All categories

3 years ago

Can someone PLEASE HELP ME!!!!!!!!

Physics

2 answers:

Tamiku [17]3 years ago

8 0

Answer:

Explanation:

sweet [91]3 years ago

7 0

<h3>WATER</h3>

Explanation:

<h2>#CARRYINGTOLEARN:)</h2>

You might be interested in

Light of wavelength 500 nm is incident perpendicularly from air on a film 10-4cm thick and of refractive index 1.375. Part of th

Marysya12 [62]

Answer

given,

wavelength (λ)= 500 n m

thickness of film= 10⁻⁴ cm

refractive index = μ = 1.375

distance traveled is double which is equal to 2 x 10⁻⁴ cm

a) Number of wave

$N = \dfrac{d}{\mu\lambda}$

$N = \dfrac{2 \times 10^{-6}}{1.375\times 500 \times 10^{-9}}$

N = 2.91

N = 3

b) phase difference is equal to

Reflection from the first surface has a 180° (½λ) phase change.

There is no phase change for the 2nd surface reflection and there is no phase difference for the 2nd wave having traveled an exact whole number of waves.

net phase difference = $180^0\times \dfrac{3}{2}$

= 270°

6 0

3 years ago

A 4-79 permalloy solenoid coil needs to produce a minimum inductance of 1.1 . If the maximum allowed current is 4 , how many tur

daser333 [38]

The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,

The magnetic field H is given as,

$H = \frac{nI}{l}$

Here,

n = Number of turns of the coil

I = Current that flows in the coil

l = Length of the coil

From the above equation, the number of turns of the coil is,

$n = \frac{Hl}{I}$

The magnetic field is again given by,

$H = \frac{B}{\mu_t}$

Where the minimum inductance produced by the solenoid coil is B.

We have to obtain n, that

$n = \dfrac{\frac{B}{\mu_t}l}{I}$

Replacing with our values we have that,

$n = \dfrac{\frac{1.1Wb/m^2 }{200000}(2m)}{4mA}$

$n = \dfrac{(\frac{1.1Wb/m^2 }{200000})(\frac{10^4 guass}{1Wb/m^2})(2m)}{4mA(\frac{10^{-3}A}{1mA})}$

$n = 27.5 \approx 28$

Therefore the number of turn required is 28Truns

4 0

3 years ago

Forces are needed to make a car do which of these?

zalisa [80]

All of the above.

if your car is on a steep hill it needs force to stop moving and to speed up

8 0

2 years ago

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 c

seraphim [82]

Answer:

Flow rate 2.34 m3/s

Diameter 0.754 m

Explanation:

Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.

The area at the well head is

$A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2$

So the volume flow rate along the pipe is

$\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s$

We can use the similar logic to find the cross-section area at the refinery

$A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2$

The radius of the pipe at the refinery is:

$A_r = \pi r^2$

$r^2 =A_r/\pi = 0.446/\pi = 0.141$

$r = \sqrt{0.141} = 0.377m$

So the diameter is twice the radius = 0.38*2 = 0.754m

6 0

3 years ago

A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At

yaroslaw [1]

The electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.

To determine σ:

σ = Q/A

Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:

σ = Q/d²

Make this substitution in the equation for E:

E = Q/(2ε₀d²)

We see that E is inversely proportional to the square of d:

E ∝ 1/d²

The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:

$E_{new} = E/4$

4 0

2 years ago