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Contact [7]
3 years ago
6

If you place 48 cm object 15 m in front of a convex mirror of focal length 5 m, what is the magnification of the image?

Physics
1 answer:
Arisa [49]3 years ago
7 0

Answer:

m = 0.25

Explanation:

Given that,

Object distance, u = -15cm

Height of the object, h = 48

Focal length, f = cm

We need to find the magnification of the image.

Let v is the image distance. Using mirror's equation.

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\=\dfrac{1}{5}-\dfrac{1}{(-15)}\\\\=3.75\ cm

Magnification,

m=\dfrac{-v}{u}\\\\=\dfrac{-3.75}{-15}\\\\=0.25

Hence, the magnification of the image is 0.25.

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A 6 N and a 10 N force act on an object. The moment arm of the 6 N force is 0.2 m. If the 10 N force produces five times the tor
Levart [38]

Answer:

The moment arm is 0.6 m

Explanation:

Given that,

First force F_{1}=6\ N

Second force F_{2}=10\ N

Distance r = 0.2 m

We need to calculate the moment arm

Using formula of torque

\tau=Force\times lever\ arm

So, Here,

\tau_{2}=5 \tau_{1}

We know that,

The torque is the product of the force and distance.

Put the value of torque in the equation

F_{2}\times d_{2}=5\times F_{1}\times r_{1}

r_{2}=\dfrac{5\times F_{1}\times r_{1}}{F_{2}}

Where, F_{1}=First force

F_{1}=First force

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Put the value into the formula

r_{2}=\dfrac{5\times6\times0.2}{10}

r_{2}=0.6\ m

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Monica [59]
Blank 1: mass
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Which option lists a form of kinetic energy followed by a form of potential
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A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20 m/s by a 5620 N braking force actin
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the distance traveled by the car is 42.98 m.

Explanation:

Given;

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time of motion of the car, t = 2.5 s

The decelaration of the car is calculated as follows;

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a = -F/m

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The distance traveled by the car is calculated as follows;

s = ut + ¹/₂at²

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