Can someone help me in the 26th one pleaseeee

A cave diver enters a long underwater tunnel. When her displacement with respect to the entry point is 20 m, she accidentally dr

Sladkaya [172]

Answer:

displacement vector is -16i

Explanation:

Given data:

Displacement relative to entry of tunnel is 20 m

Distance from entry of tunnel at which she notice about camera missing is 6 m

swim distance is 10 m

total distance from entry of tunnel after she swin back is 20+ 6 = 26 m

After covering 10 m to find a camera she will be at 26 -10 = 16 m

If direction out from tunnel is taken as positive then her displacement vector with respect to entry of tunnel is - 16i

where i is unit positive vector.

7 0

3 years ago

For which planet is the length of the plants day longer than the planets year

leonid [27]

I think Jupiter
Make as a abRainlist

6 0

3 years ago

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

How could you increase the potential energy of the apple?

VMariaS [17]

By putting an apple up on high ground. (That is 1 example)
This creates a higher gravitational force, and when it falls down it would have a lot of kinetic energy. But, if put on low ground, it wouldn't have enough potential energy to increase the kinetic energy.

7 0

3 years ago

Read 2 more answers

Two speakers face each other, and they each emit a sound of wavelength λ. One speaker is 180∘ out of phase with respect to the o

kolbaska11 [484]

The question ask to calculate and choose among the following is the distance of the microphone from the left - most speaker in order to pick up the loudest sound and the best answer would be letter C. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications

4 0

3 years ago

Can someone help me in the 26th one pleaseeee​

Can someone help me in the 26th one pleaseeee