How to find the energy of a wave

At one point in space, the electric potential energy of a 15 nC charge is 42 μJ . Part A) What is the electric potential at this

Anastasy [175]

Answer:

Part A:

$\rm 2.8\times 10^3\ Volts.$

Part B:

$\rm 5.6\times 10^{-5}\ J.$

Explanation:

Potential energy of charge at the given point, $\rm U=42\ \mu J=42\times 10^{-6}\ J.$

Charge, $\rm q=15\ nC = 15\times 10^{-9}\ C.$

The potential energy at a point due to a charge is defined as

$\rm U=qV$ .

<em>where</em>,

V = electric potential at that point.

Therefore,

$\rm V=\dfrac{U}{q}=\dfrac{42\times 10^{-6}}{15\times 10^{-9}}=2.8\times 10^3\ Volts.$

Now, if the charge at that point is replaced with $\rm q_1 = 20\ nC = 20\times 10^{-9}\ C.$ , then the electric potential energy at that point is given by

$\rm U=q_1V = 20\times 10^{-9}\times 2.8\times 10^3=5.6\times 10^{-5}\ J.$

5 0

3 years ago

A wildlife researcher is tracking a flock of geese. The geese fly 4.0km due west, then turn toward the north by 40 degrees and f

BaLLatris [955]

Answer:

(a). The distance from the initial position is 6.68 km.

(b). The magnitude of their displacement is 7.05 km.

Explanation:

Given that,

Geese fly 4.0 km due west, then turn to north by 40° and fly another 3.5 km.

(a). We need to calculate the distance from the initial position

Using formula of distance

$D=AB+BC\cos\theta$

Put the value into the formula

$D=4.0+3.5\cos40$

$D=6.68\ km$

(b). We need to calculate the magnitude of their displacement

Using formula of displacement

$AC=\sqrt{CD^2+AD^2}$

$AC=\sqrt{(3.5\sin40)^2+(6.68)^2}$

$AC=7.05\ Km$

Hence, (a). The distance from the initial position is 6.68 km.

(b). The magnitude of their displacement is 7.05 km.

8 0

3 years ago

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You should be extra careful during the hours of sunrise, sunset, and nighttime because

skelet666 [1.2K]

Poor visibility, difficulties in colour perception, lessened colour contrast vision due to darker shadows and reduced peripheral vision, these are the reasons why one has to be extra careful while driving during hours of sunrise, sunset and night time.

<h3><u> Explanation: </u></h3>

Sunrise, sunset and night time are parts of the day with minimal or absolutely no presence of sunlight. To safely navigate roads, we require enough light in order to detect presence of other vehicles, signs and pedestrians. Less sunlight during sunrise and sunset light the sky but makes the roads and vehicles have a darker, less bright view. The contrast between colours is the least, making it difficult to identify objects and see clearly.

A rising or a setting sun can also lead to glares on the driver’s view and thus obstruct it. Since a change in ambient light is observed, our eyes need to adjust with this change and this isn’t spontaneous. Night time driving has headlight glares from approaching vehicles and reduced surrounding visibility. The eyes switching for vision adaptability from dark to bright light if vehicles approach and pass by is not a quick action. Hence the driver’s vision is compromised in every such case and this may lead to accidents.

7 0

4 years ago

What is the mass of an object if a net force of 18 N causes it to accelerate at 15 m/s^2?

trasher [3.6K]

Answer:

1.2 kg

Explanation:

F = 18N

a = 15 m/s^2

F = ma

18 = 15m

m = 1.2 kg

3 0

3 years ago

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What is the Energy Transformation of an electric radiant heater?

bagirrra123 [75]

Answer:

energy transfer electric to heat

8 0

3 years ago