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igor_vitrenko [27]
3 years ago
7

Why does the lens need to be thicker for viewing nearby objects?

Physics
1 answer:
Maurinko [17]3 years ago
5 0

Answer: To focus on a near object – the lens becomes thicker, this allows the light rays to refract (bend) more strongly. To focus on a distant object – the lens is pulled thin, this allows the light rays to refract slightly.

Explanation:

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State the condition for maximum current to be drawn from the cell
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Use the equation I=V/R where I is current and V is the voltage plus R is the resistance so when voltage is the highest and resistance is lowest the current is the highest
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Vapor pressure is related to the temperature of the liquid. user: in an open system, the vapor pressure is equal to the _____. i
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Answer

-Directly;  outside air pressure

Vapor pressure is directly related to the temperature of the liquid. user: in an open system, the vapor pressure is equal to the outside air pressure.

Explanation;

-As the temperature of a system increases, the average kinetic energy of the molecules increases in both the liquid and gas phases.

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3 years ago
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Which characteristics of rocks are used in classification? Check all that apply.
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3 years ago
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What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a
Paha777 [63]
E = mgh +  \frac{1}2} m v^{2} + \frac{1}{2} I \omega^{2} = mgh +  \frac{1}2} m  r ^{2}   \omega ^{2}  + \frac{1}{2} I \omega^{2}

for a solid cylinder:  I =  \frac{1}{2} m r^{2}
for a hollow cylinder: I = mr^{2}

I will look at the case of a hollow cylinder:

E = mgh + I \omega ^{2} = constant \\ \\ I =  \frac{mgh}{  \omega^{2} }

That is as far as i get.


7 0
4 years ago
Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above
poizon [28]

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

U_{g,1} = U_{g,2} + W_{dis}

Where:

U_{g,1}, U_{g,2} - Initial and final gravitational potential energy, measured in joules.

W_{dis} - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

U = m \cdot g \cdot y

Where:

m - Mass, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

y - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

W_{dis} = f \cdot \Delta s

Where:

f - Friction force, measured in newtons.

\Delta s - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s

f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}

If m = 1000\,kg, g = 9.807\,\frac{m}{s^{2}}, y_{1} = 40\,m, y_{2} = 25\,m and \Delta s = 400\,m, then:

f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}

f = 367.763\,N

The magnitude of the frictional force between the car and the track is 367.763 N.

7 0
3 years ago
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