What is the voltage in an electric current?

What is seafloor spreading

liubo4ka [24]

Answer:

A

Explanation:

The process by which molten materials add new oceanic crust to the ocean floor.

7 0

3 years ago

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:

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

<u>Where</u>:

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:

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

<u>Where</u>:

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

It continues to fly along the same horizontal arc but increases its speed at the rate of 1.34 m/s 2 . Find the magnitude of acce

shepuryov [24]

This Question is not complete

Complete Question:

a. A hawk flies in a horizontal arc of radius 11.3 m at a constant speed of 5.7 m/s. Find its centripetal acceleration.

Answer in units of m/s2

b. It continues to fly along the same horizontal arc but increases its speed at the rate of 1.34 m/s2. Find the magnitude of acceleration under

these new conditions.

Answer in units of m/s2

Answer:

a. 2.875m/s²

b. 3.172m/s²

Explanation:

a. The formula for centripetal acceleration = (speed²) ÷ radius

Centripetal acceleration = (5.7m/s)²÷ 11.3m

Centripetal acceleration = 2.875m/s²

b. Magnitude of acceleration can be calculated by finding the sum of the vectors for the both the centripetal acceleration and the increase in the speed rate.

Centripetal acceleration ( acceleration x) = 2.875m/s²

Increase in the speed rate ( acceleration n) = 1.34m/s²

Magnitude of acceleration = √a²ₓ + a²ₙ

=√( 2.875m/s²)²+ (1.34m/s²)²

= √ 10.06m/s²

= 3.172m/s²

3 0

3 years ago

Read 2 more answers

I need hlep please help a brother out

SCORPION-xisa [38]

Answer:

a

Explanation:

4 0

2 years ago

ILL MARK BRAINLIEST

77julia77 [94]

Answer: B

Longitudinal wave

Explanation:

Transverse waves have crests and troughs

Longitudinal waves have compressions and rarefactions. A compression is where the density of the wave medium is highest. While a rarefaction is where the density of the wave medium is lowest.

Since sound wave is a longitudinal wave. And longitudinal wave exists apart from sound, we can therefore conclude that it's a longitudinal wave in spring.

8 0

2 years ago