A) What is the meaning of specific heat capacity ?

A charge of 1.5 µC is placed on the plates of a parallel plate capacitor. The change in voltage across the plates is 36 V. How m

belka [17]

Answer:

Energy stored in the capacitor is $U=2.7\times 10^{-5}\ J$

Explanation:

It is given that,

Charge, $q=1.5\ \mu C=1.5\times 10^{-6}\ C$

Potential difference, V = 36 V

We need to find the potential energy is stored in the capacitor. The stored potential energy is given by :

$U=\dfrac{1}{2}q\times V$

$U=\dfrac{1}{2}\times 1.5\times 10^{-6}\times 36$

U = 0.000027 J

$U=2.7\times 10^{-5}\ J$

So, the potential energy is stored in the capacitor is $U=2.7\times 10^{-5}\ J$ . Hence, this is the required solution.

4 0

3 years ago

Read 2 more answers

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago

Some people wish that we lived in a recollapsing universe that would eventually stop expanding and start contracting. For this t

Monica [59]

Answer:

(B) Dark energy does not exist and there is much more matter than current evidence suggests.

Explanation:

The repulsive force which is accelerating expansion of the universe is called as dark energy. Most of matter present in the universe is the dark matter of about eighty five percent.

So, a collapsing universe would not have the dark energy and there is more matter which is not the dark matter. This theory is rejected because expansion of the universe is observable.

4 0

3 years ago

hi :) is it true that zero acceleration means the object can be moving at constant velocity or at rest?

nirvana33 [79]

Answer:

No

Explanation:

Please let me know if my answer is correct

7 0

3 years ago

Read 2 more answers

(i) Calculate the energy, in electron volts, of a photon whose frequency is (a) 620 THz}, (b) 3.10GHz , and(c) 46.0 MHz

DedPeter [7]

The energy in electron volts of the photons that has the following frequencies is as follows:

620 THz = 2.564eV
3.10GHz = 1.28 × 10-⁵eV
46.0 MHz = 1.902 × 10-⁷eV

<h3>How to calculate energy?</h3>

The energy of a photon can be calculated using the following formula:

E = hf

Where;

E = energy
h = Planck's constant (6.626 × 10-³⁴ J/s)
f = frequency

First, we convert the frequencies to hertz as follows;

620THz = 6.2 × 10¹⁴Hz
3.10GHz = 3.1 × 10⁹Hz
46.0MHz = 4.6 × 10⁷Hz

E = 6.626 × 10-³⁴ × 6.2 × 10¹⁴ = 2.564eV
E = 6.626 × 10-³⁴ × 3.1 × 10⁹ = 1.28 × 10-⁵eV
E = 6.626 × 10-³⁴ × 4.6 × 10⁷ = 1.902 × 10-⁷eV

Learn more about energy of a photon at: brainly.com/question/2393994

#SPJ1

6 0

1 year ago

A) What is the meaning of specific heat capacity ?​

A) What is the meaning of specific heat capacity ?