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3 years ago

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An object is placed between two plain mirrors which make angle 45 degree with each other. Distance between all adjacent images a

re equal to 20 cm. Find distance between the object and each other mirror. How many images are formed.

1 answer:

marysya [2.9K]3 years ago

3 0

I have a formula here that might just help you solve the problem on your own:

The number of images depends on the angle between the two mirrors. The number of images formed in two plane mirrors inclined at an angle A to each other is given by the formula:

Number of images = 360<span>/A - 1.
</span>
I hope my guide has come to your help. God bless and have a nice day ahead!

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A worker exerts a pulling force on a box. The worker exerts this force by attaching a rope to the box and pulling on the rope so

shusha [124]

Answer:

<em>The magnitude of the pulling force is 20.66% of the gravitational force acting on the box</em>

Explanation:

<u>Accelerated Motion</u>

The net force exerted on a body is the (vector) sum of all forces applied to the body. The net force can be decomposed in its rectangular components and the dynamics of the body can be studied in each direction x,y separately.

Let's start off by calculating the acceleration the worker gives to the box when pulling it. The distance traveled by the box initially at rest in a time t at an acceleration a is given by

$\displaystyle x=\frac{at^2}{2}$

Solving for a

$\displaystyle a=\frac{2x}{t^2}$

$\displaystyle a=\frac{2\cdot 10}{5.12^2}$

$a=0.763\ m/s^2$

Now we analyze the geometric of the forces applied to the box. Please refer to the free body diagram provided below.

The forces in the y-axis must be in equilibrium since no movement takes place there, thus, being g the acceleration of gravity:

$T_y+N=m.g$

There Ty is the vertical component of the tension of the rope, N is the normal force, and m is the mass of the box

The decomposition of T gives us

$T_y=Tsin\theta$

$T_x=Tcos\theta$

Solving the above equation for N

$Tsin\theta+N=m.g$

$N=m.g-Tsin\theta\text{..........[1]}$

Now for the x-axis, there are two forces acting on the box, the x-component of the tension and the friction force Fr. Those forces are not equilibrated, thus acceleration is produced:

$Tcos\theta-F_r=m.a$

Recalling that

$F_r=\mu N$

$Tcos\theta-\mu N=m.a$

Replacing N from [1]

$Tcos\theta-\mu (m.g-Tsin\theta)=m.a$

Operating

$Tcos\theta-\mu m.g+\mu Tsin\theta=m.a$

Solving for T

$T(cos\theta+\mu sin\theta)=m.a+\mu m.g$

$\displaystyle T=\frac{m.a+\mu m.g}{cos\theta+\mu sin\theta}$

$\displaystyle T=m\frac{a+\mu g}{cos\theta+\mu sin\theta}$

We don't know the value of m, thus we'll plug in the rest of the data

$\displaystyle T=m\frac{0.763+0.10\cdot 9.8}{cos36.8^o+0.10 sin36.8^o}$

$T=2.0252m$

Dividing by the weight of the box m.g

$T/(m.g)=2.0252/9.8=0.2066$

Thus, the magnitude of the pulling force is 20.66% of the gravitational force acting on the box

7 0

3 years ago

A 110 n force and a 55 n force both act on an object point p. the 110 n force acts at 90 degrees. the 55 n force acts at 0 degre

ss7ja [257]

<span>To determine the magnitude and the direction of the resultant force, we assume that the forces are in XY coordinate plane and the angles that are given are from the x axis.
</span>
<span>The 110 N force is said to act at 90 deg which means it is along the Y axis. The </span><span>55 N force is said to act at 0 deg which means it is along the X axis. so, a right angle is made by the two forces. Thus, the
</span>
<span>X component of the resultant force = 55 N </span>
<span>Y component of the resultant force = 110 N </span>
<span>Magnitude of the resultant force would be calculated as follows:
R = √(Fx^2 + Fy^2)
R = √(55^2 + 110^2) </span>
<span>R = √(15125) </span>
<span>R = 123 N </span>
<span>The resultant force would have its terminal side in the x-axis. We calculate angle θ as follows:
</span>
<span>tan θ = Fy/Fx </span>
<span>tan θ = 110 N /55 N = 2
</span>θ = arctan(2)
θ <span>= 63.4 degrees
</span>Therefore, the m<span>agnitude of the resultant force is 123 N and the direction would be at an angle of 63.4 degrees.</span>

5 0

4 years ago

Which action do researchers take to make advances in science?

Ugo [173]

<span>b. They share their experimental results with other researchers</span>

6 0

3 years ago

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Which of the following refers to the increasing size of deserts worldwide? A. disintegration B. desertification C. decomposition

algol [13]

the answer is B. desertification

7 0

4 years ago

Read 2 more answers

The velocity of sound on a particular day outside is 331 meters/second. What is the frequency of a tone if it has a wavelength o

Leya [2.2K]

Answer:
frequency = 5.52 * 10² Hz

Explanation:
the equation that relates velocity, frequency and wavelength is:
velocity = frequency * wavelength

We are given that:
velocity = 331 m/sec
wavelength = 0.6 m

Substitute with the givens in the equation to get the frequency as follows:
velocity = frequency * wavelength
331 = frequency * 0.6
frequency = 331 / 0.6
frequency = 5.52 * 10² Hz

Hope this helps :)

4 0

3 years ago