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almond37 [142]
3 years ago
11

When determining a region's climate, the two main variables considered are the average what? ( They don't have the subject "Scie

nce" here. )
Physics
1 answer:
shutvik [7]3 years ago
3 0
Temperature and precipitation. (They should put a science button in) :-)
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Describe how the three methods of thermal energy transfer may take place within the iguana’s enclosure.
sasho [114]

Answer:

Heat can travel from one place to another in three ways: Conduction, Convection and Radiation. Both conduction and convection require matter to transfer heat.  Conduction is the transfer of heat between substances that are in direct contact with each other. Thermal energy is transferred from hot places to cold places by convection. Radiation is a method of heat transfer that does not rely upon any contact between the heat source and the heated object as is the case with conduction and convection. Heat can be transmitted through empty space by thermal radiation often called infrared radiation.

Explanation:

6 0
2 years ago
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A 10 N force is used to push a 30 kg box across the floor, moving it a distance of 20 m. There is no friction. What is the magni
tekilochka [14]
Work=Force • Distance
W= 10N • 20m
W= 200 J

5 0
3 years ago
As a mercury atom absorbs a photon of
Allisa [31]
The energy absorbed by photon is 1.24 eV.
This is the perfect answer.
8 0
2 years ago
A 60 kg sprinter has a momentum of +600 kg-m/s when he crosses the finish
MakcuM [25]

Answer:

10 ms⁻¹

Explanation:

The amount of momentum that an object has is dependent upon two factors

  • mass of the moving object  
  • speed of motion

In terms of an equation,

Momentum (P) = Mass(m)×velocity(v)

                     P = m×v

                 600 = 60 × v ⇒ v = 10 ms⁻¹

3 0
3 years ago
A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica
Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed \omega =40rad/s

mass of m_2=0.8 kg dropped at r=0.3 m from center

let \omega _2 be the final angular velocity of cylinder

Conserving Angular momentum

L_1=L_2

\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2

\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2

26.01\times 40=26.082\times \omega _2

\omega _2=39.88 rad/s

3 0
3 years ago
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