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DaniilM [7]
3 years ago
6

When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is direc

ted:_____
a. into the page.
b. toward the left
c. toward the right
d. toward the bottom of the page.
e. toward the top of the page.
f. out of the page.
Physics
1 answer:
Serga [27]3 years ago
8 0

Answer: F

Out of the page.

Explanation:

For an electron with a charge of -e, the magnitude of the force on it is F = BeV

Where

F = force on the electron

e = charge ( electrons )

V = velocity

B = magnetic field

F is the force acting on all the electrons in a wire which gives rise to the F = BIL

Where

I = current

L = length of the wire

The force F is always at the right angle to the particle's velocity and its direction can be found using the left hand rule.

When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed out of the page.

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A) In the case of the Boundary Thickness Layer we use the given formula,

\delta = \frac{4.91x}{\sqrt{Re}}

We know as well that,

Re = Número de Reynolds = \frac{U*x}{\upsilon}

Where,

U = velocity

\upsilon = kinematic viscosity

For water, kinematic viscosity, \upsilon = 1.21*10^{-5} ft^2 /s

So, 500,000 = \frac{ 17x}{(1.21*10^{-5})}

x = 0.355 ft

d = \frac{4.91*0.355}{\sqrt {500000}}

d = 0.002465 ft = 0.029in

B) For flat plate boundary layer. Given the Critical Reynolds Number.= 5*10^5 we know that is equal to Re above.

Thus, x = 0.355 ft

C. Wall shear stress,

\tau = \mu*\sqrt{ U^3 / (2*\nu*x) }

For water, dynamic viscosity, \nu = 2.344*10^-5 lbf-s/ft^2

\tau = 2.344*10^-5 \sqrt {17^3 / (2*1.21*10^{-5}*0.355)}

\tau = 0.5605 lbf/ft^2

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3 years ago
What is the correct definition of a fracture?
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The correct definition of a fracture is break in the bone

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A closely wound search coil has an area of 3.21 cm2, 120 turns, and a resistance of 58.7 O. It is connected to a charge-measurin
Alexxx [7]

Answer:

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

\epsilon = \frac{NBA}{t}

Where,

\epsilon  =electromotive force

N = Number of loops

B = Magnetic field

A = Area

t= Time

For Ohm's law we now that,

V = IR

Where,

I = Current

R = Resistance

V = Voltage (Same that the electromotive force at this case)

In this system we have that the resistance in series of coil and charge measuring device is given by,

R = R_c + R_d

And that the current can be expressed as function of charge and time, then

I = \frac{q}{t}

Equation Faraday's law and Ohm's law we have,

V = \epsilon

IR = \frac{NBA}{t}

(\frac{q}{t})(R_c+R_d) = \frac{NBA}{t}

Re-arrange for Magnetic Field B, we have

B = \frac{q(R_c+R_d)}{NA}

Our values are given as,

R_c = 58.7\Omega

R_d = 45.5\Omega

N = 120

q = 3.53*10^{-5}C

A = 3.21cm^2 = 3.21*10^{-4}m^2

Replacing,

B = \frac{(3.53*10^{-5})(58.7+45.5)}{120*3.21*10^{-4}}

B = 0.095T

Therefore the magnetic field in the System is 0.095T

3 0
3 years ago
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