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liq [111]
3 years ago
12

Where in the world does average daily temperature exceed the annual temperature range?

Physics
1 answer:
padilas [110]3 years ago
6 0
The average daily annual temperature range is exceeded in places such as deserts. Deserts tend to be hotter due to the lack of water contained within them. This means that all of the sun's thermal radiation heats up the ground and air. These factors cause deserts to have above average temperatures.
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If ball 4 has a mass of 2 kg and it is 5m high, what will be its gravitational potential energy? (g=10 N/kg) *
Nastasia [14]

Explanation:

Gravitational potential energy

= mgh

= (2kg)(10N/kg)(5m)

= 100J.

3 0
3 years ago
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Now review the difference between the temperatures on Earth and Mars. Also look at their distances from the sun.
aksik [14]

Answer: Use Question cove you can get it faster you can get the answer faster! ;) hope this helps ;) but yeah use that and answer is done right away

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2 years ago
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A referee will toss up the ball between to opponents.what is this called
OlgaM077 [116]
This happens in basketball. It is known as "jump ball".
5 0
3 years ago
The wheels of an automobile are locked as it slides to a stop from an initial speed of 30.0 m/s. If the coefficient of kinetic f
Amiraneli [1.4K]

Answer:

  x = 76.5 m

Explanation:

Let's use Newton's second law at the point of contact between the wheel and the floor.

     fr = m a

     fr = miy N

     N-W = 0

     N = W

     μ mg = m a

     a = miu g

    a = 0.600 9.8

    a = 5.88 m / s²

Having the acceleration we can use the kinematic relationships to find the distance

     v_{f}² = v₀² + 2 a x

    v_{f} = 0

    x = -v₀² / 2 a

Acceleration opposes the movement by which negative

   x = - 30²/2 (-5.88)

   x = 76.5 m

8 0
3 years ago
A convex security mirror has a radius of curvature of 12.0 cm. What is the magnification of a pare 3.0 m from the mirror?
Makovka662 [10]

Answer:

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length f=\frac{R}{2}=\frac{12}{2}=6cm

Now for mirror we know that \frac{1}{f}=\frac{1}{u}+\frac{1}{v}

So \frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}

16.66-0.333=\frac{1}{v}

v = 0.750 m

Now magnification of the mirror is m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025

5 0
3 years ago
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