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yarga [219]
3 years ago
9

A rock is dropped in a pond, causing circular ripples. The radius increased with a rate 1 foot per second. When the radius is 4

feet, the area is increasing by 8π square feet per second.1. True2. False
Physics
1 answer:
Brrunno [24]3 years ago
5 0

Answer:

true

Explanation:

Area = πr²

Differentiating in respect to r

we get,

dA/dr = 2πr

using chain rule,

\frac{dA}{dt} = \frac{dA}{dr} .\frac{dr}{dt}  = 2\pi \frac{dr}{dt}

since r = 4feet

so ,

2π(4) = 8π

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A thin, uniformly charged insulating rod has a linear charge density λ = 3 nC/m and lies along the x axis from x = 1m to x = 3m.
Contact [7]

Answer:

A) V_A = 11.93~V

B) The vector definition of E-field is

\vec{E} = -1.13\^x + 2.41\^y

where magnitude is E = 2.66 N/m.

Explanation:

The potential of a uniformly charged rod can be found by the method of integration. We will first choose an infinitesimal part on the rod. We will compute the potential of this part at point A. Then we will integrate this potential over the entire rod.

We will use the following formula for electric potential:

V = \frac{1}{4\pi \epsilon_0}\frac{Q}{r}

Let us choose the infinitesimal part a distance 'x' from the origin. Then the distance between this point and point A is

r = \sqrt{x^2+4^2}

The infinitesimal length is 'dx', and the potential of this length is dV. Let's apply the formula:

dV = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{\sqrt{x^2 + 4^2}}

Here, the charge Q is equal to the charge density multiplied by the length. Q = λdx

Now we have to integrate this infinitesimal potential over the rod:

V = \int\limits^3_1 {dV} \, dx = \frac{1}{4\pi \epsilon_0}\int\limits^3_1 {\frac{\lambda}{\sqrt{x^2 + 16}} \, dx

By using an integral table, this can be calculated:

V = \frac{3\times 10^{-9}}{4\pi\epsilon_0}\ln(|\sqrt{x^2+16}+x|)\left \{ {{x=3} \atop {x=1}} \right. \\V = 11.93~V

B) The electric field can be found by a similar approach, but a different formula:

\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\^r

Let's apply this formula to the infinitesimal part we have chosen.

dE_x = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\cos(\theta)\\dE_y = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\sin(\theta)

By the geometry sine and cosine terms can be found:

\sin(\theta) = \frac{4}{\sqrt{x^2+16}}\\\cos(\theta) = \frac{x}{\sqrt{x^2 + 16}}

The x- and y-components of the E-field can be found separately by integrating the infinitesimal parts over the entire rod.

E_x = \int\limits^3_1 {dE_x} \, dx = \frac{\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{x}{(x^2+16)^{3/2}}} \, dx  = 1.13(-\^x)\\E_y = \int\limits^3_1 {dE_y} \, dx = \frac{4\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{1}{(x^2+16)^{3/2}}} \, dx  = 2.41(\^y)

So, the final E-field is

\vec{E} = -1.13\^x + 2.41\^y

The magnitude of the E-field is

E = 2.66 N/m

6 0
3 years ago
Describe at least three measures that ensure safety and prevent injury during a workout.
KonstantinChe [14]
Water - helps with dehydration.
instruction - help you understand how you need to do the workout.
trainers - wether it's an online trainer or a gym trainer they always say how to be safe while working out.
5 0
3 years ago
Read 2 more answers
If you apply 3 times the force for 4 times the distance, how much does your work increase?
masha68 [24]
12 increase When force is multiplied by the distance over which it is applied, that amount is given ... force * distance = work
4 0
3 years ago
Formula:
pishuonlain [190]

Answer:

<h2>42.5 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 50 × 0.85

We have the final answer as

<h3>42.5 N</h3>

Hope this helps you

5 0
3 years ago
Which following options are examples of matter? A) light B) Molecules C) Feelings D) people
yaroslaw [1]
All except C, feelings are not physical.
6 0
4 years ago
Read 2 more answers
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