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VikaD [51]
4 years ago
14

Consider a cloudless day on which the sun shines down across the United States. If 2073 kJ of energy reaches a square meter ( m

2 ) of the United States in one hour, how much total solar energy reaches the entire United States per hour
Physics
1 answer:
Mnenie [13.5K]4 years ago
5 0

The total amount of energy per hour is 2.039\cdot 10^{16} kJ

Explanation:

In this problem we are told that the amount of energy reaching a square meter in the United States per hour is

E_1 = 2073 kJ

The total surface area of the United States is

A=9.834\cdot 10^6 km^2

And converting into squared metres,

A=9.834\cdot 10^6 \cdot 10^6 = 9.834\cdot 10^{12} m^2

Therefore, the total energy reaching the entire United States per hour is given by:

E=AE_1 = (9.834\cdot 10^{12})(2073)=2.039\cdot 10^{16} kJ

Learn more about energy and power:

brainly.com/question/7956557

#LearnwithBrainly

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Why do we warm up when we wear warm clothes? Explain the hiccup from the point of view of physics.
bazaltina [42]
Well your body would be trapping heat when your cold your blood contricts to help hold in heat so it would help keep your body tempture up and keep heat from escaping
5 0
4 years ago
Consider a metal single crystal oriented such that the normal to the slip plane and slip direction are at angles of 60° and 35°,
Vedmedyk [2.9K]

Answer:

19.5324 MPa

Explanation:

Information provided

Angle between the normal to the slip plane with tensile axis, \alpha=60^{o}


Angle by slip direction with tensile axis, \beta=35^{o}

Critical resolved shear stress, \tau_{c}=8 MPa

Applied stress \sigma=12 MPa

Shear stress at slip plane

\tau=\sigma cos\alpha cos\beta

\tau=12cos60^{o}cos35^{o}=4.915 MPa

\tau hence crystal won’t yield

Applied stress, \sigma for crystal to yield is given by

\sigma=\frac {\tau_{c}}{cos\alpha cos\beta}

\sigma=\frac {8}{cos60cos35}=19.53239342 MPa

\sigma=19.5324 MPa


7 0
3 years ago
How long must it take for a communications satellite to make one complete orbit around the earth? (such an orbit is said to be g
vitfil [10]

The time period of the communications satellites must be 24 hours to make one complete orbit around the earth.

An object that has been placed into orbit in space on purpose is known as a satellite or artificial satellite. Most spacecraft, with the exception of passive satellites, contain a means of generating electricity for the electronics they carry, such as solar cells or radioisotope thermoelectric generators.

A communications satellite is a man-made spacecraft that uses a transponder to relay and amplify radio telecommunication signals. It establishes a channel of communication between a source transmitter and a receiver situated at various points on Earth.

Most satellites that orbit the Earth do so between 160 and 2,000 kilometers above the surface. Because of how closely the satellites are orbiting the Earth, this orbital regime is known as low Earth orbit or LEO. The normal orbital period for satellites in low Earth orbit (LEO) is between 90 and 2 hours.

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5 0
2 years ago
A motorist drives north for 36.0 minutes at 96.0 km/h and then stops for 15.0 minutes. He then continues north, traveling 130 km
Over [174]

Answer:

a) d=187.6km

b)v=61.5km/h

Explanation:

First we convert our minutes to hours so we work always in the same units.

36min=36min(\frac{1h}{60min})=0.6h

15min=15min(\frac{1h}{60min})=0.25h

Where we used the fact that 1 hour are 60 min, thus the multiplying factor is equal to 1 (not altering the time, just changing the units).

a) On the first part the motorist travels a distance d_1=v_1t_1=(96km/h)(0.6h)=57.6km, and on the second part he travels d_2=130km.

The total displacement is d=d_1+d_2=57.6km+130km=187.6km

b) The average velocity is the relation between the total displacement and the time taken to cover it. Our total time is t=0.6h+0.25h+2.2h=3.05h, thus we have:

v=\frac{187.6km}{3.05h}=61.5km/h

8 0
3 years ago
N april 1974, steve prefontaine completed a 10-km race in a time of 27 min, 43.6 s. suppose \"pre\" was at the 7.15-km mark at a
aliya0001 [1]

solution:speed is 7.99km ,u=\frac{7990}{(25\times60)}=5.33m/s\\
now let the final velocity after 60s from 7.99km is v.\\
v-u =at\\v=5.33+a\times60\\  =60s+5.33\\now distance travelled during this time will be\\
s_{2}=vt\\
v\times103.6\\
=(60a+5.33)\times103.6\\
=551.84+6216am\\
now total distance travelled  is \\  also\\
s_{1}=ut+(\frac{1}{2})at^{2}      =5.33\times60+(0.5)9\times3600\\
=1800a+319.6m\\
Now the remaining time to complete rest race was\\
t=(27\times60+43.6)60-60(25\times60)\\
=103.6s\\Now, distance travelled during this time will be\\
s_{2}=vt\\
v\times103.6\\
=(60a+5.33)\times103.6\\
=551.84+6216am\\
now total distance travelled  is \\
s_{1}+s_{2}+7990=10000\\
1800a+319.6+551.84+6216a=2010\\
8016a=2010-871.44\\
8016a=1138.56\\
a=0.142m/s^2

6 0
3 years ago
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