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Temka [501]
2 years ago
5

Why is nuclear energy able to be used for practical purposes? the reactions are controlled to regulate energy output. the reacti

ons are uncontrolled for maximum energy output. all of the products are fed back into the reaction to keep it going. all of the products are immediately removed to inhibit more reactions.
Physics
1 answer:
Sonbull [250]2 years ago
3 0

Nuclear energy is is used practically for many energy requirements because the reactions are controlled to regulate energy output

<h3>What is nuclear energy?</h3>

Nuclear energy is the energy reaelesed during nuclear relations.

Nuclear energy is a clean form of energy which is available in enormous amounts.

Nuclear energy is is used practically for many energy requirements because the reactions are controlled to regulate energy output.

Therefore, the control of nuclear reactions make the use if nuclear energy possible.

Learn more about nuclear reactions at: brainly.com/question/25387647

#SPJ4

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Calculate the acceleration of an airplane that starts at rest and reaches a speed 45m/s in 9 seconds
sasho [114]

I believe the acceleration would be 5m/s

All you would need to do is divide the final speed by the time it took to get there. I am only about 80 sure this answer is correct, so take my advise only if you feel comfortable.

3 0
3 years ago
Read 2 more answers
A transformer has two sets of coils, the primary with N1 = 160 turns and the secondary with N2 = 1400 turns. The input rms volta
vovikov84 [41]

To solve the problem it is necessary to apply the concepts related to the voltage in a coil, through the percentage relationship that exists between the voltage and the number of turns it has.

So things our data are given by

N_1 = 160

N_2 = 1400

\Delta V_{1rms} = 62V

PART A) Since it is a system in equilibrium the relationship between the two transformers would be given by

\frac{N_1}{N_2} = \frac{\Delta V_{1rms}}{\Delta V_{2rms}}

So the voltage for transformer 2 would be given by,

\Delta V_{2rms} = \frac{N_2}{N_1} \Delta V_{1rms}

PART B) To express the number value we proceed to replace with the previously given values, that is to say

\Delta V_{2rms} = \frac{N_1}{N_2} \Delta V_{1rms}

\Delta V_{2rms} = \frac{1400}{160} 62V

\Delta V_{2rms} = 1446.66V

7 0
3 years ago
A 10 kg migratory swan cruises at 20m/s. A calculation that takes into ac-count the necessary forces shows that this motion requ
IgorC [24]

Answer:

Part A:

Distance=864000 m=864 km

Part B:

Energy Used=ΔE=8638000 Joules

Part C:

\frac{\triangle m}{m}=0.004998=0.49985\%

Explanation:

Given Data:

v=20m/s

Time =t=12 hours

In Secs:

Time=12*60*60=43200 secs

Solution:

Part A:

Distance = Speed**Time

Distance=v*t

Distance= 20*43200

Distance=864000 m=864 km

Part B:

Energy Used=ΔE= Energy Required-Kinetic Energy of swans

Energy Required to move= Power Required*time

Energy Required to move=200*43200=8640000 Joules

Kinetic Energy=\frac{1}{2}mv^2

K.E\ of\ Swans=\frac{1}{2} *10*(20)^2=2000\ Joules

Energy Used=ΔE=8640000 -2000

Energy Used=ΔE=8638000 Joules

Part C:

Fraction of Mass used=Δm/m

For This first calculate fraction of energy used:

Fraction of energy=ΔE/Energy required to move

ΔE is calculated in part B

Fraction of energy=8638000/8640000

Fraction of energy=0.99977

Kinetic Energy=\frac{1}{2}mv^2

Now, the relation between energies ratio and masses is:

\frac{\triangle E}{E}=\frac{\triangle m}{2m}v^2

\frac{\triangle m}{m}=\frac{2}{v^2} *\frac{\triangle E}{E}\\\frac{\triangle m}{m}=\frac{2}{20^2} *0.99977

\frac{\triangle m}{m}=0.004998=0.49985\%

3 0
3 years ago
Calculate the displacement of an object at 2.0 seconds when thrown straight up with an
Licemer1 [7]

Answer:

<h3>30m</h3>

Explanation:

Velocity is the change of rate of displacement with respect to time.

velocity = displacement/time

Given

initial velocity = 15 m/s.

time taken =2 secs

Required

Displacement of the object

From the formula;

Displacement = Velocity * time

Displacement = 15 * 2

Displacement = 30m

<em>Hence the displacement of the object is 30m</em>

8 0
3 years ago
In general, an organism will be more likely to develop phobias of __________.
BARSIC [14]

Answer:

a) dangers faced during natural circumstances

7 0
2 years ago
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