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3 years ago

7

Figure 24-40 shows a thin rod with a uniform charge density of 2.40 μC/m. Evaluate the electric potential (in V) at point P if d

= D = L/5.00.

1 answer:

LuckyWell [14K]3 years ago

7 0

Can you send a picture???

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Help, please<br> and explain

11Alexandr11 [23.1K]

We could tell a force is acting on an object if the object is being pushed, pulled, or moved in any way

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3 years ago

An action force is 40 N to the right. The reaction force must be: A. 20 N left B. 40 N left C. 20 N right D. 40 N right

Sergeeva-Olga [200]

Answer:

The answer to your question is letter B.

Explanation:

To answer this question, we must remember the third law of motion of Newton that states that For every action, there is an equal and opposite reaction.

Then, if the action force is 40 N to the right, the reaction force must be 40 N to the left.

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3 years ago

Oil at 150 C flows slowly through a long, thin-walled pipe of 30-mm inner diameter. The pipe is suspended in a room for which th

chubhunter [2.5K]

Answer:

<em>1.01 W/m</em>

Explanation:

diameter of the pipe d = 30 mm = 0.03 m

radius of the pipe r = d/2 = 0.015 m

external air temperature Ta = 20 °C

temperature of pipe wall Tw = 150 °C

convection coefficient at outer tube surface h = 11 W/m^2-K

From the above,<em> we assumed that the pipe wall and the oil are in thermal equilibrium</em>.

area of the pipe per unit length A = $\pi r ^{2}$ = $7.069*10^{-4}$ m^2/m

convectional heat loss Q = Ah(Tw - Ta)

Q = 7.069 x 10^-4 x 11 x (150 - 20)

Q = 7.069 x 10^-4 x 11 x 130 = <em>1.01 W/m</em>

4 0

3 years ago

Part of understanding the physical effects on Mars, we must understand

Ber [7]

Answer:

so easy add the subtract then multiplay the add

Explanation:

8 0

3 years ago

Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a −80-nC charge at x = −4 m on

disa [49]

Answer:

V = 48 Volts

Explanation:

Since we know that electric potential is a scalar quantity

So here total potential of a point is sum of potential due to each charge

It is given as

$V = V_1 + V_2 + V_3$

here we have potential due to 50 nC placed at y = 6 m

$V_1 = \frac{kQ}{r}$

$V_1 = \frac{(9\times 10^9)(50 \times 10^{-9})}{\sqrt{6^2 + 8^2}}$

$V_1 = 45 Volts$

Now potential due to -80 nC charge placed at x = -4

$V_2 = \frac{kQ}{r}$

$V_2 = \frac{(9\times 10^9)(-80 \times 10^{-9})}{12}$

$V_2 = -60 Volts$

Now potential due to 70 nC placed at y = -6 m

$V_3 = \frac{kQ}{r}$

$V_3 = \frac{(9\times 10^9)(70 \times 10^{-9})}{\sqrt{6^2 + 8^2}}$

$V_3 = 63 Volts$

Now total potential at this point is given as

$V = 45 - 60 + 63 = 48 Volts$

3 0

3 years ago