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Anastaziya [24]
3 years ago
8

Define power and discuss how to determine it.

Physics
1 answer:
serious [3.7K]3 years ago
4 0

there are different types of power so ill show you all the examples of power.


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Which type of graph is best suited to show the breakdown of how this teacher weights each category ?
MArishka [77]

Answer:

pie graph?

Explanation:

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3 years ago
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If chilled coke and hot tea are <br> kept together tea cools down but ko gets warm why<br>​
olga55 [171]
Tea gets cool beacuse of heat tranfer through convenction
Heat transfers from a area of hot region to cold.
The tea is hot in this case, due to taht factor the heat will move towards teh cooler region which is coke and will make the coke warm


Hope this helped!
5 0
3 years ago
A car accelerates from rest for 8.0 s, and reaches a speed of 66 m/s. How far will the
PSYCHO15rus [73]

Answer: C

Explanation:

Find the acceleration using this kinematic equation:

v_f=v_i+at\\66=0+a*8\\a=8.25m/s^2

Now use this kinematic equation to find the displacement:

v_f^2=v_i^2+2a*d\\66^2=0^2+2(8.25)d\\4356=16.5d\\d=264m

3 0
3 years ago
Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude
DENIUS [597]

Answer:

The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

Explanation:

Given;

initial velocity of proton, v_p_i = 3 x 10⁵ m/s

distance moved by the proton, d = 3.5 m

electric field strength, E = 120 N/C

The kinetic energy of the proton at the end of the motion is calculated as follows.

Consider work-energy theorem;

W = ΔK.E

W =K.E_f - K.E_i

where;

K.Ef is the final kinetic energy

W is work done in moving the proton = F x d  = (EQ) x d = EQd

K.E_f =EQd + \frac{1}{2}m_pv_p_i^2

m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C

K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5   \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\

K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J

Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

3 0
3 years ago
Anyone know a GOOD show on Netflix please kid shows pleaseeeee
Afina-wow [57]

Answer:

like horror? or action haha

Explanation:

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3 years ago
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