PHYSICS COLLEGE

What is the critical angle for light traveling from crown glass (n = 1.52) into water (n = 1.33)?42 degrees
48 degrees
57 degrees
61 degrees

Answers

Answer 1

Answer:

Answer:

61 degrees, I just did the test.

Explanation:

Answer 2

Answer:

Answer: 61 degrees

Explanation:

I just did the question and got it right

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The pressure in a section of horizontal pipe with a diameter of 2.5 cm is 139 kPa. Water ï¬ows through the pipe at 2.9 L/s. If the pressure at a certain point is to be reduced to 101 kPa by constricting a section of the pipe, what should be the diameter of the constricted section? The acceleration of gravity is 9.81 m/s2 . Assume laminar nonviscous ï¬ow.
A thin Nichrome wire connected to an ammeter surrounds a region of time-varying magnetic flux, and the ammeter reads 13 amperes. If instead of a single wire we use a coil of thin Nichrome wire containing 23 turns, what does the ammeter read?

During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of 176 N on the wire. The left section of the wire makes an angle of 12.5° relative to the horizontal and sustains a tension of 413 N. Find the (a) magnitude and (b) direction (as an angle relative to horizontal) of the tension that the right section of the wire sustains.

Answers

Answer:

a. 12.12°

b. 412.04 N

Explanation:

Along vertical axis, the equation can be written as

T_1 sin14 + T_2sinA = mg

T_2sinA = mg - T_1sin12.5 ....................... (a)

Along horizontal axis, the equation can be written as

T_2×cosA = T_1×cos12.5 ......................... (b)

(a)/(b) given us

Tan A = (mg - T_1sin12.5) / T_1 cos12.5

= (176 - 413sin12.5) / 413×cos12.5

A = 12.12 °

(b) T2 cosA = T1 cos12.5

T2 = 413cos12.5/cos12.12

= 412.04 N

Answer:

Magnitude - 11.83 Degree

Direction - 422.42 N

Explanation:

Given data:

Downward force on wire 176 N

Angle made by left section of wire 12.5 degree with horizontal

Tension force = 413 N

From figure

Applying quilibrium principle at point A

The vertical and horizontal force is 0

then we have

$Tcos\theta = 413 N$ ........1

$176 = 413 sin 12.5 + Tsin\theta$ .......2

$Tsin\theta = 176 - 89.39 = 86.6$ .......3

divide equation 3 by 1

we get

$\theta = tan^(-1) (0.2096)$

$theta = 11.83^o$ ...........4

from equation 3 and 4

$T = (86.6)/(sin 11.83)$

T = 422.42 N

A force of 40 N is applied in a direction perpendicular to the end of a 9 m long bar that pivots about its other end. Find the torque that this force produces about the pivot point. magnitude

Answers

Answer:

360 Nm

Explanation:

Torque: This is the force that tend to cause a body to rotate or twist. The S.I unit of torque is Newton- meter (Nm).

From the question,

The expression of torque is given as

τ = F×d......................... Equation 1

Where, τ = Torque, F = force, d = distance of the bar perpendicular to the force.

Given: F = 40 N, d = 9 m

Substitute into equation 1

τ = 40(9)

τ = 360 Nm.

Answer:

360Nm

Explanation:

Torque is defined as the rotational effect of a force. The magnitude of a torque τ, is given by;

τ = r F sin θ

Where;

r = distance from the pivot point to the point where the force is applied

F = magnitude of the force applied

θ = the angle between the force and the vector directed from the point of application to the pivot point.

From the question;

r = 9m

F = 40N

θ = 90° (since the force is applied perpendicular to the end of the bar)

Substitute these values into equation (i) as follows;

τ = 9 x 40 sin 90°

τ = 360Nm

Therefore the torque is 360Nm

A person on a rocket traveling at 0.47 c (with respect to the Earth) observes a meteor come from behind and pass her at a speed she measures as0.47 c.How fast is the meteor moving with respect to the Earth?

Answers

The concept to solve this problem is related to the relativistic physics for which the speed of the object in different frames of reference is related. This concept is called Velocity-addition formula

and can be written as,

$u = (v+u')/(1+(vu')/(c^2))$

Where,

u = Velocity of a body within a Lorentz Frame

v = Velocity of a second frame

u'= The transformed velocity of the body within the second frame

c = speed of light

Replacing we have to

$u = (v+u')/(1+(vu')/(c^2))$

$u = (0.47c+0.47c)/(1+(0.47c*0.47c)/(c^2))$

$u = 0.769c$

$u \approx 230'700.000m/s$

Therefore the meteor moving with respect to the Earth to 230'700.000m/s

A laser (electromagnetic wave) has the maximum electric field strength of 1.0x1011 V/m. What is the force the laser applies on a mirror (totally reflective) of 5.0 mm2 area? A. 2.76 x105N B. 1.21 x106N C. 1.94 x106N D.4.43 x105 N E. 7.82 x104N

Answers

Answer:

The correct option is D

Explanation:

From the question we are told that

The maximum electric field strength is $E = 1.0 *10^(11) \ V/m$

The area is $A = 5.0 \ mm^2 = 5.0 *10^(-6) \ m^2$

Generally the force the laser applies is mathematically represented as

$F = \epsilon_o * E ^2 * A$

Here $\epsilon_o = 8.85*10^(-12) C/(V \cdot m)$

$F = 8.85*10^(-12) * (1.0 *10^(11)) ^2 * 5.00*10^(-6 )$

=> $F = 4.43 *10^(5) \ N$

Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as he can and jump onto the tire at full speed. The tire has a mass of 10.0 kg and hangs 3.50 m straight down from a tree branch. Ezra stands back 10.0 m and accelerates to a speed of 3.62 m/s before jumping onto the tire swing. (a) How fast are Ezra and the tire moving immediately after he jumps onto the swing? m/s (b) How high does the tire travel above its initial height?

Answers

Answer:

a) $v=5.6725\,m.s^(-1)$

b) $h= 1.6420\,m$

Explanation:

Given:

mass of the body, $M=20\,kg$

mass of the tyre, $m=10\,kg$

length of hanging of tyre, $l=3.5m$

distance run by the body, $d=10m$

acceleration of the body, $a=3.62m.s^(-2)$

(a)

Using the equation of motion :

$v^2=u^2+2a.d$ ..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

$u=0m.s^(-1)$

putting the values in eq. (1)

$v^2=0^2+2* 3.62 * 10$

$v=8.5088\,m.s^(-1)$

Now, the momentum of the body just before the jump onto the tyre will be:

$p=M.v$

$p=20* 8.5088$

$p=170.1764\,kg.m.s^(-1)$

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

$(M+m)* v'=p$

$(20+10)* v'=170.1764$

$v'=5.6725\,m.s^(-1)$

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

$(1)/(2) (M+m).v'^2=(M+m).g.h$

$(1)/(2) (20+10)* 5.6725^2=(20+10)* 9.8* h$

$h\approx 1.6420\,m$

above the normal hanging position.

Which of the following sets of characteristics describe what we know about the outer planets? (2 points)Lowest daily average temperature, smaller in size and few or no moons.
Many moons, smaller in size and a ring system.
Rocky surface, closest to the sun and larger in size,
Gaseous composition, larger size and many moons.

Answers

Gaseous composition, larger size and many moons describe about the outer planets.

What are outer planets?

Jupiter, Saturn, Uranus, and Neptune are the four outer planets. They are all gas giants consisting primarily of hydrogen and helium. Their interiors are liquid and contain thick gaseous outer layers. Numerous moons and planetary rings consisting of dust and other particles are present on every one of the outer planets.

To know more about outer planets refer to :

brainly.com/question/18392639

#SPJ2

Answer: D

Explanation: Gaseous composition, larger size and many moons

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