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Westkost [7]
3 years ago
9

Which of the following changes would increase the focal length of a lens? a) decrease rodius of curvature of surfoce c) increase

index of refraction of lens material. d) use red colored lens material b) increase radius of curvature of surface
Physics
1 answer:
anastassius [24]3 years ago
8 0

Answer:

Increase the radius of curvature of surface.

Explanation:

Radius of curvature is the measurement that is two times the focal length, for a given lens. It lies on either side of the lens.

Focal length is the distance which is half of the radius of curvature. Radius of curvature is a measure of the radius of the circle . Focal length is the distance between the center of curvature of the lens and the point where all the rays are brought to a focus for a distant object.

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A fuel oil tank is an upright cylinder, buried so that its circular top is 14 feet beneath ground level. the tank has a radius o
Ierofanga [76]

Say we have a cylinder that has a height of dx, we see that the cylinder has a volume of: <span>

<span>Vcylinder = πr^2*h = π(5)^2(dx) = 25π dx

Then, the weight of oil in this cylinder is: 

Fcylinder = 50 * Vcylinder = (50)(25π dx) = 1250π dx. 

Then, since the oil x feet from the top of the tank needs to travel x feet to get the top, we have: 

Wcylinder = Force x Distance = (1250π dx)(x) = 1250π x dx. 

<span>Integrating from x1 to x2 ft gives the total work to be: (x1 = distance from top liquid level to ground level; x2 = distance from bottom liquid level to ground level)</span>

<span>W = ∫ 1250π x dx  
<span>W = 1250π ∫ x dx
W = 625π * (x2 – x1)</span></span></span></span>

<span>x2 = 14 ft  + 15 ft = 29 ft</span>

x1 = 14 ft + 1 ft = 15 ft

<span>
W = 625π * (29^2 - 15^2) 
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3 0
4 years ago
I don’t get science at all my teacher doesn’t explain anything and I was out of school can anyone explain this to me
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3 years ago
A snowball starting at rest rolls down a hill and reaches 5 m/s. If the hill is
Lostsunrise [7]

Answer:

The acceleration of the snowball is 0.3125

Explanation:

The initial speed of the snowball up the hill, u = 0

The speed the snowball reaches, v = 5 m/s

The length of the hill, s = 40 m

The equation of motion of the snowball given the above parameters is therefore;

v² = u² + 2·a·s

Where;

a = The acceleration of the snowball

Plugging in the values, we have;

5² = 0² + 2 × a × 40

∴ 2 ×  40 × a  = 5² = 25

80 × a = 25

a = 25/80 = 5/16

a = The acceleration of the snowball = 5/16 m/s².

The acceleration of the snowball = 5/16 m/s² = 0.3125 m/s² .

4 0
3 years ago
An object thrown straight up a distance ymaxymax. After tt seconds, it falls back and is caught again, just as it reaches the he
Mazyrski [523]

Answer:

Avg.velocity=(Δy/ Δt)  =(net displacement/ total time for journey)

Δy = 0

Δt = t

so avg. velocity = 0/t =0

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Explanation:

Since there is no net displacement from the original position,velocity is zero. Claudia is right!

while it covered some distance in time t so its speed is not as qouted by Hossien

6 0
4 years ago
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