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3 years ago

When the radius of curvature of the lens increases, what happens to the focal length of the lens

Physics

1 answer:

Basile [38]3 years ago

4 0

Answer

Equation of lens law

$\dfrac{1}{f}=(n -1 )(\dfrac{1}{R_1}-\dfrac{1}{R_2})$

where f is the focal length of the lens

n is the refractive index of the lens

R_1 and R_2 is the radius of curvature of the lens.

From the above expression we can see that focal length of the lens is directly proportional to radius of curvature.

hence, when the radius of curvature of the lens increases the focal length of the lens will also increase.

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This problem has been solved!

RideAnS [48]

Answer:

Option (a)

Explanation:

We will discard options that don't fit the situation:

Option b: Incorrect since if the driver "hits the gas" then velocity is augmenting and it's not constant.

Option c and d: Incorrect since the situation doesn't give us any information that could be related directly to the terrain or movement direction.

Option a: Correct. At stage 1 we can assume the driver was going at constant speed which means acceleration is constantly zero. At stage 2 we can assume the driver augmented speed linearly, this is, with constant positive acceleration. At stage 3 we can assume the driver slowed the speed linearly, with constant negative acceleration.

6 0

3 years ago

Block A, with a mass of 4.0 kg, is moving with a speed of 2.0 m/s while block B , with a mass of 8.0 kg, is moving in the opposi

Anon25 [30]

Consider velocity to the right as positive.

First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right

Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left

Total momentum of the system is
P = m₁v₁ + m₂v₂
= 4*2 + 8*(-3)
= -16 (kg-m)/s

Let v (m/s) be the velocity of the center of mass of the 2-block system.

Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
v = -1.333 m/s

Answer:
The center of mass is moving at 1.33 m/s to the left.

5 0

3 years ago

8–4. The tank of the air compressor is subjected to an internal pressure of 90 psi. If the internal diameter of the tank is 22 i

slamgirl [31]

Answer:

The stress S = 1935 [Psi]

Explanation:

This kind of problem belongs to the mechanical of materials field in the branch of the mechanical engineering.

The initial data:

P = internal pressure [Psi] = 90 [Psi]

Di= internal diameter [in] = 22 [in]

t = wall thickness [in] = 0.25 [in]

S = stress = [Psi]

Therefore

ri = internal radius = (Di)/2 - t = (22/2) - 0.25 = 10.75 [in]

And using the expression to find the stress:

$S=\frac{P*D_{i} }{2*t} \\replacing:\\S=\frac{90*10.75 }{2*0.25} \\S=1935[Psi]$

In the attached image we can see the stress σ1 & σ2 = S acting over the point A.

4 0

3 years ago

a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left. afterwards, the 282 kg car

Vaselesa [24]

The momentum of the 155 kg car afterwards is 469.7 kg m/s to the right

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum of the system is conserved before and after the collision, so we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$