PHYSICS COLLEGE

How do lenses and mirrors compare in their interactions with light? A. Lenses spread apart light; mirrors do not.
B. Lenses reflect light; mirrors do not.
C. Lenses refract light; mirrors do not.
D. Lenses focus light; mirrors do not.​

Answers

Answer 1
Answer:

This question involves the concepts of reflection and refraction.

The comparison of lenses and mirrors in their interaction with light is "C. Lenses refract light; mirrors do not.".

LENSES AND MIRRORS

When it comes to the interaction with light, the key difference between lenses and mirrors is the difference of refraction and reflection. Reflection means the complete rebound of the light rays after striking on a surface without any absorption or transmission. On the other hand, refraction is the  bending of light rays, while passing through a medium, without any rebound or absorption.

Lenses are tansparent from both sides, so they refract the light rays. While, mirrors are coated opaque from one side, so they reflect back the light rays.

Learn more about reflection and refraction here:

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Answer 2
Answer:

Answer:

C. lenses refract light; mirrors do not

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Could you please solve it with shiwing the full work1-

A high powered projectile is fired horizontally from the top of a cliff at a speed of 638.6 m/s. Determine the magnitude of the velocity (in m/s) after 5 seconds.

Take gravitational acceleration to be 9.81 m/s2.

2-

A man throws a ball with a velocity of 20.9 m/s upwards at 33.2° to the horizontal. At what vertical distance above the release height (in metres) will the ball strike a wall 13.0 m away ?

Take gravitational acceleration to be 9.81 m/s2.



3-

A particle is moving along a straight path and its position is defined by the equation s = (1t3 + -5t2 + 3) m, where t is measured in seconds. Determine the average velocity (in m/s) of the particle when t = 5 seconds.

4-

A particle has an initial speed of 26 m/s. The particle undergoes a deceleration of a = (-9t) m/s2, where t is measured in seconds. Determine the distance (in metres) the particle travels before it stops. When t = 0, s = 0.

Answers

Answer:

1.V= 640.48 m/s :total velocity in t= 5s

2. Y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. v =25m/s

4. s= (-1.5t³+26t ) m

Explanation:

1. Parabolic movement in the x-y plane , t=5s

V₀=638.6 m/s=Vx  :Constant velocity in x

Vy=V₀y +gt= 0+9.8*5  = 49 m/s : variable velocity in y

v=\sqrt{v_(x) ^(2) +v_(y) ^(2) }

v=\sqrt{ 638.6^(2) +49 ^(2) }

V= 640.48 m/s : total velocity in t= 5s

2. v_(ox) =v_(o) cos33.2=20.9*cos33,2= 17.49 m/s

v_(oy)=v_(o)*sin33,2 =20.9*sin33,2=11.44 m/s

x=v₀x*t

13=v₀x*t

13=17.49*t

t=13/17.49=0.743s : time for 13.0 m away

th=v₀y/g=11.44/9.8= 1,17s :time for maximum height

at t=0.743 sthe ball is going up ,then g is negative

y=v₀y*t - 1/2 *g¨*t²

y=11.44*0.743 -1/2*9.8*0.743²

y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. s = (1t3 + -5t2 + 3) m

v=3t²-10t=3*25-50=75-50=25m/s

at t=0, s=3 m

at t=5s s=5³-5*5²+3

4.  a = (-9t) m/s2

a=dv/dt=-9t

dv=-9tdt

v=∫ -9tdt

v=-9t²/2 + C1 equation (1)

in t=0  , v₀=26m/s ,in the equation (1) C1= 26

v=-9t²/2 + 26=ds/dt

ds=( -9t²/2 + 26)dt

s= ∫( -9t²/2 + 26)dt

s= -9t³/6+26t+C2 Equation 2

t = 0, s = 0 , C2=0

s= (-9t³/6+26t ) m

s= (-1.5t³+26t ) m

A wire is wrapped around a piece of iron, and then electricity is run through the wire. What happens to the iron?

Answers

Search ResultsBy simply wrapping wire that has an electrical current running through it around a nail, you can make an electromagnet. When the electric current moves through a wire, it makes a magnetic field. ... You can make a temporary magnet by stroking apiece of iron or steel (such as a needle) along with a permanent magnet.

Hope This Helps!

At what distance from a long straight wire carrying acurrentof 5.0A is the magnitude of the magnetic field due to the
wireequal to the strength of the Earth's magnetic field of about
5.0 x10^-5 T?

Answers

Answer:

The distance is 2 cm

Solution:

According to the question:

Magnetic field of Earth, B_{E} = 5.0* 10^(- 5) T

Current, I = 5.0 A

We know that the formula of magnetic field is given by:

B = \farc{\mu_(o)I}{2\pi d}

where

d = distance from current carrying wire

Now,

d = (\mu_(o)I)/(2\pi B)

d = (4\pi* 10^(- 7)* 5.0)/(2\pi* 5.0* 10^(- 5))

d = 0.02 m 2 cm

Which two types of energy does a book have as it falls to the floor

Answers

Answer:

kinetic and potential energy

Explanation:

A farmer is using a rope and pulley to lift a bucket of water from the bottom of a well. the farmer uses a force f1=57.5 n to pull the bucket of water upwards. the total mass of the bucket of water is f2= 3.9kg. -Calculate how much work Wg in J gravity does on the bucket filled with water as the farmer lifts it up the well.
-Calculate the net work Wnet in J done on the bucket of water by the two forces F1 and Fg.

Answers

Final answer:

To calculate the work done by gravity on the bucket of water as it is lifted up the well, multiply the weight of the bucket by the lifting distance. The net work done on the bucket by the force applied by the farmer and gravity is the sum of the work done by both forces. The net work is represented by the equation Wnet = W1 + Wg.

Explanation:

To calculate how much work gravity does on the bucket filled with water as the farmer lifts it up the well, we need to multiply the force of gravity (weight) by the vertical distance the bucket is lifted. The equation for work is W = Fd, where W is the work done, F is the force, and d is the distance. In this case, the force of gravity is the weight of the bucket, which can be calculated by multiplying the mass by the acceleration due to gravity (9.8 m/s^2).

So, the work done by gravity (Wg) on the bucket is Wg = Fg * d = (m * g) * d = (3.9 kg * 9.8 m/s^2) * d = 38.22 d Joules.

To calculate the net work done on the bucket by the two forces, we can use the equation Wnet = W1 + Wg, where W1 is the work done by force F1 and Wg is the work done by gravity. Since force F1 and the displacement (lifting distance) are both vertical, the work done by F1 is given by W1 = F1 * d.

Therefore, the net work done on the bucket by forces F1 and gravity is Wnet = F1 * d + Fg * d = (57.5 N) * d + (3.9 kg * 9.8 m/s^2) * d = (57.5 N + 38.22 d) Joules.

Learn more about Work done by forces here:

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A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30.0 m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity of the stone just before it hits the ground? Ignore any effects of air resistance.

Answers

Answer:

vf = 30 m/s : (the magnitude of the velocity of the stone just before it hits the ground)

Explanation:

Because the stone moves with uniformly accelerated movement we apply the following formulas:

vf²=v₀²+2*g*h Formula (1)

Where:  

h: displacement in meters (m)  

v₀: initial speed in m/s

vf: final speed in m/s  

g: acceleration due to gravity in m/s²

Free fall of the stone

Data

v₀ =  10 m/s

vf =  30.0 m/s

g = 9,8 m/s²

We replace data in the formula (1) to calculate h:

vf²=v₀²+2*g*h

(30)² = (10)² + (2)(9.8)*h

(30)²- (10)²= (2)(9.8)*h

h =( (30)²- (10)²) /( 2)(9.8)

h = 40.816 m

Semiparabolic movement of the stone

Data

v₀x =  10 m/s

v₀y =  0 m/s

g = 9.8 m/s²

h= 40.816 m

We replace data in the formula (1) to calculate vfy :

vfy² = v₀y² + 2*g*h

vfy² = 0 + (2)(9.8)( 40.816)

v_(fy)=√(2*9.8*40.816) = 28.284 (m)/(s)

v_(f)=\sqrt{v_(ox)^2+v_(fy)^2}=√((10)^2+(28.284)^2) = 30(m)/(s)

The magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

The given parameters;

initial vertical velocity of the stone, v_y_0 = 10 m/s

final vertical velocity of the stone, v_y_f = 30 m/s

The height traveled by the stone before it hits the ground is calculated as;

v_y_f^2 = v_y_0^2 + 2gh\n\nh = (v_y_f^2- v_y_0^2)/(2g) \n\nh = ((30)^2 - (10)^2)/(2* 9.8) \n\nh = 40.82 \ m

If the the stone is projected horizontally with initial velocity of 10 m/s;

the initial vertical velocity = 0

Final vertical velocity of the stone is calculated as follow;

v_y_f^2 = v_y_0^2 + 2gh\n\nv_y_f^2 = 0 + 2* 9.8* 40.82\n\nv_y_f^2 = 800.07\n\nv_y_f = √(800.07) \n\nv_y_f = 28.28 \ m/s

The horizontal velocity doesn't change.

the final horizontal velocity, v_x_f = initial horizontal velocity = 10 m/s

The resultant of the final velocity of the stone before it hits the ground;

v _f= √(v_x_f^2 + v_y_f^2) \n\nv_f = √(10^2 + 28.28^2) \n\nv_f= 29.99 \ m/s \approx 30 \ m/s

Thus, the magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

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