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VARVARA [1.3K]
3 years ago
14

As part

Physics
1 answer:
umka21 [38]3 years ago
8 0

Answer:

Part a)

a = 3.68 m/s^2

Part b)

a = 11.8 m/s^2

Explanation:

Part a)

For force conditions of two blocks we will have

m_1g - T = m_1 a

T - m_2g = m_2 a

now from above equations we have

(m_1 - m_2) g = (m_1 + m_2) a

a = \frac{m_1 - m_2}{m_1 + m_2} g

now we know that

m_1 = \frac{908}{9.8} = 92.65 kg

m_2 = \frac{412}{9.8} = 42 kg

now from above equation we have

a = \frac{92.65 - 42}{92.65 + 42}(9.8)

a = 3.68 m/s^2

Part b)

When heavier block is removed and F = 908 N is applied at the end of the string then we have

F - mg = ma

908 - 412 = 42 a

a = 11.8 m/s^2

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parallel-plate air capacitor is made from two plates 0.070 m square, spaced 6.3 mm apart. What must the potential difference bet
Rom4ik [11]

Answer:

V = 576 V

Explanation:

Given:

- The area of the two plates A = 0.070 m^2

- The space between the two plates d = 6.3 mm

- Te energy density u = 0.037 J /m^3

Find:

- What must the potential difference between the plates V?

Solution:

- The energy density of the capacitor with capacitance C and potential difference V is given as:

                               u = 0.5*ε*E^2

- Where the Electric field strength E between capacitor plates is given by:

                               E = V / d

Hence,

                               u = 0.5*ε*(V/d)^2

Where, ε = 8.854 * 10^-12

                               V^2 = 2*u*d^2 / ε

                               V = d*sqrt ( 2*u / ε )

Plug in values:

                               V = 0.0063*sqrt ( 2 * 0.037 / (8.854 * 10^-12) )

                               V = 576 V

4 0
3 years ago
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun
AlekseyPX

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and  i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561  × 10¹⁹ electrons per second = 1.561  × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

8 0
3 years ago
If r1 < r2 < r3, and if these resistors are connected in series in a circuit, which one dissipates the most power
Vesnalui [34]
All three have the same current, so that is not a factor. Wattage (power) is E*I or i^2 R. The higher the resistance, the more power dissipated. The answer is R3 because it has the highest resistance.
R3 <<<< ===== answer.
6 0
3 years ago
What is the wavelength of a wave with a frequency of 466 Hz and a speed of
Volgvan

Answer:

<h3>The answer is option B</h3>

Explanation:

The wavelength of a wave can be found by using the formula

\lambda =  \frac{c}{f} \\

where

c is the speed of the wave

f is the frequency

From the question

c = 343 m/s

f = 466 Hz

We have

\lambda =  \frac{343}{466} \\  =  0.73605150...

We have the final answer as

<h3>0.74 m</h3>

Hope this helps you

4 0
3 years ago
Why neon is in the same group as helium?
guajiro [1.7K]
Neon and helium are both gases that fall under the Noble Gases section in the periodic table of elements.

Hope this helped to answer your question. :D

5 0
3 years ago
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