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4 years ago

60 POINTS ANSWER CORRECTLY

Physics

2 answers:

ohaa [14]4 years ago

5 0

Answer:

C ) 1.53

Explanation:

The critical angle of a material is given by the formula

$sin c = \frac{1}{n}$

where

c is the critical angle

n is the refractive index

This formula is valid if the second medium is air (which is the case of the problem).

In this problem, we know the critical angle:

$c=40.8^{\circ}$

Therefore we can rearrange the equation to find the refractive index:

$n=\frac{1}{sin c}=\frac{1}{sin 40.8^{\circ}}=1.53$

nexus9112 [7]4 years ago

3 0

Answer:

The answer is "C"

Explanation:

I just took the test and got it correct.

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A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period 2.68 s, at what ampli

fredd [130]

Answer:

Part a)

$A = 1.78 m$

Part b)

$f = 2 rev/s$

Explanation:

Part A)

As we know that time period of the motion is given as

$T = 2.68 s$

so we have

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{2.68}$

$\omega = 2.34 rad/s$

now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

$mg = m\omega^2 A$

so we have

$A = \frac{g}{\omega^2}$

$A = \frac{9.81}{2.34^2}$

$A = 1.78 m$

Part b)

Now if the amplitude of the SHM is 6.23 cm

and now at this amplitude if object will lose the contact then in that case again we have

$mg = m\omega^2 A$

$g = \omega^2 (0.0623)$

$\omega = 12.5 rad/s$

so now we have

$2\pi f = 12.5$

$f = 2 rev/s$

3 0

3 years ago

Which of these instruments is listed with the wrong type of lens that it uses?

VikaD [51]

The wrong type of lens-Microscope, concave

Explanation:

A microscope Basically uses two convex lenses to magnify an object, or specimen.

There are 2 lenses in a microscope

Object Lens:The lens that is closer to the object
Eyepiece:The lens that is closer to the eye

Both the object lens and the eyepiece, is a convex lens.

3 0

3 years ago

A 100kg cannon at rest contains a 10kgcannonball. when fired,

slega [8]

Answer:

9 m/s

Explanation:

mass of cannon, M = 100 kg

mass of cannon ball, m = 10 kg

velocity of cannon ball, v = 90 m/s

Let the recoil velocity of cannon is V.

Us ethe conservation of linear momentum, as no external force is acting on the system, so the linear momentum of the system is conserved.

Momentum before the firing = momentum after the firing

M x 0 + m x 0 = M x V + m x v

0 = 100 x V + 10 x 90

V = - 9 m/s

Thus, the recoil velocity of cannon is 9 m/s.

4 0

3 years ago

A 0.26 kg rod of length 80 cm is suspended by a frictionless pivot at one end. It is held horizontal and released.

Daniel [21]

Answer:

a) $a_{center} = 7.38 ~m/s^2$

b) $a_{end} = 14.77 ~m/s^2$

c) $v_{center} = 2.43~m/s$

Explanation:

a) Immediately after the rod is released, the rod is still horizontal but now subject to gravity. Since one end of the rod is fixed, then the weight of the rod applies a torque. Then by Newton's Second Law, the acceleration can be found.

$\tau =I\alpha$

where I is the moment of inertia of the rod with respect to its fixed end, and α is the angular acceleration.

The net torque of the rod is

$\vec{\tau} = \vec{r} \times \vec{F}\\\tau = rF\sin(90) = rF$

where r is the distance from center of the mass to the fixed end, so r = 0.4 m.

The weight of the rod is w = mg = 0.26 x 9.8 = 2.54 N.

So the net torque is τ = 1.01 Nm.

The moment of inertia of the rod is

$I = \frac{1}{3}mL^2 = \frac{1}{3}(0.26)(0.8)^2 = 0.055~kg m^2$

So, the Newton's Second Law yields

$\tau = I\alpha\\\alpha = \frac{\tau}{I} = \frac{1.01}{0.055} = 18.47$

The relation between angular acceleration and linear acceleration is a = αr

So, the linear acceleration of the rod is

$a = \alpha r = 7.38~m/s^2$

b) Using the same relationship between angular acceleration and linear acceleration, the linear acceleration of the end of the rod can be found.

$a = \alpha L = 14.77~m/s^2$

c) The conservation of energy can be used to find the velocity when the rod is vertical.

$K_1 + U_1 = K_2 + U_2\\0 + mg(L/2) = \frac{1}{2}I\omega^2 + 0\\(0.26)(9.8)(0.4) = \frac{1}{2}(0.055)\omega^2\\\omega = 6.08~rad/s$

The linear velocity is v = ωr, so

v = 2.43 m/s.

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3 years ago

The sum of potential and kinetic energies in the particles of a substance is called

olganol [36]

Internal energy.

Explanation:

In any substance/object, the particles inside it (atoms/molecules) constantly move in random directions and with random speeds (this motion is called Brownian motion). As a result, the particles have some kinetic energy (which is proportional to the temperature of the substance). Moreover, the particles interact with each other due to the presence of electrostatic intermolecular forces, and as a result, the particles also have some potential energy.

The sum of the kinetic energies and potential energies of the particles in a substance is called internal energy.

6 0

4 years ago

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