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2 years ago

An object 10mm in height is located 20cm from a crown glass spherical surface whose power is +10.00DS. Locate the image

Physics

1 answer:

Tasya [4]2 years ago

7 0

The image is present at 20cm from the crown glass spherical surface.

To find the answer, we need to know about the lens formula.

<h3>What's the lens formula?</h3>

It's (1/V)-(1/U)= (1/f)
V= image distance from the lens, U= object distance, f= focal length of the lens

<h3>What's the image distance, if object is present at 20cm from crown glass of power 10DS?</h3>

Focal length (f)= 1/ power = 1/10 = 0.1 m
U= -20cm = -0.2m (-ve sign due to sign convection)
(1/V)-(-1/0.2)= (1/0.1)

=> (1/V)+5=10

=> 1/V= 5

=> V=0.2m = 20cm

Thus, we can conclude that the image is present at 20cm.

Learn more about the lens formula here:

brainly.com/question/2098689

#SPJ1

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A Roman centurion fires off a vat of burning pitch from

EastWind [94]

Answer:

(a) 1.11sec

(b) 14.37m/s

Explanation:

U = 18m/s, A = 37°, g = 9.8m/s^2

(a) t = UsinA/g = 18sin37°/9.8 = 18×0.6018/9.8 = 1.11sec

(b) Ux = UcosA = 18cos37° = 18×0.7986 = 14.37m/s

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4 years ago

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When the speed of electrons striking the anode of an x-ray tube increases, the _____ of the emitted x-rays increases.

Yuki888 [10]

Cathody Ray electrons striking the anode of an x-ray tube increases

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4 years ago

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Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about the amounts o

likoan [24]

Answer:

1) At the highest point of the building.

2) The same amount of energy.

3) The kinetic energy is the greatest.

4) Potential energy = 784.8[J]

5) True

Explanation:

Question 1

The moment when it has more potential energy is when the ball is at the highest point in the building, that is when the ball is at a height of 40 meters from the ground. It is taken as a point of reference of potential energy, the level of the soil, at this point of reference the potential energy is zero.

$E_{p} = m*g*h\\E_{p} = 2*9.81*40\\E_{p} = 784.8[J]$

Question 2)

The potential energy as the ball falls becomes kinetic energy, in order to be able to check this question we can calculate both energies with the input data.

$E_{p}=m*g*h\\ E_{p} = 2*9.81*20\\ E_{p} = 392.4[J]\\$

And the kinetic energy will be:

$E_{k}=0.5*m*v^{2}\\ where:\\v = velocity = 19.8[m/s]\\E_{k}=0.5*2*(19.8)^{2}\\ E_{k}=392.04[J]$

Therefore it is the ball has the same potential energy and kinetic energy as it is half way through its fall.

Question 3)

As the ball drops all potential energy is transformed into kinetic energy, therefore being close to the ground, the ball will have its maximum kinetic energy.

$E_{k}=E_{p}=m*g*h = 2*9.81*40\\ E_{k} = 784.8[J]\\ E_{k} = 0.5*2*(28)^{2}\\ E_{k} = 784 [J]$

Question 4)

It can be easily calculated using the following equation

$E_{p} =m*g*h\\E_{p}=2*9.81*40\\E_{p} =784.8[J]$

Question 5)

True

The potential energy at 20[m] is:

$E_{p}=2*9.81*20\\ E_{p}= 392.4[J]\\The kinetic energy is:\\E_{k}=0.5*2*(19.8)^{2} \\E_{k}=392[J]$

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3 years ago

As the wavelength increases, the frequency? A. decreases and energy decreases B. increases and energy increases C. decreases and

const2013 [10]

For the same wave, the product product of

(wavelength) times (frequency)

is always the same number. (It happens to be the speed of the wave.)

So if one of them changes, the other one has to change in the opposite
direction, in order to keep their product constant.

For electromagnetic waves, higher frequency means higher energy.
I'm not sure about mechanical waves just now.

8 0

4 years ago

Two charges that are 1 meter apart repel each other with a force of 2 N. If the distance between the charges is increased to 2 m

Savatey [412]

Answer:

b) 0.5 N

Explanation:

From coulomb's law,

F = kq'q/r².................... Equation 1

Where F =force of repulsion between the charges, q' = first charge, q = second charge, r = distance between the charges, k = proportionality constant.

q'q = Fr²/k........................... Equation 2

Given: F = 2 N, r = 1 m, k = 9.0×10⁹ Nm²/C²

Substituting into equation 2

q'q = 2(1)²/(9.0×10⁹)

q'q = 2/9.0×10⁹ C².

If the distance between the charges is increased to 2 meters,

r = 2 m, q'q = 2/9.0×10⁹ C².

Substitute into equation 1

F = 9.0×10⁹(2/9.0×10⁹)/2²

F = 2/4

F = 1/2 = 0.5 N.

The right option is b) 0.5 N

5 0

4 years ago