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FrozenT [24]
3 years ago
14

Show how the alternative definition of power, found in your book, can be derived by substituting the definitions of work and spe

ed into the standard
definition of power, P = W /delta t
Physics
2 answers:
Harman [31]3 years ago
8 0

Let us consider body moves a distance S due to the force F.

Hence the work by the body W = FS

If the force is not along the direction of displacement,then the work by a body for travelling a distance S will be -

                                       W=[ Fcos\theta]*S  where    Fcos\theta is the component of the force along the direction of displacement.

                                  Hence\ W= FScos\theta

                                                        = F.S

As per the question the power P is given as -

                                                  P=\frac{W}{\delta t}

                                                         =\frac{F.S}{\delta t}

                                                         = F.\frac{S}{\delta t}

                                                         = \ F.V

Hence alternative definition of power P = F.V


zimovet [89]3 years ago
5 0
Power formula is P= W/ Delta t, or we know that W= saclar product of D and F, D=length of distance traveled, F= the applied force, it is W = //D// //F// cos(D, F), when we substitute, we get P=[ //D// //F// cos (D, F)]/ deltat, or //D// = V x delta t, P= [V x delta t x //F// cos(D, F)] / delta t, and then P= V//F//cos(D, F), or V and D have the same direction, so cos(D, F)= cos(V, F), finally  P= V. F (or scalar product of V and F)
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Determine the energy required to accelerate an electron from (a) 0.500c to 0.900c and (b) 0.900c to 0.990c
Lera25 [3.4K]

Answer:

0.582 MeV

2.45 MeV

Explanation:

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(a) 0.500c to 0.900c

W=\left(\frac{1}{\sqrt{1-\frac{v_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}\right)E_r\\\Rightarrow W=\left(\frac{1}{\sqrt{1-0.9^2}}-\frac{1}{\sqrt{1-0.5^2}}\right)0.511\\\Rightarrow W=0.582\ MeV

Energy required is 0.582 MeV

(b) 0.900c to 0.990c

W=\left(\frac{1}{\sqrt{1-\frac{v_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}\right)E_r\\\Rightarrow W=\left(\frac{1}{\sqrt{1-0.99^2}}-\frac{1}{\sqrt{1-0.9^2}}\right)0.511\\\Rightarrow W=2.45\ MeV

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3 years ago
Plants appear green because they do not absorb the green wavelengths of light. What happens to those green light waves when they
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it is reflected all the other colors such as red is absorbed

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A parallel plate capacitor has a capacitance of 6.78 μF and it is connected to a 12V battery. What is the charge on the capacito
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Answer:

8.136×10⁻⁵ J

Explanation:

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Q = Cv................ Equation 1

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From the question,

Given: C = 6.78μF = 6.78×10⁻⁶ F, v = 12 V

Substitute these values into equation 1

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Answer:

T = 37.5 N

Explanation:

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now when it will reach to the height of the peg then its speed is given as

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v_f = 2.6 m/s

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