How can we teach our kitty to behave properly

The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70

MrRa [10]

Answer:

a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

C = $\epsilon_o \frac{A}{d}$

C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

C = $\frac{Q}{\Delta V}$

Q₀ = C ΔV

Q₀ = 2.62 10⁻¹² 8.70

Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

C₁ = 1.04 10⁻¹² F

C₁ = Q₀ / ΔV₁

ΔV₁ = Q₀ / C₁

ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

ΔV₁ = 21.9 V

b) initial stored energy

U₀ = $\frac{Q_o}{ 2C}$

U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)

U₀ = 99.2 10⁻¹² J

c) final stored energy

U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)

U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

W = U_f - U₀

W = (249.2 - 99.2) 10⁻¹²

W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0

3 years ago

You are trying to overhear a most interesting conversation, but from your distance of 10.0 m , it sounds like only an average wh

Alexandra [31]

Answer:

r₂=0.1 m

Explanation:

Given that

r₁= 10 m , β₁ = 20 dB

At r₂ ,β₂= 60 dB

As we know that intensity level of sound given as

$\beta =10\ log\dfrac{I}{10^{-12}}$

$\beta _1=10\ log\dfrac{I_1}{10^{-12}}$

$20=10\ log\dfrac{I_1}{10^{-12}}$

10² x 10⁻¹² = I₁

I₁=10⁻¹⁰ W/m²

$\beta _2=10\ log\dfrac{I_2}{10^{-12}}$

$60=10\ log\dfrac{I_1}{10^{-12}}$

10⁶ x 10⁻¹² = I₂

I₂ = 10⁻⁶ W/m²

I₁=10⁻¹⁰ W/m²

P = I A

P=Power ,I =Intensity ,A=Area

$\dfrac{I_1}{I_2}=\dfrac{r^2_2}{r^2_1}$

$\dfrac{10^{-10}}{10^{-6}}=\dfrac{r^2_2}{10^2}$

r₂=0.1 m

4 0

3 years ago

Two 2.0 g plastic buttons each with + 40 nC of charge are placed on a frictionless surface 2.0 cm (measured between centers) on

EleoNora [17]

Answer:

a. There are three potential energy interaction. b. 2.16 m/s c. 2.16 m/s d. 0 m/s

Explanation:

a. There are three potential energy interaction.

Let the charges be q₁ = +40 nC, q₂ = +250 nC and q₃ = + 40 nC and the distances between them be q₁ and q₂ is r, the distance between q₂ and q₃ is r and the distance between q₁ and q₃ is r₁ = 2r respectively. So, the potential energies are

U₁ = kq₁q₂/r, U₂ = kq₁q₃/2r and U₃ = kq₂q₃/r

U = U₁ + U₂ + U₃ = kq₁q₂/r + kq₁q₃/2r + kq₂q₃/r (q₁ = q₃ = q and q₂ = Q)

U = kqQ/r + kq²/2r + kqQ/r = qk/r(2Q + q/2)

b. To calculate the final speed of the left 2.0 g button, the potential energy = kinetic energy change of the particle.

ΔU = -ΔK

0 - qk(2Q + q/2)/r = -(1/2mv² - 0). Since the final potential at infinity equals zero and the initial kinetic energy is zero.

So qk(2Q + q/2)/r = -1/2mv²

v = √[2qk(2Q + q/2)/mr] where m = 2.0 g r = 2.0 cm

substituting the values for the variables,

v = √[2 × 40 × 10⁻⁹ × 9 × 10⁹(2 × 250 × 10⁻⁹ + 40 × 10⁻⁹/2)/2 × 10⁻³ × 2 × 10⁻²]

v = √[360(500 × 10⁻⁹ + 20 × 10⁻⁹)/2 × 10⁻⁵]

v = √[720(520 × 10⁻⁹)/4 × 10⁻⁵] = 2.16 m/s

c. The final speed of the right 2.0 g button is also 2.16 m/s since we have the same potential energy in the system

d.

Since the net force on the 5.0 g mass is zero due to the mutual repulsion of the charges on the two 2.0 g masses, its acceleration a = 0. Since it starts from rests u = 0, its velocity v = u + at.

Hence,

v = u + at = 0 + 0t = 0 m/s

8 0

3 years ago

A 57-gram tennis ball moving at 70 miles per hour is hit with a 110-gram tennis racquet moving at 60 miles per hour. Which state

Whitepunk [10]

Answer:

D. Because mass and energy are both conserved, the total amounts of mass and energy are the same before and after impact.

Explanation:

As we know that, the energy in motion is Kinetic Energy mathematically given as:

$KE=\frac{1}{2} m.v^2$

Now, according to the law of conservation of energy:

$\frac{1}{2} m_1.u_1^2+\frac{1}{2} m_2.u_2^2=\frac{1}{2} m_1.v_1^2+\frac{1}{2} m_2.v_2^2$

where:

$m_1\ \&\ m_2$ mass of racquet and ball respectively and $u_1\ \&\ u_2$ are their respective initial velocities.

$v_1\ \&\ v_2$ are the respective final velocities.

Also the law of conservation of momentum is applicable in this case:

$m_1.u_1+ m_2.u_2= m_1.v_1+m_2.v_2$

In this case the velocity of the lighter mass will get increases in the final condition.

5 0

3 years ago

In 1985, during the construction of a skyscraper in Austin, Texas, the remains of a mastodon were unearthed. If only 1/64 of the

Art [367]

Answer:

34,380 years

Explanation:

The remaining amount of a radioactive isotope is found as the product of the original amount by (1/2) raised to the number of half-lives elapsed. The formla is:

M = M₀ × (1/2)ⁿ

Where M is the remaining amoun, M₀ i s the initial amount, and n is the number of half-lives.

Here, 1/64 of the original carbon-14 remained, meaning that M/M₀ = 1/64.

Then, you can substitute in the equation and solve:

(1/64) = (1/2)ⁿ

(1/2⁶) = (1/2ⁿ)

2⁶ = 2ⁿ

6 = n

Then, 6 half-lives elapsed since the mastodon died and the remains were dated.

Then, you must multiply 6 by the half-life time:

6 × 5730 years = 34,380 years ← answer

8 0

3 years ago

How can we teach our kitty to behave properly​

How can we teach our kitty to behave properly