Who developed the universal law of gravitation

A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.155 Î¼C on each plate. The plates have a

butalik [34]

Answer:

The potential difference between the plates is 596.2 volts.

Explanation:

Given that,

Capacitance $C=260\ pF$

Charge $q=0.155\ \mu\ C$

Separation of plates = 0.313 mm

We need to calculate the potential difference between the plates

Using formula of potential difference

$V= \dfrac{Q}{C}$

Where, Q = charge

C = capacitance

Put the value into the formula

$V=\dfrac{0.155\times10^{-6}}{260\times10^{-12}}$

$V=596.2\ volts$

Hence,The potential difference between the plates is 596.2 volts.

7 0

3 years ago

Please help on this one?

olganol [36]

the object distance of both lenses are positive.

6 0

3 years ago

A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to

gizmo_the_mogwai [7]

Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

$v=at+v_o$ where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

$v=at+v_o$

$(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})$

$1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t$

$\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}$

$10\text{ [s]}=t$

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

<u>Verification:</u>

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is $a=\dfrac{100 [\frac{m}{s}]}{1[s]}$ , the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

So, indeed, after 10 seconds, the velocity reaches 1000 m/s

5 0

2 years ago

Whats 6 3/7 ×1 5/9.

aniked [119]

6 3/7 * 1 5/9
45/7 * 14/9
630/63
10

7 0

3 years ago

Read 2 more answers

The speed of light in vacuum is defined to be c = 299,792,458 m/s = 1 μ0ε0 . The permeability constant of vacuum is defined to b

Radda [10]

Explanation:

It is given that,

The speed of light in vacuum is, c = 299,792,458 m/s

The permeability constant of vacuum is, $\mu_o=4\pi\times 10^{-7}\ N.s^2/C^2$

Let $\epsilon_o$ is the permittivity of free space. The relation between $\mu_o,\epsilon_o\ and\ c$ is given by :

$c=\dfrac{1}{\sqrt{\mu_o\epsilon_o}}$

$\epsilon_o=\dfrac{1}{c^2u_o}$

$\epsilon_o=\dfrac{1}{(299792458\ m/s)^2\times 4\pi\times 10^{-7}\ N.s^2/C^2}$

$\epsilon_o=8.85\times 10^{-12}\ C^2/N.m^2$

Hence, this is the required solution.

3 0

3 years ago

Read 2 more answers