1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
IrinaVladis [17]
3 years ago
5

A plane mirror of circular shape with radius r=20cm is fixed to the ceiling. A bulb is to be placed on the axis of the mirror. A

circular area of r=1m on the floor is to be illuminated after reflection of light from mirror.The height of the room is 3m.What is maximum distance from the centre of the mirror and the bulb so that the required area is illuminated?

Physics
1 answer:
KIM [24]3 years ago
5 0

Answer:

0.75 m

Explanation:

Let's call the distance between the bulb and the mirror x.

The bulb and the length of the mirror form a triangle.  The mirror and the illuminated area on the floor form a trapezoid.  If we extend the lines from the mirror edge to the reflected image of the bulb, we turn that trapezoid into a large triangle.  This triangle and the small triangle are similar.  So we can say:

x / 0.4 = (3 + x) / 2

Solving for x:

2x = 0.4 (3 + x)

2x = 1.2 + 0.4 x

1.6 x = 1.2

x = 0.75

So the bulb should located no more than 0.75 m from the mirror.

You might be interested in
Physics students study a piano being pulled across a room on a rug. They know that when it is at rest, it experiences a gravitat
Vanyuwa [196]
The static frictional force is greater than the kinetic frictional force, so the static frictional force is greater than 1200 N.
5 0
3 years ago
Read 2 more answers
What is the massof the largest ruby?
alexandr402 [8]
I think the answer is 2283g
4 0
3 years ago
How do you find acceleration due to gravity with time and height given?
Feliz [49]

Here, height is given which will be the distance for a freely falling object.

The velocity will be

v=\text{ }\frac{h}{t}

and the acceleration will be

a=\frac{v}{t}

In this way, the formula works.

3 0
1 year ago
Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi
kozerog [31]

Answer:

a) (0, -33, 12)

b) area of the triangle : 17.55 units of area

Explanation:

<h2>a) </h2>

We know that the cross product of linearly independent vectors \vec{A} and \vec{B} gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors \vec{A} = \vec{P}-\vec{Q} and \vec{B} = \vec{R} - \vec{Q} are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)

\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)

\vec{A} \times  \vec{B} = (A_y B_z - B_y A_z) \  \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}

\vec{A} \times  \vec{B} = ( (-1) * 3 - 1 * (-3) ) \  \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}

\vec{A} \times  \vec{B} = ( - 3 + 3 ) \  \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}

\vec{A} \times  \vec{B} = 0 \  \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}

\vec{A} \times  \vec{B} =(0, -33, 12)

<h2>B)</h2>

We know that \vec{A} and \vec{B} are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

|\vec{A} \times  \vec{B} | = 2 * area_{triangle}

so:

\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}

\sqrt{1233} = 2 * area_{triangle}

35.114= 2 * area_{triangle}

17.55 \ units \  of \ area =  area_{triangle}

5 0
2 years ago
En que parte de la isla de cuba tuvieron lugar los principales acontecimientos de la guerra de independencia?
DedPeter [7]

Answer:

El 24 de febrero de 1895, por órdenes de Martí se levantan 35 aldeas en el Oriente de Cuba en lo que se ha dado en llamar el Grito de Baire.

Entre los lugares están Ibarra, Guantánamo y Manzanillo.

Explanation:

8 0
3 years ago
Other questions:
  • A baseball is traveling (+20 m/s) and is hit by a bat. It leaves the bat traveling (-30 m/s). What is the change in the velocity
    9·2 answers
  • What creates an ionic bond?
    12·2 answers
  • Farmers during the dust bowl of the 1930s watched their soil be carried away by wind what process removed the soil from the plac
    13·2 answers
  • A young lady can paddle a canoe in a lake 7.7 m/s. She paddles downstream in a river whose current is 12.4 m/s. What is the comb
    10·1 answer
  • A group of college students eager to get to Florida on a spring break drove the 710-mi trip with only minimum stops. They comput
    15·1 answer
  • 15. Use the chemical equation below to determine how many moles of ammonia
    6·1 answer
  • A flask with a tap has a volume of 200cm3 when full of air, tye flask has a mass of 30.98g. the flask is connected to vacuum pum
    6·1 answer
  • A common black ant discovers a piece of bread 85 cm east of the entrance of her nest. If the ant carries 10 bits of bread back t
    13·1 answer
  • Which statement is correct about the equation for work?
    10·1 answer
  • Una carga eléctrica de 120 Coulomb pasa uniformemente por la sección transversal de un hilo conductor durante un minuto. La inte
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!