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8_murik_8 [283]
3 years ago
12

Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the

m has charge −q. What is the magnitude of the electric force on the −q charge due to the three q charges?

Physics
1 answer:
nydimaria [60]3 years ago
6 0

Answer:F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

F_1=\frac{kq(-q)}{(L\sqrt{2})^2}

where  L\sqrt{2}=Distance between the two charges

F_1=-\frac{kq^2}{2L^2}

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

F_2=\frac{kq(-q)}{(L)^2}

The magnitude of force by both the  charge is same but at an angle of 90^{\circ}

thus combination of two forces at 2 and 3 will be

F'=\sqrt{2}\frac{kq^2}{2L^2}

Now it will add with force due to 1 charge

Thus net force will be

F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

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Answer:

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An internal combustion engine uses fuel, of energy content 44.4 MJ/kg, at a rate of 5 kg/h. If the efficiency is 28%, determine
Alika [10]
Efficiency =  Power Output / Power Input

Power Input = Rate of Energy input = 44.4 MJ/kg * 5 kg/h

 = 222 MJ/h

But 1 hour = 3600seconds

222 MJ/h = 222 MJ/3600s =  0.061667 MW              J/s = Watts

Power input = 0.061667 MW = 61 667 W

From  Efficiency =  Power Output / Power Input

   28% =  Power Output / 61667

   Power Output = 0.28 * 61667

   Power Output = 17266.76 W

    Power Output ≈ 17 267 W

   Rate of heat rejection = Power Input - Power Output

                                   =  61667 - 17267 = 44400 W

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The 100-m dash can be run by the best sprinters in 10.0 s. A 66-kg sprinter accelerates uniformly for the first 45 m to reach to
irga5000 [103]

(a) 154.5 N

Let's divide the motion of the sprinter in two parts:

- In the first part, he starts with velocity u = 0 and accelerates with constant acceleration a_1 for a total time t_1 During this part of the motion, he covers a distance equal to s_1 = 45 m, until he finally reaches a velocity of v_1 = u + a_1t_1. We can use the following suvat equation:

s_1 = u t_1 + \frac{1}{2}a_1t_1^2

which reduces to

s_1 = \frac{1}{2}a_1 t_1^2 (1)

since u = 0.

- In the second part, he continues with constant speed v_1 = a_1 t_1, covering a distance of d_2 = 55 m in a time t_2. This part of the motion is a uniform motion, so we can use the equation

s_2 = v_1 t_2 = a_1 t_1 t_2 (2)

We also know that the total time is 10.0 s, so

t_1 + t_2 = 10.0 s\\t_2 = (10.0-t_1)

Therefore substituting into the 2nd equation

s_2 = a_1 t_1 (10-t_1)

From eq.(1) we find

a_1 = \frac{2s_1}{t_1^2} (3)

And substituting into (2)

s_2 = \frac{2s_1}{t_1^2}t_1 (10-t_1)=\frac{2s_1}{t_1}(10-t_1)=\frac{20 s_1}{t_1}-2s_1

Solving for t,

s_2+2s_1=\frac{20 s_1}{t_1}\\t_1 = \frac{20s_1}{s_2+2s_1}=\frac{20(45)}{55+2(45)}=6.2 s

So from (3) we find the acceleration in the first phase:

a_1 = \frac{2(45)}{(6.2)^2}=2.34 m/s^2

And so the average force exerted on the sprinter is

F=ma=(66 kg)(2.34 m/s^2)=154.5 N

b) 14.5 m/s

The speed of the sprinter remains constant during the last 55 m of motion, so we can just use the suvat equation

v_1 = u +a_1 t_1

where we have

u = 0

a_1  =2.34 m/s^2 is the acceleration

t_1 = 6.2 s is the time of the first part

Solving the equation,

v_1 = 0 +(2.34)(6.2)=14.5 m/s

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3 years ago
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