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Nezavi [6.7K]
3 years ago
12

An extended period when temperatures are well below average is known as a/an

Physics
2 answers:
sasho [114]3 years ago
6 0
Depends on the "period". If weeks, months, or years, then it's a cold wave. If centuries, then you've got yourself an ice age.
Alecsey [184]3 years ago
3 0
It would be a cold wave. 
Ice ages occur when the temperatures are extremely cold.(long-term)  
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If Matthew was traveling into space from Earth, which of these would he be able to reach first? A) Sun B) Venus C) Alpha centaur
const2013 [10]

C.) Alpha Centauri

Explanation:

Due to it being the closest planetary system to earth.

About 4.367 light years away.

3 0
3 years ago
A 23.0 kg iron weightlifting plate has a volume of 2920 cm3 . what is the density of the iron plate in g/cm3?
yanalaym [24]
The first thing you should know for this case is that density is defined as the quotient between mass and volume:
 D = M / V
 In addition, you should keep in mind the following conversion:
 1Kg = 1000g
 Substituting the values we have:
 D = (23.0 * 1000) / (2920) = 7.88 g / cm ^ 3
 answer
 the density of the iron plate is 7.88 g / cm ^ 3
8 0
3 years ago
Select the correct answer.
Effectus [21]
A. logic, would be your answer i believe!
4 0
2 years ago
When you stand outside your are standing in the, Stratosphere. Troposphere. Mesosphere no
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8 0
3 years ago
A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul
zepelin [54]

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}

\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}

\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 +  \omega^2}

\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2}  }

\mathbf{\omega^2=\dfrac{39.24 }{2}}

\mathbf{\omega=\sqrt{19.62 } \ rad/sec}

\mathbf{\omega=4.429 \ rad/sec}

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

4 0
2 years ago
Read 2 more answers
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