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3 years ago

Anyone know this... pls help its for a grade

Physics

1 answer:

hodyreva [135]3 years ago

3 0

His average speed is 45 miles an hour

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A 5.00-V battery charges the parallel plates in a capacitor, with a plate area of 865 mm2 and an air-filled separation of 3.00 m

Westkost [7]

Answer:

W = 3.21x10⁻¹¹ J

Explanation:

The work required to separate the plates can be calculated using the following equation:

$W = U_{2} - U_{1} = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2})$

Where:

U₂: is the final stored energy

U₁: is the initial stored energy

C₂: is the final capacitance

C₁: is the initial capacitance

V₁: is the initial potential difference = 5.00 V

V₂: is the final potential difference

The initial and final capacitance is:

$C_{1} = \epsilon_{0}*\frac{A}{d_{1}}$

Where:

ε₀: is the vacuum permittivity = 8.85x10⁻¹² C²/(N*m²)

d: is the initial distance = 3.00 mm = 3.00x10⁻³ m

A: is the plate area = 865 mm² = 8.65x10⁻⁴ m²

$C_{1} = \epsilon_{0}*\frac{A}{d_{1}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 \cdot 10^{-3} m} = 2.55 \cdot 10^{-12} F$

Similarly, C₂ is:

$C_{2} = \epsilon_{0}*\frac{A}{d_{2}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 + 3.00 \cdot 10^{-3} m} = 1.28 \cdot 10^{-12} F$

Now, V₂ can be calculated by finding the initial charge (q₁):

$q_{1} = C_{1}V_{1} = 2.55 \cdot 10^{-12} F*5.00 V = 1.28 \cdot 10^{-11} C$

Since, q₁ is equal to q₂, V₂ is:

$V_{2} = \frac{q_{2}}{C_{2}} = \frac{1.28 \cdot 10^{-11} C}{1.28 \cdot 10^{-12} F} = 10 V$

Finally, we can find the work:

$W = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2}) = \frac{1}{2}(1.28 \cdot 10^{-12} F*(10 V)^{2} - 2.55 \cdot 10^{-12} F(5.00 V)^{2}) = 3.21 \cdot 10^{-11} J$

Therefore, the work required to separate the plates is 3.21x10⁻¹¹ J.

I hope it helps you!

4 0

3 years ago

Is energy released or absorbed during the formation of a solution?

kodGreya [7K]

Energy can be released and absorbed during the formation of a solution, not one or the other. When a solute interacts with the solvent, energy is absorbed so the solvent can overcome the intermolecular bonds of the solute and energy is released, most commonly, in the form of heat, light, or a gaseous byproduct.

3 0

3 years ago

Hat processes lead to glacial erosion ?

azamat

Yes......................

8 0

3 years ago

Aliya deposited half as much money in a savings account earning 1.9% simple interest as she invested in a money market account t

yanalaym [24]

Let A be the amount of money that Aliya deposited in the savings account. Since A is half as much as money as she invested in a money market account, then the amount that she invested in the market account is 2A.

Express the interest that Aliya earned in terms of A. Set it equal to the amount of $297.60 and then solve for A.

Since the savings account gives 1.9% simple interest, the total amount of interest that she will earn from the savings account is 1.9% of A, which is equal to:

$\frac{1.9}{100}\times A$

Since the money market account gives 3.7% simple interest, the total amount of interest that she will earn from the money market account, is 3.7% of 2A, which is equal to:

$\frac{3.7}{100}\times2A$

Add both interests in terms of A and simplify the expression:

$\begin{gathered} \frac{1.9}{100}\times A+\frac{3.7}{100}\times2A \\ =\frac{1.9}{100}\times A+\frac{7.4}{100}\times A \\ =(\frac{1.9}{100}+\frac{7.4}{100})\times A \\ =\frac{1.9+7.4}{100}\times A \\ =\frac{9.3}{100}\times A \end{gathered}$

The expression (9.3/100)*A represents the total interest after one year. Then:

$\begin{gathered} \frac{9.3}{100}\times A=297.60 \\ \Rightarrow A=\frac{100}{9.3}\times297.60 \\ \Rightarrow A=\frac{100\times297.60}{9.3} \\ \Rightarrow A=\frac{29760}{9.3} \\ \Rightarrow A=3200 \end{gathered}$

Use the value of A to find the amount that was invested in the money market account:

$2A=2\times3200=6400$

Therefore, Aliya deposited 3200 in a savings account and 6400 in a money market account.

3 0

1 year ago

A point charge q1 = 3.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 10 m.

UkoKoshka [18]

Answer:

a) 1.6 mN b) -1.6 mN c) -1.6 mN d) 1.6 mN

Explanation:

The electrostatic force between 2 point charges, obeys the Coulomb's Law, that can be expressed as follows:

F₁₂ = k*q₁*q₂/(r₁₂)² (in magnitude)

The direction of the force, is along the line that joins the charges (along the x axis) and as q₁ and q₂ are of the same sign, aims away from both charges.

a) So, for the force on q₂, we have:

F₁₂ = 9*18*10⁻⁵ N = 1.6 mN (positive as it is aiming in the positive x direction)

b) The force on q1, according to Newton's 3rd Law, is just equal and opposite to the one on q2:

F₂₁ = (-9*18*10⁻⁵) N = -1.6 mN (towards the negative x direction, away from q1)

c) If q₂ were -6.0 μC, the force will be the same in magnitude, but as now both charges have different signs, they wil attract each other, so the direction of the forces will be exactly the opposite to the first case:

F₁₂ = -1.6 mN (going towards the origin, where q₁ is located)

F₂₁ = 1.6 mN (going in the positive x direction, towards q₂)

6 0

4 years ago

Read 2 more answers

Anyone know this... pls help its for a grade​

Anyone know this... pls help its for a grade