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Dima020 [189]
3 years ago
5

What name is used for a high-power variable resistor?

Physics
2 answers:
alexandr1967 [171]3 years ago
6 0
The Answer B:Potentiometer
const2013 [10]3 years ago
3 0
The answer would be C: Rheostat.  :)
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1. What types of elements does an ionic bond occur between?
sweet [91]

Answer:

Ionic bonds form when a nonmetal and a metal exchange electrons, while covalent bonds form when electrons are shared between two nonmetals. An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions.

Explanation:

6 0
3 years ago
The magnetic field at the center of a wire loop of radius , which carries current , is 1 mT in the direction (arrows along the w
Citrus2011 [14]

Complete Question

 The complete question is shown on the first uploaded image

Answer:

The magnetic field is B_{net} = \frac{1}{4}  * mT

And the direction is  -\r k

Explanation:

      From the question we are told that

                 The magnetic field at the center is B = 1mT

Generally magnetic field is mathematically represented as

              B = \frac{\mu_o I}{2R}

We are told that it is equal to 1mT

So

                B = \frac{\mu_o I}{2R} = 1mT

From the first diagram we see that the effect of the current flowing in the circular loop is  (i.e the magnetic field generated)

                         \frac{\mu_o I}{2R} = 1mT

 This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)

                   B_1 = \frac{1}{2} \frac{\mu_o I}{2R}

and  for the larger semi-circular loop  is

                 B_2 = \frac{1}{2} \frac{\mu_o I}{2 * (2R)}

Now a closer look at the second diagram will show us that the current in the semi-circular loop are moving in the opposite direction

    So the net magnetic field would be

                   B_{net} = B_1 - B_2

                        =  \frac{1}{2} \frac{\mu_o I}{2R} -\frac{1}{2} \frac{\mu_o I}{2 * (2R)}

                        =\frac{\mu_o I}{4R} -\frac{\mu_o I }{8R}

                        =\frac{\mu_o I}{8R}

                        = \frac{1}{4} \frac{\mu_o I}{2R}

Recall  \frac{\mu_o I}{2R} = 1mT

    So  

             B_{net} = \frac{1}{4}  * mT

Using the Right-hand rule we see that the direction is into the page which is -k

3 0
3 years ago
On the moon, the acceleration due to gravity is one-sixth that of earth. That is gmoon = gearth /6 = (9.8 m/s2 )/6 = 1.63 m/s2 .
MAVERICK [17]
The pendulum would differ from 300 inches
8 0
3 years ago
A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round
Svetllana [295]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s

v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s

Now we find the centripetal acceleration which is given by

a_c=\frac{v^2}{r}

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2

we also have a tangential acceleration, which is given by

a_t = \frac{v-u}{t}

where

t = 17.0 s

Substituting values,

a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2

6 0
3 years ago
Read 2 more answers
You run away from a plane mirror at 2.30 m/s. At what speed does your image move away from you?
ivanzaharov [21]

Answer:

4.60m/s

.............

5 0
3 years ago
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