All categories

3 years ago

6300 divided by 32.8

Physics

2 answers:

Ilya [14]3 years ago

7 0

please mark this answer as brainlist

Anna71 [15]3 years ago

5 0

The answer is 192.0731707317 and if you round it it’s going to be 196

You might be interested in

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

<u>Distance of the center from each corners</u> $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

<u>For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:</u>

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

<u>Force by each of the charges at the remaining corners:</u>

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

<u> Now, net electric field in the vertical direction:</u>

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

<u>Now, net electric field in the horizontal direction:</u>

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

What two quantities must stay the same in order for an object to have a constant velocity? A) The speed and kinetic energy must

GarryVolchara [31]

the answer is c) the speed and direction of travel must be constant

7 0

4 years ago

Help ASAP ITS DUE IN 30 MIN

Katena32 [7]

can't read it, need larger picture

5 0

4 years ago

A force of 25N acts on a mass of 5.0kg, initially at rest. Calculate the distance travelled before achieving a velocity of 20m/s

spin [16.1K]

Answer:

40m

Explanation:

let's calculate the acceleration first

force = mass × acceleration

rearranging to find acceleration:

acceleration = force ÷ mass

force = 25N, mass = 5.0kg

acceleration = 25 ÷ 5 = 5ms^-2

we can now use the formula v^2 = u^2 + 2as where v = final velocity, u = initial velocity, a = acceleration and s = distance

rearranging v^2 = u^2 + 2as the distance is

s = (v^2 - u^2) ÷ 2a

v = 20, u = 0, a = 5

s = (20^2 - 0^2) ÷ (2 × 5) = 40m

the distance is 40m

6 0

2 years ago

Water waves of wavelength 4m are produced by two generators S1 and S2.

Vaselesa [24]

A is not the correct answer because the amplitude and oscillation and there is 1/2 A in oscillation

8 0

2 years ago