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Sindrei [870]
3 years ago
15

a spherical mirror produces a magnification of -1 on a screen placed at a distance of 40cm from the mirror i) write the type of

mirror ii) What is the focal length of the mirror
Physics
1 answer:
Darya [45]3 years ago
8 0

Answer:

f =-20 cm

Explanation:

Given that,

The magnification of a spherical mirror, m = -1

The image distance, v = 40 cm (for negative magnification)

The magnification of a concave mirror is negative. The mirror showing -1 magnification is a concave mirror.

Let f be the focal length of the mirror. We know that,

m=\dfrac{-v}{u}\\\\-1=\dfrac{-v}{u}\\\\v=u

Object distance, u = -40 cm

Using mirror's formula i.e.

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}\\\\f=\dfrac{1}{\dfrac{1}{-40}+\dfrac{1}{-40}}\\\\f=-20\ cm

So, the focal length of the mirror is 20 cm.

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What are the wavelengths of electromagnetic wave in free space that have the following frequencies?
Ne4ueva [31]

Explanation:

Given that,

(a) Frequency, f_1=4\times 10^{19}\ Hz

All electromagnetic wave moves with the speed of light. It is given by :

c=f\lambda

\lambda_1=\dfrac{c}{f_1}\\\\\lambda_1=\dfrac{3\times 10^8}{4\times 10^{19}}\\\\\lambda_1=7.5\times 10^{-12}

(b) Frequency, f_2=5.5\times 10^{1=9}\ Hz

All electromagnetic wave moves with the speed of light. It is given by :

c=f\lambda

\lambda_2=\dfrac{c}{f_2}\\\\\lambda_2=\dfrac{3\times 10^8}{5.5\times 10^{9}}\\\\\lambda_2=0.054\ m

Hence, this is the required solution.

7 0
3 years ago
Suppose that an object travels from one point in space to another. Make a comparison between the magnitude of the displacement a
Rashid [163]

Answer:

- Distance is a scalar quantity, defined as the total amount of space covered by an object while moving between the final position and the initial position. Therefore, it depends on the path the object has taken: the distance will be minimum if the object has travelled in a straight line, while it will be larger if the object has taken a non-straight path.

- Displacement is a vector quantity, whose magnitude is equal to the distance (measured in a straight line) between the final position and the initial position of the object. Therefore, the displacement does NOT depend on the path taken, but only on the initial and final point of the motion.

If the object has travelled in a straight path, then the displacement is equal to the distance. In all other cases, the distance is always larger than the displacement.

A particular case is when an object travel in a circular motion. Assuming the object completes one full circle, we have:

- The distance is the circumference of the circle

- The displacement is zero, because the final point corresponds to the initial point

3 0
3 years ago
Choose the correct words from the box to complete the definition of resistance.
zzz [600]

Answer:

Opposition of passing a electric circuit

6 0
3 years ago
A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C
mario62 [17]

Answer:

3.62 m  and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}

\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}

\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}

x = - 1.4 m

6 0
4 years ago
Based on its location on the periodic table which element would be most likely to form a negative ion
blagie [28]

Answer:

Since Fluorine is very electronegative, it can easily absorb the electrons of other elements. Since it sucks up electron, this gives Fluorine an excess electron thus making it a negative ion F-.

Explanation:

because it is

6 0
3 years ago
Read 2 more answers
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