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
pantera1 [17]
3 years ago
11

Dr. Kirwan is preparing a slide show that he will present to the executive board at tonight's committee meeting. He places a 3.5

0-cm slide behind a lens of 20.0 cm focal length in the slide projector. A) How far from the lens should the slide be placed in order to shine on a screen 6.00 m away? B) How wide must the screen be to accommodate the projected image?​
Physics
1 answer:
lisov135 [29]3 years ago
8 0

Answer:

A) d_o = 20.7 cm

B) h_i = 1.014 m

Explanation:

A) To solve this, we will use the lens equation formula;

1/f = 1/d_o + 1/d_i

Where;

f is focal Length = 20 cm = 0.2

d_o is object distance

d_i is image distance = 6m

1/0.2 = 1/d_o + 1/6

1/d_o = 1/0.2 - 1/6

1/d_o = 4.8333

d_o = 1/4.8333

d_o = 0.207 m

d_o = 20.7 cm

B) to solve this, we will use the magnification equation;

M = h_i/h_o = d_i/d_o

Where;

h_o = 3.5 cm = 0.035 m

d_i = 6 m

d_o = 20.7 cm = 0.207 m

Thus;

h_i = (6/0.207) × 0.035

h_i = 1.014 m

You might be interested in
A small car with mass of 0.800 kg travels at a constant speed
Alexandra [31]

Answer:

The equation of equilibrium at the top of the vertical circle is:

\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}

The speed experimented by the car is:

\frac{N}{m}+g=\frac{v^{2}}{R}

v = \sqrt{R\cdot (\frac{N}{m}+g) }

v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}

v\approx 9.302\,\frac{m}{s}

The equation of equilibrium at the bottom of the vertical circle is:

\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}

The normal force on the car when it is at the bottom of the track is:

N=m\cdot (\frac{v^{2}}{R}+g )

N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)

N=21.690\,N

7 0
3 years ago
Hello please help i’ll give brainliest
Rashid [163]

Answer:

i believe the answer would be fault lines

Explanation:

6 0
3 years ago
1. Take the speed at which the body moves as 10 m/s 2. Calculate the distance travelled by the body every second 3. Take the tim
exis [7]

Answer:

a)this graph is also a line     b) in both cases we have a uniform movement

Explanation:

In this exercise we have a uniform movement

     v = d / t

     d = v t

in the table we give some values ​​to make the graph

       t (s)    d (m)

        1         10

        2        20

        3         30

In the attached we can see the graph that is a straight line

we have another vehicle at v = 50 me / S

t (s)     d (m)

1         50

2        100

3         150

this graph is also a line

b) in both cases we have a uniform movement

3 0
4 years ago
What is the relative size and composition of the universe, a galaxy, and a solar system?
slava [35]

The solar system is smaller than a galaxy. Galaxies are composed of solar systems. The solar system is composed of planets revolving around a star (such as our sun). To put into perspective, there are about hundreds of billions of stars in our galaxy (the milky way). On the other hand, billions of galaxies make up the universe.

The speed of light is constant which is 299 792 458 m / s. The time is taken for light to reach an object can be used to measure the distance between objects in space. This is done by multiplying the time by the speed of light.

Distance in space is so vast and therefore impractical to use the basic SI units of meters and kilometers we use here on earth. Therefore the use of astronomical units and light years is used. Light years is the distance that light can cover in a year in the vacuum of space, while astronomical units is the distance between solar systems.

6 0
2 years ago
At a time when mining asteroids has become feasible, astronauts have connected a line between their 3740-kg space tug and a 5690
Gwar [14]

Answer:

61.4 s

Explanation:

The distance d₁ traveled by the asteroid:

d_1=\frac{1}{2}a_1t^2

The distance d₂ traveled by the space ship:

d_2=\frac{1}{2}a_2t^2

The total distance d:

d=d_1+d_2=\frac{1}{2}(a_1+a_2)t^2

Solving for time t:

t=\sqrt{\frac{2d}{a_1+a_2}}=\sqrt{\frac{2d}{F(\frac{1}{m_1}+\frac{1}{m_2})}}=\sqrt{\frac{2dm_1m_2}{F(m_1+m_2)}}

3 0
3 years ago
Read 2 more answers
Other questions:
  • Find the kinetic energy of a block twice as massive (20kg) moving at the same speed
    14·1 answer
  • The temperature scale which starts at absolute zero is the _____.
    8·2 answers
  • What are three physical of aluminum foil
    7·1 answer
  • Why does your hair stand up when you go down a slide?
    6·1 answer
  • What does a thermostat do if it gets too cool? ____________________________________ 2. what does a thermostat do if it gets too
    13·1 answer
  • Convert 35 centimeters to inches
    13·1 answer
  • What is the power rating of a light bulb that transfers 120J of energy in 2seconds
    11·1 answer
  • Explain the difference between a high tide and a low tide.
    10·1 answer
  • What is constant of a spring if there is 150 j when stretched 0.25 m?
    10·1 answer
  • When light enters water, it bends. what does the amount of bending depend on?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!