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Nuetrik [128]
3 years ago
13

Hello I need the formula for getting the focal length of a concave mirror​

Physics
2 answers:
marissa [1.9K]3 years ago
7 0
The focal length of a concave mirror is negative and that of a convex mirror is positive. You can also prove the same by using the mirror formula (1/f = 1/v +1/u).

Let's see how,

Since we know object is always placed on the left side or direction opposite to the incidence ray of the mirror. Therefore object distance will always be negative.

u = -u

v = -v (Image distance is negative, since image formed by concave mirrors are generally on the left side or direction opposite to the incidence ray)

Using mirror formula,

1/f = 1/v +1/u

or, 1/f = (-u-v)/uv

or, f = -uv/(u+v)

As you can see, The value of f is negative here…

However v is not always negative in concave mirror, when object is placed between focus and pole, you will get a virtual image on the right side of the mirror, so in this case image distance will be positive. But even in this case focal length will be negative. How?

Just use the mirror formula again, and you will know how.
mars1129 [50]3 years ago
6 0

Answer:

f= -9cm, v= -6cm. The focal length of a concave mirror can be estimated by focusing a distant object on a screen. Parallel rays from a distant object converge at the focal plane of the mirror. The distance between the mirror and the screen is the estimated focal length of the concave mirror.

Explanation:

Hope this help!

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Force X has a magnitude of 1260 ​pounds, and Force Y has a magnitude of 1530 pounds. They act on a single point at an angle of 4
weeeeeb [17]

Answer:

Fe= 2579.68 P

α= 24.8°

Explanation:

Look at the attached graphic

we take the forces acting on the x-y plane and applied at the origin of coordinates

FX = 1260 P , horizontal (-x)

FY = 1530  P , forming 45° with positive x axis

x-y components FY

FYx= - 1530*cos(45)° = - 1081.87 P

FYy= -  1530*sin(45)° = - 1081.87 P

Calculation of the components of net force (Fn)

Fnx= FX + FYx

Fnx= -1260 P -1081.87 P

Fnx= -2341.87 P

Fny=FYy

Fny= -1081.87 P

Calculation of the components of equilibrant force (Fe)

the x-y components of the  equilibrant force are equal in magnitude but in the opposite direction to the net force components:

Fnx= -2341.87 P, then, Fex= +2341.87 P

Fny=  -1081.87 P P, then, Fex= +1081.87 P

Magnitude of the equilibrant (Fe)

F_{n} = \sqrt{(F_{nx})^{2} +(F_{ny})^{2}  }

F_{e} =\sqrt{(2341.87)^{2}+(1081.87)^{2}  }

Fe= 2579.68 P

Calculation of the direction of  equilibrant force (α)

\alpha =tan^{-1} (\frac{F_{ny} }{F_{nx} } )

\alpha =tan^{-1} (\frac{1081.87 }{2341.87} )

α= 24.8°

Look at the attached graphic

6 0
3 years ago
An astronaut lands on another planet and wishes to determine the acceleration due to gravity.the astronaut measure a period of 0
jarptica [38.1K]

Answer:

1.927  m/s^2

Explanation:

period = 2 pi  sqrt ( l/g)  

3.2   =  2 pi sqrt (.5/g) =1.927 m/s^2

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How do you measure the volume of irregular object? explain with diagram​
masya89 [10]

Answer:

by using graphical meythod

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Consider the resistances of short and open circuits. Fill in the blanks: The voltage across a short circuit will always be _____
aleksandr82 [10.1K]

Answer:

a) Zero

b) power source

Explanation:

According to Ohm's law, the voltage dropped in a resistance is proportional to the current flow and the resistor opposing to it.

V=I*R

For the case of a short circuit, the resistance tends to zero, so the voltage will tend to zero too.

In the case of the open circuit, the resitance will tend to infinity, because we cannot obtain an infite voltage, it will be limited by the power source.

3 0
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A moon orbits a planet every 42 hours with a mean orbital radius of .002819 AU. The mass of the moon is 8.932 x 1022 kg. Using N
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Answer:

The mass of the planet  is 1.9407\times10^{27}\ kg

Explanation:

Given that,

Time period = 42 hours = 151200 sec

Orbital radius = 0.002819 AU = 421716397.5 m

Mass of moon m=8.932\times10^{22}\ kg

We need to calculate the mass of the planet

Using Kepler’s third law

T^2\propto a^3

T^2=\dfrac{4\pi^2}{G(M+m)}\times a^3

Where, a = orbital radius

T = time period

G = gravitational constant

M = mass of moon

m = mass of planet

Put the value into the formula

(151200)^2=\dfrac{4\pi^2}{6.673\times10^{-11}(8.932\times10^{22}+m)}\times(421716397.5)^3

(8.932\times10^{22}+m)=\dfrac{4\pi^2}{6.673\times10^{-11}}\times\dfrac{(421716397.5)^3}{(151200)^2}

(8.932\times10^{22}+m)=1.94087\times10^{27}

m=1.94087\times10^{27}-8.932\times10^{22}

m=1.9407\times10^{27}\ kg

Hence, The mass of the planet  is 1.9407\times10^{27}\ kg

8 0
4 years ago
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