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4 years ago

why does the value of capacitance of a capacitor increases in parallel combination but not in series??

1 answer:

5 0

It is easiest to consider problems like this by thinking exclusively about parallel plate capacitors for which $C \equiv \frac{Q}{V} =\kappa \epsilon_0 \frac{A}{d}$ where Q is the charge separated (+Q on one plate, -Q on the other), V is the voltage difference between the plates, A is the area of each plate, and d is the separation between the plates.

When capacitors are connected in parallel, the voltage across each capacitor is the same. But with two capacitors, it will require more charge to reach the voltage V than it would with just one capacitor. In fact, if capacitor 1 requires charge

Read more: brainly.com/question/8898885

8 0

2 years ago

Can an object have increasing speed while its acceleration is decreasing?

alexdok [17]

The best option is C. This is due to friction.

6 0

3 years ago

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago

(a) find the magnitude of the gravitational force between a planet with mass 7.50 3 1024 kg and its moon, with mass 2.70 3 1022

anzhelika [568]

<span>The magnitude of the gravitational force between two bodies is the product of their masses divided by the square of the distance between them. So we have F = M1*M2 / r^2. M1 = 7.503 * 10e24 and M2 = 2.703 * 10e22 and r= 2.803 * 10e8; r^2 = 5.606 *10e16. So we have 7.503 *2.703 *10^(24+22) = 20.280 * 10^(46). Then we divide our answer by 5.606 * 10e16 which is the distance ; then we have 3.6175 * 10 e (46- 16) = 3.6175 * 10e30. To find the acceleration we use Newton's second law F = ma. F is 3.6175 * 10e30 and M is 7.503 * 10e24 so a = F/M and then we have 3.6175/7.503 * 10e (30-24) = 0.48 * 10e6. Similarly for moon, we have a = 3.6715/2.703 * 10e(30-22). = 1.358 * 10e8</span>

4 0

3 years ago

According to the law of conservation of mass, in a chemical reaction, the mass of the reactant will always be different from the

Lostsunrise [7]

That is not true.

<span>According to the law of conservation of mass, in a chemical reaction, the mass of the reactants will always be the same as the mass of the products.</span>

6 0

3 years ago