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3 years ago

Workin

Physics

1 answer:

max2010maxim [7]3 years ago

8 0

Answer:

Soru okunmuyor keşke fotoğrafını çekip atsaydın öyle atarsan sorunu çözerim

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Help me, i don't know the answer

8_murik_8 [283]

Explanation:

A) Bb BB

B) 50%

A) 50%

B) b. b

B. Bb. Bb

b. bb. bb

4 0

3 years ago

If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is

leva [86]

Answer:

31.831 Hz.

Explanation:

Given:

$\rm y = 0.02\sin(30x-200 t).$

The vertical displacement of a wave is given in generalized form as

$\rm y = A\sin(kx -\omega t).$

where,

A = amplitude of the displacement of the wave.
k = wave number of the wave = $\dfrac{2\pi }{\lambda}.$
$\lambda$ = wavelength of the wave.
x = horizontal displacement of the wave.
$\omega$ = angular frequency of the wave = $\rm 2\pi f$ .
f = frequency of the wave.
t = time at which the displacement is calculated.

On comparing the generalized equation with the given equation of the displacement of the wave, we get,

$\rm A=0.02.\\k=30.\\\omega =200.\\$

therefore,

$\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.$

It is the required frequency of the wave.

3 0

3 years ago

________ is the average kinetic energy of each atom. radiation temperature potential

solong [7]

I think it's temperature

7 0

3 years ago

Un tubo cilindrico hueco de cobre mide 3 m de longitud tienen un diametro exterior de 4cm y un diametro interior de 2 cm¿cuanto

Irina-Kira [14]

Answer:

W = 9.93 10² N

Explanation:

To solve this exercise we must use the concept of density

ρ = m / V

the tabulated density of copper is rho = 8966 kg / m³

let's find the volume of the cylindrical tube

V = A L

V = π (R_ext ² - R_int ²) L

let's calculate

V = π (4² - 2²) 10⁻⁴ 3

V = 1.13 10⁻² m³

m = ρ V

m = 8966 1.13 10⁻²

m = 1.01 10² kg

the weight of the tube

W = mg

W = 1.01 10² 9.8

W = 9.93 10² N

4 0

3 years ago

A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the

Sergio039 [100]

Answer:

W = 1.06 MJ

Explanation:

- We will use differential calculus to solve this problem.

- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.

- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.

- Now develop and expression of Force required:

F = p*V*g

F = 1000*(2*0.5*x*8*dx)*g

F = 78480*x*dx

- Now, the work done is given by:

W = F.s

- Where, s is the distance from top of hose to the differential volume:

s = (5 - x)

- We have the work as follows:

dW = 78400*x*(5-x)dx

- Now integrate the following express from 0 to 3 till the tank is empty:

W = 78400*(2.5*x^2 - (1/3)*x^3)

W = 78400*(2.5*3^2 - (1/3)*3^3)

W = 78400*13.5 = 1058400 J

5 0

3 years ago