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mars1129 [50]
3 years ago
13

fins the powee of a convex lens which forms a real and inverted image of magnification -1 of an object placed at a distance from

its optical centre​
Physics
1 answer:
anzhelika [568]3 years ago
7 0

Explanation:

hope i am correct

hii

how are you

You might be interested in
Determine the thrust produced if 1.5 x 10^3 kg of gas exits the combustion chamber each second, with a speed of 4.00 x 10^3 m/s.
ozzi

Answer:

The thrust is 6\times 10^6\ N

Explanation:

Given that,

Mass of gas, m=1.5\times 10^3\ kg

The rate at which the gas is expelling, \dfrac{dv}{dt}=4\times 10^{3}\ m/s

We need to find the thrust produced by the gas.

We know that force is equal to the rate of change of momentum. So,

F=\dfrac{p}{t}

Also, p = mv

F=\dfrac{mv}{t}

So,

F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N

So, the thrust is 6\times 10^6\ N

3 0
3 years ago
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
Alex places 2 cubes side-by-side on a ramp made of wood. Cube #1 is ice and Cube #2 is wood
zmey [24]

Answer:

Explanation:

The sandpaper block did not move because the forces of friction and gravity were balanced.

6 0
3 years ago
An object is thrown upward from the edge of a tall building as with a velocity of 10m/s. where will the object be 3s after it is
KonstantinChe [14]
20 m is the answer, hope that helps
6 0
3 years ago
This is a Science question pls help give 15 coins for answer
Vlad1618 [11]

Answer:

<u>Conservation</u>: using less water

<u>Xeriscaping</u>: replanting your yard with plants that do not require great amounts of water

<u>Desalination</u>: process of removing salt from water so that it can be used for consumption

<u>Water Budget</u>: finite amount of usable water available

<u>Potable</u>: water that is safe to use a drink

7 0
3 years ago
Read 2 more answers
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