A light bulb has a resistance of 100 ohms. If a current of 1.2 amps is going through it, calculate

A vector d has a magnitude of2.6 m and points north. What are the magnitudes and directions of the vectors (a) - d, (b) d/2.0, (

AleksandrR [38]

Solution :

A vector is defined as an element that has magnitude of some measure and direction.

It is given there is vector 'd' which has magnitude 2.6 m and its direction is towards north.

a). -d

The magnitude of the vector '-d' is 2.6 m and its direction is reversed, i.e. its direction is towards south.

b). d/2.0

The magnitude of the vector 'd/2.0' is 1.3 m and its direction is towards north.

c). - 2.5d

The magnitude of the vector increases by 2.5 times i.e. 2.5 x 2.6 = 6.5 m and the direction is towards south.

d). 5.0d

The magnitude of the vector increases by 5 times i.e. 5 x 2.6 = 13 m and the direction is towards north.

4 0

3 years ago

A negative charge is moving from point A to point B to the RIGHT in an area where there is a constant electric field also pointi

Aneli [31]

Answer:

Explanation:

In the electric field pointing right , force on the electron will be towards left . Since charge is moving towards right so it is doing some work against the force . Hence its potential energy will increase. Whenever a work is done against a force , the work done is converted into potential energy in conservative field. Electric field is also a conservative field.

3 0

4 years ago

The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti

deff fn [24]

Answer:

Explanation:

Given

side of square shape $a=5\ cm$

Electric flux $\phi =3\times 10^{-5}\ N.m^2/C$

Permittivity of free space $\epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}$

Flux is given by

$\phi =EA\cos \theta$

where E=electric field strength

A=area

$\theta$ =Angle between Electric field and area vector

$E=\frac{\phi }{A\cos (0)}$

$E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}$

$E=0.012\ N/C$

and Electric field by a uniformly charged sheet is given by

$E=\frac{\sigma }{2\epsilon_0}$

where $\sigma$ =charge density

$=\frac{\sigma }{\epsilon_0}$

$\sigma =0.012\times 8.85\times 10^{-12}$

$\sigma =2.12\times 10^{-13}\ C/m^2$

5 0

3 years ago

If the net force acting on a stationary object is zero, then the object will A. remain at rest. B. accelerate in the direction o

ra1l [238]

The object will remain at rest

Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

Here net force is zero

An object with a net force of zero acting on it will remain at rest, if initially at rest, or it will maintain a constant velocity.

F=MA

Force F=0

Than Acceleration A=0

With zero Acceleration stationary object will remain at rest.

Hence Remain at rest is the correct answer

Learn more about Newton's law of motion here

brainly.com/question/25545050

#SPJ4

6 0

2 years ago

When none of the magnetic domains in a material will stay aligned, the material is called __________________?

Kisachek [45]

The material is called paramagnetic or non magnetic material. Opposite to this are materials whose magnetic domains are aligned and possess asmagnetism which is due to its ferrous or iron content.

7 0

3 years ago