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GaryK [48]
3 years ago
7

A parallel-plate capacitor with plates of area 540 cm2 is charged to a potential difference V and is then disconnected from the

voltage source. When the plates are moved 0.6 cm farther apart, the voltage between the plates increases by 100 V. What is the charge Q on the positive plate of the capacitor?
Physics
1 answer:
notsponge [240]3 years ago
5 0

Answer: 0.8 nC

Explanation: In order to explain this problem we have top use the expression for the capacitor of parallel plates, which is given by:

C=εo*A/d where A and d are the area and the separation of the plates.

Then we have

V=Q/C so ΔV= Q* Δd/(εo*A)

Q= 100* 0.054*8.85*10^-12/( 0.06)=0.8 nC

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he "escape velocity from Earth (the speed required to escape Earth's gravity) is 2.5 x 10 miles per hour. What is this speed in
Natasha_Volkova [10]

Answer: 11.17\ \text{ m/s}

Explanation:

Given : The escape velocity : v=2.5\times10\text{ miles per hour}

We know that 1 mile = 1609 meters  (approx)

and 1 hour= 3600 seconds

To convert escape velocity 2.5 x 10 miles per hour into m/s , we need to multiply it by 1609.34 and divide it by 3600.

Thus, the escape velocity in m/s is given by :-

v=2.5\times10\times\dfrac{1609}{3600}\\\\=11.1736111111\approx11.17\text{ m/s}

Hence, the speed in m/s = 11.17

4 0
3 years ago
Can someone help with these?
Flauer [41]

6.

Basically, all of these questions use F = ma

m = 80 kg

a = 1.62

Weight = F = 80*1.62 = 129.6

The closest answer is B

7.

<em><u>Step One</u></em>

Find the mass here on earth

m = F/a

F = 40 kg

a = 9.2

F = 40/9.2 = 4.34

Now take everything to the moon

F = 20 kg

a = ??

m = 4.34

a = F/m = 20/4.34 = 4.6 m/s^2

8

m = 40 kg

F = 20 N

a = ??

a = F/m = 20N/40kg = 1/2 m/s^2

Comment

All of these depend on F = m*a.  None but the first one talk about vertical forces where gravity would play a part. Moving horizontally means that there is no gravitational force if there is no friction.  a = 9.8 has nothing to do with the problem.

4 0
3 years ago
A 5.0-kg object is pulled along a horizontal surface at a constant speed by a 15-n force acting 20° above the horizontal. How mu
puteri [66]

Answer:

84.6 J

Explanation:

The work done by the force is given by

W=Fd cos \theta

where

W is the work done

F = 15 N is the force applied

d = 6.0 m is the displacement

\theta=20^{\circ} is the angle between the force's direction and the displacement

Substituting the numbers into the equation, we find

W=(15 N)(6.0 m)cos 20^{\circ} =84.6 J

3 0
3 years ago
Air that enters the pleural space during inspiration but is unable to exit during expiration creates a condition called a. open
sertanlavr [38]

Answer:

The correct answer is d. tension pneumothorax.

Explanation:

The increasing build-up of air that is in the pleural space is what we call the tension pneumothorax and this happens due to the lung laceration that lets the air to flee inside the pleural space but it does not return.

4 0
3 years ago
Part C: Quantitative Problems when vf is not 0
Alina [70]

Answer:

(a)

\triangle v=-8\ m/s\\\triangle mv=-56\ kg.m/s

(b)

1120 N

Explanation:

Change in velocity, \triangle v is given by subtracting the initial velocity from the final velocity and expressed as \triangle v= v_f -v_i

Where v represent the velocity and subscripts f and i represent final and initial respectively. Since the ball finally comes to rest, its final velocity is zero. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by

\triangle v=0-8=-8\ m/s

To find m\triangle v then we substitute 7 kg for m and -8 m/s for \triangle v therefore \triangle\ v=7 Kg\times -8 m/s=-56\ Kg.m/s

(b)

The impact force, F is given as the product of mass and acceleration. Here, acceleration is given by dividing the change in velocity by time ie

a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}

Substituting t with 0.05 s then a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}=\frac {-8}{0.05}=-160 m/s^{2}

Since F=ma then substituting m with 7 Kg we get that F=7*-160=-1120 N

Therefore, the impact force is equivalent to 1120 N

3 0
3 years ago
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