What is the resistance of a 600 W kettle that draws a current of 5.0 A? please answer with steps

A parallel-plate vacuum capacitor has 8.60 J of energy stored in it. The separation between the plates is 3.80 mm . If the separ

creativ13 [48]

Answer:

Part a)

$E = \frac{8.60}{2.62} = 3.28 J$

Part b)

$E = 2.62(8.60) = 22.5 J$

Explanation:

As we know that the energy of capacitor when it is not connected to potential source is given as

$U = \frac{Q^2}{2C}$

As we know that initial energy is given as

$8.60 = \frac{Q^2}{2C}$

now we know that capacitance of parallel plate capacitor is given as

$C = \frac{\epsilon_0A}{d}$

now the new capacitance when distance is changed from 3.80 mm to 1.45 mm

$C' = \frac{Cd}{d'}$

$C' = \frac{C(3.80)}{1.45}$

$C' = 2.62 C$

Now the new energy of the capacitor is given as

$E = \frac{Q^2}{2(2.62C)}$

$E = \frac{8.60}{2.62} = 3.28 J$

Part b)

Now if the voltage difference between the plates of capacitor is given constant

now the energy energy of capacitor is

$U = \frac{1}{2}CV^2$

$8.60 = \frac{1}{2}CV^2$

now when capacitance is changed to new value then new energy is given as

$E = \frac{1}{2}C'V^2$

$E = \frac{1}{2}(2.62C)V^2$

$E = 2.62(8.60) = 22.5 J$

6 0

3 years ago

The cable of the 1800kg elevator cab in Fig. 8−56 snaps when the cab is at rest at the first floor, where the cab bottom is a di

drek231 [11]

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

a ) The speed of the cab just before it hits the spring,

Ki + Pi = K final + P final + W

1 / 2 m vo² + m g hi = 1 / 2 m v² + m g h + f d

0 + ( 1800 * 9.8 * 3.7 m ) = ( 1 / 2 * 1800 * v² ) + 0 + ( 4400 * 3.7 )

v = 7.4 m / s

b ) The maximum distance x that the spring is compressed,

Ki + Pi = K final + P final + W + Fs

1 / 2 m vo² + m g x = 1 / 2 m v² + m g h + f d + 1 /2 k x²

( 1/2 * 1800 * 7.4² ) + ( 1800 * 9.8 * x ) =0 + 0+ ( 4400 * x ) + ( 1/2 * 1800 * x² )

75000 x² + 13420 x - 50625 = 0

x = 0.9 m

c ) The distance that the cab will bounce back up the shaft,

Ki + Pi + Fs = K final + P final + W

1 / 2 m vo² + m g x + 1 /2 k x² = 1 / 2 m v² + m g h + f d

0 + 0 + ( 1 / 2 * 0.15 * 0.9² ) = 0 + ( 1800 * 9.8 * h ) + ( 4400 * h )

h = 2.8 m

Therefore,

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

To know more about law of conservation of energy

brainly.com/question/12050604

#SPJ4

3 0

8 months ago

Help with 5 and the question below. BRAINLIEST FOR THE CORRECT ANSWER! Urgent Physics help!

krok68 [10]

Sound level at distance of 15 m is given as 20 dB

so intensity at this distance is given as

$L = 10 Log\frac{I}{I_0}$

$20 = 10 Log{I}{10^{-12}}$

$I_1 = 10^{-10}W/m^2$

now if we move closer to some some distance the sound level is now 50 dB

now the intensity is given as

$L = 10 Log\frac{I}{I_0}$

$50 = 10Log\frac{I}{10^{-12}}$

$I_2 = 10^{-7}W/m^2$

now we know that

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

$\frac{10^{-10}}{10^{-7}} = \frac{r_2^2}{15^2}$

$r_2 = 47 cm$

so now the distance from friend must be 47 cm

8 0

2 years ago

What is the purpose of an experiment design?

Eddi Din [679]

An experimental design is used to assign variables for testing. In contrast to a control design where nothing is changed, the experimental design allows you to test various new inputs to see how they would vary from the original results.

5 0

3 years ago

If an object absorbs all colors but red, we see

julia-pushkina [17]

[I researched for you, since I am not in that particular level to know that knowledge yet. I assure this is accurate info :)]

The answer is A, red.
"Remember, the color you see is light REFLECTING off the surface of that object. If all colors are absorbed in to the surface EXCEPT red, red must be reflected, and you'll see red." - Yahoo User @Chap

8 0

3 years ago