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3 years ago

Point R is 0.4 m from a 2C charge. Find the electric ﬁeld at point R.

1 answer:

8 0

We know, E = 1/4π∈ * q/r²
Here, 1/4π∈ = 9 × 10⁹
q = 2 C
r = 0.4

Substitute their values,
E = 9 × 10⁹ * 2/(0.4)²
E = 9 × 10⁹ * 2/0.16
E = 9 × 10⁹ * 12.5
E = 112.5 × 10⁹ N/C

So, Your Final Answer would be 1.125 × 10¹¹ N/C

Hope this helps!

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Chegg Given that the mean radius of the Moon’s orbit is 3.84 x 108 m and its period is 2.36 x 106 sec, at what altitude above th

alukav5142 [94]

Answer:

The altitude of geostationary satellite is $3.58\times10^{7}\ m$

Explanation:

Given that,

Radius of moon's orbit $r=3.84\times10^{8}\ m$

Time period $T=2.36\times10^{6}\ sec$

We need to calculate the orbital radius of geostationary satellite is

Using formula of time period

$T=\sqrt{\dfrac{4\pi^2}{GM}a^3}$

$a=((\dfrac{GM}{4\pi^2})T^2)^{\dfrac{1}{3}}$

Where, G = gravitational constant

M = Mass of earth

T = time period of geostationary satellite orbit

Put the value in to the formula

$a=((\dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{4\times\pi^2})\times(86160)^2)^{\dfrac{1}{3}}$

$a=4.217\times10^{7}\ m$

We need to calculate the altitude of geostationary satellite

Using formula of altitude

$h = a-R_{e}$

Where, R = radius of earth

a = radius of geostationary satellite

Put the value into the formula

$h =4.217\times10^{7}-6.38\times10^{6}$

$h =35790000\ m$

$h=3.58\times10^{7}\ m$

Hence, The altitude of geostationary satellite is $3.58\times10^{7}\ m$

4 0

4 years ago

The transmission of heat requiring the movement of a liquid or a gas is

AysviL [449]

Answer:

convection

Explanation:

7 0

4 years ago

To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3

iren [92.7K]

Answer:

$372.3 J/^{\circ}C$

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

$P=VI=(3.6)(2.6)=9.36 W$

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

$E=Pt=(9.36)(350)=3276 J$

Finally, the change in temperature of an object is related to the energy supplied by

$E=C\Delta T$

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

$\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C$ is the change in temperature

Solving for C, we find:

$C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C$

5 0

3 years ago

A plate drops onto a smooth floor and shatters into three pieces of equal mass.Two of the pieces go off with equal speeds v at r

Firlakuza [10]

Answer:

Speed of the this part is given as

$v_3 = \sqrt2 v$

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

Explanation:

As we know by the momentum conservation of the system

we will have

$P_1 + P_2 + P_3 = P_i$

here we know that

$P_1 = P_2$

the momentum of two parts are equal in magnitude but perpendicular to each other

so we will have

$P_1 + P_2 = \sqrt{P^2 + P^2}$

$P_1 + P_2 = \sqrt2 mv$

now from above equation we have

$P_3 = -(P_1 + P_2)$

$mv_3 = -(\sqrt 2 mv)$

$v_3 = \sqrt2 v$

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

6 0

3 years ago

A piston/cylinder contains 2 kg of water at 20◦C with a volume of 0.1 m3. By mistake someone locks the piston, preventing it fro

morpeh [17]

Answer:

Final temperature = 250.11 °C

Final volume = 0,1 m3.

Process work = 0

Explanation:

The specific volume in the initial state is: v = 0.1m3/2 kg = 0.05 m3/kg.

This volume is located between the volumes as saturated liquid and saturated steam at 20 °C. For this reason the water is initially in a liquid vapor mixture. As the piston was blocked the volume remains constant and the process is isometric, also known as isocoric process, so the final temperature will be the water temperature at a saturated steam of v=0.05m3/kg, which is obtained by using steam tables for water, by linear interpolation. As follows, using table A-4 of the Cengel book 7th Edition:

v=0.05 m3/kg

v1=0.057061 m3/kg

T1=242.56°C

v2=0.049779 m3/kg

T2=250.35°C

T= $\frac{T2-T1}{v2-v1} x(v-v1)+T1=\frac{250.35°C-242.56°C}{0.049779m3/kg-0.057061m3/kg}x(0.05m3/kg-0.057061m3/kg)+242.56°C=250.11°C$

The process work is zero because there is no change in volume during heating:

W=PxΔv=Px0=0

where

W=process work

P=pressure

Δv=change of volume, is zero because the piston was blocked so the volume remains constant.

7 0

3 years ago