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4 years ago

We know we have exerted of force even when we have done no work this is called _____

Physics

1 answer:

zheka24 [161]4 years ago

5 0

Answer: The correct answer is zero work done.

Explanation:

Work is said to be done when the object moves through a distance when the force is applied to the object.

If the object does not move a distance even the force is exerted on the object then the work done is zero in this case.

Therefore, when the force is exerted even when no work is done then this is called zero work done.

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A layer of oil (n = 1.38) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid.

Vinil7 [7]

Answer:

Refractive index of unknown liquid = 1.56

Explanation:

Using Snell's law as:

$n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence ( 65.0° )

${\theta_r}$ is the angle of refraction ( 53.0° )

${n_r}$ is the refractive index of the refraction medium (unknown liquid, n=?)

${n_i}$ is the refractive index of the incidence medium (oil, n=1.38)

Hence,

$1.38\times {sin65.0^0}={n_r}\times{sin53.0^0}$

Solving for ${n_r}$ ,

Refractive index of unknown liquid = 1.56

4 0

3 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

3 years ago

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz. What is the period of the crystal's motion?

Elan Coil [88]

Answer:

Time period, $T=3.05\times 10^{-5}\ s$

Explanation:

Given that,

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz

We need to find the period of the crystal's motion. The relationship between the frequency and the time period is given by :

$T=\dfrac{1}{f}$

T is the time period of the crystal's motion.

Time period is given by :

$T=\dfrac{1}{32768}$

$T=3.05\times 10^{-5}\ s$

So, the time period of the crystal's motion is $3.05\times 10^{-5}\ s$ . Hence, this is the required solution.

8 0

3 years ago

Macronutrients include carbohydrates, proteins, lipids, and _____.

melamori03 [73]

<span>Macronutrients include carbohydrates, proteins, lipids, and fats.</span>

8 0

3 years ago

3. A coil of 100 turns encloses an area of 100 cm2. It is placed at an angle of 700 with a

sasho [114]

Explanation:

Given that,

Number of turns in the coil, N = 100

Area of the coil, A = 100 cm² = 0.01 m²

It is placed at an angle of 70°.

Magnetic field, B = 0.1 Wb/m²

We need to find the magnetic flux through the coil and the emf is induced in the coil after 10⁻³ s.

Magnetic flux is given by :

$\phi =BA\cos\theta\\\\\text{For N turns},\\\phi =NBA\cos\theta \\\\\phi=100\times 0.1\times 0.01\times \cos(70)\\\\=0.034\ Wb$

So, the magnetic flux through the coil is 0.1 Wb.

Emf induced in the coil is :

$\epsilon=\dfrac{-d\phi}{dt}\\\\=\dfrac{0.034}{10^{-3}}\\\\=34\ V$

So, 34V of emf is induced in the coil.

7 0

3 years ago