Which state of matter has a definite volume but a variable shape?

Power of sunlight on Earth The Sun emits about 3.9 * 1026 J of electromagnetic radiation each second. (a) Estimate the power tha

ratelena [41]

Answer:

a)6.34 x $10^{7}$ W/m²

b)1.37 x $10^{3$ W/m²

c) see explanation.

Explanation:

a)The relation of intensity'I' of the radiation and area 'A' is given by:

I= P/A

where P= power of sunlight i.e 3.9 x $10^{26}$ J

and the area of the sun is given by,

A= 4π $R_{sun}$ => 4π $(\frac{1.4*10^{9} }{2} )^{2}$

A=6.15 x $10^{18}$ m²

$I_{sun}$ = 3.9 x $10^{26}$ / 6.15 x $10^{18}$ => 6.34 x $10^{7}$ W/m²

b) First determine the are of the sphere in order to determine intensity at the surface of the virtual space

A= 4π $R$

Now R= 1.5 x $10^{11$ m

A= 4π x 1.5 x $10^{11$ =>2.83 x $10^{23$ m²

The power that each square meter of Earths surface receives

$I_{earth$ = 3.9 x $10^{26}$ /2.83 x $10^{23$ =>1.37 x $10^{3$ W/m²

c) in part (b), by assuming the shape of the wavefront of the light emitted by the Sun is a spherical shape so each point has the same distance from the source i.e sun on the wavefront.

6 0

3 years ago

Sasha is interviewing people for a study he is doing. Before he starts the interview, Sasha explains the purpose of the study an

Alex73 [517]

D.

Sasha is informing participants about the study and making sure they wish to participate.

5 0

3 years ago

If enough heat was REMOVED from B, it would change into ____________.

Sindrei [870]

It would change into a gas.

5 0

3 years ago

The aurora is caused when electrons and protons, moving in the earth's magnetic field of ≈5.0×10−5T, collide with molecules of t

Vika [28.1K]

Answer:

$r=0.341m$

Explanation:

The circular movement is produced by the magnetic field:

$F=q*v*B$

This force is a centripetal force because the electron moves in a plane perpendicular to the magnetic field. For a centripetal Force, that produces a circular orbit with a radius r:

$F=m*a=m*v^{2}/r$

If we solve these two equations in order to find r, with the mass and charge of a electron:

$qvB=mv^{2}/r\\r=mv/(qB)=9,1*10^{-31}*3*10^{6}/(1.6*10^{-19}*5*10^{-5})=0.341m$

3 0

3 years ago

Ayden and Steven are playing catch with a football. Ayden throws the football at a velocity of 15 m/s with an angle of 60° above

Pachacha [2.7K]

1) 1.33 s

2) 8.6 m

3) 19.9 m

Explanation:

1)

The motion of the football is a projectile motion, which consists of two independent motions:

- A uniform motion (=constant velocity) along the horizontal direction

- A uniformly accelerated motion (=constant acceleration) along the vertical direction

The hang time of the football is the time it takes for the football to reach its maximum height. It is given by:

$t=\frac{u_y}{g}$

where

$u_y = u sin \theta$ is the initial vertical velocity of the ball, where

u = 15 m/s is the initial velocity

$\theta=60^{\circ}$ is the angle of projection of the ball

$g=9.8 m/s^2$ is the acceleration due to gravity

Substituting, we find the hang time:

$t=\frac{u sin \theta}{g}=\frac{(15)(sin 60^{\circ})}{9.8}=1.33 s$

2)

The motion of the ball along the vertical direction is a uniformly accelerated motion, so we can use the suvat equation:

$s=u_y t - \frac{1}{2}gt^2$

where:

s is the vertical displacement

$u_y = u sin \theta$ is the initial vertical velocity

t is the time

$g=9.8 m/s^2$ is the acceleration due to gravity

In this part, we want to find the maximum height, so the height reached by the ball when the time is

t = 1.33 s

Therefore, by substituting the values in the equation, we can find the maximum height:

$s=usin \theta t-\frac{1}{2}gt^2=(15)(sin 60^{\circ})(1.33)-\frac{1}{2}(9.8)(1.33)^2=8.6 m$

3)

Here we want to find the horizontal range covered by the ball during its flight.

The horizontal range for a projectile is given by the equation

$d=\frac{u^2 sin(2\theta)}{g}$

where

u is the initial velocity

$\theta$ is the angle of projection

g is the acceleration due to gravity

For the ball in this problem we have:

u = 15 m/s

$\theta=60^{\circ}$

$g=9.8 m/s^2$

Substituting into the equation, we find the horizontal distance covered by the ball:

$d=\frac{(15)^2sin(2\cdot 60^{\circ})}{9.8}=19.9 m$

6 0

4 years ago