Ask question

Ask question

All categories

2 years ago

9

A 80-cm-wide object seen through an optical device appears inverted and 50 cm wide. What is the magnification of the optical dev

ice?
A. 1.3
B. 0.63
C. -0.63
D. 2.5
E. -1.3

1 answer:

Arte-miy333 [17]2 years ago

6 0

The magnification is the ratio between the size of the image $h_i$ and the size of the original object $h_o$ :
$M= \frac{h_i}{h_o}$
The sign of the magnification contains another piece of information: if it's positive, then the image is upright, if it's negative, the image is inverted. In our problem, the image is inverted, so the magnification will have a negative sign, and it will be
$M=(-) \frac{50 cm}{80 cm}=-0.63$

You might be interested in

Vaccine ASAP don't got much time

djyliett [7]

Answer:

i answered this question not that long ago....it's D

Explanation:

7 0

2 years ago

Read 2 more answers

Which experiment has a non-uniform probability model? Select two answers. A. Tossing a fair coin one time B. Randomly guessing t

DiKsa [7]

Answer:

C and D

Explanation:

A uniform probability model is a probabilistic model characterized by a uniform probability density function, or uniform distribution.

In common language, a uniform probability distribution means that all possible outcomes in the probability space have the same probability of occurrence.

So:

- A fair toss of coin every possible outcome (H,T) has probability 0.5. It is modeled by by a uniform discrete distribution.

- Randomly selected answer to an MCQ with four options would have probability of success 0.25 for every MCQ. It is modeled by by a uniform discrete distribution.

- Spinning a spinner with sections that are different sizes, each section would have different probabilities proportional to the coverage area on the. It is modeled by a non-uniform discrete distribution

- Pulling a red marble out of a bag with 6 red marbles, 3 green marbles, and 1 yellow marble. Each successive time a red marble is drawn the probability decreases. Hence, non uniform distribution.

- Spinning a spinner on which all sections are the same size. Each section would have similar probabilities proportional to the coverage area on the. It is modeled by a uniform discrete distribution .

5 0

2 years ago

50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t

r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

$PE=mgh$

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

$KE=\frac{1}{2}mv^2$

where v is the speed of the object

When the ball is sitting on the top of the building, we have

$h=40 m$ , therefore the potential energy is not zero
$v=0$ , since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

$h=20 m$

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

$E=PE+KE=const.$

At the top of the building,

$E=PE_{top}$

While halfway through the fall,

$PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}$

And the mechanical energy is

$E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}$

which means

$KE_{half}=\frac{E}{2}$

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

The height of the ball relative to the ground is now zero: $h=0$ . This means that the potential energy of the ball is zero: $PE=0$

The kinetic energy, instead, is not zero: in fact, the ball has gained speed during the fall, so $v\neq 0$ , and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

4)

The potential energy of the ball as it sits on top of the building is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

h = 40 m is the height of the building, where the ball is located

Substituting the values, we find the potential energy of the ball at the top of the building:

$PE=(2)(9.8)(40)=784 J$

5)

The potential energy of the ball as it is halfway through the fall is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

h = 20 m is the height of the ball relative to the ground

Substituting the values, we find the potential energy of the ball halfway through the fall:

$PE=(2)(9.8)(20)=392 J$

6)

The kinetic energy of the ball halfway through the fall is given by

$KE=\frac{1}{2}mv^2$

where

m = 2 kg is the mass of the ball

v = 19.8 m/s is the speed of the ball when it is halfway through the fall

Substituting the values into the equation, we find the kinetic energy of the ball when it is halfway through the fall:

$KE=\frac{1}{2}(2)(19.8)^2=392 J$

We notice that halfway through the fall, half of the initial potential energy has converted into kinetic energy.

7)

The kinetic energy of the ball just before hitting the ground is given by

$KE=\frac{1}{2}mv^2$

where:

m = 2 kg is the mass of the ball

v = 28 m/s is the speed of the ball just before hitting the ground

Substituting the values into the equation, we find the kinetic energy of the ball just before hitting the ground:

$KE=\frac{1}{2}(2)(28)^2=784 J$

We notice that when the ball is about to hit the ground, all the potential energy has converted into kinetic energy.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0

2 years ago

An athlete of mass 70kg runs at a velocity of 6ms5-1 calculate the kinetic energy of the athlete.

Molodets [167]

KE=1/2 mv²
= 1/2 × 70 × (6)²
= 1260

8 0

2 years ago

Use the information and diagram to answer the following question.

Mumz [18]

Answer:b

Explanation:

Because I don’t know

8 0

2 years ago