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Luda [366]
3 years ago
7

A battery establishes a voltageV on a parallel-plate capacitor. After the battery is disconnected, the capacitor doesn't loss ch

arge however the distance between the plates is doubled and the area of the and the voltage between the plates plates is quadrupled. Accordingly, the capacitance
A. increases; decreases.
B. decreases; increases.
C. increases; increases.
D. decreases; decreases
E. stays the same; stays the same.
Physics
1 answer:
Digiron [165]3 years ago
7 0

Answer:

E

Explanation:

I me and my friends did a project on this

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Three boards, each 50 mm thick, are nailed together to form a beam that is subjected to a 1200-N vertical shear. Knowing that th
lisabon 2012 [21]

Answer:

Using equation

qs=F

q=VQ/I   V=1200

q=998.45 N

qs=F

s=F/q

s=700/998.45

spacing=s=0.701 m

7 0
3 years ago
Read 2 more answers
(50 POINTS) please answer each question:
Advocard [28]

Answer:

i know the first one.

Explanation:

the distance from earth in light years is 6 trillion miles.

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3 years ago
Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo
Elena-2011 [213]

a) 3.14 \cdot 10^{-4} s

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

a)

The position of the tip of the lever at time t is described by the equation:

y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t] (1)

The generic equation that describes a wave is

y(t)=A sin (\frac{2\pi}{T} t) (2)

where

A is the amplitude of the wave

T is the period of the wave

t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}

Therefore, the period is

T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s

b)

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when sin(...)=1, and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

c)

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

\lambda=2L

where

\lambda is the wavelength of the wave

L is the length of the string

In this problem,

L = 5.00 m is the length of the string

Therefore, the wavelength is

\lambda =2(5.00)=10.0 m

d)

The amplitude of a wave is the magnitude of the maximum displacement of the wave, measured relative to the equilibrium position.

In this problem, we can easily infer the amplitude of this wave by looking at eq.(1).

y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]

And by comparing it with the general equation of a wave:

y(t)=A sin (\frac{2\pi}{T} t)

In fact, the maximum displacement occurs when the sine part is equal to 1, so when

sin(\frac{2\pi}{T}t)=1

which means that

y(t)=A

And therefore in this case,

y=0.500 cm

So, this is the displacement.

6 0
3 years ago
Two charges repel each other with a force. If the magnitude of both charges double, but the distance between them remains consta
-Dominant- [34]

Answer:

C.

Explanation:

4 0
3 years ago
In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.8 m/s at an
tensa zangetsu [6.8K]
The equation to be used here is the trajectory of a projectile as written below:

y = xtanθ +/- gx²/2v²(cosθ)²
where
y is the vertical distance
x is the horizontal distance
θ is the angle of trajectory or launch angle
g is 9.81 m/s²
v is the initial velcity

Since the angle is below horizontal, let's use the minus equation. Substituting the values:

- 0.8 m = xtan15° - (9.81 m/s²)x²/2(4.8 m/s)²(cos15°)²
Solving for x,
x = 2.549 m

However, we only take half of this distance because it was specified that the distance asked before bouncing. Hence, the horizontal distance is equal to 1.27 m.
5 0
3 years ago
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