Please help on this one?

A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it

Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface $\Phi$ it is possible to calculate the magnitude $\frac{\Phi}{A}=B\ cos \theta$ .

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux $\Phi$ is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is ( $m^{2}$ ). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with $\theta$ being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

$\Phi=B\ A\ cos\theta$

We are told the values of $\Phi$ and B, then we can calculate the magnitude

$\frac{\Phi}{A}=B\ cos\theta$

3 0

2 years ago

What new tool did geologists use to estimate the changing flow rates of lava over the duration of this eruption? Hint: The answe

Vladimir [108]

Answer:

Option b, pothographs from drones.

Explanation:

the USGS (U.S. Geological Survey) decided to make photographic captures from drones to the volcanic surfaces, which allowed through observations to understand things like the characteristics of the lava, the height of the volcanic plumes (among others).

Podemos ver en el siguiente enlace un ejemplo de fotografía tomada desde un dron al Kilauea.

https://www.usgs.gov/media/images/k-lauea-volcano-drone-over-lava-channel

4 0

2 years ago

Which metals have similar chemical properties? Check all that apply.

Alika [10]

Answer:

yes

Explanation:

3 0

2 years ago

Read 2 more answers

We wrap a light, nonstretching cable around a 8.00 kg solid cylinder with diameter of 30.0 cm. The cylinder rotates with negligi

antoniya [11.8K]

Answer:

h = 16.67m

Explanation:

If the kinetic energy of the cylinder is 510J:

$Kc=510=1/2*Ic*\omega c^2$

$\omega c=\sqrt{510*2/Ic}$

Where the inertia is given by:

$Ic=1/2*m_c*R_c^2=1/2*(8)*(0.15)^2=0.0225kg.m^2$

Replacing this value:

$\omega c=106.46rad/s$

Speed of the block will therefore be:

$V_b=\omega_c*R_c=106.46*0.15=15.969m/s$

By conservation of energy:

Eo = Ef

Eo = 0

$Ef = 510+1/2*m_b*V_b^2-m_b*g*h$

So,

$0 = 510+1/2*m_b*V_b^2-m_b*g*h$

Solving for h we get:

h=16.67m

3 0

3 years ago

You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, w

Vesna [10]

Answer:

$\ \text{m/s}$

Explanation:

$u_1$ = Velocity of one lump = $3x+3y-3z$

$u_2$ = Velocity of the other lump = $-4x+0y-4z$

m = Mass of each lump = $30\ \text{g}$

The collision is perfectly inelastic as the lumps stick to each other so we have the relation

$mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=\ \text{m/s}$

The velocity of the stuck-together lump just after the collision is $\ \text{m/s}$ .

4 0

2 years ago