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3 years ago

Real images can be upright or inverted.

1 answer:

Nadusha1986 [10]3 years ago

5 0

Real images can be either upright or inverted. Real images can be magnified in size, reduced in size or the same size as the object. Real images can be formed by concave, convex and plane mirrors. Real images are not virtual; thus you could never see them when sighting in a mirror.

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What is the difference between the hegelian right, center, and left? islamic?

Marta_Voda [28]

The main difference between the two is that Enlightenment rationalism dwells in abstract inwardness. and it is only through this echo that German Christianity is ...

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3 years ago

A sample contains 36 g of a radioactive isotope. How much radioactive isotope remains in the sample after 3 half-lives?

Gre4nikov [31]

Answer:

Option "C": "4.5 g"

Explanation:

N0 = 36 g, Let half-life is T.

t = 3 T, n is number of half lives = t / T = 3

By using the decay law of radioactivity

N / N0 = (1 / 2)^n

where

"N0" be the "initial amount"

"N" be the "amount left"

"n" be the "number of half-lives"

N / 36 = (1/2)^3

N / 36 = 1 / 8

N = 36 / 8 = 4.5 g

3 0

3 years ago

Read 2 more answers

A horizontal force of 400 N is exerted on a 2.0-kg ball as it rotates (at

frutty [35]

Answer:

the speed of the ball is 10 m/s

Explanation:

Given;

magnitude of exerted force, F = 400 N

mass of the ball, m = 2 kg

radius of the circle, r = 0.5

The speed of the ball is calculated by applying centripetal force formula;

$F = \frac{mv^2}{r} \\\\v^2 = \frac{Fr}{m}\\\\v = \sqrt{\frac{Fr}{m}}\\\\ v = \sqrt{\frac{400*0.5}{2}}\\\\v = 10 \ m/s$

Therefore, the speed of the ball is 10 m/s

6 0

3 years ago

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A crate of 45.2-kg tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and observe that th

iragen [17]

Answer:

μ = 0.725

Explanation:

This problem refers to Newton's second law.

F = ma

Let's write the equations on each axis

Y Axis

N-W = 0

N = W

N = mg

X axis

F-fr = ma

With the body not started moving its acceleration is zero

F-fr = 0

F = fr

The friction force equation is

fr = μ N

fr = μ m g

Let's replace and calculate

F = μ m g

μ = F / mg

μ = 321 /45.2 9.8

μ = 0.725

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3 years ago

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pickupchik [31]

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

$T^2 = \frac{4\pi^2}{GM}a^3$