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klio [65]
3 years ago
12

A point charge is placed 3 m from a 4 uC charge. What is the strength of the electric field on the point charge at this

Physics
2 answers:
brilliants [131]3 years ago
4 0

Answer:

4,000N/C

Explanation:

the information we have is:

distance:  r=3m

charge: q= 4\mu C

since the prefix \mu is equal to 1x10^{-6}

the charge is: q=4x10^{-6}C

------------------

To calculate the strength of the electric field we use the following formula:

E=\frac{kq}{r^2}

where q is the charge, r is the distance and k is Coulomb's constant

k=9x10^9Nm^2/C^2

substituting all the known values into the formula:

E=\frac{(9x10^9Nm^2/C^2)(4x10^{-6}C)}{(3m)^2} \\E=\frac{36,000Nm^2/C}{9m^2} \\E=4,00N/C

The strength of the electric field on the point charge at the distance of 3m is 4,000 N/C.

mars1129 [50]3 years ago
3 0

Answer:4000v/m

Explanation:

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a toy propeller fan with a moment of inertia of .034 kg x m^2 has a net torque of .11Nxm applied to it. what angular acceleratio
Harman [31]

Answer:

The  angular acceleration is  \alpha  = 3.235 \ rad/s ^2

Explanation:

From the question we are told that

    The moment of inertia is  I  =  0.034\ kg \cdot m^2

     The  net torque is  \tau  =  0.11\ N \cdot m

Generally the net torque is mathematically represented as

           \tau =  I  *  \alpha

Where \alpha is the angular acceleration so  

        \alpha  =  \frac{\tau }{I}

substituting values

         \alpha  =  \frac{0.1 1}{ 0.034}

        \alpha  = 3.235 \ rad/s ^2

6 0
3 years ago
As 390 g of hot milk cools in a mug it transfers 30 000 j of heat to the environment. whats is the temperature change of the mil
Fed [463]
You have to use the specific heat equation. 

Q = cmΔT where Q is the energy, c is specific heat, m is mass, and ΔT is change in temp.

So we can substitute our variables into the equation.

30000J = (390g)(3.9J*g/C)ΔT

Solving for ΔT, we get:

30000J/[(390g)*(3.9J*g/C) = ΔT

ΔT = 19.72386588C

I'm assuming the temperature is C, since it was not specified.

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liquid helium has a very low boiling point, 4.2 k, as well as a very low latent heat of vaporization, 2.00 104 j/kg. if energy i
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4.80 \times 10^3 \text { seconds }  long does it take to boil away 2.40 kg of the liquid.

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Latent heat of vapourization $L=2.00 \times 10^4 \mathrm{~J} / \mathrm{kg}$

Power of electrical heater $P=30 \mathrm{w}$

mass of liquid is $m=2.40 \mathrm{~kg}$

amount of heat required to boil

$$\begin{aligned}&Q=m L \\&Q=2.40 \times 2 \times 10^4 \mathrm{~J} \\&Q=4.80 \times 10^4 \mathrm{~J}\end{aligned}$$

Power $p=\frac{\text { work }}{\text { time }}=\frac{\text { Energy }}{\text { Time }}$

$$\begin{aligned}P &=\frac{Q}{t} \\\text { tine } t &=\frac{Q}{P}=\frac{4.80 \times 10^4 \mathrm{~J}}{10} \\t &=4.80 \times 10^3 \text { seconds }\end{aligned}$$

The heat or energy that is absorbed or released during a substance's phase shift is known as latent heat. It could go from a solid to a liquid or from a liquid to a gas, or vice versa. Enthalpy, a characteristic of heat, is connected to latent heat.

The heat that is used or lost as matter melts and transitions from a solid to a fluid form at a constant temperature is known as the latent heat of fusion.

Due to the fact that during softening the heat energy anticipated to transform the substance from solid to fluid at air pressure is the latent heat of fusion and that the temperature remains constant during the process, the "enthalpy" of fusion is a latent heat. The enthalpy change of any quantity of material during dissolution is known as the latent heat of fusion.

For learn more about Latent heat of vaporization, visit: brainly.com/question/14980744

#SPJ4

3 0
1 year ago
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