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2 years ago

Which of the following represents an image that is located behind a mirror?

Physics

2 answers:

9966 [12]2 years ago

3 0

Answer:

i think its -di........

Marina CMI [18]2 years ago

3 0

Explanation:

There are some conventions while solving the problems based on mirrors. All parameters are taken like x-y coordinate system.

All the measurements are done from the optical centre of the mirror.

$d_i$ is the image distance

$d_o$ is the object distance

$-d_i$ is the image distance which is formed behind the mirror.

$-d_o$ is the object distance when an object is placed behind the mirror.

So, $-d_i$ shows the image that is located behind the mirror. Hence, this is the required solution.

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Based on the free-body diagram, the net force acting on this firework is

Strike441 [17]

0N. The net force acting on this firework is 0.

The key to solve this problem is using the net force formula based on the diagram shown in the image. Fnet = F1 + F2.....Fn.

Based on the free-body diagram, we have:

The force of gases is Fgases = 9,452N

The force of the rocket Frocket = -9452

Then, the net force acting is:

Fnet = Fgases + Frocket

Fnet = 9,452N - 9,452N = 0N

4 0

2 years ago

Read 2 more answers

We can determine the velocity of a wave when given the frequency and the

Kobotan [32]

Hello.

The answer is: D. wavelength

This is correct because frequency x wavelength = speed

Have a nice day

3 0

2 years ago

Read 2 more answers

Suppose a plane accelerates from rest for 30 s, achieving a takeoff speed of 80 m/s after traveling a distance of 1200 m down th

Margaret [11]

Answer:

300 m

Explanation:

The train accelerate from the rest so u = 0 m/sec

Final speed that is v = 80 m/sec

Time t = 30 sec

The distance traveled by first plane = 1200 m

We know the equation of motion $S=ut+\frac{1}{2}at^2$ where s is distance a is acceleration and u is initial velocity

Using this equation for first plane $1200=0\times 30+\frac{1}{2}a30^2$

$a=2.67\frac{m}{sec^2}$

As the acceleration is same for both the plane so a for second plane will be 2.67 $\frac{m}{sec^2}$

The another equation of motion is $v^2=u^2+2as$ using this equation for second plane $40^2=0+2\times 2.67\times s$

s = 300 m

5 0

2 years ago

The histogram below shows the number of downloads of a song over time.

Allisa [31]

Given data:

It is a graphical display where the data is grouped in to ranges

A diagram consists rectangles, whose area is proportional to frequency of a variable and whose width is equal to the class interval.

It is an accurate representation of the distribution of numerical data.

From Figure:

Each box in the graph (small rectangle box) is assumed to be one download. So, in the graph the time between 8 p.m to 9 p.m, the number of downloads are 8.75 approximately (because the last box is incomplete, therefore 8 complete boxes and 9th is more than half).

So, We conclude that the total number of downloads are approximately 9 in the time span of 8 p.m. to 9 p.m.

7 0

2 years ago

A mass weighting 16 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 20 lbs when the

const2013 [10]

Answer:

The equation for the object's displacement is $u(t)=0.583cos11.35t$