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4 years ago

6

The difference in electric potential between a thunder cloud and the ground is 2.08 x 10^8 V. Electrons move from the ground whi

ch is at a lower potential to the cloud which is at a higher potential. Determine the change in electric potential energy of one of the electrons that move to the cloud.

1 answer:

Mandarinka [93]4 years ago

7 0

Answer:

3,3*10^-11 J

Explanation:

A=Uq=2,08*10^8 V * 1,6*10^-19 C=3,3*10^-11J

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When an object is lifted 12 meters off the ground, it gains a certain amount of potential energy. If the same object is lifted 2

alexandr402 [8]

I think the answer is twice as much

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2 years ago

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

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3 years ago

What is the same about each type of electromagnetic wave?

xenn [34]

There all diffrent speeds

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4 years ago

What would the velocity of an object be if it was on a 1.75m

JulijaS [17]

Answer:

T = 4 sec / 2 = 2 sec period of revolution

S = 2 pi R = 2 * pi * 1.75 m = 11 m

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3 years ago

A 720 kg roller-coaster starts off from Location A. Assuming friction does not impede the car's motion, what will be the change

vaieri [72.5K]

We know, Potential Energy = m * g * h
Here, mass & gravity would be same, but their height will change so it will be:

ΔU = U₂ - U₁
ΔU = mgh₂ - mgh₁
ΔU = mg (h₂ - h₁)

Hope this helps!

7 0

3 years ago

Read 2 more answers