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lisov135 [29]
3 years ago
11

How can heat be transferred across empty space?

Physics
2 answers:
erica [24]3 years ago
7 0
By thermal radiation 
yawa3891 [41]3 years ago
4 0
The answer is: Heat can<span> be </span>transmitted<span> though </span>empty space<span> by thermal radiation. Thermal radiation (often called infrared radiation) is a type electromagnetic radiation (or light). Radiation is a form of energy transport consisting of electromagnetic waves traveling at the speed of light.</span>
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What is the role of electrical forces in nuclear fission
77julia77 [94]
The electrical forces pulls nucleus apart
8 0
3 years ago
How much work does it take to lift 345 boxes to a height of 6.00 m of each box has a mass of 7.89 kg
Art [367]

Given the value of the mass of each boxes, the work done in lifting the boxes to the given height is 1.6 × 10⁵J.

<h3>Work done</h3>

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

W = F × d

Where F is force applied or Weight and d is distance

Also Force = Weight = mass × acceleration due to gravity.

Since gravity is acting on the boxes as it been lift

W = Weight × height from ground level

W = mg × d

Where m is mass of the boxes, g is accelration due to gravity( g = 9.8m/s² ) and d is distance from ground level.

Given the data in the question;

  • Since each box has a mass of 7.89 kg
  • Mass of the 345 boxes = 345 × 7.89 kg = 2722.05kg
  • Distance or height d = 6.0m
  • Work done W = ?

To determine the work done, we substitute our values into the expression above.

W = mg × d

W = 2722.05kg × 9.8m/s² × 6.0m

W = 160056.5kgm²/s²

W = 160056.5J

W = 1.6 × 10⁵J

Therefore,  Given the value of the mass of each boxes, the work done in lifting the boxes to the given height is 1.6 × 10⁵J.

Learn more about work done here: brainly.com/question/26115962

3 0
2 years ago
You are sitting 3 m away from you friend who is watching a cartoon on his phone. How will the sound itensity change if your frie
Zinaida [17]

Answer:

Decreases by $3.6 \times 10^{-3}$ times

Explanation:

The intensity of a sound is defined as the energy of the sound that is flowing in an unit time through the unit area which is in the direction that is perpendicular to the direction of the sound waves movement.

The intensity of energy is described by the inverse square law. It states that the intensity varies inversely with the distance square of the distance.

In other words, the sound intensity decreases as inversely proportional to the squared of the distance.  i.e. $\frac{1}{r^2}$

In the context when the distance was 3 m, the intensity of the sound was = $\frac{1}{9}$

But when the distance became 6 cm or 0.06 m, the sound intensity decreases by =  $\frac{1}{0.06^2}$

                       = $3.6 \times 10^{-3}$ times

3 0
3 years ago
A 3250 N car is pushed a distance of 35 m the power was 11375 J, how long did it take?
irakobra [83]

Answer:

10.8s

Explanation:

Given parameters:

Force on the car  = 3250N

Distance  = 35m

Power  = 11375W

Unknown:

Time taken = ?

Solution:

To solve this problem;

 Power is the rate at which work is done

         Power = \frac{work done }{time}  

  Work done  = force x distance  = 3250 x 35  = 123200J

Now;

          11375  = \frac{123200}{t}  

           11375t  = 123200  

                   t  = 10.8s

5 0
3 years ago
a fan acquires a speed of 180 rpm in 4s, starting from rest. calculate the speed of the fan at the end of the 5th second startin
KengaRu [80]

Answer:

225 rpm

Explanation:

The angular acceleration of the fan is given by:

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

where

\omega_f is the final angular speed

\omega_i is the initial angular speed

\Delta t is the time interval

For the fan in this problem,

\omega_i = 0\\\omega_f = 180 rpm\\\Delta t=4 s

Substituting,

\alpha = \frac{180-0}{4}=45 rpm/s

Now we can find the angular speed of the fan at the end of the 5th second, so after t = 5 s. It is given by:

\omega' = \omega_i + \alpha t

where

\omega_i = 0\\\alpha = 45 rpm/s\\t = 5 s

Substituting,

\omega' = 0 + (45)(5)=225 rpm

7 0
3 years ago
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