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4 years ago

8

A capacitor stores 7.6 × 10–11 C when the electric potential difference between the plates is 5.8 V. What is the electric potent

ial energy stored in the capacitor?

1 answer:

AlekseyPX4 years ago

8 0

Energy stored in the capacitor= 2.2 x 10⁻¹⁰ J

Explanation:

The energy stored in the capacitor is given by

E= 1/2 q V

E= energy stored

q= charge= 7.6 x 10⁻¹¹ C

V= potential difference=5.8 V

so energy stored= E= 1/2 (7.6 x 10⁻¹¹) (5.8)

Energy stored= 2.2 x 10⁻¹⁰ J

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The results of a dart game were precise but not accurate. The accepted value of the game was the center of the dartboard. Which

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Answer:

The darts hit very different areas of the board.

Explanation:

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3 years ago

Read 2 more answers

What is the wavelength in nanometers of light when the energy is 1. 91 × 10^6 j for a mole of photons?.

JulijaS [17]

The wavelength in nanometers of light when the energy is 1. 91 × 10^6 j for a mole of photons is <u>62. 8 nm.</u>

Wavelength is the distance among the same points (adjacent crests) within the adjoining cycles of a waveform signal propagated in space or along a cord. In wi-fi structures, this period is typically specified in meters (m), centimeters (cm), or millimeters (mm).

The wavelength is the distance between wave crests, and it is going to be the same for troughs. The frequency is the variety of vibrations that skip over a given spot in one 2nd, and it's far measured in cycles consistent with the second (Hz) (Hertz).

Frequency is the ratio of pace and wavelength in relation to speed. In comparison, wavelength refers to the ratio of pace and frequency. Audible sound waves are characterized by way of a frequency range of 20 to 20 kHz. In contrast, the variety of wavelengths of visible light is from four hundred to seven hundred nm.

<u>calculation:-</u>

*E=hc/λ

1.91 × 10^6 J = (6.62610⁻³⁴) (3.00*10⁸) / λ

λ= (6.62610⁻³⁴) (3.00*10⁸) / 1.91 × 10⁶ J

λ= 1.0410⁻³¹× 10⁻⁹ × 6.022*10²³

=<u> 62. 8 nm </u>

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1 year ago

Figure P2.23 is a somewhat simplified velocity graph for Olympic sprinter Carl Lewis starting a 100 m dash. Estimate his acceler

ehidna [41]

A) Acceleration in part A: $6.1 m/s^2$

B) Acceleration in part B: $2.7 m/s^2$

C) Acceleration in part C: $1.5 m/s^2$

Explanation:

A)

The picture of the problem is missing: find it in attachment.

The acceleration of a body is equal to the rate of change of its velocity:

$a=\frac{v-u}{\Delta t}$

where

v is the final velocity

u is the initial velocity

$\Delta t$ is the time it takes for the velocity to change from u to v

In part A of the race, we have:

u = 0

v = 5.5 m/s (estimate)

$\Delta t = 0.9 - 0 = 0.9 s$

So the acceleration is

$a=\frac{5.5-0}{0.9}=6.1 m/s^2$

B)

In part B of the race, we have:

u = 5.5 m/s is the initial velocity (estimate)

v = 9.5 m/s is the final velocity (estimate)

$\Delta t = 2.4 - 0.9 = 1.5 s$ is the time interval between the two points considered

Therefore, using the equation for the acceleration, we can find the acceleration in part B:

$a=\frac{9.5-5.5}{1.5}=2.7 m/s^2$

C)

In part C of the race, we have:

u = 9.5 m/s is the initial velocity (estimate)

v = 11 m/s is the final velocity (estimate)

$\Delta t = 3.4 - 2.4 = 1 s$ is the time interval between the two points considered

And therefore, the acceleration in part C of the race is:

$a=\frac{11-9.5}{1}=1.5 m/s^2$

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3 years ago

A boy whirls a stone in a horizontal circle of radius 1.8 m and at height 1.8 m above level ground. The string breaks, and the s

umka2103 [35]

Answer:

Acceleration is $148.33\ m/s^{2}$

Solution:

As per the question:

Radius of the circle, R = 1.8 m

Height above the ground, h = 1.8 m

Horizontal distance, x = 9.9 m

Now,

The magnitude of the centripetal acceleration can be calculated as:

$a_{c} = \frac{v^{2}}{R}$

where

v = velocity

R = radius

$a_{c}$ = centripetal acceleration

Now, if we consider the vertical component of motion only, then considering the initial velocity, 'u' = 0, from kinematic eqn:

$h = ut + \frac{1}{2}gt^{2}$

$h = 0.t + \frac{1}{2}gt^{2}$

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2\times 1.8}{9.8}}$

t = 0.606 s

Now, for the horizontal component of velocity:

x = vt

$v =\frac{x}{t}$

$v =\frac{9.9}{0.606} = 16.34\ s$

Now, we know that the centripetal acceleration is given by:

$a_{c} = \frac{16.34^{2}}{1.8} = 148.33\ m/s^{2}$

6 0

3 years ago

A glass rod that has been charged to +12.0 nC touches a metal sphere. Afterward, the rod's charge is +8.0 nC. a) What kind of ch

SpyIntel [72]

Answer:

Explanation:

charge transferred to sphere = 12 - 8 = 4 nC

a )

The positive charge is decreased on glass rod , that means electrons must have entered the glass rod from the sphere which reduces its positive charge.

b )

charge transferred = 4 nC = 4 x 10⁻⁹ C

charge on one electron = 1.6 x 10⁻¹⁹ C

No of electrons transferred

= Total charge transferred / charge on one electron

= 4 x 10⁻⁹ / 1.6 x 10⁻¹⁹

= 2.5 x 10¹⁰ .

4 0

3 years ago