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Juliette [100K]

3 years ago

7

What is the strength of the electric field in a region where the electric potential is constant?

1 answer:

Roman55 [17]3 years ago

4 0

Answer:

Where the electric potential is constant, the strength of the electric field is zero.

Explanation:

As a test charge moves in a given direction, the rate of change of the electric potential of the charge gives the potential gradient whose negative value is the same as the value of the electric field. In other words, the negative of the slope or gradient of electric potential (V) in a direction, say x, gives the electric field (Eₓ) in that direction. i.e

Eₓ = - dV / dx ----------(i)

From equation (i) above, if electric potential (V) is constant, then the differential (which is the electric field) gives zero.

<em>Therefore, a constant electric potential means that electric field is zero.</em>

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A wave has a speed of 360 m/s. It has a frequency of 20hz what is its wavelength (include correct unit)

abruzzese [7]

Answer:

18m

Explanation:

v=frequency × wavelenght

wavelength=v/f

wavelength=360÷20

=18m

7 0

3 years ago

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A meteorite strikes the Moon at the speed of 11 km/s. It has KE of 60,500,000 J. What will be its mass?

Vikentia [17]

KE=1/2 mv²
60,500,000 = 1/2 m ( 11,000 m/s )²
60,500,000= 1/2 ( 11,000 )² m
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8 0

3 years ago

Algebraicallycalculate the rise time and fall timevalues for a sine wave at an arbitrary frequency f0(expressed in Hz). Assume t

Blababa [14]

Answer:

185.95 μsec

Explanation:

Given that:

Sine wave = $Amsinw_{o}t$

$V_{min}=-V_{max}$

Sinusodial signal $f_{o}$ = 50 kHz = $5*10^4Ht$

Rise time is said to be defined as the time needed for a pulse to rise from 10% - 90% of maximum rate.

∴

$t_1$ = 10% = 0.1 Am

$t_2$ =90% = 0.9 Am

Using sine wave for $t_1$ ; we have:

$Amsinw_{o}t_1$ = 0.1 Am

$sinw_{o}t_1$ = 0.1

$w_{o}t_1 = sin^{-1}(0.1)$

$w_{o}t_1$ = 5.7392

$t_1=\frac{5.7392}{2 \pi f_0}$

$t_1=\frac{0.9134}{ f_0}$

Using sine wave for $t_2$ ; we have:

$Amsinw_{o}t_2$ = 0.9 Am

$sinw_{o}t_2 = 0.9$

$w_ot_2$ = $sin^{-1}(0.9)$

$w_ot_2$ = 64.158

$t_2$ = $\frac{64.1581}{2 \pi f_o}$

$t_2$ = $\frac{10.211}{f_o}$

Change in rise time $t_r$ = $t_2$ - $t_1$

$t_r$ = $\frac{10.211}{f_o}-\frac{0.9134}{ f_0}$

$t_r$ = $\frac{10.211-0.9134}{f_o}$

$t_r$ = $\frac{9.2976}{f_o}$

since; $f_{o}$ = 50 kHz = $5*10^4Ht$

$t_r$ = $\frac{9.2976}{5*10^4}$

$t_r$ = 1.85952 × 10⁻⁴

$t_r$ = 185.952 × 10⁻⁶ sec

$t_r$ = 185.95 μ sec

∴ The rise in time in (μ sec) for the sinosodial signal at 50 kHz = 185.95 μ sec

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3 years ago

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Ivenika [448]

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3 years ago

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Llustration 2: Aman can run a distance of 100 m in 20 seconds. Find the speed of Aman in m/s.

lawyer [7]

Answer:

$\boxed {\boxed {\sf 5 \ meters/second}}$

Explanation:

Speed is equal to distance over time.

$s=\frac{d}{t}$

The distance is 100 meters and the time is 20 seconds.

$d= 100 \ m \\t= 20 \ s$

Substitute the values into the formula.

$s=\frac{100 \ m }{20 \ s}$

Divide.

$s= 5 \ m/s$

Aman's speed is 5 meters per second.

3 0

3 years ago