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3 years ago

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Point charges of 28.0 µC and 42.0 µC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

ectric field zero? m (from the smaller charge) (b) What is the magnitude (in N/C) and direction of the electric field halfway between them? magnitude N/C direction

1 answer:

olga_2 [115]3 years ago

6 0

Answer: a) electric field will be zero at zero meters apart

b) for smaller charge q, E = 6.048X10^6N/m towards away from the charge,

for bigger charge Q, E = 4.032X10^6N/m

Explanation:

Detailed explanation and calculation is shown in the image below.

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A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found

kati45 [8]

Answer:

1) $\rm q_2$ is<u> positive.</u>

<u></u>

2) $\rm q_2=4.56\times 10^{-10}\ C.$

Explanation:

<h2><u>Part 1:</u></h2>

<u></u>

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., $\rm q_2$ is <u>positive.</u>

<u></u>

<h2><u>Part 2:</u></h2>

<u></u>

<u>Given:</u>

Mass of the balloon, m = 0.00275 kg.
Charge on the balloon, $\rm q_1 = -3.50\times 10^{-8}\ C.$
Distance between the rod and the balloon, d = 0.0640 m.
Acceleration due to gravity, $\rm g = 9.81\ m/s^2.$

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, $\rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.$

The magnitude of the electrostatic force on the balloon due to the rod is given by

$\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.$

$\rm \dfrac{1}{4\pi \epsilon_o}$ is the Coulomb's constant.

For the elecric force and the weight to be balanced,

$\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.$

3 0

3 years ago

An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,

insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

$Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}$

Where

$t_{hf}$ = Temperature at the inside of the pipe = 300 °C

$t_{f}$ = Temperature at the outside of the pipe = 20 °C

r₁ =internal radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

$k_A$ = 15 W/m·K

$k_B$ = 0.038 W/m·K

$h_{hf}$ = 75 W/m²·K

$h_{cf}$ = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

$Q = \frac{t_A-t_B}{R_T}$ Where $R_T$ for the air film on the pipe outer surface is given by

$R_T= \frac{1}{\alpha A}$

where A =area of the outside of the pipe

= $\frac{1}{10*2\pi*0.07*1 }$ = 0.227 K/W

Therefore

131.32 W = $\frac{t_A-20}{0.227}$ which gives

$t_A$ = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

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3 years ago

IN THIS FORMULA FOR WATER WHAT DOES THE SUBSCRIPT 2 INDICATE

masha68 [24]

The chemical formular for water is H2O.
The H aspect of the formula stands for hydrogen gas and the subscript 2 which is attached to the H symbol signifies that two atoms of hydrogen are joined together, that is two atom of hydrogen are present.
The chemical formula of water indicates that, two atom of hydrogen react with one atom of oxygen to form one molecule of water.
In chemical formulae, subscripts are normally used to indicate the number of atoms that are present in a molecule.

7 0

3 years ago

Which of the following scenarios will cause the current to stop flowing within a circuit?

Lelechka [254]

C is correct. Only a closed circuit can carry current, opening a switch disconnects the circuit and stops current from flowing. Remember, all current flows in loops, if you can't make a loop out of it, current can't flow, and opening a switch, assuming no other paths for current exist, will completely stop the current.

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3 years ago

Read 2 more answers

a car traveling at 24 m/s starts to decelerate steadily. It comes to a complete stop in 6 seconds what is its acceleration?

Irina18 [472]

We can solve for the acceleration by using a kinematic equation. First we should identify what we know so we can choose the correct equation.

We are given an original velocity of 24 m/s, a final velocity of 0 m/s, and a time of 6 s. We and looking for acceleration (a) in m/s^2.

The following equation has everything we need:

$v_f=v_i + at$

So plug in the known values and solve for a:

0 = 24 + 6a

-24 = 6a

a = -4 m/s^2

8 0

3 years ago