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brilliants [131]

1 year ago

8

If the surface temperature of that person's skin is 30∘C (that's a little lower than healthy internal body temperature becaus

e your skin is usually a little colder than your insides), what is the total power that person will radiate? For now, assume the person is a perfect blackbody (so ϵ=1).

1 answer:

anastassius [24]1 year ago

5 0

Answer:

$E=477.92\ W.m^{-2}$

Explanation:

Given that:

Absolute temperature of the body, $T=273+30=303\ K$

emissivity of the body, $\epsilon=1$

<u>Using Stefan Boltzmann Law of thermal radiation:</u>

$E=\epsilon. \sigma.T^4$

where:

$\sigma =5.67\times 10^{-8}\ W.m^{-2}.K^{-4}$ (Stefan Boltzmann constant)

Now putting the respective values:

$E=1\times 5.67\times 10^{-8}\times 303^4$

$E=477.92\ W.m^{-2}$

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A 2400-kg satellite is in a circular orbit around a planet. the satellite travels with a constant speed of 6670 m/s. the radius

Ad libitum [116K]

The gravitational force exerted on the satellite is called the centrifugal force, the force keeping it orbiting to the planet. Its formula is F= mass times the square of the velocity all over the radius.Thus,

F = 2400 * 6670^2 * (1/8.92x10^6)
F = 11,970 N

I hope I was able to help you. Have a good day.

4 0

2 years ago

Read 2 more answers

Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0

1 year ago

Un cuerpo se mueve en línea recta segun la ecuación x=10+20t-4.9t2 (x está expresado en metros y t en segundos). ¿Cuál es la lon

lora16 [44]

Answer:

La longitud del camino recorrido es de 25.9 [m]

Explanation:

Se reemplaza el valor de tiempo en segundos en la ecuación dada de desplazamiento

x=10+20*(3) - 4.9*(3)^2

x= 25.9 [metros]

4 0

1 year ago

Read 2 more answers

The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0. Exp

olganol [36]

Answer:

Ft

Explanation:

We are given that

Initial velocity=u=0

We have to find the magnitude of p of the momentum of the particle at time t.

Let mass of particle=m

Applied force=F

Acceleration, $a=\frac{F}{m}$

Final velocity , $v=a+ut$

Substitute the values

$v=0+\frac{F}{m}t=\frac{F}{m}t$

We know that

Momentum, p=mv

Using the formula

$p=m\times \frac{F}{m}t=Ft$

6 0

2 years ago

A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder h

Anestetic [448]

Answer:

Speed of 1.83 m/s and 6.83 m/s

Explanation:

From the principle of conservation of momentum

$mv_o=m(v_1 + v_2)$ where m is the mass, $v_o$ is the initial speed before impact, $v_1$ and $v_2$ are velocity of the impacting object after collision and velocity after impact of the originally constant object

$5m=m(v_1 +v_2)$

Therefore $v_1+v_2=5$

After collision, kinetic energy doubles hence

$2m*(0.5mv_o)=0.5m(v_1^{2}+v_2^{2})$

$2v_o^{2}=v_1^{2} + v_2^{2}$

Substituting 5 m/s for $v_o$ then

$2*(5^{2})= v_1^{2} + v_2^{2}$

$50= v_1^{2} + v_2^{2}$

Also, it’s known that $v_1+v_2=5$ hence $v_1=5-v_2$

$50=(5-v_2)^{2}+ v_2^{2}$

$50=25+v_2^{2}-10v_2+v_2^{2}$

$2v_2^{2}-10v_2-25=0$

Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then $v_2=6.83 m/s$

Substituting, $v_1=-1.83 m/s$

Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s

6 0

1 year ago