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2 years ago

Which combination of units can be used to express the magnetic field?

Physics

1 answer:

Zolol [24]2 years ago

8 0

Answer:

The magnetic field unit in the International System is the tesla (T). A tesla is defined as the magnetic field that exerts a force of 1 N (newton) on a load of 1 C (coulomb) that moves at a speed of 1 m / s within the field and perpendicular to the field lines

Explanation:

Magnetic induction or magnetic flux density (B), is the magnetic flux that causes a diffusion charge in motion for each unit of normal area to the direction of the flow. It is also called the magnetic field strength.

The unit of magnetic flux density in the International System of Units is the tesla (T).

The tesla (symbol T), is the magnetic induction unit (or magnetic flux density) of the International System of Units (SI). It is defined as a uniform magnetic induction that, normally distributed over a surface of one square meter, produces through this surface a total magnetic flux of a weber.

<u>Equivalences: </u>

1 T = 1 Wb · m-2 = 1 kg · s-2 · A-1 = 1 kg · C-1 · s-1

A Tesla is also defined as the induction of a magnetic field that exerts a force of 1 N (newton) on a load of 1 C (coulomb) that moves at a speed of 1 m / s within the field and perpendicular to the lines of magnetic induction.

1 T = 1 N · s · m-1 · C-1

Basic Unit in the Cegesimal System of Units (CGS): Gauss (G)

A gauss (G) is a magnetic field unit of the Cegesimal System of Units (CGS). A gauss (G) is defined as a maxwell per square centimeter.

1 gauss = 1 maxwell / cm2

A gauss is equivalent to 10-4 tesla:

1 T = 10,000 G

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An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's

ivolga24 [154]

1) Focal length

We can find the focal length of the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}$ (1)
where
f is the focal length
$d_o$ is the distance of the object from the mirror
$d_i$ is the distance of the image from the mirror

In this case, $d_o = 18 cm$ , while $d_i=-4 cm$ (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
$\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}$
from which we find
$f=- \frac{36}{7} cm=-5.1 cm$

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
$r=2f=2 \cdot 5.1 cm=10.2 cm$

3 0

2 years ago

A proton of mass mp is released from rest just above the lower plate and reaches the top plate with speed vp. An electron of mas

vodka [1.7K]

Answer:

$v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

Explanation:

You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:

$F=qE$

$F_p=F_e$

where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.

by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

$m_pa_p=qE\\\\a_p=\frac{v_p^2}{2d}\\\\\frac{m_pv_p^2}{2d}=qE$

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.

By doing the same for the electron you obtain:

$\frac{m_ev_e^2}{2d}=qE$

we can equals these expressions for both proton and electron, because the forces qE are the same:

$\frac{m_pv_p^2}{2d}=\frac{m_ev_e^2}{2d}\\\\v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

4 0

2 years ago

There are devices to put in a light socket that control the current through a lightbulb, thereby increasing its lifetime. Which

Dmitrij [34]

Answer: B

Explanation:

Limiting the maximum current through the bulb. This will help in preserving or improving the bulb's lifetime and also this won't have an effect on the brightness of the bulb as brightness is affected by the average value. Although brightness is a factor of current, reducing the maximum current won't have any bearing on the average current the bulb is getting.

4 0

2 years ago

A boy standing on a 19.6 meter tall bridge sees a motorboat approaching the bridge at a constant speed. When the boat is 27 mete

azamat

Answer:

A. 12 m/s

Explanation:

Let’s remember that the definition of velocity is the variation of position of an object respect with to time. We know that the boy dropped the stone when the boat was 27 meters from the bridge and the stone hit the water 3 meters in front of the boat. So, the Boat must have traveled x=27 m-3m=24 m. The next step is calculating the amount of time that took the boat to make that travel; coincidentally, it is the same time that takes the stone to reach the water.

The equation that describes the motion of the stone is:

y = y_0 + v_0 * t+1/2 * a * t^2

The boy drops the stone from rest, so we can say that v_0=0. We can fixate the reference line on top of the bridge, so y_0=0 as well. The equation will be then:

-19,6 m = -1/2 * 9,8 m/s^2 * t^2

t^2= -(19,6 m)/(-4,9 m/s^2) = 4,012 s^2

t=√(4,012 s^2) = 2,003 s

Knowing the time that takes the stone to reach the water, that is the same that time that the boat uses to travel the 24 meters. The velocity of the boat is:

v = ∆x/∆t = (27 m-3 m)/(2,003 s-0s) = 11,9816 m/s ≈ 12 m/s

Have a nice day! :D

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2 years ago

A 0.20-kg object attached to the end of a string swings in a vertical circle (radius = 80 cm). at the top of the circle the spee

gayaneshka [121]

Answer:

Tension in the string at this position: 3.1 N.

Explanation:

Convert the radius of the circle to meters:

$r = 80\;\text{cm} = 0.80\;\text{m}$ .

What's the net force on the object?

The object is in a circular motion. As a result,

$\displaystyle \Sigma F = \frac{m\cdot v^{2}}{r}$ ,

where

$\Sigma F$ is the net force on the object,
$m$ is the mass of the object,
$v$ is the velocity of the object, and
$r$ is the radius of the circular motion.

For this object,

$\displaystyle \Sigma F = \frac{0.20\times {4.5}^{2}}{0.80} = 5.0625\;\text{N}$ .

The output unit of net force should be standard if the unit for mass, velocity, and radius are all standard. The net force shall always point towards the center. In this case the net force points downwards.

What are the forces on this object?

There are two forces on the object at this moment:

Weight, $W$ , which points downwards. $W = m\cdot g = 0.20\times 9.81 = 1.962\;\text{N}$ .
Tension, $T$ , which also points downwards. The size of the tension force needs to be found.

What's the size of the tension force?

Gravity and tension points in the same direction. The size of their resultant force is the sum of the two forces. In other words,

$\Sigma F = T + W$ .

$T = \Sigma F - W = 5.0625 - 1.962 = 3.1$ .

All three values in this question are given with two sig. fig. Round the value of $T$ to the same number of significant figures.

4 0

2 years ago