Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of 20C

Answers

Answer 1

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

[tex]Q =\frac{KA(\delta T)}{L}[/tex]

[tex]Q =\frac{25.87*1*20}{1}[/tex]

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

Answer 2

Answer: Rate of heat transfer q = 517.4W

Explanation:

Thermal conductivity is a material property that describes its ability to conduct heat. Thermal conductivity is the quantity of heat transmitted through a unit thickness of a material, in a direction normal to a surface of unit area,due to a unit temperature gradient under steady state conditions. It can be shown mathematically using the equation below;

Q/t = kA(∆T)/d

q = kA(∆T)/d

q = Q/t = rate of heat transfer

Q = total amount of heat transfer

t = time

k = thermal conductivity of material

A = Area

d= distance

For the case above, the material used is still air.

k for still air at 20°C and 1bara = 0.02587W/mK

A = 1m^2

d = 1mm = 0.001m

∆T = 20°C = 20K

q = 0.02587×1×20/0.001

q = 517.4W

Therefore, the rate of heat conduction through the still air is 517.4W


Related Questions

The bellow of a territorial bull hippopotamus is measured at 106 dB above the threshold of hearing. What is the sound intensity? Hint: The threshold of human hearing is I0 = 1.00 x 10-12 W/m2.

Answers

Answer:

0.03981 W/m²

Explanation:

I = Sound intensity

[tex]\beta[/tex] = Intensity level = 106 dB

[tex]I_0[/tex] = Threshold of human hearing = [tex]10^{-12}\ W/m^2[/tex]

Intensity of sound is given by

[tex]\beta=10log\dfrac{I}{I_0}\\\Rightarrow 106=10log\dfrac{I}{10^{-12}}\\\Rightarrow \dfrac{106}{10}=log\dfrac{I}{10^{-12}}\\\Rightarrow 10^{\dfrac{106}{10}}=\dfrac{I}{10^{-12}}\\\Rightarrow I=10^{\dfrac{106}{10}}\times 10^{-12}\\\Rightarrow I=10^{-1.4}\\\Rightarrow I=0.03981\ W/m^2[/tex]

The sound intensity is 0.03981 W/m²

Final answer:

The sound intensity is approximately 1.00 x 10^-4 W/m^2.

Explanation:

The sound intensity can be calculated using the formula:

I = I0 * 10^(dB/10)

where I is the sound intensity, I0 is the threshold of human hearing (given as 1.00 x 10^-12 W/m^2), and dB is the decibel level above the threshold. In this case, the decibel level is 106 dB. Plugging in the given values into the formula:

I = (1.00 x 10^-12 W/m^2) * 10^(106/10)

Simplifying the equation:

I ≈ 1.00 x 10^-4 W/m^2

The sound intensity is approximately 1.00 x 10^-4 W/m^2.

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Two radioactive nuclei A and B are present in equal numbers to begin with. Three days later, there are 4.04 times as many A nuclei as there are B nuclei. The half-life of species B is 1.37 days. Find the half-life of species A (in days).

Answers

Answer:

The half-life of A is 17.1 days.

Explanation:

Hi there!

The half-life of B is 1.73 days.

Let´s write the elapsed time (3 days) in terms of half-lives of B:

1.37 days = 1 half-life B

3 days = (3 days · 1 half-life B / 1.37 days) = 2.19 half-lives B.

After 3 days, the amount of A in terms of B is the following:

A = 4.04 B

The amount of B after 3 days can be expressed in terms of the initial amount of B (B0) and the number of half-lives (n):

B after n half-lives = B0 / 2ⁿ

Then after 2.19 half-lives:

B = B0 /2^(2.19)

In the same way, the amount of A can also be expressed in terms of the initial amount and the number of half-lives:

A = A0 / 2ⁿ

Replacing A and B in the equation:

A = 4.04 B

A0 / 2ⁿ = 4.04 · B0 / 2^(2.19)

Since A0 = B0

A0 / 2ⁿ = 4.04 · A0 / 2^(2.19)

Dividing by A0:

1/2ⁿ = 4.04 / 2^(2.19)

Multipliying by 2ⁿ and dividing by  4.04 / 2^(2.19):

2^(2.19) / 4.04 = 2ⁿ

Apply ln to both sides of the equation:

ln( 2^(2.19) / 4.04) = n ln(2)

n = ln( 2^(2.19) / 4.04) / ln(2)

n = 0.1756

Then, if 3 days is 0.1756 half-lives of A, 1 half-life of A will be:

1 half-life ·(3 days / 0.1756 half-lives) = 17.1 days

The half-life of A is 17.1 days.

In a certain region of space, the electric field is zero. From this fact, what can you conclude about the electric potential in this region? (a) It is zero. (b) It does not vary with position. (c) It is positive. (d) It is negative. (e) None of those answers is necessarily true.

Answers

Answer:

It is constant.

Explanation:

As we know that the electric field is a change in the electric potential so,

E = -ΔV

E = 0 ∵ V = constant

As we know that the electric field is zero. Which means that the electric potential is constant and it's not changing which results in the zero electric field.

Final answer:

In a region of space where the electric field is zero, it does not necessarily mean that the electric potential in that region is zero. The electric potential can still have a non-zero value even if the electric field is zero.

Explanation:

In a certain region of space where the electric field is zero, it does not necessarily mean that the electric potential in that region is zero. The electric potential can still have a non-zero value even if the electric field is zero. This is because electric potential is determined by the distribution of charges and their distances from the region in consideration. While the electric field measures the force experienced by a charge, the electric potential is related to the work required to move a charge in a region with an electric field.



Therefore, the correct answer is (e) None of those answers is necessarily true.

A wheel with a 0.10-m radius is rotating at 35 rev/s. It then slows uniformly to 15 rev/s over a 3.0-s interval.

a. What is the angular acceleration of a point on the wheel?
b. how to do and explain will give lifesaver!

Answers

Answer:

The angular acceleration of a point on the wheel is [tex]41.89\ rad/s^2[/tex] and it is decelerating.

Explanation:

It is given that,

Radius of the wheel, r = 0.1 m

Initial angular velocity of the wheel, [tex]\omega_i=35\ rev/s=219.91\ rad/s[/tex]

Final angular velocity of the wheel, [tex]\omega_f=15\ rev/s=94.24\ rad/s[/tex]

Time, t = 3 s

We need to find the angular acceleration of a point on the wheel. It is given by the rate of change of angular velocity divided by time taken. It is given by :

[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}[/tex]

[tex]\alpha =\dfrac{(94.24-219.91)\ rad/s}{3\ s}[/tex]

[tex]\alpha =-41.89\ rad/s^2[/tex]

So, the angular acceleration of a point on the wheel is [tex]41.89\ rad/s^2[/tex] and it is decelerating. Hence, this is the required solution.

Final answer:

The angular acceleration of a point on the wheel can be found using the formula: angular acceleration = change in angular velocity / time interval. Given the initial and final angular velocities and the time interval, we can calculate the angular acceleration.

Explanation:

The angular acceleration of a point on the wheel can be found using the formula:

angular acceleration = change in angular velocity / time interval

Given that the initial angular velocity is 35 rev/s, the final angular velocity is 15 rev/s, and the time interval is 3.0 s, we can substitute these values into the formula:

angular acceleration = (15 rev/s - 35 rev/s) / 3.0 s

Simplifying the equation, we get:

angular acceleration = -20 rev/s / 3.0 s

Converting rev/s to rad/s, we have:

angular acceleration = -20 rev/s ×(2π rad/rev) / 3.0 s

angular acceleration ≈ -41.89 rad/s²

Therefore, the angular acceleration of a point on the wheel is approximately -41.89 rad/s².

The acceleration due to gravity on the moon is about 5.4 ft/s2 . If your weight is 150 lbf on the earth:
What is your mass on the moon, in slugs.

Answers

Answer:

4.662 slugs

Explanation:

Your mass on the moon should always be the same as any planet you are on (due to law of mass conservation), only your weight be different as gravitational acceleration is different on each planet.

If you weight 150 lbf on Earth, and gravitational acceleration on Earth is 32.174 ft/s2. The your mass on Earth is

m = W / g = 150 / 32.174 = 4.662 slugs

which is also your mass on the moon.

A 0.10 kg meter stick is held perpendicular to a vertical wall by a 2.5 mm string going from the wall to the far end of the stick.
(A) Find the tension in the string.
(B) If a shorter string is used, will its tension be greater than, less than, or the same as that found in part (A)?
(C) Find the tension in a 2.0 m string.

Answers

Explanation:

a)

Sum of moments = 0 (Equilibrium)

T . cos (Q)*L = m*g*L/2

[tex]cos Q = \frac{\sqrt{(2.5^2 - L^2) } }{2.5}[/tex]

[tex]T*\frac{\sqrt{(2.5^2 - L^2) } }{2.5} * L = 0.1 *9.81*L/2[/tex]

[tex]T = \frac{2.4525}{\sqrt{(2.5)^2 - L^2} }[/tex]

b) If the String is shorter the Q increases; hence, Cos Q decreases which in turn increases Tension in the string due to inverse relationship!

c)

[tex]T = \frac{1.962}{\sqrt{(2)^2 - L^2} }[/tex]

The rotational equilibrium condition allows finding the responses for the tension of the rope are:

   A) T = 0.53 N

   B) If the rope shortens the tension increase..

   C) T = 0.57 N

Newton's second law for rotational motion gives a relationship between the torque, the moment of inertia and the angular acceleration of the bodies, in the special case that the acceleration is zero, it is called the equilibrium condition.

                 

                  Σ τ = 0

Torque is defined as the vector product of the force and the distance, its modulus is:

                τ = F rsin θ  

where τ is the torque, F the force, r the distance and tea the angle between the force and the distance, it is a product (r sin θ ) it is called the perpendicular distance or arm.

In the attached we have a free body diagram of the system, let's apply the equilibrium condition,

Let's use trigonometry to descompose the force.  

                cos θ = [tex]\frac{T_x}{T}[/tex]  

               sin  θ = [tex]\frac{T_y}{T}[/tex]  

               Tₓ = T cos θ

               [tex]T_y[/tex] = T sin  θ

They indicate that the length of the bar is x = 1 m and the length of the cable is L = 2.5 m, let's find the angle

             cos  θ = [tex]\frac{x}{L}[/tex]  

              θ = cos⁻¹ [tex]\frac{x}{L}[/tex]  

              θ = cos⁻¹ [tex]\frac{1}{2.5}[/tex]

              θ = 66.4º

A) let's set our pivot point at the junction with the wall and the anti-clockwise direction of rotation is positive.

                 W [tex]\frac{L}{2}[/tex]  - Ty L = 0

                 W [tex]\frac{L}{2}[/tex]  = (T sin 66.4) L

                 [tex]T = \frac{mg}{s sin 66.4} \\ \\T = \frac{0.1 \ 9.8}{2 \ sin66.4}[/tex]  

                 T = 0.53 N

B) how the tension changes as the length of the string changes.

               T = [tex]\frac{mg}{2 sin \theta}[/tex]  

we can see that the change of the tension occurs by changing the value of the sine function.

           sin θ = [tex]\frac{y}{L_o}[/tex]

Let's use the Pythagorean theorem to find the opposite leg.

         L² = x² + y²

         y = [tex]\sqrt{L^2 - x^2 }[/tex] = [tex]L \ \sqrt {1^2 + (\frac{x}{L})^2 }[/tex]

Let's substitute.

         sin θ =  [tex]\sqrt{1 - \frac{x^2}{L^2} }[/tex]

if we use a binomial expansion.

          [tex](1 - a) ^{0.5} = 1 - \frac{1}{2} a + ...[/tex]

Let's substitute.

            sin  θ  = 1 -  [tex]\frac{1}{2} \ \frac{x}{L}[/tex]    

We can see that when the value of the length decreases the value of the sine decreases and this term is in the denominator of the expression of the tension, therefore the tension must increase.

C) the length of the rope is L = 2 m

           sin θ =  [tex]\sqrt{1 - (\frac{1}{2})^2 }[/tex]  

           sin θ = 0.866

           T = [tex]\frac{mg}{2sin \theta}[/tex]  

            T = [tex]\frac{0.1 \ 9.8}{2 \ 0.866}[/tex]  

            T =0.57 N

In conclusion, using the rotational equilibrium condition we can find the answers for the tension of the rope are:

   A) T = 0.53 N

   B) If the rope shortens the tension increase.

   C) T = 0.57 N

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A 3-in-thick slab is 12 in wide and 15 ft long. Thickness of the slab is reduced in three steps in a hot rolling operation. In each step, thickness is reduced by 20% and width increases by 3%. If the entry speed of the slab in the first step is 40 ft/min, and roll speed is the same for the three steps, determine (a) length and (b) exit velocity of the slab after the final reduction.

Answers

Answer:

[tex]L_f=26.8108 ft[/tex]

Part B:

For Final Reduction

[tex]v_f=48.5436ft/min[/tex]

Explanation:

Part A:

At each step 0.8 (100-20)% of thickness is left

Final Thickness t_f:

[tex]t_f=(0.80)(0.80)(0.80)*3\\t_f=1.536 in[/tex]

Width increases by 0.03 in each step so (100+3)%=1.03

Final Width w_f:

[tex]w_f=(1.03)(1.03)(1.03)*12\\w_f=13.1127 in[/tex]

Conservation of volume:

[tex]t_ow_oL_o=t_fw_fL_f\\L_f=\frac{t_ow_oL_o}{t_fw_f} \\L_f=\frac{3*12*(15*12)}{1.536*13.1127}\\L_f=321.730 in\\L_f=26.8108 ft[/tex]

Part B:

[tex]t_ow_ov_o=t_fw_fv_f[/tex]

At First reduction exit Velocity:

[tex]v_f=\frac{t_ow_oL_o}{t_fw_f} \\v_f=\frac{3*12*(40)}{0.8*3*1.03*12}\\v_f=48.5439ft/min[/tex]

At 2nd Reduction:

[tex]v_f=\frac{0.8*3*1.03*12*40}{0.8^2*3*1.03^2*12}\\v_f=48.5436ft/min[/tex]

For Final Reduction:

[tex]v_f=\frac{0.8^2*3*1.03^2*12*40}{0.8^3*3*1.03^3*12}\\v_f=48.5436ft/min[/tex]

If the momentum of a 1000 kg car travelling at 10 m/s was transferred completely to a 20.0 kg traffic barrier, what would the final speed of the barrier be
a. 40 m/s
b. 50 m/s
c. 250 m/s
d. 500 m/s

Answers

Answer:

d. 500 m/s

Explanation:

Momentum: This is the product of the mass of a body to its velocity, The S.I unit of momentum is kgm/s.

Mathematically, momentum can be expressed as,

M = mv....................... equation 1

Where M = momentum, m = mass, v = velocity

deduced from the question,

Momentum of the car = momentum of the barrier.

MV = mv ............................. Equation 1

Where M = mass of the car, V = velocity of the car, m = mass of the barrier, v = velocity of the barrier.

making v the subject of the equation,

v = MV/m........................ Equation 2

Given: M = 1000 kg, V = 10 m/s, m = 20 kg.

Substitute into equation 2

v = 1000(10)/20

v = 500 m/s.

Hence the speed of the barrier = 500 m/s

The right option is d. 500 m/s

A free electron and a free proton are released in identical electric fields.

(i) How do the magnitudes of the electric force exerted on the two particles compare?

It is millions of times greater for the electron.

It is thousands of times greater for the electron.

They are equal.

It is thousands of times smaller for the electron.

It is millions of times smaller for the electron.


(ii) Compare the magnitudes of their accelerations.

It is millions of times greater for the electron.

It is thousands of times greater for the electron.

They are equal.It is thousands of times smaller for the electron.

It is millions of times smaller for the electron.

Answers

Answer:

i) They are equal . ii) It is thousands of times greater for the electron.

Explanation:

i) By definition, the electric field is the electric force per unit charge. If the field is the same, the force will depend on the value of the charge under the influence of the field.

As the magnitude of  the charge of the electron and the proton are the same, we conclude that the electric force on both must be equal in magnitude.

ii) The acceleration on both particles must meet the Newton´s 2nd Law, so, if the forces are equal in magnitude (neglecting any other external interaction), the acceleration will only depend on the mass of both particles, according this general expression:

a = F/m

As the mass of the electron is approximately two thousands times smaller than the proton´s, it concludes that the acceleration on the electron must be thousands of times greater for the electron.

Final answer:

The magnitude of the electric force exerted on the electron is millions of times greater than that exerted on the proton. The magnitudes of their accelerations are equal.

Explanation:

(i) The magnitude of the electric force exerted on the electron is millions of times greater than that exerted on the proton. This is because the electron has a much smaller mass compared to the proton, and the electric force depends on the charge and mass of the particle.



(ii) The magnitudes of their accelerations are equal. The acceleration of a particle in an electric field depends only on the magnitude of the electric field and the mass of the particle. Since both the electron and proton experience the same electric field, their accelerations will be the same.

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Suppose you hear a clap of thunder 16.2 seconds after seeing the associated lightning stroke. The speed of sound waves in air is 343-m/s and the speed of light in air is 3.00 x 10^8-m/s. How far are you from the lightning stroke in both meters and miles? Assume that the light reaches you instantaneously.

Answers

Answer:

D = 3.45 mile

Explanation:

given,

time taken by the sound wave, t = 16.2 s

speed of sound, v = 343 m/s

speed of the light = 3 x 10⁸ m/s

distance where lightning stroke

 we know,

 distance = speed x time

  D = 343 x 16.2

  D = 5556.6 m

 1 m = 0.000621371 mile

5556.6 m = 5556.6 x 0.0006214

D = 3.45 mile

the distance lightning strike is 3.45 mile away from the observer.

Final answer:

To calculate the distance from a lightning stroke: use the time delay between seeing the lightning and hearing the thunder, and multiply it by the speed of sound (343 m/s). In this case, the lightning is approximately 5562.6 meters or about 3.46 miles away.

Explanation:

Your distance from a lightning stroke can be calculated by knowing the speed of sound and the time it takes for the sound from the lightning stroke to reach you. You said you hear the thunder sound 16.2 seconds after seeing the lightning. The speed of sound in air is 343 m/s. Therefore, the distance can be calculated using the formula: distance = speed * time.

In this case, distance = 343 m/s * 16.2s = 5562.6 meters or about 5.56 kilometers.

When this is converted to miles (since 1 mile is about 1609.34 meters), it is approximately 3.46 miles away. Therefore, based on the time difference between the flash and the thunder, you are about 5562.6 meters or 3.46 miles away from the lightning strike.

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If you were designing a room in a house, where would be the better place to put a heater, near the floor or near the ceiling? Why? Consider your answer in the context of convection.1. Near the ceiling, because the air warmed by the heater would quickly spread down and across the room, replacing the cold air.2. Near the floor, because the heater would warm the air close to it. After being heated, the now-warm air would rise, and be replaced by cool air, which the heater would warm. The cycle would continue until the room is heated.3. Near the ceiling, because the heater would warm the air close to it. After being heated, the now-warm air would descend, and be replaced by cool air, which the heater would warm. The cycle would continue until the room is heated.4. Near the floor, because the air warmed by the heater would quickly spread up and across the room, replacing the cold air.

Answers

Answer: The correct explanation is 2.

Explanation: The warm air is less dense (it expands) and thus it is lighter than the cold air so it will rise up to the floor. Therefore, when you place the heater on the floor it will warm the cold air which would then rise and be replaced by more cold air which would again get warm and rise and so on until the room is heated. This means that the correct explanation is 2.

On the other hand, if you put the heater at the ceiling, it will warm the cold air near the ceiling which would stay up there (it is lighter than the cold air under it). This means that the only way for the heat to spread from this ceiling level warm air to the lower levels is via conduction which is slow.  

Final answer:

The heater should be placed near the floor due to convection, where warm air rises after being heated, circulates, and evenly distributes throughout the room as it cools and descends. Therefore, option 2 is the correct answer.

Explanation:

Proper Placement of a Heater in a Room

When considering the placement of a heater in a room in the context of convection, the better option is to place the heater near the floor. This is because, as per the principles of convection, warm air will rise after being heated by the heater. The now-warm air, having become less dense, will rise to the ceiling, creating a convective loop that circulates the warm air throughout the room. As the air cools down near the ceiling and outside walls, it contracts, becoming denser, and subsequently sinks back to the floor, where it will be heated again by the heater. This cycle continues, leading to an efficient distribution of heat throughout the room. Therefore, option 2 is correct: placing the heater near the floor allows it to warm the air close to it, after which the warm air rises and is replaced by cool air that the heater warms again, maintaining a continuous cycle of heating.

Find the equivalent resistance Req between terminals a and b if terminals c and d are open and again if terminals c and d are shorted together. (Round the final answers to two decimal places.) With terminals c-d open, Req = Ω. With terminals c-d shorted, Req = Ω.

Answers

Answer:

351 ohm

720 ohm

Explanation:

When c and d are open:

Terminals c and d are open.. If you  redraw the circuit as below, you can see that the two resistors in the first  column are in parallel as, they are connected together at both pairs of terminals  (due to the short).

Hence, we have a pair of parallel resistors:

Req1 = (R1*R2)/ (R1 + R2) = 360*540/(360+540) = 216 ohms

Req2 = (R3*R4)/ (R3 + R4) = 180*540/(180+540) = 135 ohms

Now these two sets are  in series with another Hence,

Req = Req1 + Req2 = 216 + 135 = 351 ohms

Answer: 351 ohms

When c and d are shorted:

The current will flow through the least resistant path naturally from resistors R3 and R1 or R4.

Both of these resistor lie in a single path placing the resistors in series to one another, hence

Req = R3 + R1 = 180 + 540 = 720 ohms

Answer:720 ohms

Final answer:

To find the equivalent resistance Req between terminals a and b, we can follow a series of steps to simplify the circuit. The final equivalent resistance can be calculated by adding the resistors in series. With terminals c and d open, Req is 12.22 ohms, and with terminals c and d shorted, Req is 22.00 ohms.

Explanation:

To find the equivalent resistance between terminals a and b, we need to consider the circuit shown in Figure 10.15. To simplify the circuit, we can apply a series of steps to reduce it to a single equivalent resistance. Step 1 involves reducing resistors R3 and R4 in series. Step 2 involves reducing resistors R2 and the equivalent resistance R34 in parallel. Finally, Step 3 involves reducing resistor R1 and the equivalent resistance R234 in series. The final equivalent resistance Req can be calculated by adding the resistors in series.

Following these steps, the equivalent resistance Req between terminals a and b with terminals c and d open is 12.22 ohms. With terminals c and d shorted together, the equivalent resistance Req is 22.00 ohms.

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The momentum of a type of bumper car at an amusement park should not exceed 2100 kg×m/s to ensure the safety of the visitors to the park. If each bumper car has a mass of 780 kg, will the ride be safe if the bumper cars are limited to a top speed of 3 m/s

Answers

Answer:

No, the ride will not be safe according to the standards mentioned in the question because the value of momentum exceeds.

Explanation:

Given:

magnitude of safe momentum of bumper cars, [tex]p=2100\ kg.m.s^{-1}[/tex]mass of each bumper car, [tex]m=780\ kg[/tex]top velocity of bumper car, [tex]v=3\ m.s^{-1}[/tex]

Now, we find the momentum of the car using the given mass and speed:

[tex]p'=m.v[/tex]

[tex]p'=780\times 3[/tex]

[tex]p'=2340\ kg.m.s^{-1}[/tex]

Therefore the ride will not be safe according to the standards mentioned in the question because the value of momentum exceeds.

Consider n equal positive charged particles each of magnitude Q/n placed symmetrically around a circle of radius a. Calculate the magnitude of the electric field at a point a distance x from the center of the circle and on the line passing through the center and perpendicular to the plane of the circle. (Use any variable or symbol stated above along with the following as necessary: ke.) (b) Now consider a ring of radius a that carries a uniformly distributed positive total charge Q. Recall the calculation of the electric field at point a point a distance x from the center of the ring and on the line passing through the center and perpendicular to the plane of the ring. Explain why the result in part (a) is identical to the result for the ring.

Answers

Answer:

To make it easier to Understand, consider the circle to be in the usual 3 dimensional x-z plane (the "horizontal plane") and the point of measurement C to be at a distance p (instead of x to avoid confusion with the x-axis) on the y-axis (the "vertical direction").

The answer for a and b is the same.  This is because a the horizontal component of a charged portion of a uniform ring is canceled by the opposite portion of the ring since they are placed at an equal distance from the position in which the electric field is being measured or applied, just as are the opposing point charges described in part a.

Explanation:

A.)  Using Coulomb's law of point charges, each charge on the circle would exert a field Ec at C given by:

(1)  Ec   =   Ke * (Q / n) / d²

where:

Ke is Coulomb's constant,

Q / n  is the magnitude of the charge, and

d   =   the distance between the charge and the point of measurement C, with   d²    =    a²  +  c²  

Since the charges are in a circle in the x-z  plane, all force components in both the x- and the z-directions are canceled by symmetry; the vertical force (that in the y-direction) is the only component that does not cancel.  

Therefore the resultant vector Ecy points up (+y-direction) and has a magnitude of:

(2)  Ecy   =   Eq  *  sin(theta)

=  (Ke * (Q / n) / d²)  *  (c  /  d)

Then, summing the forces from all the charges, the magnitude of the total electric field is given by:

(3)  Ey   =   n * Ecy

=   n * [ (Ke * (Q / n) / d²)  *  (c  /  d) ]

=   c * Ke * Q  / d^3

B.) The equation is the same. This is because both the x- and z-components (the two planar components) of a charged portion of a uniform ring are canceled by the opposite portion of the ring,  since they lie at an equal distance but opposite direction from c.  

This is the same way the opposing point charges described in part A behave.

Final answer:

In both scenarios described, because of symmetry, the electric fields generated by opposing charge pairs cancel each other out, resulting in a net electric field of zero at the specified point.

Explanation:

The magnitude of the electric field in the scenario described is given by Coulomb's law, which states that the electric field E due to a charge Q at a distance r from a point is given by E = ke*(Q/r^2), where ke is Coulomb's constant.

In this case, the magnitude of each charge is Q/n and we have n charges symmetrically distributed along the circle. Because we are calculating the electric field at a point on the line passing through the center of the circle and perpendicular to the plane of the circle, each pair of charges on the circle will produce an electric field that is symmetric and thus its contribution will be canceled out by another pair's contribution. The final resultant electric field will hence be zero.

(b) Now let's consider a ring of radius a that carries a uniformly distributed positive total charge Q. Because the charge is symmetrically distributed, the electric field at any point on the line passing through the center and perpendicular to the plane of the ring will be sum of electric fields due to all small charged particles which make up this ring, and due to symmetry of the distribution, it will add up to zero.

Therefore, the results in part (a) and (b) are identical because in both cases, due to symmetry, the electric fields generated by opposing charge pairs cancel each other out, resulting in a net electric field of zero at the specified point.

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During launches, rockets often discard unneeded parts. A certainrocket starts from rest on the launch pad and accelerates upward ata steady 3.25 m/s^2. When it is 230 m above the launch pad, it discards a usedfuel canister by simply disconnecting it. Once it is disconnected,the only force acting on the canister is gravity (air resistancecan be ignored).

A) How high is the rocket when thecanister hits the launch pad, assuming that the rocket does notchange its acceleration?
B) What total distance did the canistertravel between its release and its crash onto the launchpad?

Answers

Answer:

915.69549 m

306.1968 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 3.25 m/s²

g = Acceleration due to gravity = 9.81 m/s²

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 3.25\times 230+0^2}\\\Rightarrow v=38.665\ m/s[/tex]

Height is given by

[tex]h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{38.665^2}{2\times 9.81}\\\Rightarrow h=76.1968\ m[/tex]

Total distance the canister has to travel is 230+76.1968 = 306.1968 m

Time to reach the max height

[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{g}\\\Rightarrow t=\dfrac{0-38.665}{-9.81}\\\Rightarrow t=3.9413\ s[/tex]

Time taken to reach the ground is

[tex]s=ut+\frac{1}{2}gt^2\\\Rightarrow 306.1968=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{306.1968\times 2}{9.81}}\\\Rightarrow t=7.9\ s[/tex]

Total time taken would be 3.9413+7.9 = 11.8413 seconds

[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=38.665\times 11.8413+\dfrac{1}{2}\times 3.25\times 11.8413^2\\\Rightarrow s=685.69549\ m[/tex]

The rocket is 685.69549+230 = 915.69549 m from the launchpad.

The total distance is 230+76.1968 = 306.1968 m

At the instant when the electron is 4.40 cmcm from the wire and traveling with a speed of 6.10×104 m/sm/s directly toward the wire, what is the magnitude of the force that the magnetic field of the current exerts on the electron?

Answers

Answer:

The force that the magnetic field of the current exerts on the electron is [tex]2.30\times10^{-19}\ N[/tex]

Explanation:

Given that,

Distance = 4.40 cm

Speed [tex]v= 6.10\times10^{4}\ m/s[/tex]

Suppose a long, straight wire carries a current of 5.20 . An electron is traveling in the vicinity of the wire.

We need to calculate the magnetic field of the current exerts on the electron

Using formula of force

[tex]F=qv\times B[/tex]

[tex]F=qv\times\dfrac{\mu_{0}I}{2\pi r}[/tex]

Put the value into the formula

[tex]F=1.6\times10^{-19}\times6.10\times10^{4}\times\dfrac{4\pi\times10^{-7}\times5.20}{2\pi\times4.40\times10^{-2}}[/tex]

[tex]F=2.30\times10^{-19}\ N[/tex]

Hence, The force that the magnetic field of the current exerts on the electron is [tex]2.30\times10^{-19}\ N[/tex]

If you wanted to move an electron from the positive to the negative terminal of the battery, how much work W would you need to do on the electron

Answers

Answer:

W = qV

Explanation:

Let V be the potential difference of the battery, and q be the charge on the electron.

The work done in moving a charge through a potential difference V =?

Work = force * distance

Work = F.r

F = qE

But E = V / r

F = q. V / r

Work ( W ) = (qV/ r ) * r

Work = qV

Therefore the force required to move a charge through a potential difference V = qV

Final answer:

The amount of work needed to move an electron from the positive to the negative terminal of a battery using the equation W = qV, where q is the charge of the electron and V is the voltage of the battery is  1.92 x 10-18 joules.

Explanation:

In order to move an electron from the positive to the negative terminal of a battery, you would need to do work on the electron.

The amount of work, W, can be calculated using the equation W = qV, where q is the charge of the electron and V is the voltage of the battery.

The charge of an electron is approximately 1.6 x 10-19 coulombs.

For example, if the voltage of the battery is 12 volts, the work done on the electron would be:
W = ([tex]1.6[/tex]×[tex]10^{-19}[/tex][tex]C[/tex]) × ([tex]12[/tex] [tex]v[/tex]) = [tex]1.92[/tex]×[tex]10^{-18}[/tex] joules.

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What would you expect the barometric pressure outside of a jet flying at 9.6 km to be?

Answers

Answer:

The pressure is expected to decrease because as we move up in the atmosphere the is lesser mass of atmosphere weight above us.

Explanation:

The atmospheric pressure is the the force exerted by the layers of the atmosphere above a point. Being a fluid and due to the gravity of earth there is a pressure due to the weight of the atmospheric columns above us.

So, at a height h above the earth surface in the atmosphere the pressure will decrease and this can be calculated as:

[tex]P_h=101325\times (1-2.25577\times 10^{-5}\times h)^{5.25588}[/tex]

putting h=9.6 meter

[tex]P_h=101209.7268\ Pa[/tex]

At an elevation of 9.6 km, the pressure is approximately 254 millibars.

The pressure is inversely proportional to elevation, higher the elevation lower will be the pressure while on the other hand, lower the elevation higher will be the pressure. At an elevation of 10 km pressure drops to 265 millibars then the elevation of 9.6 km, the pressure is approximately 254 millibars.

The atmospheric pressure at sea level is about 1,014 millibars which is equal to the pressure of 14.7 pounds/inch2 so we can conclude that at the elevation of 9.6 km, the pressure is approximately 254 millibars.

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A thin cylindrical shell and a solid cylinderhave the same mass and radius. The two arereleased side by side and roll down, withoutslipping, from the top of an inclined planethat is 1.3 m above the ground.Find the final linear velocity of the thincylindrical shell.The acceleration of gravity is9.8 m/s2

Answers

Answer:

v = 5.05m/s

Explanation:

H = 1.3m

initial velocity = 0

final velocity = v = ?

g =9.8 m/s^2

we apply the conservation of energy; all potential energy is comletely converting to kinetic energy

[tex]mgh = \frac{mv^{2}}{2}[/tex]

mass is same; it cancels out

[tex]v =\sqrt{2gh} = \sqrt{2(9.81)(1.3)}[/tex]

v = 5.05m/s

Answer:

the final linear velocity is 3.56931 m/s

Explanation:

the solution is in the attached Word file

The 160-lblb crate is supported by cables ABAB, ACAC, and ADAD. Determine the tension in these wire

Answers

Final answer:

The question pertains to the physics discipline, dealing with calculating the tension in supporting cables using principles of static equilibrium and considering the load's weight and the angles of attachment.

Explanation:

The question involves determining the tension in cables that support a structure or an object. This type of problem is common in physics, specifically in the area of mechanics, and it involves understanding forces and how they are distributed within a system. When multiple cables support a load, tensions in each cable can be found using principles of static equilibrium, where the sum of forces in each direction (horizontal and vertical) equals zero. In addition to the weight of the supported object, angles at which cables are attached can play a crucial role in determining individual tensions.

A piece of tape is pulled from a spool and lowered toward a 190-mg scrap of paper. Only when the tape comes within 8.0 mm is the electric force magnitude great enough to overcome the gravitational force exerted by Earth on the scrap and lift it.
Determine the magnitude and direction of the electric force exerted by the tape on the paper at this distance

Answers

Final answer:

The magnitude of the electric force exerted by the tape on the paper at a distance of 8.0 mm is 3.72 x 10^-7 C. The direction of the force is repulsive as like charges repel each other.

Explanation:

When the tape comes within 8.0 mm of the scrap of paper, the electric force magnitude is great enough to overcome the gravitational force exerted by Earth on the scrap and lift it. To determine the magnitude and direction of the electric force exerted by the tape on the paper at this distance, we can use the equation for the electric force:

F = k * (Q1 * Q2) / r^2

where F is the magnitude of the force, k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.

We know that the gravitational force is equal to the weight of the paper:

Fg = m * g

where Fg is the gravitational force, m is the mass of the paper (190 mg = 0.190 g), and g is the acceleration due to gravity (9.8 m/s^2).

At the point where the tape comes within 8.0 mm of the paper, the electric force is equal to the gravitational force:

F = Fg

k * (Q1 * Q2) / r^2 = m * g

We can solve this equation for the magnitude of the charge Q1 or Q2:

Q1 * Q2 = (m * g * r^2) / k

Substituting the known values, we get:

Q1 * Q2 = (0.190 g * 9.8 m/s^2 * (8.0 mm)^2) / (8.99 x 10^9 Nm^2/C^2)

Q1 * Q2 = 1.383 x 10^-12 C^2

To find the magnitude of the charge, we can assume that Q1 and Q2 are equal, so:

Q1 = Q2 = sqrt(1.383 x 10^-12 C^2) = 3.72 x 10^-7 C

The magnitude of the electric force exerted by the tape on the paper at this distance is 3.72 x 10^-7 C. The direction of the force is repulsive because like charges repel each other.

Calculate the period of a ball tied to a string of length 0.468 m making 3.8 revolutions every second. Answer in units of s. Your answer must be within ± 2.0%

Answers

Answer:

0.26315 s

Explanation:

The frequency of the ball tied to a string system is 3.8 rev/s.

That means in one second the ball will complete 3.8 revolutions.

The time period will be the reciprocal of this frequency

[tex]T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{3.8}\\\Rightarrow T=0.26315\ s[/tex]

The time period is 0.26315 s

It can be also solved in the following way

[tex]1\ s=3.8\ rev\\\Rightarrow 1\ rev=\dfrac{1}{3.8}\ s\\\Rightarrow 1\ rev=0.26315\ s[/tex]

The time period is 0.26315 s

A positron is an elementary particle identical to an electron except that its charge is . An electron and a positron can rotate about their center of mass as if they were a dumbbell connected by a massless rod.What is the orbital frequency for an electron and a positron 1.50 apart?

Answers

Explanation:

According to the energy conservation,

          [tex]F_{centripetal} = F_{electric}[/tex]

            [tex]\frac{mv^{2}}{r} = \frac{kq^{2}}{d^{2}}[/tex]

           [tex]v^{2} = \frac{kq^{2}r}{d^{2}m}[/tex]

                 = [tex]\frac{9 \times 10^{9} N.m^{2}/C^{2} \times 1.6 \times 10^{-19} C \times 0.75 \times 10^{-9} m}{(1.50 \times 10^{-9}m)^{2} \times 9.11 \times 10^{-31} kg}[/tex]

                = [tex]8.430 \times 10^{10} m^{2}/s^{2}[/tex]

             v = [tex]\sqrt{8.430 \times 10^{10} m^{2}/s^{2}}[/tex]

                = [tex]2.903 \times 10^{5} m/s[/tex]

Formula for distance from the orbit is as follows.

               S = [tex]2 \pi r[/tex]

                  = [tex]2 \times 3.14 \times 0.75 \times 10^{-9} m[/tex]

                  = [tex]4.71 \times 10^{-9} m[/tex]

Now, relation between time and distance is as follows.

                T = [tex]\frac{S}{v}[/tex]

       [tex]\frac{1}{f} = \frac{S}{v}[/tex]

or,           f = [tex]\frac{v}{S}[/tex]          

                = [tex]\frac{2.903 \times 10^{5} m/s}{4.71 \times 10^{-9} m}[/tex]      

                = [tex]6.164 \times 10^{13} Hz[/tex]

Thus, we can conclude that the orbital frequency for an electron and a positron that is 1.50 apart is [tex]6.164 \times 10^{13} Hz[/tex].

A laser beam is incident at an angle of 33.0° to the vertical onto a solution of cornsyrup in water.(a) If the beam is refracted to 24.84° to the vertical, what is the index ofrefraction of the syrup solution?(b) Suppose the light is red, with wavelength 632.8 nm in a vacuum.Find its wavelength in the solution.(c) What is its frequencyin the solution?(d) What is its speed in the solution?

Answers

Answer:

1.29649

488.08706 nm

[tex]6.14644\times 10^{14}\ Hz[/tex]

231715700.28346 m/s

Explanation:

n denotes refractive index

1 denotes air

2 denotes solution

[tex]\lambda_0[/tex] = 632.8 nm

From Snell's law we have the relation

[tex]n_1sin\theta_1=n_2sin\theta_2\\\Rightarrow n_2=\dfrac{n_1sin\theta_1}{sin\theta_2}\\\Rightarrow n_2=\dfrac{1\times sin33}{sin24.84}\\\Rightarrow n_2=1.29649[/tex]

Refractive index of the solution is 1.29649

Wavelength is given by

[tex]\lambda=\dfrac{\lambda_0}{n_2}\\\Rightarrow \lambda=\dfrac{632.8}{1.29649}\\\Rightarrow \lambda=488.08706\ nm[/tex]

The wavelength of the solution is 488.08706 nm

Frequency is given by

[tex]f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{488.08706\times 10^{-9}}\\\Rightarrow f=6.14644\times 10^{14}\ Hz[/tex]

The frequency is [tex]6.14644\times 10^{14}\ Hz[/tex]

[tex]v=\dfrac{c}{n_2}\\\Rightarrow v=\dfrac{3\times 10^8}{1.29469}\\\Rightarrow v=231715700.28346\ m/s[/tex]

The speed in the solution is 231715700.28346 m/s

Final answer:

The index of refraction can be found using Snell's law. The wavelength of red light within the solution can be calculated from the index and its frequency remains constant. The speed of light in the solution is the product of its frequency and the calculated wavelength.

Explanation:

Solution to the Laser Beam Question

To answer the student's question regarding refraction of a laser beam in a corn syrup solution, we can use Snell's law, which relates the incident angle, refracted angle, and indices of refraction of the two media. We can also calculate the wavelength, frequency, and speed of light within the solution using the principles of wave optics.

(a) According to Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n2 is the index of refraction of the corn syrup solution, θ1 is the incident angle, 90° - 33.0° = 57.0°, and θ2 is the refracted angle, 90° - 24.84° = 65.16°. Since the laser light is passing from air (n1=1) to the solution, we have 1 * sin(57.0°) = n2 * sin(65.16°). Solving for n2 gives us the index of refraction of the syrup solution.

(b) The wavelength of light in the solution is given by λ' = λ0/n, where λ0 is the wavelength in a vacuum, and n is the index of refraction. Inserting the red light's wavelength (632.8 nm) in vacuum and the obtained index of refraction will yield the wavelength in the solution.

(c) The frequency of light does not change when it enters another medium, so we use the vacuum frequency calculated by f = c/λ0, and c is the speed of light in a vacuum.

(d) The speed of light in the solution, v, can be found using the equation v = f * λ', which uses the frequency found in part (c) and the wavelength in the solution from part (b).

If I time the movement of waves for 1 second, and calculate that 100 waves passed through in that 1 second, what have I calculated?

A.crest
B.amplitude
C.wavelength.
D.frequency.
E.none of the above

Answers

Answer:

D.frequency.

Explanation:

The number of times a wave passes through a particular point in a time of 1 second is called the frequency.

Amplitude is the maximum height the wave reaches from the reference axis of the wave

Wavelength is the distance between the two upper (crest) or lower (troughs) points of a wave.

Crest is the top most part of a wave.

Hence, the question is referring to frequency.

A projectile of mass 5 kg is fired with an initial speed of 176 m/s at an angle of 32◦ with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 2 kg and 3 kg . The 3 kg fragment lands on the ground directly below the point of explosion 4.1 s after the explosion. The acceleration due to gravity is 9.81 m/s 2 . Find the magnitude of the velocity of the 2 kg fragment immediatedly after the explosion. Answer in units of m/s.

Answers

Answer:

v1 = 377.98 m/s

Explanation:

m = 5 Kg

v0 = 176 m/s

v0x = v0*Cos 32° = 176 m/s*Cos 32° = 149.256 m/s

m1 = 2 Kg

m2 = 3 Kg

t = 4.1 s

g = 9.81 m/s²

Before  the explosion

pix = m*v0x = 5 Kg*149.256 m/s = 746.282 Kgm/s

piy = 0

After the explosion

pfx = m1*v1x

knowing that pix = pfx

we have

746.282 = 2*v1x

v1x = 373.14 m/s

v2y = g*t

pfy = m1*v1y + m2*v2y

pfy = 2*v1y + 3*(9.81*4.1)

pfy = 2*v1y + 120.663

knowing that piy = pfy = 0

we have

0 = 2*v1y + 120.663

v1y = 60.33 m/s

Finally we apply

v1 = √(v1x² + v1y²)

v1 = √(373.14² + 60.33²)

v1 = 377.98 m/s

Final answer:

To find the magnitude of the velocity of the 2 kg fragment immediately after the explosion, we use the principle of conservation of momentum. By setting the momentum before the explosion equal to the momentum after the explosion, we can solve for the velocity of the 2 kg fragment. The magnitude of the velocity is approximately 56.8 m/s.

Explanation:

Projectile Motion

To find the magnitude of the velocity of the 2 kg fragment immediately after the explosion, we need to use the principle of conservation of momentum. Since there are no external forces acting on the system, the total momentum before the explosion is equal to the total momentum after the explosion.

Before the explosion, the momentum of the projectile can be calculated using the formula:

momentum = mass * velocity

After the explosion, the momentum of the 2 kg fragment can be calculated as:

momentum = mass * velocity

Setting the two equations equal to each other, we can solve for the velocity of the 2 kg fragment.

Plugging in the given values:

Initial velocity of the projectile = 176 m/s
Initial angle = 32°
Mass of the 2 kg fragment = 2 kg
Mass of the 3 kg fragment = 3 kg
Time after explosion = 4.1 s
Acceleration due to gravity = 9.81 m/s^2

Solving the equation, we find that the magnitude of the velocity of the 2 kg fragment immediately after the explosion is approximately 56.8 m/s.

The gravitational force exerted on a solid object is 5.00N. When the object is suspended from a spring scale and submerged completely in water the scale reads 3.50N. Find the density of the object.

Answers

Answer:

3333.33 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of a body to its volume.

The unit of density is kg/m³.

From Archimedes principle,

R.d = W/U = D/D'

Where R.d = relative density, W = weight of the object in air, u = upthrust in water, D = Density of the object, D' = Density of water.

W/U = D/D'

making D the subject of the equation

D = D'(W/U).......................... Equation 1

Given: W = 5.0 N, U = 5.0 -3.5 = 1.5 N, D' = 1000 kg/m³

Note: U = lost in weight = weight in air - weight in water

Substitute into equation 1

D = 1000(5/1.5)

D = 3333.33 kg/m³

Thus the density of the object = 3333.33 kg/m³

The first step is to calculate the buoyant force that the water exerts on the solid object. The buoyant force can be found by subtracting the scale reading from the gravitational force, or 5.00N - 3.50N = 1.50N.

Next, we find the volume of the water displaced by the solid. The buoyant force is equal to the weight of the fluid displaced, so we can use the formula F = ρf * V * g, where ρf is the fluid density, V is the water volume displaced, g is the acceleration due to gravity and F is the buoyant force.

We're given that the density of water is 9.8 * 1000 N/m³ and we've just calculated the buoyant force. Thus, we have 1.50N = 9.8 * 1000 * V * 9.81, which simplifies to V = 1.5 / (9.8 * 1000) = 0.0001530612244897959 m³.

Now we calculate the object's mass using the equation, m = F / g, where F is gravitational force and g is acceleration due to gravity. Substituting the given values, we get m = 5.00 / 9.81 = 0.509683995922528 kg.

Finally, we find the density of the object using the formula ρ = m / V, where m is the mass and V is the volume. Substituting the values calculated earlier, we get ρ = 0.509683995922528 / 0.0001530612244897959 = 3329.935440027183 kg/m³.

Therefore, the density of the object is approximately 3330 kg/m³.

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A charge of + 3.00 μC is located at the origin, and a second charge of −2.00 μC is located on the x−y plane at the point (30.0 cm, 20.0 cm). Determine the electric force exerted by the −2.00 μC charge on the 3.00 μC charge.

Answers

Answer:

[tex]\vec{F} = -0.34\^x - 0.22\^y\\|\vec{F}| = -0.41~N[/tex]

Explanation:

The electric force between two point charges can be calculated by Coulomb's Law:

[tex]\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r[/tex]

We have to calculate the distance between two points; (0,0) and (0.3 m, 0.2 m).

[tex]r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(0.3)^2 + (0.2)^2} = 0.36~m[/tex]

Now we can apply Coulomb's Law

[tex]F = \frac{1}{4\pi\epsilon_0}\frac{(3\times 10^{-6})(-2\times 10^{-6})}{(0.36)^2} = -0.41~N[/tex]

The minus sign in front of the force means that the force is attractive.

The direction of the force can be calculated as follows:

[tex]F_x = F\cos(\theta)\\F_y = F\sin(\theta)[/tex]

where θ is the angle between F and the x-axis. This angle can be calculated by the triangle with edges 0.3 m, 0.2 m, and 0.36 m.  

So, sin(θ) = 0.2/0.36 = 0.55 and cos(θ) = 0.3/0.36 = 0.83.

Finally,

[tex]F_x = -0.41 \times 0.83 = -0.34~N\\F_y = -0.41 \times 0.55 = -0.22~N[/tex]

Final answer:

An electric force of approximately 0.0137 Newtons is exerted by the -2.00 μC charge on the 3.00 μC charge. The direction is attractive, implying that the force pulls the 3.00 μC charge towards the -2.00 μC charge.

Explanation:

The subject of this query pertains to the concept of electric force in Physics. Given the position coordinates and charge values, we can calculate the electric force between the two charges using Coulomb's Law, which states that the force between two charges is equal to the absolute value of the product of the charges, divided by the square of the distance between them, multiplied by the electrostatic constant (k = 8.99 x 10^9 N m²/C²).

First, determine the distance between the charges. Using the Pythagorean theorem, the distance is √(0.30² + 0.20²) = 0.36 m. Now plug the charge values (Q1=3.00 μC = 3.00 x 10^-6 C, Q2=-2.00 μC = -2.00 x 10^-6 C), and the distance (r=0.36 m) into Coulomb's Law (F=k|Q1*Q2|/r²).

The absolute value of the electric force would therefore be approximately |-2 x 8.99 x 10^9 N m²/C² x 3.00 x 10^-6 C x -2.00 x 10^-6 C / (0.36 m)²|, or about 0.0137 N (Newtons). Because force is a vector quantity, to find the direction of the force we need to consider the signs of the charges. Since they have opposite signs, the force is attractive, hence, the -2.00μC charge exerts a force towards itself on the +3.00 μC charge.

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If A is the amplitude of a mass on an oscillating spring, then in one period the mass travels a distance of ___________.

a. 0
b. A
c. 2A
d. 4A

Answers

Answer:

d. 4A

Explanation:

Given that

Amplitude = A

We know that the distance cover by a particle before coming the the rest from the mean point in the oscillation motion is known as amplitude.

The distance cover by particle from 1 - 2 = A

The distance cover by particle from 2 - 1 = A

The distance cover by particle from 1 - 3 = A

The distance cover by particle from 3-4 = A

Therefore the total distance cover by particle = A+A+A+A = 4 A

Therefore the total distance = 4 A

That is why the answer will be 4 A.

d. 4A

While punting a football, a kicker rotates his leg about the hip joint. The moment of inertia of the leg is 3.75 kg⋅m² and its rotational kinetic energy is 175 J.
(a) What is the angular velocity of the leg?
(b) What is the velocity of the tip of the punter's shoe if it is 1.05 m from the hip joint?

Answers

Answer:

9.6609 rad/s

10.143945 m/s

Explanation:

I = Moment of inertia = 3.75 kgm²

K = Kinetic energy = 175 J

r = Radius = 1.05 m

Kinetic energy is given by

[tex]K=\dfrac{1}{2}I\omega^2\\\Rightarrow \omega=\sqrt{\dfrac{2K}{I}}\\\Rightarrow \omega=\sqrt{\dfrac{2\times 175}{3.75}}\\\Rightarrow \omega=9.6609\ rad/s[/tex]

The angular velocity of the leg is 9.6609 rad/s

Velocity is given by

[tex]v=r\omega\\\Rightarrow v=1.05\times 9.6609\\\Rightarrow v=10.143945\ m/s[/tex]

The velocity of the tip of the punters shoe is 10.143945 m/s

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