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3 years ago

Good evening everyone, State the law of conservation of energy .Thank u

Physics

2 answers:

miskamm [114]3 years ago

5 0

Answer:

The law of conservation of energy states that energy can neither be created, nor destroyed only converted from one form of energy to another. This means a system always has the same amount of energy unless it is added from the outside.

Explanation:

Hope this helps :)

disa [49]3 years ago

3 0

Energy can neither be created nor be destroyed but can be converted from one form to another .

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A metal sphere of radius 11 cm has a net charge of 2.8 × 10–8 C. (a) What is the electric field at the sphere's surface? (b) If

harina [27]

Answer:

Therefore,

a) $E=20.82\ kN/C$

b) $V= 2290\ Volt$

c) distance from the sphere's surface = 1.8 cm

Explanation:

Given:

Radius, r = 11 cm = 0.11 m

Charge, $Q = 2.8\times 10^{-8}\ C$

To Find:

a) electric field at the sphere's surface = ?

b) If V = 0 at infinity, what is the electric potential at the sphere's surface = ?

c) At what distance from the sphere's surface has the electric potential decreased by 320 V = ?

Solution:

Electric field at the sphere's surface is given as,

$E=\dfrac{k\times Q}{r^{2}}$

Where,

E = Electric Field,

$k = Coulomb's\ constant = 9\times 10^{9}$

Q = Charge

r = Radius

Substituting the values we get

$E=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{(0.11)^{2}}=20.82\ kN/C$

Now, Electric Potential at point surface is given as,

$V=\dfrac{k\times Q}{r}$

Substituting the values we get

$V=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{0.11}=2.29\ kV$

For distance from the sphere's surface has the electric potential decreased by 320 V,

So V becomes,

V = 2290 - 320 = 1970 Volt, then r =?

$V=\dfrac{k\times Q}{r}$

Substituting the values we get

$r=\dfrac{k\times Q}{V}\\\\r=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{1970}=0.128\ m$

Therefore the distance from the sphere's surface,

$d = 12.8 - 11 = 1.8\ cm$

Therefore,

a) $E=20.82\ kN/C$

b) $V= 2290\ Volt$

c) distance from the sphere's surface = 1.8 cm

7 0

4 years ago

At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average power of approximately 3.00

alexira [117]

Energy of a wave:
E = nhc/λ
3000 = (n x 6.63 x 10⁻³⁴ x 3 x 10⁸)/(510 x 10⁻⁹)
n = 7.69 x 10 ²¹ photons per second per meter²
2.70 cm² = 2.70/10,000 m²
= 2.7 x 10⁻⁴
Photons per second = 7.69 x 10 ²¹ x 2.7 x 10⁻⁴
= 2.08 x 10¹⁸ photons per second

4 0

3 years ago

Classical conditioning requires _____.

LekaFEV [45]

I would say A classical conditioning definitely demands a pairing of two stimuli
hope i helped

5 0

3 years ago

Read 2 more answers

Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven t

Bond [772]

Answer:

$\frac{D_{jet}}{D_{prop}}=2.865$

Explanation:

Given data

Speed of jet Vjet=1190 km/h

Speed of prop driven Vprop=595 km/h

Height of jet 7.5 km

Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

$D=\frac{1}{2}CpAv^2\\$

The ratio between the drag force on the jet to the drag force on prop-driven transport is then given by:

$\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\ \frac{D_{jet}}{D_{prop}}=2.865$

4 0

3 years ago

A tuning fork of frequency 254 Hz and an open orang pipe of slightly lower frequency are at 15oC. When

katovenus [111]

The temperature of the air in the open orang pipe has been altered by 18.73° C

The frequency of an open orang pipe is estimated by using the formula:

$\mathbf{f = \dfrac{v}{2L}}$

Then, the combination of the frequency of the tuning fork and the open orang pipe is:

$\mathbf{254 - \dfrac{v}{2L} }$

These combinations of frequency produce 4 beats per sound.

i.e.

$\mathbf{254 - \dfrac{v}{2L} =4}$

$\mathbf{ \dfrac{v}{2L} = 254-4 }$

$\mathbf{ \dfrac{v}{2L} = 250 ----(1)}$

When it is altered, the beats first diminish and increase again by 4.

i.e.

$\mathbf{ \dfrac{v'}{2L} = 254+4 }$

$\mathbf{ \dfrac{v'}{2L} = 258 --- (2) }$

If we equate both equations (1) and (2) together, we have:

$\mathbf{\dfrac{v'}{v}= \dfrac{258}{250}}$

However, from our previous knowledge, we understand that the velocity of an object varies directly proportional to the square root of its temperature.

Hence;