PHYSICS COLLEGE

Light bulb 1 operates with a filament temperature of 2700 K whereas light bulb 2 has a filament temperature of 2100 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs.

Answers

Answer 1

Answer:

Answer:

0.3659

Explanation:

The power (p) is given as:

P = AeσT⁴

where,

A =Area

e = transmittivity

σ = Stefan-boltzmann constant

T = Temperature

since both the bulbs radiate same power

P₁ = P₂

Where, 1 denotes the bulb 1

2 denotes the bulb 2

thus,

A₁e₁σT₁⁴ = A₂e₂σT₂⁴

Now e₁=e₂

⇒A₁T₁⁴ = A₂T₂⁴

$(A_1)/(A_2) =(T_(2)^(4))/(T_(1)^(4))$

substituting the values in the above question we get

$(A_1)/(A_2) =(2100_(2)^(4))/(2700_(1)^(4))$

$(A_1)/(A_2) }$ =0.3659

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What is the magnitude and direction of the electric field atradiaConsider a coaxial conducting cable consisting of a conductingrod of radius R1 inside of a thin-walled conducting shell of radius 2(both are infinite length). Suppose the inner rod hasradiusR1= 1.3 mm and outer shell has radiusR2= 10R1Ifthe net charge density on the center rod isq1= 3.4×10−12C/mand the outer shell isq2=−2q1,a.)What is the magnitude and direction of the electric field atradial distancer= 5R1from the center rod

Answers

Answer:

E = 9.4 10⁶ N / C, The field goes from the inner cylinder to the outside

Explanation:

The best way to work this problem is with Gauss's law

Ф = E. dA = qint / ε₀

We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.

The flow on the faces is zero, since the field goes in the radial direction of the cylinders.

The area of the cylinder is the length of the circle along the length of the cable

dA = 2π dr L

A = 2π r L

They indicate that the distance at which we must calculate the field is

r = 5 R₁

r = 5 1.3

r = 6.5 mm

The radius of the outer shell is

r₂ = 10 R₁

r₂ = 10 1.3

r₂ = 13 mm

r₂ > r

When comparing these two values we see that the field must be calculated between the two housings.

Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is

λ = q / L

Qint = λ L

Let's replace

E 2π r L = λ L /ε₀

E = 1 / 2piε₀ λ / r

Let's calculate

E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3

E = 9.4 10⁶ N / C

The field goes from the inner cylinder to the outside

A body with initial velocity 8.0 m/s moves along a straight line with constant acceleration and travels640 m in 40 s. For the 40 s interval, find (a) the average velocity, (b) the final velocity, and (c) the
acceleration.

Answers

Answer:

(a) The average velocity is 16 m/s

(b) The acceleration is 0.4 m/s^2

(c) The final velocity is 24 m/s

Explanation:

Constant Acceleration Motion

It's a type of motion in which the velocity (or the speed) of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, final speed is calculated as follows:

$v_f=v_o+at\qquad\qquad [1]$

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+(a.t^2)/(2)\qquad\qquad [2]$

(a) The average velocity is defined as the total distance traveled divided by the time taken to travel that distance.

We know the distance is x=640 m and the time taken t= 40 s, thus:

$\displaystyle \bar v=(x)/(t)=(640)/(40)=16$

The average velocity is 16 m/s

Using the equation [1] we can solve for a:

$\displaystyle a=(v_f-v_o)/(t)$

$\displaystyle a= 2(x-v_ot)/(t^2)$

Since vo=8 m/s, x=640 m, t=40 s:

$\displaystyle a= 2(640-8\cdot 40)/(40^2)=0.4$

The acceleration is 0.4 m/s^2

(b) The final velocity is calculated by [1]:

$v_f=8+0.4\cdot 40$

$v_f=8+16=24$

The final velocity is 24 m/s

Final answer:

The average velocity is 16 m/s, the final velocity is 8.0 m/s + (acceleration * 40 s), and the acceleration can be found by solving the equation 640 m = (8.0 m/s * 40 s) + (0.5 * acceleration * (40 s)^2.

Explanation:

To find the average velocity, we use the formula: average velocity = total displacement / total time. In this case, the total displacement is 640 m and the total time is 40 s, so the average velocity is 640 m / 40 s = 16 m/s.

To find the final velocity, we can use the formula: final velocity = initial velocity + (acceleration * time). In this case, the initial velocity is 8.0 m/s and the time is 40 s. Since the question states that it moves with constant acceleration, we can assume that the acceleration is the same throughout the 40 s interval. Therefore, the final velocity is 8.0 m/s + (acceleration * 40 s).

To find the acceleration, we can use the formula: total displacement = (initial velocity * time) + (0.5 * acceleration * time^2). In this case, the total displacement is 640 m, the initial velocity is 8.0 m/s, and the time is 40 s. Solving for acceleration, we have 640 m = (8.0 m/s * 40 s) + (0.5 * acceleration * (40 s)^2).

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Students have four identical, hollow, uncharged conducting spheres, W, X, Y, and Z.Sphere Z is given a positive charge of +40 C. Sphere Z is touched first to sphere W, then sphere X, and finally to sphere Y. What is the resulting charge on sphere Y?

a. +5 με

b. +10 μC

c. +20 μC

d. +40 με

Answers

Explanation:

because they made contact that means their new force will be the same

Final answer:

Sphere Z is initially charged with +40 C. When it is touched to three other spheres, the charge is evenly distributed among them. The resulting charge on sphere Y is +10 μC.

Explanation:

The initial charge on sphere Z is +40 C. When sphere Z is touched to sphere W, the charge is evenly distributed between the two spheres, resulting in each sphere having a charge of +20 C. Then, when sphere Z is touched to sphere X, the total charge is evenly distributed between all three spheres, resulting in each sphere having a charge of +13.33 C. Finally, when sphere Z is touched to sphere Y, the total charge is evenly distributed between all four spheres, resulting in each sphere having a charge of +10 C. Therefore, the resulting charge on sphere Y is +10 μC (option b).

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An athlete swings a 6.50-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.900 m at an angular speed of 0.700 rev/s. (a) What is the tangential speed of the ball

Answers

Answer:

v = 3.951 m/s

Explanation:

Given that,

Mass of a ball, m = 6.5 kg

Radius of the circle, r = 0.9 m

Angular speed of the ball, $\omega=0.7\ rev/s=4.39\ rad/s$

Let v is the tangential speed of the ball. It is given in terms of angular speed is follows :

$v=r\omega\n\nv=0.9* 4.39\n\nv=3.951\ m/s$

So, the tangential speed of the ball is 3.951 m/s.

The mass of a string is 20 g and it has a length of 3.2 m. Assuming that the tension in the string is 2.5 N, what will be the wavelength of a travelling wave that is created by a sinusoidal excitation of this string with a frequency of 20 Hz. Provide the wavelength in units of m. Please note: You do not include the units in your answer. Just write in the number.

Answers

Answer:

The wavelength of the wave is 1 m

Explanation:

Given;

mass of the string, m = 20 g = 0.02 kg

length of the string, L = 3.2 m

tension on the string, T = 2.5 N

the frequency of the wave, f = 20 Hz

The velocity of the wave is given by;

$v = \sqrt(T)/(\mu) {}$

where;

μ is mass per unit length = 0.02 kg / 3.2 m

μ = 6.25 x 10⁻³ kg/m

$v = \sqrt{(T)/(\mu) } \n\nv = \sqrt{(2.5)/(6.25*10^(-3)) } \n\nv = 20 \ m/s$

The wavelength of the wave is given by;

λ = v / f

λ = (20 m/s )/ (20 Hz)

λ = 1 m

Therefore, the wavelength of the wave is 1 m

If a single circular loop of wire carries a current of 61 A and produces a magnetic field at its center with a magnitude of 1.70 10-4 T, determine the radius of the loop.