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GrogVix
1 month ago
7

An ideal parallel-plate capacitor consists of a set of two parallel plates of area A separated by a very small distance d. When

this capacitor is connected to a battery that maintains a constant potential difference between the plates, the energy stored in the capacitor is U0. If the separation between the plates is doubled, how much energy is stored in the capacitor?a. Uo/2 b. Uo c. Uo/4 d. 4Uo e. 2Uo
Physics
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Richard needs to fly from san diego to halifax, nova scotia and back in order to give an important talk about mathematics. on th
Ostrovityanka [3204]

As the plane heads toward Halifax, the wind speed supports the flight path

resulting in an overall improved speed

Conversely, during the return trip, the wind will resist the plane's motion, decreasing the net speed

The total journey lasts 13 hours

of which 2 hours was dedicated to the mathematics discussion

Consequently, the total flight time is 13 - 2 = 11 hours

Now we apply the formula to calculate the time for traveling to Halifax

t_1 = \frac{d}{v + 50}

Time needed to return

t_2 = \frac{d}{v - 50}

Let’s look at the total time

T = t_1 + t_2

11 = \frac{d}{v - 50} + \frac{d}{v + 50}

Here d = 3000 miles

11 = \frac{3000}{v - 50} + \frac{3000}{v + 50}

3.67 * 10^{-3} = \frac{2v}{v^2 - 2500}

v^2 - 2500 = 545.45v

By solving the derived quadratic equation

v = 550 mph

the plane's speed calculates to 550 mph

3 0
3 months ago
Read 2 more answers
Three point charges are arranged on a line. Charge q3=+5.00x10^-9C and is at the origin. Charge q2=-3.00x10^-9C and is at x=+4.0
Sav [3153]

Answer:

The result is 0.750 NC

Explanation:

I am also taking the same test

5 0
2 months ago
Read 2 more answers
The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
Sav [3153]
The masses of particle A, B, and C are given, with all three particles aligned linearly. The distances between them are noted. The gravitational forces are attractive, compounding when acting in the same direction. The effects on each particle are formulated based on their distances.
3 0
3 months ago
Two carts are involved in an elastic collision. Cart A with mass 0.550 kg is moving towards Cart B with mass 0.550 kg, which is
Sav [3153]

Answer:

A) v_b=0.8\ m.s^{-1} denotes the resultant velocity of cart B post-collision.

B) KE_A=0.176\ J

C) KE_B=0\ J

D) ke_a=0\ J

E) ke_B=0.176\ J

F) Yes, kinetic energy remains conserved in this situation because both colliding bodies have identical mass.

G) Yes, momentum is conserved in every elastic collision.

Explanation:

Given:

  • mass of car A, m_a=0.55\ kg
  • mass of car B, m_b=0.55\ kg
  • initial velocity of car A, u_a=0.8\ m.s^{-1}
  • final velocity of car A, v_a=0\ m.s^{-1}

A)

The question mentions the cars experience an elastic collision:

By applying momentum conservation principles:

m_a.u_a+m_b.u_b=m_a.v_a+m_b.v_b

0.55\times 0.8+0.55\times 0=0.55\times 0+0.55\times v_b

v_b=0.8\ m.s^{-1} denotes the resulting velocity of cart B after collision.

B)

Initial kinetic energy of cart A:

KE_A=\frac{1}{2} m_a.u_a^2

KE_A=0.5\times 0.55\times 0.8^2

KE_A=0.176\ J

C)

Initial kinetic energy of cart A:

KE_B=\frac{1}{2} \times m_b.u_b^2

KE_B=0.5\times 0.55\times 0^2

KE_B=0\ J

D)

The final kinetic energy of cart A:

ke_A=\frac{1}{2} m_a.v_a^2

ke_a=0.5\times 0.55\times 0^2

ke_a=0\ J

E)

The final kinetic energy of cart B:

ke_B=\frac{1}{2} m_b.v_b^2

ke_B=0.5\times 0.55\times 0.8^2

ke_B=0.176\ J

F)

Yes, kinetic energy is conserved in this case due to both masses being identical in the collision.

G)

Indeed, momentum is consistently conserved in elastic collisions.

5 0
2 months ago
A charge of 4.9 x 10-11 C is to be stored on each plate of a parallel-plate capacitor having an area of 150 mm2 and a plate sepa
Keith_Richards [3271]

Answer:2.53*10^-10F

Explanation:

C=£o£r*A/d

Where £ represents the permittivity constant

£o= 8.85*10^-12f/m

£r=6.3

A=150mm^2=0.015m^2

d=3.3mm= 0.0033m

C=8.85*10^-12*6.3*0.015/0.0033

C=8.85*6.3*10^-12*0.015/0.0033

C=55.755*0.015^-12/0.003

C=8.36/3.3*10^-13+3

C=2.53*10^-10F

7 0
3 months ago
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