A sonar signal leaves a submarine, travels through

Which of these forces are acting within a nucleus? Check all that apply.

zavuch27 [327]

Answer:

Excuse me what?

Explanation:

5 0

3 years ago

Four charges of equal magnitude q = 2.16 µC are situated as shown in the diagram below. If d = 0.88 m, find the electric potenti

Triss [41]

Answer:

<em>The total potential (magnitude only) is 11045.45 V</em>

Explanation:

<u>Electric Potential </u>

The total electric potential at location A is the sum of all four individual potentials produced by the charges, including the sign since the potential is a scalar magnitude that can be computed by

$\displaystyle V=\frac{kq}{r}$

Where k is the Coulomb's constant, q is the charge, and r is the distance from the charge. Let's find the potential of the rightmost charge:

$\displaystyle V_1=\frac{9\cdot 10^{9}\times -2.16\cdot 10^{-6}}{0.88}=-22090.91\ V$

The potential of the leftmost charge is exactly the same as the above because the charges and distances are identical

$V_2=-22090.91\ V$

The potential of the topmost charge is almost equal to the above computed, is only different in the sign:

$V_3=+22090.91\ V$

The bottom charge has double distance and the same charge, thus the potential's magnitude is half the others':

$\displaystyle V_4=\frac{9\cdot 10^{9}\times 2.16\cdot 10^{-6}}{1.76}=+11045.45 \ V$

The total electric potential in A is

$V=-22090.91\ V-22090.91\ V+22090.91\ V+11045.45 \ V$

$V=-11045.45 \ V$

The total potential (magnitude only) is 11045.45 V

8 0

3 years ago

Most of the currents identified have a circular shape. This is because as the air and water move between the equator to the pole

Morgarella [4.7K]

I believe it is true . Somebody correct me if I’m wrong!

5 0

2 years ago

A constant amount of charge passes through a conductor. How is current affected if the same amount of charge passes in less time

baherus [9]

Answer:

B. The current increases.

Explanation:

As we know that rate of flow of charge through the conductor is known as electric current

So we have

$i = \frac{q}{t}$

here we know that charge Q flowing through the conductor is constant while the time in which it passes through it is decreased

so we can say that the ratio of charge and time will increase

so here we have

$i = increased$

So correct answer will be

B. The current increases.

4 0

3 years ago

Mike is standing on the roof of a building looking at the roof of the neighboring building that is 15 meters away and 10 meters

amid [387]

Answer:

Part a)

$t = 1.65 s$

Part b)

$x = 40.4 m$

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

$v = 27.3 m/s$

direction of velocity is given as

$[tex]\theta = 26.35 degree$

Explanation:

Part a)

acceleration due to gravity on this planet is 3/4 times the gravity on earth

So the acceleration due to gravity on this new planet is given as

$a = \frac{3}{4}(9.81)$

$a = 7.36 m/s^2$

now the vertical displacement covered by the canister is given as

$y = 10 m$

now by kinematics we have

$y = \frac{1}{2}gt^2$

$10 = \frac{1}{2}(7.36)t^2$

$t = 1.65 s$

Part b)

Horizontal speed of the canister is given as

$v_x = 24.5 m/s$

now the distance moved by it

$x = v_x t$

$x = 24.5 (1.65)$

$x = 40.4 m$

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

Final velocity in X direction will remains the same

$v_x = 24.5 m/s$

final velocity in Y direction

$v_y = v_i + at$

$v_y = 0 + (7.36)(1.65)$

$v_y = 12.14 m/s$

now magnitude of velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{24.5^2 + 12.14^2}$

$v = 27.3 m/s$

direction of velocity is given as

$\theta = tan^{-1}\frac{v_y}{v_x}$

$\theta = tan^{-1}\frac{12.14}{24.5}$

$[tex]\theta = 26.35 degree$

6 0

3 years ago