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3 years ago

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A 100-m-long wire carrying a current of 4.0 A will be accompanied by a magnetic field of what strength at a distance of 0.050 m

from the wire? (magnetic permeability in empty space µ0 = 4π ´ 10-7 T×m/A)

1 answer:

solong [7]3 years ago

3 0

Answer:

1.6 x 10^-5 T

Explanation:

i = 4 A

r = 0.05 m

The magnetic field due to long wire at a distance r is given by

$B = \frac{\mu _{0}}{4\pi }\times \frac{2i}{r}$

B = 10^-7 x 2 x 4 / 0.05

B = 1.6 x 10^-5 T

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When you cook a marshmallow on a metal poker tool over an open flame, energy is transferred. Identify the three different ways t

ss7ja [257]

When we cook a marshmallow on a metal poker tool over an open flame, there are three ways in which heat energy is transferred: Conduction, convection, and radiation.

<h3>Heat energy transfer</h3>

Heat transfer is the natural transfer of heat from an object with a higher temperature to an object with a lower temperature. Heat transfer can occur in three ways, namely conduction, convection, and radiation.

Conduction occurs when heat flows from a place with a high temperature to a place with a lower temperature using a fixed heat-conducting medium. Heat transfer from the open flame to the marshmallows via direct fire contact with the marshmallows is an example of conduction.
Convection is the transfer of heat by means of a stream in which the intermediate substance also moves. If the particles move and cause heat to propagate, convection will occur. The hot air rising from the flames burning the marshmallows is an example of convection.
Radiation is heat transfer without a medium. Radiation can also usually be accompanied by light. The direct transfer of heat from the flame to the marshmallow in the form of waves is an example of radiation.

Learn more about heat transfer here: brainly.com/question/16055406

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3 0

10 months ago

The precision of a laboratory instrument is ± 0.05 g. The accepted value for your measurement is 7.92 g. Which measurements are

skelet666 [1.2K]

Answer:

7.89 7.91

Explanation:

The ranges of measurement lie between 7.92-0.05 and 7.92+0.05

7.87g and 7.97g

3 0

2 years ago

Read 2 more answers

A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit

Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

$y = \frac{1}{2} at^2+v_0t+y_0$

Where,

a = acceleration

t = time

$v_0$ = Initial velocity

$y_0$ = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

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

x = Displacement

t = time

With the data we have we can find the time as well

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

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

$t = \frac{18.6}{34}$

$t = 0.547s$

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

$y = \frac{1}{2} at^2+v_0t+y_0$

$y = \frac{1}{2} gt^2+0+0$

$y = \frac{1}{2} gt^2$

$y = \frac{1}{2} 9.8*0.547^2$

$y = 1.46m$

Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.

6 0

2 years ago

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.

4 0

2 years ago

If displacement covered by a particle is zero then distance cover by it<br>

sleet_krkn [62]

Answer:

When displacement is zero, the particle may be at rest, therefore, distance travelled = 0.

Again, when displacement is zero, the final position matches with the initial position after some time, but the distance travelled will not be zero.

7 0

2 years ago