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3 years ago

In an electric motor, electrical energy is converted into

Physics

1 answer:

N76 [4]3 years ago

5 0

Answer:

idrk im jus here for the points $\lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n$ ∈ω≠≠

Explanation:

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A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between

Oduvanchick [21]

Answer:

Net electric field, $E_{net}=91406.24\ N/C$

Explanation:

Given that,

Charge 1, $q_1=7.5\ nC=7.5\times 10^{-9}\ C$

Charge 2, $q_2=-2.9\ nC=-2.9\times 10^{-9}\ C$

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

$E_1=\dfrac{kq_1}{r^2}$

$E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}$

$E_1=65917.96\ N/C$

Electric field due to charge 2 is given by :

$E_2=\dfrac{kq_2}{r^2}$

$E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}$

$E_2=25488.28\ N/C$

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

$E_{net}=E_1+E_2$

$E_{net}=65917.96+25488.28$

$E_{net}=91406.24\ N/C$

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0

4 years ago

A ball player catches a ball 3.55 s s after throwing it vertically upward. with what speed did he throw it

Hatshy [7]

Answer:

i believe he threw the ball up at about 18 mph

Explanation:

5 0

3 years ago

The monomer used as the building block in polyethylene is? Ethene, ethane, monoethane, amino acids?

LuckyWell [14K]

<span>it's ethene and not an amino acid!!! amino acids form proteins as their polymers. </span>

6 0

4 years ago

How much work is done if you push a 200 N box across a floor with a force of 50 N for a distance of 20m

vovikov84 [41]

<h2>Answer: 1000 J</h2>

The Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.

It should be noted that it is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter:

$1J=(1N)(1m)=Nm$

Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:

$W=(F)(d)$ (1)

When they are not parallel, both directions form an angle, let's call it $\alpha$ . In that case the expression to calculate the Work is:

$W=Fdcos{\alpha}$ (2)

For example, in order to push the 200 N box across the floor, you have to apply a force along the distance $d$ to overcome the resistance of the weight of the box (its 200 N).

In this case both <u>(the force and the distance in the path) are parallel</u>, so the work $W$ performed is the product of the force exerted to push the box $F=50N$ by the distance traveled $d$ . as shown in equation (1).