle and electric forces depends on the magnitude of the charge and distance of separation.

Explanation:

The magnetic force on a free moving charge depends on the velocity of the charge and the magnetic field and direction of the force is given by the right hand rule. While gravitational depends on the mass and distance of the moving particle and electric forces depends on the magnitude of the charge and distance of separation.

The magnetic force is given by the charge times the vector product of velocity and magnetic field. While gravitational force is given by the square of the particle mass divided by the square its distance of separation. Also electric forces is given by the square of the charge magnitude divided by the square its distance separation.

Answer:

[tex]5.8\cdot 10^{-5}T[/tex]

Explanation:

The magnetic field produced by a current-carrying wire is:

[tex]B=\frac{\mu_0 I}{2\pi r}[/tex]

where

[tex]\mu_0=4\pi \cdot 10^{-7} H/m[/tex] is the vacuum permeability

I is the current in the wire

r is the distance from the wire

The direction of the field is given by the right-hand rule: the thum is placed in the same direction of the current, and the other fingers "wrapped" around the thumb gives the direction of the field.

First of all, here we are finding the magnetic field strength at a point 12.0cm from one wire and 13.6cm from the other: since the two wires are 6.50 cm apart, this means that the point at which we are calculating the field is located either on the left or on the right of both wires. So, by using the right-hand rule for both, we see that both fields go into the same direction (because the currents in the two wires have same direction).

Therefore, the net magnetic field will be the sum of the two magnetic fields.

For wire 1, we have:

[tex]I_1=18.5 A\\r_1=12.0 cm =0.12 m[/tex]

So the field is

[tex]B_1=\frac{(4\pi \cdot 10^{-7})(18.5)}{2\pi(0.12)}=3.1\cdot 10^{-5} T[/tex]

For wire 2, we have:

[tex]I_2=18.5 A\\r_2=13.6 cm=0.136 m[/tex]

So the  field is

[tex]B_2=\frac{(4\pi \cdot 10^{-7})(18.5)}{2\pi(0.136)}=2.7\cdot 10^{-5} T[/tex]

Therefore, the total magnetic field is

[tex]B=B_1+B_2=3.1\cdot 10^{-5}+2.7\cdot 10^{-5}=5.8\cdot 10^{-5}T[/tex]