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The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the

FrozenT [24]

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

$B=\frac{2\pi*a }{u*I}$

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

$I=\frac{2\pi*a*B }{u}$ Formula(2)

$B=0.2*10^{-5} T = 0.2*10^{-5} \frac{weber}{m^{2} }$

a =8cm=0.08m

$u=4*\pi *10^{-7} \frac{Weber}{A*m}$

We replace the known information in the formula (2)

$I=\frac{2\pi*0.08*0.2*10^{-5} }{4\pi *10x^{-7} }$

I=0.8 A

Answer: The electric current in the wire is 0.8 A

4 0

3 years ago

Joe and Jim start at rest at the starting line of a race track. Both runners accelerate at the same rate, but Joe accelerates fo

eduard

Answer:

Vf₂ = 2 Vf₁

It shows that final speed of Joe is twice the final speed of Jim.

Explanation:

First, we analyze the final speed of Jim by using first equation of motion:

Vf₁ = Vi + at

where,

Vf₁ = final speed of Jim

Vi = initial speed of Jim = 0 m/s

a = acceleration of Jim

t = time of acceleration for Jim

Therefore,

Vf₁ = at ---------------- equation (1)

Now, we see the final speed of Joe. For Joe the parameters will become:

Vf = Vf₂

Vi = 0 m/s

a = a

t = 2t

Therefore,

Vf₂ = 2at

using equation (1):

<u>It shows that final speed of Joe is twice the final speed of Jim.</u>

3 0

3 years ago

The buoyant force on an object fully submerged in a liquid depends on (select all that apply)

Natasha_Volkova [10]

2. The object's volume.
3. The density of the liquid.
Remember what the buoyant force is. It's the lifting force caused by the displacement of a fluid. I'm using the word fluid because it can be either a liquid or gas. For instance a helium balloon floats due to the buoyant force exceeding the mass of the balloon. So let's look at the options and see what's correct.
1. Object's mass
* This doesn't affect the buoyant force directly. It can have an effect if the object's mass is lower than the buoyant force being exerted. Think of a boat as an example. The boat is floating on the top of the water. If cargo is loaded into the boat, the boat sinks further into the water until the increased buoyant force matches the increased mass of the boat. But if the density of the object exceeds the density of the fluid, then increasing the mass of the object will not affect the buoyant force. So this is a bad choice.
2. The object's volume.
* Yes, this directly affects the buoyant force. So this is a good choice.
3. The density of the liquid.
* Yes, this directly affects the buoyant force. You can drop a piece of iron into water and it will sink. You could also drop that same piece of iron into mercury and it will float. The reason is that mercury has a much higher density than water. So this is a good choice.
4. Mass of the liquid
* No. Do not mistake mass for density. As a mental exercise, imagine the buoyant force on a small piece of metal dropped into a swimming pool. Now imagine the buoyant force on that same piece of metal dropped into a lake. In both cases, the buoyant force is the same, yet the lake has a far greater mass of water than the swimming pool. So this is a bad choice.

8 0

3 years ago

A runner whose mass is 54 kg accelerates from a stop to a speed of 7 m/s in 3 seconds. (A good sprinter can run 100 meters in ab

Darya [45]

Answer:

a. $F=126N$

b. $E_K=1323J$

Explanation:

Given:

$m=54kg$

$v=7 m/s$

$t= 3s$

The runner force average to find given the equations

a.

$F=m*a$

$a=\frac{v}{t}$

$F=m*\frac{v}{t}=54kg*\frac{7m/s}{3s}$

$F=126N$

b.

Work done by the system by this force so

$W=F*d$

$W=E_K$

$E_K=\frac{1}{2}*m*v^2$

$E_K=\frac{1}{2}*54kg*(7m/s)^2$

$E_K=1323J$

6 0

3 years ago

On a straight road (taken to be in the x direction) you drive for an hour at 60 km per hour, then quickly speed up to 120 km per

Luda [366]

Answer:

The average velocity is 180 km/hr

Explanation:

Given;

initial velocity, u = 60 km per hour

final velocity, v = 120 km per hour

initial time = 1 hour

final time = 2 hour

Initial position = 60 km/h x 1 hour = 60 km

final position = 120 km/h x 2 hour = 240 km

The average velocity is given by;

$V_{avg} = \frac{Final \ position\ - \ Initial \ position}{final \ time\ - \ initial \ time}\\\\V_{avg} = \frac{240km \ - \ 60km}{2hr\ - \ 1hr} \\\\V_{avg} = \frac{180 \ km}{1hr} \\\\V_{avg}= 180 \ km/hr$

Therefore, the average velocity is 180 km/hr

3 0

3 years ago