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3 years ago

7

What are the variables in Gay-Lussac's law? pressure and volume pressure, temperature, and volume pressure and temperature volum

e, temperature, and moles of gas

2 answers:

Nataly [62]3 years ago

8 0

Answer:

the answer is C on edge.

Explanation:

Yanka [14]3 years ago

3 0

Answer:

pressure and temperature (assuming volume is constant)

Explanation:

You might be interested in

What is uniform motion ,?

bogdanovich [222]

Answer:

Uniform Motion and Non Uniform Motion

Uniform Motion. In Physics, uniform motion is defined as the motion, wherein the velocity of the body travelling in a straight line remains the same.

Explanation:

5 0

2 years ago

Read 2 more answers

Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is

Xelga [282]

Answer:

$7.3114*10^{-5}V$

Explanation:

To give a solution to the exercise, it is necessary to consider the concepts related to magnetic flux and Faraday's law of induction. Faraday's law states that the voltage induced in a closed circuit is directly proportional to the speed with which the magnetic flux that crosses any surface with the circuit as an edge changes over time.

It is represented under the equation,

$\epsilon = N \frac{\Delta\Phi}{\Delta t}$

Where,

$\epsilon$ is the induced electromotive force

N = Number of loops

$\Delta t$ = Time

$\Delta\Phi$ = Magnetic Flux

For definition the change in magnetic flux is:

$\Delta \Phi = \Delta B A cos\phi$

Where,

B= Magnetic field

Substituting at the first equation we have

$\epsilon = N \frac{\Delta B A Cos\phi}{\Delta t}$

$\epsilon = N \frac{(B_2-B_1) (\pi r^2) Cos\phi}{\Delta t}$

Our values are given by,

N = 1 turn

$B_2 = 1T$

$B_1 = 0T$

r = 1.6mm

$\phi = 0\°$

$\Delta t = 100ms$

Replacing,

$\epsilon = (1) \frac{(1-0) (\pi (1.6*10^{-3})^2) Cos(0)}{110*10^{-3}}$

$\epsilon = 7.311*10^{-5}V$

<em>Therefore the magnitud of the induced emf around a horizontal circle of tissue is $7.3114*10^{-5}V$ </em>

8 0

3 years ago

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.

snow_lady [41]

Complete Question

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)

Answer:

The velocity is $v = 3.79 *10^{5} \ m/s$

Explanation:

From the question we are told that

The magnitude of the electric field is $E = 144 \ kV /m = 144*10^{3} \ V/m$

The magnetic field is $B = 0.38 \ T$

The force due to the electric field is mathematically represented as

$F_e = E * q$

and

The force due to the magnetic field is mathematically represented as

$F_b = q * v * B * sin(\theta )$

Now given that it is perpendicular , $\theta = 90$

=> $F_b = q * v * B * sin(90)$

=> $F_b = q * v * B$

Now given that it is not deflected it means that

$F_ e = F_b$

=> $q * E = q * v * B$

=> $v = \frac{E}{B }$

substituting values

$v = \frac{ 144 *10^{3}}{0.38 }$

$v = 3.79 *10^{5} \ m/s$

7 0

3 years ago

To model this process, assume two charged spherical conductors are connected by a long conducting wire and a 1.20-mC charge is p

il63 [147K]

Answer:

Part a: The electric potential of each sphere is 1.35x10⁸V

Part b: The electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

Explanation:

As the complete question is not given, the similar question is attached herewith. The values are used as indicated in the given question

Let r_1 = 6 cm=0.06 m

r2 = 2 cm = 0.02 m

Q = 1.2 mC

Let q1 and q2 are the charges on each sphere.

q1 + q2 = 1.2 mC -------(1)

In the equilibrium, V1 = V2

k*q1/r1 = k*q2/r2

q1/0.06 = q2/0.02

q1/q2 = 0.06/0.02

q1/q2 = 3 ---------(2)

On solving equation 1 and 2

we get

q1 = 0.9 mC

q2 = 0.3 mC

So

V1 = k*q1/r1 = (9*10^9*0.9*10^-3)/0.06 = 1.35*10^8 Volts

V2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02 = 1.35*10^8 Volts

So the electric potential of each sphere is 1.35x10⁸V

Part b

Now the electric potential is given as

E1 = k*q1/r1^2 = 9*10^9*0.9*10^-3/0.06^2 = 2.25*10^9 N/C

E2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02^2 = 6.75*10^9 N/C

So the electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

7 0

3 years ago

1. Which of the following is an example of deduction? a. Hector calls the weather service to find out if the temperature outside

Ann [662]

Answer:

b

Explanation:

she made a conclusion based off of her own observation, and isnt dumb enough to call a weather station

5 0

3 years ago