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Gwar [14]
3 years ago
15

How much electrical energy is used by a 75 W laptop that is operating 12 minutes?

Physics
1 answer:
LiRa [457]3 years ago
4 0
75 watts means 75 joules per second. 12 minutes is (12 x 60) = 720 seconds. 75 j/sec x 720sec = 54,000 joules.
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A object with a mass of 1.5 kg is lifted from the ground to a height of 0.22 m what is the objects potential energy
svet-max [94.6K]

Answer:

<h2>3.3 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 1.5 × 10 × 0.22

We have the final answer as

<h3>3.3 J</h3>

Hope this helps you

5 0
3 years ago
How can you be both at rest and also moving at 100,000 km/h at the same time
4vir4ik [10]

You could be lying completley still on your bed, and all though it seems you are at rest, you are moving along with the earth around the sun and hence are motion. This is why 'being at rest' is more of a relative term. Hope this helps!

7 0
3 years ago
What is the Kinetic Energy of a 1200 kg object that is moving with a speed of 24 m/s
SVEN [57.7K]

Kinetic Energy = 1/2mv^2

m= 1200kg

v= 24 m/s

KE = 1/2 (1200kg)(24m/s)^2 = 345,600 N

8 0
3 years ago
A positron is an elementary particle identical to an electron except that its charge is . An electron and a positron can rotate
denis23 [38]

Explanation:

According to the energy conservation,

          F_{centripetal} = F_{electric}

            \frac{mv^{2}}{r} = \frac{kq^{2}}{d^{2}}

           v^{2} = \frac{kq^{2}r}{d^{2}m}

                 = \frac{9 \times 10^{9} N.m^{2}/C^{2} \times 1.6 \times 10^{-19} C \times 0.75 \times 10^{-9} m}{(1.50 \times 10^{-9}m)^{2} \times 9.11 \times 10^{-31} kg}

                = 8.430 \times 10^{10} m^{2}/s^{2}

             v = \sqrt{8.430 \times 10^{10} m^{2}/s^{2}}

                = 2.903 \times 10^{5} m/s

Formula for distance from the orbit is as follows.

               S = 2 \pi r

                  = 2 \times 3.14 \times 0.75 \times 10^{-9} m

                  = 4.71 \times 10^{-9} m

Now, relation between time and distance is as follows.

                T = \frac{S}{v}

       \frac{1}{f} = \frac{S}{v}

or,           f = \frac{v}{S}          

                = \frac{2.903 \times 10^{5} m/s}{4.71 \times 10^{-9} m}      

                = 6.164 \times 10^{13} Hz

Thus, we can conclude that the orbital frequency for an electron and a positron that is 1.50 apart is 6.164 \times 10^{13} Hz.

7 0
3 years ago
Define the term Distance.​
Aloiza [94]

Distance is defined as the space between two points in space.

<h3>What is distance?</h3>

Distance is defined as the space between two points in space.

The unit of distance is metres or kilometres.

Distance can be measured directly usind instruments such as metre rule and tape rule.

The formula for calculating distance is given below:

Distnce = velocity × time.

Therefore, distance is defined as the space between two points in space.

Learn more about about distance at: brainly.com/question/4931057

7 0
3 years ago
Read 2 more answers
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