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guajiro [1.7K]
3 years ago
5

How hot can the desert get​

Physics
2 answers:
Art [367]3 years ago
8 0

Answer:

134 f

Explanation:

The hottest temperature ever reliably measured in a desert was 134 degrees F, in Death Valley of the Mojave Desert in 1913.

Volgvan3 years ago
7 0
<h2>not as hot as me.</h2><h2>jk.</h2>

Temperature. During the day, desert temperatures rise to an average of 38°C (a little over 100°F). At night, desert temperatures fall to an average of -3.9°C (about 25°F). At night, desert temperatures fall to an average of -3.9 degrees celsius (about 25 degrees fahrenheit).

Desert surfaces receive a little more than twice the solar radiation received by humid regions and lose almost twice as much heat at night. Many mean annual temperatures range from 20-25 degrees Celsius. The extreme maximum ranges from 43.5-49 degrees Celsius. Minimum temperatures sometimes drop to -18 degrees Celsius.

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Astronaut Rob leaves Earth in a spaceship at a speed of 0.960crelative to an observer onEarth. Rob's destination is a star syste
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To solve this problem we will use the mathematical definition of the light years in metric terms, from there, through the kinematic equations of motion we will find the distance traveled as a function of the speed in proportion to the elapsed time. Therefore we have to

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Then

14.4Ly = 1.36231609*10^{17} m

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Where,

v = Velocity

x = Displacement

t = Time

We have that

t = \frac{1.36231609*10^{17}}{0.96c} \rightarrow c= Speed of light

t = \frac{1.36231609*10^{17}}{0.96(3*10^8)}

t= 454105363 s (\frac{1hour}{3600s})

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5 0
3 years ago
Calculate the kinetic energy, in joules of a 1160-kg automobile moving at 19.0 m/s.
Rudiy27

Answer:

Explanation:

K.E=1/2 mv²

m=1160 kg

v=19.0 m/s

so k.e=1/2*1160*(19.0)²

K.E=1/2*1160*361

K.E=1/2*418760

K.E=209380=2.0*10^5 j

7 0
2 years ago
A boy is pedaling his bicycle at a velocity of 0.20km/ minute . How far will he travel in 2.5 hours
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problems like this A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet fi
Ymorist [56]

This question is incomplete the complete question is

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

Answer:

(a) Xs=0.459m

(b) t=0.984 s

(c) Vc=6.65 m/s

Explanation:

(a) To reach maximum distance

g=-9.8m/s^{2}\\ Vf=0\\v_{b}^{2}=v_{a}^{2}+2gx_{s} \\  x_{s}=\frac{0-(3^{2} )}{-2*9.8}\\ x_{s}=0.459m

(b) For Time

To find t we must find t1 and t2

as

t=t1+t2

For T1

t_{1}=(Vb-Va)/g \\t_{1}=(0-3)/9.8\\t_{1}=0.306s

For T2

x_{l}=Vbt+(1/2)gt_{2}^{2}\\   as\\x_{l}=x_{1}+x_{s}\\x_{l}=1.8+0.459\\x_{l}=2.259\\so\\t_{2}=\frac{2.259*2}{9.8} \\t_{2}=0.6789s

For Total Time

t=t1+t2

t=0.306+0.6789

t=0.984s

(c) To find Vc

Vc=Vb+gt2

Vc=(0)+(9.8)(0.6789)

Vc=6.65 m/s

7 0
3 years ago
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