All categories

3 years ago

What does the electric field strength tell about the electric firld?

Physics

1 answer:

Solnce55 [7]3 years ago

8 0

Answer:

Explanation:

The region around a charged particle where another charged particle experiences a force of attraction or repulsion is called electric field.

The strength of electric field is defined as the force experienced by the unit positive test charge.

E = F / q

Electric field strength is a vector quantity and it is measured in newton per coulomb.

Where, F is the force of attraction or repulsion between the two charges and q is the test charge on which the electric field strength is to be calculated.

The strength of electric field is more if the field is strong. It means more be the electric field strength at a point more be the electric field.

You might be interested in

Newton's third law states that for every action force there is an equal and opposite reaction force. An idiot in your class says

Otrada [13]

A. Because the third laws say that for every action force the is an equal and opposite reaction force

4 0

3 years ago

Read 2 more answers

Does anyone know this?!

Valentin [98]

Answer:

2 is the numerical answer.

Explanation:

Hello there!

In this case, according to the given information and formula, it is possible for us to remember that equation for the calculation of the average kinetic energy of a gas is:

$KE=\frac{3}{2} \frac{R}{N_A} T$

Whereas R is the universal gas constant, NA the Avogadro's number and T the temperature.

Which means that for the given ratio, we can obtain the value as follows:

$=\frac{\frac{3}{2} \frac{R}{N_A} T_1}{\frac{3}{2} \frac{R}{N_A} T_2} \\\\=\frac{T_1}{T_2} \\\\=\frac{500K}{250K} \\\\=2$

Regards!

8 0

3 years ago

Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t

earnstyle [38]

1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
$B= \frac{\mu_0I}{2 \pi r}$
where
$\mu_0$ is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance
$r=40.0 cm=0.40 m$ ,
which is the distance at which the other wire is located:
$B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T$

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
$F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N$

2) direction of the force:
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.

6 0

3 years ago

If 2 grams of element X combine with 4 grams of element Y to form compound XY, how many grams of element Y would combine with 46

Butoxors [25]

92 grams of Y would combine with 46 grams of X

3 0

3 years ago

Now in "real life," this automobile is cruising at 20.5 m/s (equal to 73.8 km/hr) when it is about to hit a pedestrian stuck in

algol13

Answer:

He needs 1.53 seconds to stop the car.

Explanation:

Let the mass of the car is 1500 kg

Speed of the car, v = 20.5 m/s

He will not push the car with a force greater than, $F=2\times 10^4\ N$

The impulse delivered to the object is given by the change in momentum as :

$F\times t=mv\\\\t=\dfrac{mv}{F}\\\\t=\dfrac{1500\times 20.5}{2\times 10^4}\\\\t=1.53\ s$

So, he needs 1.53 seconds to stop the car. Hence, this is the required solution.

5 0

3 years ago