Which law states that if the temperature of a gas is held

The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi

Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

$\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$

Where,

B= Magnetic Field

l = length

$\mu_0$ = Vacuum permeability

$\epsilon_0$ = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

$B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}$

Recall that the speed of light is equivalent to

$c^2 = \frac{1}{\mu_0 \epsilon_0}$

Then replacing,

$B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}$

$B = \frac{r}{2C^2} \frac{dE}{dt}$

Our values are given as

$dE = 2150N/C$

$dt = 5s$

$C = 3*10^8m/s$

$D = 0.440m \rightarrow r = 0.220m$

Replacing we have,

$B = \frac{r}{2C^2} \frac{dE}{dt}$

$B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}$

$B =5.25*10^{-16}T$

Therefore the magnetic field around this circular area is $B =5.25*10^{-16}T$

3 0

3 years ago

You are making a round trip from City A to City B and back to City A again at constant speed. At what point in the trip is your

MA_775_DIABLO [31]

Answer:

Halfway between B and A on the return leg.

Explanation:

Your average SPEED for the entire trip will equal your constant speed as the time and distance increase at proportionate rates.

Your average VELOCITY will equal your constant speed while you travel from A to B because time and displacement are increasing at proportionate rates.

When you turn around at B to return, your Displacement is now decreasing while your travel time continues to increase, so your average velocity decreases.

Lets say the distance from A to B is 90 km and your constant speed is 30 km/hr.

your average speed is 30 km/hr because you took 6 hrs to travel 180 km

We want to find your position when your average velocity is 30/3 = 10 km/hr

it took 3 hrs to go 90 km from A to B. Let t be the time lapsed since turn around

your displacement is given by d = 90 - 30(t)

and your total time of travel is t + 3 hrs

v = d/t

10 = (90 - 30t) / (t + 3)

10(t + 3) = (90 - 30t)

10t + 30 = 90 - 30t

40t = 60

t = 1.5 hrs

This will occur when you are halfway between B and A

3 0

2 years ago

At the 1 s mark, which objects have equal momentum?

Rudik [331]

C. The bowling ball and the bicycle

p = mv
P of bike = 12x5 = 60
P of rock = 2x20 = 40
P of ball = 5 x 10 = 60

3 0

3 years ago

You observe a hockey puck of mass 0.12 kg, traveling across the ice at speed 18.3 m/sec. The interaction of the puck and the ice

Galina-37 [17]

The stopping distance is 143.1 m

Explanation:

First of all, we have to find the acceleration of the hockey puck. This can be done by using Newton's second law of motion:

$\sum F =ma$

where

$\sum F = F_f = -0.14 N$ is the net force acting on the puck (the force of friction, negative because it acts in a direction opposite to the direction of motion)

m = 0.12 kg is the mass of the puck

a is the acceleration

Solving for a,

$a=\frac{\sum F}{m}=\frac{-0.14}{0.12}=-1.17 m/s^2$

The motion of the puck is a uniformly accelerated motion, therefore we can use the following suvat equation:

$v^2-u^2=2as$

where:

v = 0 is the final velocity (the puck comes to a stop)

u = 18.3 m/s is the initial velocity

$a=-1.17 m/s^2$ is the acceleration

s is the stopping distance

And solving for s, we find

$s=\frac{v^2-u^2}{2a}=\frac{0-(18.3)^2}{2(-1.17)}=143.1 m$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

6 0

3 years ago

What is the mission of the mars Land Rover, and why is it important?

GaryK [48]

Primarily to survey and map the landscape of Mars for future missions. It also takes core samples and searches for water underground. This is important for many reasons, but most obviously for the upcoming missions to mars and the colonies that follow. Hope this helps?

4 0

3 years ago