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Leno4ka [110]
3 years ago
10

You have two square metal plates with side lengths of (6.50 C) cm. You want to make a parallel-plate capacitor that will hold a

charge of (12.5 A) nC when connected to a (34.8 B) V potential difference. Determine the necessary separation in mm. Round your answer to three significant figures.
Physics
1 answer:
gtnhenbr [62]3 years ago
6 0

Answer:

The necessary separation between  the two parallel plates is 0.104 mm

Explanation:

Given;

length of each side of the square plate, L = 6.5 cm = 0.065 m

charge on each plate, Q = 12.5 nC

potential difference across the plates, V = 34.8 V

Potential difference across parallel plates is given as;

V = \frac{Qd}{L^2 \epsilon_o} \\\\d = \frac{V L^2 \epsilon_o}{Q}

Where;

d is the separation or distance between the two parallel plates;

d = \frac{VL^2 \epsilon_o}{Q} \\\\d =  \frac{34.8*(0.065)^2 *8.854*10^{-12}}{12.5*10^{-9}} \\\\d = 0.000104 \ m\\\\d = 0.104 \ mm

Therefore, the necessary separation between  the two parallel plates is 0.104 mm

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A 2000 kg car accelerates from 0 to 25 m / s in 21.0 s. How much is the average power delivered by the motor? (1 hp = 746 W)
mash [69]

Answer: 40.2 hp

Explanation:

The mass of the car is 2000kg.

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We know that force = mass*acceleration, then:

F = 2000kg*1.2m/s^2 = 2400N.

Now, the Power can be written as:

P = W/t where W is work = F*d (force per distance) and t is time.

then we have:

P = (F*d)/t and d/t  is the average velocity, and we know the velocity v = 25m/s, so the average velocity will be (1/2)*25m/s = 12.5m/s

P = F*12.5m/s

and the force is F = 2400N

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and we know that 1hp = 746 W

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A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800.0 N on him. Th
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Answer:

a) F_{fric} = 692 N

b) F_{applied} = 932 N

Explanation:

a)

According to newton's second law of motion, acceleration of an object is directly proportional to the net force acting on it. When there is no net force force acting on the body, there is no acceleration. A force is a push or a pull, and the net force ΣF is the total force, or sum of the forces exerted on an object  in all directions.

F_{net}  ∝ a

F_{net} =  ma

F_{applied} - F_{fric} = ma

Given data:

F_{applied} = 800 N

Mass = m = 90 kg

acceleration = a = 1.2 m/s²

F_{fric} = ?

800 - F_{fric} = (90)(1.2)

F_{fric} = 692 N

b)

According to newton's second law of motion,

F_{net}  ∝ a

F_{net} =  ma

F_{applied} - F_{fric} = ma

Given data:

If we assume the same friction and acceleration between player's feet and ground as calculated in part a

F_{fric} = 692 N

acceleration = a = 1.2 m/s²

We take the equal mass to the total mass of both the players because when the winning player push losing player backward, he exert force on the ground not only due to his mass but also due to the mass of losing player.

Mass = M = m₁ + m₂ = 110 kg + 90 kg

= 200 kg

F_{applied} = ?

F_{applied} - 692 N = (200)(1.2)

F_{applied} = 692 + 240

F_{applied} = 932 N

7 0
3 years ago
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