What is the resistance of a 7 W light bulb connected to a 7 V battery source. The answer key says .39 but not sure how they got that?

Answer:

The magnitude and direction of the non conservative force acting on the box is <F= 59.36 N - DUE EAST DIRECTION>.

Explanation:

m= 4 kg

Vi= 11 m/s

Vf= 1.5 m/s

d= 4m

d= Vi * t - a * t²/2

clearing a:

a= 2*(Vi * t - d)/ t²

Vf= Vi - a * t

replacing "a" and clearing t:

t= 2d/(Vf+Vi)

t= 0.64 s

found now the value of a:

a= Vi - Vf / t

a= 14.84 m/s ²

F= m * a

F= 59.36 N

Answer:

Magnetic field shall be zero at exactly in between the wires.

Explanation:

We can find the magnetic field by biot Savart law as follows

[tex]\overrightarrow{dB}=\frac{\mu _{0}I}{4\pi }\int \frac{\overrightarrow{dl}\times \widehat{r}}{r^{2}}[/tex]

For current carrying wire in positive y direction we have

[tex]\overrightarrow{dB_{1}}=\frac{\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{1}}}{r_{1}^{2}}[/tex]

Similarly for wire carrying current in -y direction we have [tex]\overrightarrow{dB_{2}}=\frac{-\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{2}}}{r_{2}^{2}}[/tex]

Thus the net magnetic field at any point in space is given by

[tex]\overrightarrow{dB_{1}}+\overrightarrow{dB_{2}}[/tex]

[tex]\frac{\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{1}}}{r_{1}^{2}}+\frac{-\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{2}}}{r_{2}^{2}}=0\\\\\Rightarrow \overrightarrow{r_{1}}=\overrightarrow{r_{2}}[/tex]

For points with same position vectors from the 2 wires we have a net zero magnetic field. These points are exactly midway between the 2 wires

Answer:

Student b is suspended at rest above the floor.

Explanation:

Student a and student b are holding the two ends of the rope which passes over a pulley.

The mass of student a = 70 kg

The mass of student b = 60 kg

The tension force acting on the rope is uniform through out the length in this case.

Force due to gravitation:

F = mg

Where,

m is the mass and g is the acceleration due to gravity

Since g is constant for both the students. Thus, F is directly proportional to the mass of the student.

Since the mass of student a is greater than student b. Thus, the force due to the weight of the student a is greater than student b.

Thus, Student a will be at the floor and Student b is suspended at rest above the floor.

Answer:

Part a)

[tex]L = 18 kg m^2/s[/tex]

Part b)

[tex]v = 0.6 m/s[/tex]

Part c)

[tex]R_{max} = 6.04 kg[/tex]

[tex]R_{min} = 5.96 kg[/tex]

Explanation:

As we know that Ferris wheel start from rest with angular acceleration

[tex]\alpha = 0.001 rad/s^2[/tex]

time taken = 2 min

so here we have its angular speed after t = 2min given as

[tex]\omega = \alpha t[/tex]

[tex]\omega = (0.001)(2\times 60)[/tex]

[tex]\omega = 0.12 rad/s[/tex]

Part a)

Angular momentum of the Penguine about the center of the wheel is given as

[tex]L = I\omega[/tex]

[tex]L = (6\times 5^2)(0.12)[/tex]

[tex]L = 18 kg m^2/s[/tex]

Part b)

tangential speed is given as

[tex]v = r\omega[/tex]

[tex]v = (5)(0.12)[/tex]

[tex]v = 0.6 m/s[/tex]

Part c)

Maximum reading of the scale at the lowest point is given as

[tex]R_{max} = \frac{m\omega^2 r + mg}{g}[/tex]

[tex]R_{max} = \frac{6(0.12^2)(5) + 6(9.81)}{9.81}[/tex]

[tex]R_{max} = 6.04 kg[/tex]

Minimum reading of the scale at the top point is given as

[tex]R_{min} = \frac{mg - m\omega^2 r}{g}[/tex]

[tex]R_{min} = \frac{6(9.81) - 6(0.12^2)(5)}{9.81}[/tex]

[tex]R_{min} = 5.96 kg[/tex]

6 ohms is the resistance of the unknown resistor

Electrical resistance, or resistance to electricity, is a force that opposes the passage of current. It acts as a gauge for the difficulty of current flow in this way. Ohms () are used to express resistance values.

When two resistors are connected in parallel, the equivalent resistance can be calculated using the following formula:

1/Req = 1/R1 + 1/R2

where Req is the equivalent resistance, R1 and R2 are the resistances of the two individual resistors.

In this case, we know the resistance of one of the resistors (R1 = 12 ohms) and we can find the current through it (I1) using Ohm's law:

I1 = V/R1 = 12 V / 12 ohms = 1 A

We also know the total current (It = 3 A) and we can find the current through the unknown resistor (I2) using Kirchhoff's current law:

It = I1 + I2

I2 = It - I1 = 3 A - 1 A = 2 A

Now we can use Ohm's law to find the resistance of the unknown resistor (R2):

R2 = V/I2 = 12 V / 2 A = 6 ohms

Therefore, the resistance of the unknown resistor is 6 ohms.

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Final answer:

The resistance of the unknown resistor is 4.0 ohms.

Explanation:

To find the resistance of the unknown resistor, we can use Ohm's Law, which states that V = IR, where V is the voltage, I is the current, and R is the resistance. In this case, the voltage across the unknown resistor is the same as the voltage across the 12-ohm resistor, which is 12V. The current through the unknown resistor is given as 3.0A.

Using Ohm's Law, we can rearrange the formula to solve for R:

R = V/I

R = 12V/3.0A

R = 4.0 ohms

Therefore, the resistance of the unknown resistor is 4.0 ohms.

Answer:

C) 7,500 m/s

Explanation:

The satellite's acceleration due to gravity equals its centripetal acceleration.

v² / r = GM / r²

Solving for velocity:

v² = GM / r

v = √(GM / r)

Given:

G = 6.67×10⁻¹¹ m³/kg/s²

M = 5.98×10²⁴ kg

r = 6.357×10⁶ m + 6.8×10⁵ m = 7.037×10⁶ m

Substituting the values:

v = √(6.67×10⁻¹¹ × 5.98×10²⁴ / 7.037×10⁶)

v = √(5.67×10⁷)

v = 7500 m/s

The force by the pitcher's hand on the ball during the acceleration phase is found using Newton's Second Law of Motion and is calculated to be 116 Newton.

The force exerted by the pitcher's hand can be found using Newtons Second Law of Motion which states that the force acting on an object is equal to the mass of the object times its acceleration.

it can be expressed as

F = m * a

Here, the mass (m) is 0.145 kg, and the acceleration (a) can be found using the formula a = Δv/Δt, where Δv is the change in velocity (40 m/s) and Δt is the change in time (50 ms or 0.05 s).

So, a = 40/0.05 = 800 m/s², and then the force F = 0.145 * 800 = 116 N. Therefore, the force of the pitcher's hand on the ball during this acceleration phase is 116 Newton.

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The force of the pitcher's hand on the ball can be calculated using Newton's second law of motion.

The force of the pitcher's hand on the ball can be calculated using Newton's second law of motion, which states that force equals mass times acceleration. In this case, the mass of the ball is 0.145 kg and the acceleration is the change in velocity divided by the time interval. The change in velocity is 40 m/s (the final velocity) minus 0 m/s (the initial velocity), and the time interval is 50 ms (or 0.05 s). Therefore, the force can be calculated as:

Force = mass × acceleration = 0.145 kg × (40 m/s - 0 m/s) / 0.05 s = 0.116 N.

So, the force of the pitcher's hand on the ball during this acceleration phase is approximately 0.116 Newtons.

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Answer: A. force with which air rushes across the vocal folds

Explanation:

The human voice is produced in the larynx, whose essential part is the glottis. This is how the air coming from the lungs is forced during expiration through the glottis, making its two pairs of vocal folds to vibrate.

It should be noted that this process can be consciously controlled by the person who speaks (or sings), since the variation in the intensity of the sound of the voice depends on the strength of the breath.

The loudness of a person's voice mainly depends on the force of airflow across their vocal folds. This causes the vocal folds to vibrate and the sound to be produced. The volume of this sound is determined by the amplitude of the resulting sound pressure wave.

The loudness of a person's voice is primarily dependent on the force with which air rushes across the vocal folds. Sound is created when air is pushed up from the lungs through the throat, causing the vocal folds to vibrate. When air flow from the lungs increases, the amplitude of the sound pressure wave becomes greater, resulting in a louder voice. Changes in pitch are related to muscle tension on the vocal cords.

Vocal cord vibration and sound pressure wave amplitude are thus key factors in determining the loudness of a person's voice.

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Answer: 80J

Explanation:

According to the first principle of thermodynamics:  

"Energy is not created, nor destroyed, but it is conserved."  

Then this priciple (also called Law) relates the work and the transferred heat exchanged in a system through the internal energy [tex]U[/tex], which is neither created nor destroyed, it is only transformed. So, in this especific case of the compressed gas:

[tex]\Delta U=Q+W[/tex]  (1)

Where:

[tex]\Delta U[/tex] is the variation in the internal (thermal) energy of the system (the value we want to find)

[tex]Q=-100J[/tex] is the heat transferred out of the gas (that is why it is negative)

[tex]W[/tex] is the work is done on the gas (as the gas is compressed, the work done on the gas must be considered positive )

On the other hand, the work done on the gas is given by:

[tex]W=-P \Delta V[/tex]  (2)

Where:

[tex]P=450kPa=450(10)^{3}Pa[/tex] is the constant pressure of the gas

[tex]\Delta V=V_{f}-V_{i}[/tex] is the variation in volume of the gas

In this case the initial volume is [tex]V_{i}=600{cm}^{3}=600(10)^{-6}m^{3}[/tex] and the final volume is [tex]V_{f}=200{cm}^{3}=200(10)^{-6}m^{3}[/tex].

This means:

[tex]\Delta V=200(10)^{-6}m^{3}-600(10)^{-6}m^{3}=-400(10)^{-6}m^{3}[/tex]  (3)

Substituting (3) in (2):

[tex]W=-450(10)^{3}Pa(-400(10)^{-6}m^{3})[/tex]  (4)

[tex]W=180J[/tex]  (5)

Substituting (5) in (1):

[tex]\Delta U=-100J+180J[/tex]  (6)

Finally:

[tex]\Delta U=80J[/tex]  This is the change in thermal energy in the compression process.

Answer:

The pressure in the chamber are of p= 0.127 atm.

Explanation:

n= 2.5 moles

T= 310 K

V= 0.5 m³ = 500 L

R= 0.08205746 atm. L /mol . K

p= n*R*T/V

p= 0.127 atm

Answers: (A)[tex]F=G\frac{M^2}{4R^2}[/tex] (B) [tex]V=\sqrt{\frac{GM}{4R}}[/tex] (C)[tex]T=4\pi R\sqrt{\frac{R}{GM}}[/tex] (D)

[tex]E=-\frac{GM^{2}}{4R}[/tex]

Explanation:

According to the law of universal gravitation:

[tex]F=G\frac{m_{1}m_{2}}{r^2}[/tex]   (1)

Where:

[tex]F[/tex] is the module of the gravitational force exerted between both bodies  

[tex]G[/tex] is the universal gravitation constant.

[tex]m_{1}[/tex] and [tex]m_{2}[/tex] are the masses of both bodies.

[tex]r[/tex] is the distance between both bodies

In the case of this binary system with two stars with the same mass [tex]M[/tex] and separated each other by a distance [tex]2R[/tex], the gravitational force is:

[tex]F=G\frac{(M)(M)}{(2R)^2}[/tex]   (2)

[tex]F=G\frac{M^2}{4R^2}[/tex]   (3) This is the gravitational force between the two stars.

Taking into account both stars describe a circular orbit and the fact this is a symmetrical system, the orbital speed [tex]V[/tex] of each star is the same. In addition, if we assume this system is in equilibrium, gravitational force must be equal to the centripetal force  [tex]F_{C}[/tex] (remembering we are talking about a circular orbit):

So: [tex]F=F_{C}[/tex]   (4)

Where [tex]F_{C}=Ma_{C}[/tex]  (5) Being [tex]a_{C}[/tex] the centripetal acceleration

On the other hand, we know there is a relation between [tex]a_{C}[/tex] and the velocity [tex]V[/tex]:

[tex]a_{C}=\frac{V^{2}}{R}[/tex]  (6)

Substituting (6) in (5):

[tex]F_{C}=M\frac{V^{2}}{R}[/tex] (7)

Substituting (3) and (7) in (4):

[tex]G\frac{M^2}{4R^2}=M\frac{V^{2}}{R}[/tex]   (8)

Finding [tex]V[/tex]:

[tex]V=\sqrt{\frac{GM}{4R}}[/tex] (9) This is the orbital speed of each star

The period [tex]T[/tex] of each star is given by:

[tex]T=\frac{2\pi R}{V}[/tex]  (10)

Substituting (9) in (10):

[tex]T=\frac{2\pi R}{\sqrt{\frac{GM}{4R}}}[/tex]  (11)

Solving and simplifying:

[tex]T=4\pi R\sqrt{\frac{R}{GM}}[/tex]  (12) This is the orbital period of each star.

The gravitational potential energy [tex]U_{g}[/tex] is given by:

[tex]U_{g}=-\frac{Gm_{1}m_{2}}{r}[/tex]  (13)

Taking into account this energy is always negative, which means the maximum value it can take is 0 (this happens when the masses are infinitely far away); the variation in the potential energy [tex]\Delta U_{g}[/tex] for this case is:

[tex]\Delta U_{g}=U-U_{\infty}[/tex] (14)

Knowing [tex]U_{\infty}=0[/tex] the total potential energy is [tex]U[/tex] and in the case of this binary system is:

[tex]U=-\frac{G(M)(M)}{2R}=-\frac{GM^{2}}{2R}[/tex]  (15)

Now, we already have the potential energy, but we need to know the kinetic energy [tex]K[/tex] in order to obtain the total Mechanical Energy [tex]E[/tex] required to separate the two stars to infinity.

In this sense:

[tex]E=U+K[/tex] (16)

Where the kinetic energy of both stars is:

[tex]K=\frac{1}{2}MV^{2}+\frac{1}{2}MV^{2}=MV^{2}[/tex] (17)

Substituting the value of [tex]V[/tex] found in (9):

[tex]K=M(\sqrt{\frac{GM}{4R}})^{2}[/tex] (17)

[tex]K=\frac{1}{4}\frac{GM^{2}}{R}[/tex] (18)

Substituting (15) and (18) in (16):

[tex]E=-\frac{GM^{2}}{2R}+\frac{1}{4}\frac{GM^{2}}{R}[/tex] (19)

[tex]E=-\frac{GM^{2}}{4R}[/tex] (20) This is the energy required to separate the two stars to infinity.

F = [tex]\frac{GM1M2}{R} \\\\[/tex]

Where F = Gravitational force that is between the two masses

M1 = mass of the first star

M2 = Mass of the second star

d= distance between the masses

G = Gravitational constant

These stars are identical so M1 = M2 = M

d = 2R

d² = 2R² = 4R²

[tex]F= \frac{GM1M2}{4R^2}[/tex]

Given that theses circles are said to move in a circular orbit, the net centripetal force

Fnet = F second law of Newton

[tex]Fnet = \frac{MV^2}{R} =\frac{GM^2}{4R^2} \\\\v^2 = \frac{GM}{4R}[/tex]

[tex]v = \sqrt[]{}\frac{{GM} }{4R}[/tex]

C. The orbital period

Period = time T

speed = distance traveled / time

distance = circumference of a circle = 2πR

V = 2πR/T

T = 2πR/V

[tex]T = 2\pi R(\sqrt{(MG)/(4R)}^{-1} \\[/tex]

[tex]T = (2\pi R\sqrt{4R} )/(\sqrt{MG} )[/tex]

[tex]T = (2\pi R(2R^{1/2} ))/\sqrt{MG}[/tex]

[tex]T = (4\pi R^{3/2}) /\sqrt{MG[/tex]

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Answer:

D) multiply the force in the direction of motion by the distance the object moved

Explanation:

To calculate work done on an object, multiply the force in the direction of motion by the distance the object moved.

Final answer:

To calculate the work done on an object, multiply the force in the direction of motion by the distance the object moved, using the formula W = F d.

Explanation:

To calculate work done on an object, you multiply the force in the direction of motion by the distance the object moved. This calculation is represented by the formula W = F d, where W is the work done on the system, F is the constant force applied parallel to the direction of motion, and d is the displacement of the object in the direction of the force. It's important to note that the force considered in this calculation is only the component that acts in the same direction as the displacement of the object.

Answer:

Work out = 28.27 kJ/kg

Explanation:

For R-134a, from the saturated tables at 800 kPa, we get

[tex]h_{fg}[/tex] = 171.82 kJ/kg

Therefore, at saturation pressure 140 kPa, saturation temperature is

[tex]T_{L}[/tex] = -18.77°C = 254.23 K

At saturation pressure  800 kPa, the saturation temperature is

[tex]T_{H}[/tex] = 31.31°C = 304.31 K

Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.

Thus, [tex]q_{reject}[/tex] = [tex]h_{fg}[/tex] = 171.82 kJ/kg

We know COP of heat pump

COP = [tex]\frac{T_{H}}{T_{H}-T_{L}}[/tex]

        = [tex]\frac{304.31}{304.31-254.23}[/tex]

         = 6.076

Therefore, Work out put, W = [tex]\frac{q_{reject}}{COP}[/tex]

                                              = 171.82 / 6.076

                                              = 28.27 kJ/kg

Final answer:

Accurately calculating the net work input for a heat pump using the reversed Carnot cycle requires additional data beyond the pressure limits and the typical phase change of R-134a. Specific temperatures, heat transfers, or detailed thermodynamic tables for R-134a are essential for this calculation.

Explanation:

To determine the net work input for a heat pump operating on the reversed Carnot cycle with R-134a as the working fluid, it's important to understand the foundational concepts of heat pumps and thermodynamics. Heat pumps transfer heat from a colder area (cold reservoir) to a warmer area (hot reservoir). When examining a reversed Carnot cycle on a P-V diagram, the area within the cycle represents work, however, since it is reversed, the location of heat transfer and work are also reversed compared to a regular heat engine. In a standard heat engine, work is output, but in a heat pump, work is input.

The net work input is calculated by considering the efficiency of the system, which is dependent on the temperatures of the cold and hot reservoirs. However, since the question does not provide specific temperatures or heat transfers but only pressures, and asks us to consider a scenario where the working fluid, R-134a changes from saturated vapor to saturated liquid during the heat-rejection process, additional information such as thermodynamic tables for R-134a or further details about the specific amounts of heat transfer would be needed to calculate the precise net work input.

Final answer:

To find the initial temperature, we can use the combined gas law. By plugging in the values of the initial and final volumes and the final temperature, we can solve for the initial temperature.

Explanation:

To solve this problem, we can use the combined gas law, which states that for an ideal gas at constant pressure, the ratio of the initial volume and the initial temperature is equal to the ratio of the final volume and the final temperature:

V1/T1 = V2/T2

Given that the initial volume is 2.70 L, the final volume is 1.75 L, and the final temperature is 17.00 °C, we can plug these values into the combined gas law to find the initial temperature:

2.70/T1 = 1.75/17.00

Solving for T1, we get T1 = 45.545 °C. Therefore, the initial temperature of the gas was approximately 45.545 °C.

Answer:

57.94°

Explanation:

we know that the expression of flux

[tex]\Phi =E\times S\times COS\Theta[/tex]

where Ф= flux

           E= electric field

           S= surface area

        θ = angle between the direction of electric field and normal to the surface.

we have Given Ф= 78 [tex]\frac{Nm^{2}}{sec}[/tex]

                          E=[tex]1.44\times 10^{4}\frac{Nm}{C}[/tex]

                          S=[tex]\pi \times 0.057^{2}[/tex]

                         [tex]COS\Theta =\frac{\Phi }{S\times E}[/tex]

 =   [tex]\frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}[/tex]

 =0.5306

 θ=57.94°

In an isothermal expansion of a monatomic ideal gas where the volume doubles, Boyle's Law implies that the pressure of the gas will decrease by a factor of 2, or in other words, will be halved.

The change in pressure for a monatomic ideal gas in an isothermal expansion where the volume doubles can be explained using Boyle's Law. Boyle's Law states that the product of the pressure and volume for a given gas sample is constant as long as the temperature is constant (P₁V₁ = P₂V₂). Therefore, if the volume of the gas doubles, the pressure would reduce by half, assuming the temperature remains constant.

Using the ideal gas equation, p = nRT/V, where 'p' is the final pressure, 'n' is the number of moles, 'R' is the gas constant, 'T' is temperature, and 'V' is volume, we find the final pressure of the gas to be Po/2, where Po is the initial pressure.

Thus, in an isothermal expansion of an ideal monatomic gas where the volume doubles, the pressure will decrease by a factor of 2, that is, it will be halved.

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Answer:

109.09°C

Explanation:

Given data:

The capacity of the car cooling system =  4.40 gallons

Ratio of water and solute = 50/50

thus,

Volume of solute = Volume of water = 4.40/ 2 gallons = 2.20 gallons

thus, the Mass of solute, M₁ = Density × Volume = 1.1 g/mL × [tex]2.20\times\frac{3785.41\ \textup{mL}}{1\ \textup{gallon}}[/tex] = 9160.69 grams

Mass of water = 0.998 g/mL × [tex]2.20\times\frac{3785.41\ \textup{mL}}{1\ \textup{gallon}}[/tex] = 8310.346 grams

molality , m = [tex]\frac{mass\ of \ solute\ \times 1000}{molar\ mass\ of\ solute\times\ mass\ of\ water}[/tex]

on substituting the values, we get

m = [tex]\frac{9160.69 \times 1000}{62.07\times8310.346}[/tex] =  17.759 m

Now, the elevation in boiling point is given as:

[tex]\Delta T_b = k_bm[/tex]

where,

[tex]k_b[/tex] is a constant = 0.512 °C/m

on substituting the values we get

Boiling point = 100°C + 17.759 × 0.512= 109.09°C

The boiling point of the solution will be equal to 109.09°C.

We can arrive at this answer as follows:

First, we must take into account that the water and solute ratio is 50/50, therefore, we must divide the capacity of the car's cooling system by 2, to know the volume of solute and solvent, which will be equal per account of the reason between them.

[tex]\frac{4.40}{2} = 2.20[/tex] gallons

[tex]Density * Volume\\1.1*(2.20*3785.41)\\9160.69 g[/tex]

[tex]0.998*(2.20*3585.41)\\8310.346 g[/tex]

[tex]Molality = \frac{M_s*1000}{molar M_S*M_W}\\\\Molality = \frac{9160.69*1000}{62.07*8310.34} \\\\Molality= 17.759 m[/tex]

[tex]\Delta T= K_b*molality\\[/tex]

We must take into account that [tex]K_b[/tex] is a constant and its value is always 0.512 °C/m.

[tex]\Delta T= 0.512*(100C + 17.759)\\\Delta T= 109.09C[/tex]

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Answer:

When the net external force is zero

Explanation:

We can answer to this question by referring to Newton's Second Law, which can be written in the following form

[tex]F=\frac{\Delta p}{\Delta t}[/tex]

where

F is the net external force acting on a system

[tex]\Delta p[/tex] is the change in momentum of the system

[tex]\Delta t[/tex] is the time interval

When the total momentum of a system is conserved (so, it does not change), its variation is zero:

[tex]\Delta p = 0[/tex]

In order to satisfy this condition, we see from the formula that we must also have

F = 0

so the net external force acting on the system must be zero.

Answer:

The temperature must the ring be heated so that the sphere can just slip through is 106.165 °C.

Explanation:

For brass:

Radius = 1.3590 cm

Initial temperature = 23.0 °C

The sphere of radius 1.3611 cm must have to slip through the brass. Thus, on heating the brass must have to attain radius of 1.3611 cm

So,

Δ r = 1.3611 cm - 1.3590 cm = 0.0021 cm

The linear thermal expansion coefficient of a metal is the ratio of the change in the length per 1 degree temperature to its length.

Thermal expansion for brass = 19×10⁻⁶ °C⁻¹

Thus,

[tex]\alpha=\frac {\Delta r}{r\times \Delta T}[/tex]

Also,

[tex]\Delta T=T_{final}-T_{Initial}[/tex]

So,

[tex]19\times 10^{-6}=\frac {0.0021}{1.3290\times (T_{final}-23.0)}[/tex]

Solving for final temperature as:

[tex](T_{final}-23.0)=\frac {0.0021}{1.3290\times 9\times 10^{-6}}[/tex]

Final temperature = 106.165 °C

Answer:

Explanation:

Given that,

Number of mole

n = 2.1mole

Temperature

T = 278 K

Initial volume

V1 = 14.5L

Work done

W = 945J

R = 8.314 J/mol•K

Work done is given as at constant temperature is

W = -P1•V1 In(V2/V1)

Now, let know the pressure using is ideal gas law

PV = nRT

P = nRT/V

P = 2.1 × 8.314 ×278 / 14.5

P = 334.74 N/L

Then,

W = -P1•V1 In(V1/V2)

945 = -334.74×14.5 In(V1/V2)

-945/(334.74×14.5) = In(V1/V2)

In(V1/V2) = -0.1947

Take exponential of both sides

V1/V2 = exp(-0.1947)

14.5/V2 = 0.823

14.5 = 0.823V2

V2 = 14.5/0.823

V2 = 17.62 L

The third option is correct

A) The magnetic field doubles as well

The relationship between rms value of the electric field and rms value of the magnetic field for an electromagnetic wave is the following:

[tex]E_{rms}=cB_{rms}[/tex]

where

E_rms is the magnitude of the electric field

c is the speed of light

B_rms is the magnitude of the magnetic field

From the equation, we see that the electric field and the magnetic field are directly proportional: therefore, if the rms value of the electric field is doubled, then the rms value of the magnetic field will double as well.

B) The intensity will quadruple

The intensity of an electromagnetic wave is given by

[tex]I=\frac{1}{2}c\epsilon_0 E_0^2[/tex]

where

c is the speed of light

[tex]\epsilon_0[/tex] is the vacuum permittivity

[tex]E_0[/tex] is the peak intensity of the electric field

The rms value of the electric field is related to the peak value by

[tex]E_{rms}=\frac{E_0}{\sqrt{2}}[/tex]

So we can rewrite the equation for the intensity as

[tex]I=c\epsilon_0 E_{rms}^2[/tex]

we see that the intensity is proportional to the square of the rms value of the electric field: therefore, if the rms value of the electric field is doubled, the average intensity of the wave will quadruple.

Answer:

The correct answer is "The rigid body can have rotational and transnational motion, as long as it's transnational and angular accelerations are equal to zero."

Explanation:

A rigid body by definition does not deform when forces act on it. In case of static equilibrium a rigid body cannot have any sort of motion while in case of dynamic equilibrium it can move but with constant velocities only thus having no acceleration weather transnational or angular.

The rigid body can move in both directions as long as its angular acceleration and transverse accelerations dosent get vanished"

The phrase "angular acceleration" refers to the temporal rate at which angular velocity changes. Consequently, = d d rotational acceleration is another name for angular acceleration.

When a body experiences linear acceleration, the force or acceleration acts simultaneously on the entire body. a straight course's rate of change in velocity per unit of time. This acceleration is linear. The rotational acceleration of an item about an axis is known as angular acceleration.

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Answer:

Explanation:

Since the equation for the illumination of an object, i.e. the brightness of the light, is inversely proportional to the square of the distance from the light source, the form of the function is:

Where x is the distance between the object and the light force, k is the constant of proportionality, and f(x) is the brightness.

Then, if you move halfway to the lamp the new distance is x/2 and the new brightness (call if F) is :

[tex] F=k(x/2)^{-2}=\frac{k}{(x/2)^2}= \frac{k}{x^2}. 4=f(x).4[/tex]

Then, you have found that the light is 4 times as bright as it originally was.

The light becomes four times brighter when you move halfway closer to a light source, due to the inverse square law.

When you move halfway closer to a light source, the illumination on the book you are reading increases dramatically because of the inverse square law for light propagation. According to this law, the brightness or intensity of light is inversely proportional to the square of the distance from the light source. If you reduce the distance to half, the new brightness becomes (1/0.5)^2 = 4 times the initial brightness.

For example, if the initial intensity of the light is I at a distance d, when you move to a distance of d/2, the intensity becomes 4I. This is because the new distance squared is (d/2)^2 which is 1/4th of the original distance squared (d^2), hence reversing the effect we get four times the initial intensity due to the inverse law.

Answer:0.0242 Nm

Explanation: The distance of 0.145 mm is very small. This will also result in a small ttorque. The question would make more sense if the distance was 0.145 m. Answering with the 0.145 mm as distance :

Changin from mm to m : 1mm=0.001m

0.145 mm = 0.000145 m

The formula for Torque is T=Fxd

Wher F is the force and d is the perpendicular distance between the centre of the bolt and the applied force.

T= 167 x 0.000145 = 0.024215 Nm

Answer:

The towns are 32,635 ft apart.

Explanation:

From the image drawn below:

AB = x ft

BC = a ft

AC = (x + a) ft

Considering triangle PCB,

tan 50° = 25000 / a

Or,

a = 25000 / tan 50°

Since tan x = 1/ cot x

a = 25000×cot 50°------------------------------------1

Considering triangle PCA,

tan 25° = 25000 / (a + x)

Or,

a + x = 25000 / tan 25°

Since tan x = 1/ cot x

a + x = 25000×cot 25° -------------------------------2

Thus, finding x from equation 1 and 2, we get:

x = 25000 (cot 25° - cot 50°)

Using cot 25° = 2.1445 and cot 50° = 0.8394, we get:

x ≈ 32,635 ft

Thus, the distance between two towns is 32,635 ft.

Answer:

32635.17 ft

Explanation:

In the diagram, AB = 25000 ft

Let C and D be the town, where CD = d (Distance between two towns)

By triangle, ABC

tan 50 = AB / AC

AC = 25000 / tan 50 = 20977.5 ft

By triangle, ABD

tan 25 = AB/AD

AD = 25000 / tan 25 = 53612.67 ft

So, the distance between two towns

d = AD - AC = 53612.67 - 20977.5 = 32635.17 ft

Thus, the distance between two towns is 32635.17 ft.

Explanation:

   Force                     Description

1. [tex]F_g[/tex]         It is also known as the weight of an object. It is the force that is exerted on an object due to its mass

2. [tex]F_p[/tex]         It is force which is exerted by a push or a pull on an object. It is also known as applied force.

3. [tex]F_f[/tex]          It is known as resistive force. It opposes the motion of an object.

4. [tex]F_n[/tex]        It is the force which is at a right angle to the surface or                              perpendicular to the surface.

Answer:

Explanation:

Fp- Force exerted by a push or pull

Ff- force that comes from a resistances

Fg- also known as an objects weight

FN- support force at a right angle to a surface. JUST DID THIS QUESTION

Answer:

[tex]\omega_f=0.0356 rev/day[/tex]

Explanation:

Given:

Angular speed [tex]\omega_1=2\ rev/day[/tex]

Radius of the solid sphere = R

Radius of the spherical shell, R' = 5.8R

now the initial moment of inertia i.e the moment of inertia of the solid sphere is given as:

[tex]I_i=\frac{2}{5}MR^2[/tex]

where, M is the mass of the solid sphere

Now, the final moment of inertia i.e the moment of inertia of the spherical shell is given as:

[tex]I_f=\frac{2}{3}MR'^2[/tex]

or

[tex]I_f=\frac{2}{3}M(5.8R)^2[/tex]

or

[tex]I_f=22.426MR^2[/tex]

Now applying the concept of conservation of angular momentum

we get

[tex]I_i\omega_i=I_f\omega_f[/tex]

substituting the values, we get

[tex]\frac{2}{5}MR^2\times 2=22.426\times MR^2\omega_f[/tex]

or

[tex]\omega_f=\frac{\frac{2}{5}\times 2}{22.426}=0.0356 rev/day[/tex]

Answer: (a)[tex]F=7(10)^{-7}N[/tex]

              (b)[tex]F=1.344(10)^{-6}N[/tex]  

              (c) The force of Jupiter on the baby is slightly greater than the the force of the father on the baby.

Explanation:

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

[tex]F=G\frac{m_{1}m_{2}}{r^2}[/tex]   (1)

Where:

[tex]F[/tex] is the module of the force exerted between both bodies

[tex]G[/tex] is the universal gravitation constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]

[tex]m_{1}[/tex] and [tex]m_{2}[/tex] are the masses of both bodies.

[tex]r[/tex] is the distance between both bodies

Knowing this, let's begin with the answers:

Using equation (1) and taking into account the mass of the father [tex]m_{1}=100kg[/tex], the mass of the baby [tex]m_{2}=4.20kg[/tex] and the distance between them [tex]r=0.2m[/tex], the force [tex]F_{F}[/tex]  exerted by the father is:

[tex]F_{F}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(100kg)(4.20kg)}{(0.2m)^2}[/tex]   (2)

[tex]F_{F}=0.0000007N=7(10)^{-7}N[/tex]   (3)

Using again equation (1) but this time taking into account the mass of Jupiter [tex]m_{J}=1.898(10)^{27}kg[/tex], the mass of the baby [tex]m_{2}=4.20kg[/tex] and the distance between Jupiter and Earth (where the baby is) [tex]r_{E}=6.29(10)^{11}m[/tex], the force [tex]F_{J}[/tex]  exerted by the Jupiter is:

[tex]F_{J}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.898(10)^{27}kg)(4.20kg)}{(6.29(10)^{11}m)^2}[/tex]   (4)

[tex]F_{J}=0.000001344N=1.344(10)^{-6}N[/tex]   (5)

Now, comparing both forces:

[tex]F_{J}=0.000001344N=1.344(10)^{-6}N[/tex]   and [tex]F_{F}=0.0000007N=7(10)^{-7}N[/tex]  we can see [tex]F_{J}[/tex] is greater than [tex]F_{F}[/tex]. However, the difference is quite small as well as the force exerted on the baby.

The gravitational force a 100 kg father exerts on a 4.20 kg baby is approximately 0.014 N, while Jupiter's gravitational force on the baby is just around 9.36 * 10^-11 N. This implies that the father's gravitational force on the baby is significantly greater than that of Jupiter.

This question involves calculating the gravitational force, which according to Newton's Law of Gravitation is given by the formula F = G*(m1*m2)/r^2, where G is the universal gravitational constant (6.674 * 10^-11 N(m/kg)^2), m1 and m2 are the masses of the two objects, and r is the distance between the centers of the two objects.

(a) For the father and the baby, substitute in the given values into the formula, F (father) = (6.674 * 10^-11)*(4.20*100)/0.200^2 = 0.014 N.

(b) For Jupiter and the baby, know that the mass of Jupiter is about 1.898 × 10^27 kg, F (Jupiter) = (6.674 * 10^-11)*(4.20*1.898 × 10^27)/(6.29 * 10^11)^2 which is approximately 9.36 * 10^-11 N.

(c) The comparison of the forces shows that the gravitational force the father exerts on the baby is considerably larger than the gravitational force Jupiter exerts on the baby, reinforcing the observation that gravity is a relatively weak force that is only noticeable when dealing with massive objects.

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The force applied perpendicular to the door exerts more torque, because all of it is used for rotation. When a force is applied at an angle, only the part of the force acting perpendicular to the door contributes to the torque.

The force that exerts more torque on the door in this case is the one applied perpendicular to the door. This is because torque is calculated as the product of the force applied and the distance from the pivot point at which it is applied. The key point here is that only the component of the force acting perpendicular to the door creates torque. When a force is applied at an angle, only the component of the force acting perpendicular to the door will create a torque, hence reducing the total torque. So, in this case, the first force applied perpendicular to the door exerts greater torque because the full magnitude of the force is acting to rotate the door.

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