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Vadim26 [7]
3 years ago
11

If a burning log is a black object with a surface area of 0.25 m2 and a temperature of 800 8c, how much power does it emit as th

ermal radiation?
Physics
1 answer:
tresset_1 [31]3 years ago
8 0

The correct answer of this question: 18789.79 watt.

EXPLANATION:

As per the question, the surface area of the log is given as A = 0.25\ m^2

The temperature of the of the surface of the log t = 800 ^0C

We know that t^0C\ =(273+t)\ K

                      ⇒ 800 degree celsius= 800+273 K

                                                           = 1073 K.

Here, K stands for kelvin scale.

We are asked to calculate the thermal power radiation .

The power is defined as the amount of energy released per unit time.

Hence, power is calculated as-

                                Power P = \sigma AT^4

                                               = 5.67\times 10^{-8} \times 0.25\times (1073)^4\ watt

                                               = 18789.79 watt.      [ans]

Here, \sigma is the Stefan-Boltzmann constant and T is the absolute temperature of the log.

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GalinKa [24]

Answer:

how large a magnetic field would you experience = 8.16 x 10∧-4T

Explanation:

I = 20KA = 20,000A

r = 4.9 m

how large a magnetic field would you experience =  u.I/2πr

how large a magnetic field would you experience = (4π x10∧-7) × 20000/2π × 4.9

how large a magnetic field would you experience = 8.16 x 10∧-4T

4 0
3 years ago
Read 2 more answers
Spiderman, whose mass is 70.0 kg, is dangling on the free end of a 12.2-m-long rope, the other end of which is fixed to a tree l
Lana71 [14]

Answer:

U = -3978.8 J

Explanation:

The work of the gravitational force U just depends of the heigth and is calculated as:

U = -mgh

Where m is the mass, g is the gravitational acceleration and h the alture.

for calculate the alture we will use the following equation:

h = L-Lcos(θ)

Where L is the large of the rope and θ is the angle.

Replacing data:

h = 12.2-12.2cos(58.4)

h = 5.8 m

Finally U is equal to:

U = -70(9.8)(5.8)

U = -3,978.8 J

5 0
3 years ago
Consider two diffraction gratings with the same slit separation. The only difference between the two gratings is that one gratin
kobusy [5.1K]

Answer:

True The grid with more slits gives more angle separation increases

True. The grating with 10 slits produces better-defined (narrower) peaks

Explanation:

Such a system can be seen as a diffraction network in this case with different number of lines per unit length, the expression for the constructive interference of a diffraction network is

      d sin θ = m λ

where d is the distance between slits or lines, m the order of diffraction and λ the wavelength.

For network with 5 slits

      d = 1/5 = 0.2

For the network with 10 slits

      d = 1/10 = 0.1

let's calculate the separation (teat) for each one

      θ = sin⁻¹ (m λ / d)

for 5 slits

     θ₅ = sin⁻¹ (m λ 5)

for 10 slits

     θ₁₀ = sin⁻¹ (m λ 10)

we can appreciate that for more slits the angle increases

the intensity of a series of slits is

       I = I₀ sin²2 (N d/2) / sin² d/2)

when there are more slits (N) the peaks have greater intensity and are more acute (half width decreases)

let's analyze the claims

False

True The grid with more slits gives more angle separation increases

False

True The expression for the intensity of the diffraction peaks the intensity of the peaks increases with the number of slits as well as their spectral width decreases

False

5 0
3 years ago
An electron follows a helical path in a uniform magnetic field of magnitude 0.340 T. The pitch of the path is 6.00 µm, and the m
dedylja [7]

Answer:

The electron's speed is 34007.35 m/s

Explanation:

It is given that,

Magnetic field, B = 0.34 T

Magnetic force on the electron, F=1.85\times 10^{-15}\ N

The electron follows a helical path. We have to find the speed of an electron. The formula for magnetic force is given by :

F=B\times q\times v

q = charge on an electron, q=1.6\times 10^{-19}\ C

v = velocity of an electron

v=\dfrac{F}{Bq}

v=\dfrac{1.85\times 10^{-15}\ N}{0.34\ T\times 1.6\times 10^{-19}\ C}

v = 34007.35 m/s

Hence, this is the required solution.

4 0
2 years ago
A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t
elena55 [62]

Answer:

The first part can be solved via conservation of energy.

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For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}

where N = 0 because we are looking for the case where the car loses contact.

mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

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Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

7 0
2 years ago
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