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Lelechka [254]
3 years ago
13

Estimate the radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 9.5 cm from the center of the bul

b. Assume that light is completely absorbed. Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
4 0

Given Information:  

Power of bulb = w = 25 W atts

distance = d = 9.5 cm = 0.095 m

Required Information:  

Radiation Pressure = ?

Answer:

Radiation Pressure =7.34x10⁻⁷ N/m²

Explanation:

We know that radiation pressure is given by

P = I/c

Where I is the intensity of radiation and is given by

I = w/4πd²

Where w is the power of the bulb in watts and d is the distance from the center of the bulb.

So the radiation pressure becomes

P = w/c4πd²

Where c = 3x10⁸ m/s is the speed of light

P = 25/(3x10⁸*4*π*0.095²)

P = 7.34x10⁻⁷ N/m²

Therefore, the radiation pressure due to a 25 W bulb at a distance of 9.5 cm from the center of the bulb is 7.34x10⁻⁷ N/m²

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An unbalanced force of 20N acts on a 4.0kg mass what is it's acceleration​
tankabanditka [31]

Hi there!

According to Newton's second law:

∑F = m · a, where:

∑F = net force (N = kgm/s²)

m = mass (kg)

a = acceleration (m/s²)

Rearrange to solve for acceleration:

F/m = a

20N / 4.0kg = 5 m/s²

4 0
3 years ago
A student is running to catch the campus shuttle bus, which is stopped at the bus stop. The student is runnign at a constant spe
Serggg [28]

Answer:

Part a)

t = 16.8 s

d = 100.8 m

Part b)

v_f = 2.86 m/s

Explanation:

Part a)

Constant speed by which the student will run is given as

v = 5 m/s

now after some time if student is going to overtake the position of bus

so here the final positions will be same

so we have

x_{bus} = x_{student}

0 + \frac{1}{2}at^2 + d = v_{student} t

\frac{1}{2}(0.170)t^2 + 60 = 5 t

0.085 t^2 - 5t + 60 = 0

so it is

t = 16.8 s

So student will run the total distance

d = vt

d = (6)(16.8)

d = 100.8 m

Part b)

Speed of bus when student reach the bus is given as

v_f = v_i + at

v_f = 0 + (0.170)(16.8)

v_f = 2.86 m/s

4 0
3 years ago
A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5
harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0
3 years ago
How does unequal solar heating lead to the Gulf Stream?
adoni [48]
Http://earthguide.ucsd.edu/virtualmuseum/virtualmuseum/OriginofGulfStream.shtml this website might help u find ur answer
8 0
3 years ago
Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current
Marrrta [24]

Answer:

Explanation:

Since the wires attract each other , the direction of current will be same in both the wires .

Let I be current in wire which is along x - axis

force of attraction per unit length between the two current carrying wire is given by

\frac{\mu_0}{4\pi}  x \frac{2 I_1\times I_2}{d}

where I₁ and I₂ are currents in the wires and d is distance between the two

Putting the given values

285 x 10⁻⁶ = 10⁻⁷ x \frac{2\times25.5\times I_2}{.3}

I₂ = 16.76 A

Current in the wire along x axis is 16.76 A

To find point where magnetic field is zero due the these wires

The point will lie between the two wires  as current is in the same direction.

Let at y = y , the neutral point lies

k 2 x  \frac{16.76}{y} = k 2 x \frac{25.5}{.3-y}

25.5y = 16.76 x .3 - 16.76y

42.26 y = 5.028

y = .119

= .12 m

3 0
3 years ago
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