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Degger [83]
3 years ago
10

The charge on a parallel plate capacitor is 7.68 C, and the plate is connected to a 9.0 V battery. What is the capacitance of th

e capacitor?
A. 8.5x10^-7 pF
B. 1.1x10^-6 pF
C. 8.5x10^5 pF
D. 6.9x10^7 pF
Physics
1 answer:
Natali5045456 [20]3 years ago
8 0

Charge on a capacitor = (capacitance) x (voltage)

7.68 Coulomb = (capacitance) x (9.0 V)

Capacitance = (7.68 Coul) / (9.0 V)

<em>Capacitance = 0.83 Farad</em>

<em></em>

Apparently, the " 10^something " after the 7.68 C got lost.

If the charge is actually 7.68 <u>micro</u>coulombs, then the answer is<em> (A) .</em>

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Blue-violet light on the visible light spectrum can cause damage to retina cells, increasing one's risk of developing age-related macular degeneration, explains Essilor.


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A shopper is pushing a cart down a grocery store aisle. Starting from rest, the shopper applies a constant force to the cart for
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A shopping cart that starts from rest, is accelerated for 4 s, moves at constant velocity for 4 s, and is decelerated for 4s until returning to rest, has an average acceleration of 0 m/s².

A shopper is pushing a cart down a grocery store aisle. The movement of the cart is:

  • It starts from rest.
  • From t = 0 s to t = 4.0 s it is accelerated with a constant force.
  • From t = 4 s to t = 8.0 s it receives just enough force to balance the friction on the cart.
  • From t = 8 s to t = 12 s it is decelerated until it comes to rest.

All in all, at the initial time (t = 0 s), the velocity is 0 m/s (rest) and at the final time (t = 12 s) the velocity is 0 m/s as well (rest). The average acceleration in that period is:

a = \frac{v_{12}-v__o}{t_{12}-t_0} = \frac{0m/m-0m/s}{12s-0s}  = 0 m/s^{2}

A shopping cart that starts from rest, is accelerated for 4 s, moves at constant velocity for 4 s, and is decelerated for 4s until returning to rest, has an average acceleration of 0 m/s².

Learn more: brainly.com/question/16274121

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2 years ago
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The peak intensity of radiation from Star Beta is 350 nm. In what spectral band is this? UV, radio waves, visible light, or infa
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Answer the following. (a) What is the surface temperature of Betelgeuse, a red giant star in the constellation of Orion, which r
bagirrra123 [75]

Answer:

(a) T = 2987.6 k

(b) T = 19986.2 k

Explanation:

The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}

where,

T = Radiated surface temperature

\lambda_{max} = peak wavelength

(a)

here,

\lambda_{max} = 970 nm = 9.7 x 10⁻⁷ m

Therefore,

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}

<u>T = 2987.6 k</u>

(b)

here,

\lambda_{max} = 145 nm = 1.45 x 10⁻⁷ m

Therefore,

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{1.45\ x\ 10^{-7}\ m}

<u>T = 19986.2 k</u>

6 0
3 years ago
Points a and b lie in a region where the y-component of the electric field is Ey=α+β/y2. The constants in this expression have t
Drupady [299]

Answer:

V_{a} - V_{b} = 89.3

Explanation:

The electric potential is defined by

         V_{b} - V_{a} = - ∫ E .ds

In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.

         V_{b} - V_{a} = - ∫ E ds

We substitute

         V_{b} - V_{a} = - ∫ (α + β/ y²) dy

We integrate

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We evaluate between the lower limit A  2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m

           V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)

            V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33

            V_{b} - V_{a} = - 89.3 V

As they ask us the reverse case

             V_{b} - V_{a} = - V_{b} - V_{a}

             V_{a} - V_{b} = 89.3

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