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QveST [7]
3 years ago
11

A black, totally absorbing piece of cardboard of area A = 1.7 cm2 intercepts light with an intensity of 8.1 W/m2 from a camera s

trobe light. What radiation pressure is produced on the cardboard by the light?
Physics
2 answers:
Ipatiy [6.2K]3 years ago
6 0

Answer:

The answer is 2.7x10⁻⁸ Pa

Explanation:

The expression for total absorption is:

Pr = I/c

Where I = intensity of light

c = velocity of light

Replacing:

Pr = 8.1/3x10⁸ = 2.7x10⁻⁸ Pa

Furkat [3]3 years ago
3 0

Answer:

2.7x10⁻⁸ N/m²

Explanation:

Since the piece of cardboard absorbs totally the light, the radiation pressure can be found using the following equation:

p_{rad} = \frac{I}{c}

<u>Where:</u>

p_{rad}: is the radiation pressure

I: is the intensity of the light = 8.1 W/m²

c: is the speed of light = 3.00x10⁸ m/s

Hence, the radiation pressure is:

p_{rad} = \frac{I}{c} = \frac{8.1 W/m^{2}}{3.00 \cdot 10^{8} m/s} = 2.7 \cdot 10^{-8} N/m^{2}

Therefore, the radiation pressure that is produced on the cardboard by the light is 2.7x10⁻⁸ N/m².

I hope it helps you!

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Q 24, 25, 26 i dont get them and need answers for it
Alexandra [31]

Answer:

24) W = 75 [J]; 25) W = 1794[J]; 26) n = 8.8 (times) or 9 (times)

Explanation:

24) This problem can be solved by means of the following equation.

DU = Q-W\\

where:

DU = internal energy difference [J]

Q = Heat transfer [J]

W = work [J]  

Since there are no temperature changes the internal energy change is equal to zero

DU = 0

therefore:

Q = W\\

The work is equal to the heat transfered, W = 75 [J].

25) The heat transfer can be calculated by means of the following equation.

Q = m*c_{p}*DT\\where:\\m = mass = 0.4[kg]\\c_{p} = specific heat = 897[J/kg*K]\\DT= 5 [C]

Q = 0.4*897*5 = 1794[J]

Work is equal to heat transfer, W = 1794[J]

26) Each time the bag falls the potential energy is transformed into heat energy, which is released into the environment. In this way the potential energy is equal to the developed heat.

E_{p}=Q\\\\E_{p}=m*g*h

where:

m = mass = 0.5[kg]

g = gravity = 9.81[m/s^2]

h = 1.5 [m]

E_{p}=0.5*9.81*1.5\\E_{p}=7.36[J]

The heat developed can be calculated by means of the following equation.

Q=m*c_{p}*DT\\Q=0.5*130*1\\Q=65[J]

The number of times will be calculated as follows

n = 65/7.36

n = 8.8 (times) or 9 (times)

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3 years ago
Which statement about atmospheric pressure near Earth’s surface is true
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6 0
3 years ago
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The eiffel tower is a steel structure whose height increases by 19.5 cm when the temperature changes from −8 to +42 °c. what is
Readme [11.4K]
 The coefficient of expansion is 13 * 10^-6 m per meter length.per oK 
The temperature difference = 42 - - 8 = 50 oC 
delta T = (42 + 273) - (-8 + 273) = 50 oK 
delta L = L * 13* 10^6 m/oK 
oK = 50 oK delta L = 19.5 cm = 19.5 cm [1m / 100 cm] = 0.195m 
So we need to find the length and it is computed by:
0.195= L * 13 * 10^-6 * 50 L = 0.195 / (13*10^-6*50) L = 300 m 
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3 years ago
What two factors determine how fast weathering occurs
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3 years ago
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
iren [92.7K]

Answer:

372.3 J/^{\circ}C

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

P=VI=(3.6)(2.6)=9.36 W

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

E=Pt=(9.36)(350)=3276 J

Finally, the change in temperature of an object is related to the energy supplied by

E=C\Delta T

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C is the change in temperature

Solving for C, we find:

C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C

5 0
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