Please fast if you solve this you will get 20 point

Agata [3.3K]

E, B, C# minor and A for the most songs

4 0

4 years ago

How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter to p

katrin [286]

Answer:

Required charge $q=2.6\times 10^{9}C$ .

$n=1.622\times 10^{10}\ electrons$

Explanation:

Given:

Diameter of the isolated plastic sphere = 25.0 cm

Magnitude of the Electric field = 1500 N/C

now

Electric field (E) is given as:

$E =\frac{kq}{r^2}$

where,

k = coulomb's constant = 9 × 10⁹ N

q = required charge

r = distance of the point from the charge where electric field is being measured

The value of r at the just outside of the sphere = $\frac{25.0}{2}=12.5cm=0.125m$

thus, according to the given data

$1500N/C=\frac{9\times 10^{9}N\times q}{(0.125m)^2}$

or

$q=\frac{0.125^2\times 1500}{9\times 10^{9}}$

or

Required charge $q=2.6\times 10^{9}C$ .

Now,

the number of electrons (n) required will be

$n=\frac{required\ charge}{charge\ of\ electron}$

or

$n=\frac{2.6\times 10^{-9}}{1.602\times 10^{-19}}$

or

$n=1.622\times 10^{10}\ electrons$

6 0

4 years ago

A hydraulic press has one piston of diameter 4.0 cm and the other piston of diameter 8.0 cm. What force must be applied to the s

yan [13]

Answer:

The force is $F_1 = 400.8 \ N$

Explanation:

From the question we are told that

The first diameter is $d_1 = 4.0 \ cm = 0.04 \ m$

The second diameter is $d_2 = 8.0 \ cm = 0.08 \ m$

Generally the first area is

$A_1 = \pi * \frac{d^2_1 }{4}$

=> $A_1 = 3.142 * \frac{0.04^2}{4}$

=> $A_1 = 0.00126 \ m^2$

The second area is

$A_2 = \pi * \frac{d^2_2 }{4}$

$A_2 = 3.142 * \frac{0.08^2}{4}$

$A_2 = 0.00503 \ m^2$

For a hydraulic press the pressure at both end must be equal .

Generally pressure is mathematically represented as

$P = \frac{F}{A}$

=>

$\frac{F_1}{A_1 } = \frac{F_2}{A_2 }$

=> $F_1 = \frac{1600}{0.00503} * 0.00126$

=> $F_1 = 400.8 \ N$

8 0

3 years ago

Someone help please.....

Westkost [7]

Answer:

0.0928km/min (4dp)

Explanation:

To find the jogger's speed in km per minute, we just need to divide the number of km jogged by the time in minutes it took to jog that distance. This will give us the distance they jogged every minute which is their speed.

4km in 32 minutes:

4/32 = 0.125km/min

2km in 22 minutes:

2/22 = 0.091 (3dp)km/min

1km in 16 minutes:

0.0625km/min

Now to find the average speed of these 3 speeds, we just add them all together and divide by how many values there are (3 values).

Average (mean) = $\frac{0.125+0.091+0.0625}{3}$

Average = 0.2785/3

Average speed of jogger = 0.0928 (4dp) km/min

Hope this helped!

8 0

3 years ago

Two point charges each have a value of 3.0 c and are separated by a distance of 4.0 m. what is the electric field at a point mid

swat32

<span>Place a test charge in the middle. It is 2cm away from each charge.
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point.
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out.
THIS IS A TRICK QUESTION.
THe electric field exactly midway between them = 0/Q = 0.
But if the point moves even slightly you need the following formula
F= (1/4Piε)(Q1Q2/D^2)
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>

4 0

3 years ago