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ki77a [65]
3 years ago
8

Plz will someone help me on 28 will give brainlest if right plus 100 points.

Physics
2 answers:
PilotLPTM [1.2K]3 years ago
7 0
B. Because it has more increments
hammer [34]3 years ago
3 0

Answer:

its B

Explanation:

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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= 4πR_{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} =><u> 6.34 x  </u>10^{7}<u>W/m²</u>

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

A= 4πR

Now R= 1.5 x 10^{11m

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 =><u>1.37 x </u>10^{3<u> W/m²</u>

<u />

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
A 100-kg running back runs at 5 m/s into a stationary linebacker. It takes 0.5 s for the running back to be completely stopped.
Elza [17]

Answer:

1000 N

Explanation:

First, we need to find the deceleration of the running back, which is given by:

a=\frac{v-u}{t}

where

v = 0 is his final velocity

u = 5 m/s is his initial velocity

t = 0.5 s is the time taken

Substituting, we have

a=\frac{0-5 m/s}{0.5 s}=-10 m/s^2

And now we can calculate the force exerted on the running back, by using Newton's second law:

F=ma=(100 kg)(-10 m/s^2)=-1000 N

so, the magnitude of the force is 1000 N.

6 0
4 years ago
Read 2 more answers
How much farther from the charge is the 2000 v equipotential surface than the 3000 v surface?
olganol [36]

Answer:

r1 -r2 = 3.75cm

Explanation:

Check the attached file for the solution

7 0
4 years ago
A uniform horizontal strut weighs 400.0 N. One end of the strut is attached to a hinged support at the wall, and the other end o
Olegator [25]

Answer:

Explanation:

Given

Weight of strut W_1=400 N

Weight of sign Board W_2=200 N

In this question angle which cable makes with horizontal is not given so

assuming \theta =30^{\circ}

From Diagram

Moment about hinge point

T\sin 30\times L=W_1\times \frac{L}{2}+W_2\times L

T\sin 30=\frac{W_1}{2}+W_2

T\sin 30=\frac{400}{2}+200

T=800 N

\sum F in x direction is zero

F_x=T\cos 30

F_x=800\cos 30=346.41 N

\sum F in Y direction is zero therefore

F_y+T\sin 30=W_1+W_2

F_y=400+200-800\cdot \sin 30

F_y=200 N

F_{net}\ at\ hinge=\sqrt{F_x^2+F_y^2}

F_{net}=399.99\approx 400 N                              

6 0
3 years ago
HELPPPP
PSYCHO15rus [73]
That’s kinda messed up ⬆️
8 0
3 years ago
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