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GrogVix [38]
3 years ago
13

Assume you are a perfect blackbody at a temperature of T = 310 K. What is the rate, in watts, at which you radiate energy? (For

the purposes of this problem, you may assume you are spherical with a mass of 70 kg and density equal to that of water)
Physics
1 answer:
madreJ [45]3 years ago
7 0

Answer:

P = 942.54 W

Explanation:

Given:

Temperature, T = 310 K

mass, m = 70 kg

Now from the stefan's law,

we have the formula

P = σAT⁴

where,

P = radiate energy

σ = Stefan's constant = 5.67 × 10⁻⁸ W/(m² K⁴)

A = Area of the body

Average surface area for a human body = 1.8 m²

T = Temperature

on substituting the values we get,

P = 5.67 × 10⁻⁸ × 1.8 × 310⁴

or

P = 942.54 W

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What happens when a candle burns?
slega [8]

Answer:

oxygen is used up is the answer

Explanation:

These vaporized molecules are drawn up into the flame, where they react with oxygen from the air to create heat, light, water vapor (H2O) and carbon dioxide (CO2).

5 0
2 years ago
A motorcycle, which has an initial linear speed of 9.7 m/s, decelerates to a speed of 4.0 m/s in 4.4 s. Each wheel has a radius
Morgarella [4.7K]

Hi there!

We can begin by solving for the linear acceleration as we are given sufficient values to do so.

We can use the following equation:

vf = vi + at

Plug in given values:

4 = 9.7 + 4.4a

Solve for a:

a = -1.295 m/s²

We can use the following equation to convert from linear to angular acceleration:

a = αr

a/r = α

Thus:

-1.295/0.61 = -2.124 rad/sec² ⇒ 2.124 rad/sec² since counterclockwise is positive.

Now, we can find the angular displacement using the following:

θ = ωit + 1/2αt²

We must convert the initial velocity of the tire (9.7 m/s) to angular velocity:

v = ωr

v/r = ω

9.7/0.61 = 15.9 rad/sec

Plug into the equation:

θ = 15.9(4.4) + 1/2(2.124)(4.4²) = 20.56 rad

6 0
3 years ago
Calculate 2 x 10^-3cm ÷ 2.5 x 10^4cm
Mamont248 [21]
Here is your answer:

First find the notations:

2×10^-3
=0002
And...
2.5×10^4=25000

Then divide:

0002÷25000=8E-9

Your answer:
=8 x 10-8
3 0
3 years ago
A cannonball is launched from the ground at an angle of 30 degrees above the horizontal and a speed of 30 m/s. Ideally (no air r
ICE Princess25 [194]
B is the answer please mark brainliest
6 0
3 years ago
Read 2 more answers
Please help me with this very easy question I just don’t get it.
diamong [38]

Answer:

150m

Explanation:

The relation of speed/time and distance/time is a derivative/integral one, as in speed is the derivative of distance (the faster you go, the faster the distance changes, duh!).

So we need to compute the integral of speed over time from 0.0s to 5.0s.

The easiest way here is to compute the area under the line (it's going to be faster than computing the acceleration and using a formula of distance based on acceleration).

The area under the line is a trapezoid with "height" 5s, and the bases 10m/s and 50m/s. Using the trapezoid area formula of h*(a + b)/2

distance = 5s * (10m/s + 50m/s) / 2 = 5s * 60m/s / 2 = 5s * 30m/s = 150m

Alternatively, we can use the acceleration formula:

a = (50m/s - 10m/s)/5s = 40m/s / 5s = 8m/s^2

distance = v0 * t + a * t^2 / 2 = 10m/s * 5s + 8m/s^2 * (5s)^2 / 2 = 50m + 8m * 25 / 2 = 50m + 100m = 150m.

5 0
2 years ago
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