HELP URGENT!!!!!!!!!!!!!!

How did some U.S. citizens oppose the Vietnam war?

creativ13 [48]

They protested by marching in the streets.

3 0

3 years ago

What will be the pressure exerterd by the 0bject if 4000n is acting on an area of 50msqure

Ivahew [28]

<h3>Given, </h3>

Force,F = 4000 N

Area,a = 50 m²

Pressure = Force/Area

★ Putting the values in the above formula,we get:

$\sf \rightarrow \: pressure = \dfrac{4000}{50}$

$\sf \rightarrow pressure = 80 \: N {m}^{ - 2}$

7 0

3 years ago

An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are inv

PolarNik [594]

Complete Question

An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are investigating the environmental effects of the accident and need to know the area of the spill. The tanker captain informs you that 18000 liters of oil have escaped and that the oil has an index of refraction of n = 1.1. The index of refraction of the ocean water is 1.33. From the deck of your ship you note that in the sunlight the oil slick appears to be blue. A spectroscope confirms that the dominant wavelength from the surface of the spill is 485 nm. Assuming a uniform thickness, what is the largest total area oil slick

Answer:

The largest total area of the oil slick $A = 8.257 *10^{9} \ m^2$

Explanation:

From the question we are told that

The volume of oil the escaped is $V = 18000 \ L$

The refractive index of oil is $n_o = 1.1$

The refractive index of water is $n_w = 1.33$

The wavelength of the light is $\lambda = 485 \ nm = 485 * 10^{-9} \ m$

Generally the thickness of the oil for condition of constructive interference between the oil and the water is mathematically represented as

$d = m *\frac{\lambda}{2n_w}$

Where is the order of interference of the light and it value ranges from 1, 2, 3,...n

It is usually take as 1 unless stated otherwise by the question

substituting value

$d = 1 * \frac{485 *10^{-9}}{2 * 1.1}$

$d = 218 nm$

The are can be mathematically evaluated as

$A = \frac{V}{d}$

Substituting values

$A = \frac{18000}{218*10^{-8}}$

$A = 8.257 *10^{9} \ m^2$

6 0

3 years ago

Solve for the length of the inclined plane if the angle equals 19.45 degrees.

mel-nik [20]

The length of the inclined plane is approximately 12 ft

The situation forms a right angle triangle.

<h3>Right triangle</h3>

Right triangle have one of its angle as 90 degrees.

Therefore,

The length of the inclined plane is the hypotenuse of the triangle. The length of the inclined plane can be found using trigonometric ratios.

height = 4 ft

angle(∅) = 19.45°

sin 19.45 = 4 / h

h = 4 / 0.33298412235

h = 12.0125847796

h = 12 ft

Therefore, the length of the inclined plane is approximately 12 ft

learn more on inclined plane:brainly.com/question/14163589?referrer=searchResults

5 0

2 years ago

A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$

4 0

3 years ago