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umka2103 [35]
3 years ago
11

A plane mirror and a concave mirror (f = 8.20 cm) are facing each other and are separated by a distance of 25.0 cm. An object is

placed between the mirrors and is 12.5 cm from each mirror. Consider the light from the object that reflects first from the plane mirror and then from the concave mirror. What is the distance of the image (di) produced by the concave mirror?
Physics
2 answers:
vaieri [72.5K]3 years ago
7 0

Answer:

Sol: - I1: Reflexión a la distancia del primer espejo del objeto = 5 cm a la derecha ==> distancia de la imagen = 5 cm a la izquierda Ahora, I2: Reflexión al espejo 2 distancia del objeto = 20-5 = 15 cm a la izquierda == > distancia de imagen = 15 cm a la derecha Ahora, I3: la imagen I2 será como objeto para el espejo 1 distancia de objeto = 15 + 20 = 35 cm a la derecha ==> distancia de imagen = 35 cm a la izquierda Ahora, I4: I1 la imagen será como objeto para el espejo 2 distancia del objeto = 5 + 20 = 25 cm a la izquierda ==> distancia de la imagen = 25 cm a la derecha Ahora, I5: la imagen I4 será como objeto para la distancia del espejo 1 ...

Explanation:

Dafna11 [192]3 years ago
6 0

Answer:

The distance of image formed by the concave mirror is at 10.495 cm.

Explanation:

Since we know that a plane mirror forms image at a distance behind the mirror at which the object is placed in front of the mirror thus we have.

Image formed by the plane mirror is at a distance of 12.5 cm behind the plane mirror.

The image formed by the plane mirror is at a distance of 12.5 + 25.0 = 37.5 cm  from the pole of the concave mirror.

Thus by  mirror formula we have

\frac{1}{v}+\frac{1}{u}=\frac{1}{f}

Using the standard sign convention we have

u = location of object = -37.5 cm

f = focal length of mirror = -8.20 cm

Thus using the given values we obtain

\frac{1}{v}+\frac{1}{-37.5}=\frac{1}{-8.2}\\\\\therefore v=-10.495cm

Thus the image is formed at a distance of 10.495 cm from the pole of mirror towards left of mirror.

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A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

Compute U if the bulb remains on for 5h

Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

7 0
3 years ago
(a)
Marta_Voda [28]

a) The momentum of the coconut is 3 kg m/s

b) At first, the air resistance is negligible, so the coconut accelerates due to the force of gravity

c) The coconut reaches its terminal velocity

Explanation:

a)

The momentum of an object is given by the equation

p=mv

where

m is the mass of the object

v is its velocity

For the coconut in this problem, we have:

m = 1.5 kg (mass)

v = 2 m/s (velocity)

Therefore, its momentum is

p=(1.5)(2)=3 kg m/s

B)

There are only two forces acting on the coconut during its fall:

  • The force of gravity, of magnitude mg (m= mass of the coconut, g = acceleration of gravity), acting downward
  • The air resistance, acting upward, whose magnitude is proportional to the speed of the coconut

During the first momentums of the fall, the speed of the coconut is still low, so the air resistance is mostly negligible, and therefore only the force of gravity is acting on the coconut. Since this force is constant, it means that the acceleration of the coconut is constant: therefore, its velocity keeps increasing during the fall, and the coconut speeds up.

C)

If the tree is very tall, the fall of the coconut lasts long, and the  speed of the coconut keeps increasing. Since the air resistance is proportional to the speed, this means that at some point, the air resistance is no longer negligible, and it starts to have some effect on the fall of the coconut. In particular, at a certain point, the air resistance will become equal (in magnitude) to the force of gravity (but opposite in direction): this means that  from this point, the acceleration of the coconut will be zero, and therefore the coconut will continue its motion at constant velocity. This velocity is called terminal velocity, and it occurs when the force of gravity is equal to the air resistance:

mg = F_r

where F_r is the air resistance.

Learn more about forces and weight:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

8 0
3 years ago
You have 1.2kg of (AU)gold (that's quite a bit of money at $1839/ounce). Using a heat source you apply 3096 J to the gold and re
melamori03 [73]

Answer:

129 J/Kg°C

Explanation:

Given :

Mass of gold, m = 1.2kg

Quantity of heat applied, Q = 3096 J

Temperature, t2 = 40°C

Temperature, t1 = 20°C

Change in temperature, dt = (40-20)°C = 20°C

Using the relation :

Q = mcdt

Where, C = specific heat capacity of gold

3096 = 1.2kg * C * 20°C

3096 J = 24kg°C * C

C = 3096 J / 24 kg°C

C = 129 J/Kg°C

7 0
3 years ago
PLEASE HELP IN ONE MINUTE
Greeley [361]
I think it’s 8 hours. I’m sorry if I’m wrong.
I just did 400 divided by 50
4 0
3 years ago
Bob walks 370 m south, then jogs 410 m southwest, then walks 370 m in a direction 28 degrees east of north.
wlad13 [49]
We will measure all angles from West, the negative x-axis and divide the journey into 3 parts:
P1 = 370y
P2 = 410cos(45)x + 410sin(45)y = 290x + 290y
P3 = 370cos(270 - 28)x + 370sin(270 - 28) = -174x - 327y

Overall displacement:
x = 290 - 174 = 116 m
y = 370 + 290 - 327 = 333 m

displacement = √(116² + 333²)
= 353 m

Direction:
tan(∅) = y/x
∅ = tan⁻¹ (333 / 116)
∅ = 70.8° from West.
5 0
3 years ago
Read 2 more answers
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