All categories

3 years ago

Creates an image that appears upside down behind the focal point

1 answer:

8 0

An image that appears upside down behind the focal point is an image that is reflected on a concave mirror. Mirrors reflect different kinds of images based on the placement of an object that is reflected towards it. There are two kinds of mirrors, concave and a convex mirrors, the latter makes objects seem smaller and farther than where it is exactly.

You might be interested in

An object weighs 15N in air and 13N when completely

anastassius [24]

Answer:

The volume of water displaced = 0.204 m³

Explanation:

From Archimedes' principle,

Upthrust = lost in weight = weight of water displaced.

U = W₁ - W₂..................... Equation 1

Where U = upthrust, W₁ = weight in air, W₂ = weight when completely submerged in water.

Given: W₁ = 15 N, W₂ = 13 N

Substituting these values into equation 1

U = 15 - 13

U = 2 N

But,

Weight of water displaced = mass of water displaced. × acceleration due to gravity.

Mass of water displaced = weight of water displaced/ acceleration due to gravity.................... Equation 2

Where: acceleration due to gravity = 9.8 m/s², weight of water displaced = 2 N

Substituting these values into equation 2

Mass of water displaced = 2/9.8

Mass of water displaced = 0.204 kg.

Also,

Volume of water displaced = mass of water displaced/ Density of water.............. Equation 3

Where Density of water = 1.0 kg/m³, mass of water displaced = 0.204 kg

Substituting these values into equation 3 above,

Volume of water displaced = 0.204/1

volume of water displaced = 0.204 m³

Therefore the volume of water displaced = 0.204 m³

8 0

2 years ago

A box is pushed 40 m by a mover. The amount of work done was 2,240 j. How much force was exerted on the box

Georgia [21]

The force exerted on the box is 56 N

Explanation:

The work done by a force on an object is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and of the displacement

For the box in this problem, we have:

W = 2240 J is the work done

d = 40 m is the displacement of the box

Assuming that the force is parallel to the displacement, $\theta=0$

Solving the equation for F, we find the force exerted on the box:

$F=\frac{W}{d cos \theta}=\frac{2240}{(40)(cos 0)}=56 N$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0

2 years ago

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

8 0

2 years ago

Read 2 more answers

1. why did aristarchus choose the time of a half (quarter) moon to make his measurements for calculating the earth-sun distance?

Stells [14]

In order to make his measurements for determining the Earth-Sun distance, Aristarchus waited for the Moon's phase to be exactly half full while the Sun was still visible in the sky. For this reason, he chose the time of a half (quarter) moon.

<h3 /><h3>How did Aristarchus calculate the distance to the Sun?</h3>

It was now possible for another Greek astronomer, Aristarchus, to attempt to determine the Earth's distance from the Sun after learning the distance to the Moon. Aristarchus discovered that the Moon, the Earth, and the Sun formed a right triangle when they were all equally illuminated. Now that he was aware of the distance between the Earth and the Moon, all he needed to know to calculate the Sun's distance was the current angle between the Moon and the Sun. It was a wonderful argument that was weakened by scant evidence. Aristarchus calculated this angle to be 87 degrees using only his eyes, which was not far off from the actual number of 89.83 degrees. But when there are significant distances involved, even slight inaccuracies might suddenly become significant. His outcome was more than a thousand times off.

To know more about how Aristarchus calculate the distance to the Sun, visit:

brainly.com/question/26241069

#SPJ4

7 0

1 year ago

What is Angular acceleration, please explain this concept. Give any equations that involve angular acceleration and explain them

Eva8 [605]

Answer:

Angular acceleration is defined as the rate of change of angular velocity of a body.

consider the attached figure as shown

It rotates with an angular velocity $\omega$

An point inside the object rotates along the path as indicated thus turning by an angle $\theta$ in time 't'

Thus we have

$\alpha =\frac{d\omega }{dt}\\\\=\frac{d}{dt}(\frac{d\theta }{dt})\\\\\therefore \alpha =\frac{d^{2}\theta }{dt^{2}}$

physically angular acceleration can be understood as the rate at which the angular speed of any object is changing with time.

8 0

3 years ago