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Digiron [165]
3 years ago
12

A small lightbulb is 1.06 m from a screen. A) If you have a convex lens with 20 cm focal length, where are the two lens location

s that will project an image of the lightbulb onto the screen?
B) What’s the magnification in the first case (smaller distance from the lightbulb)?
C) What’s the magnification in the second case (larger distance from the lightbulb)?
Physics
1 answer:
MatroZZZ [7]3 years ago
7 0

Answer:

A)Explanation:

Let object distance be u.

Image distance v = 1.06 - u

Focal length = 20 cm = .2 m

Applying lens formula

1 / v - 1 / u = 1 / f

1 /( 1.06-u)  - 1 / u = 1 / .2

5 u² -3.3u - 1.06 =0

It will have two roots

u₁ = .1636 m, u₂ = .8964 m

B ) Magnification

= u = .1636m , v = .8964m

m = .8964 / .1636 = 5.48

C ) When u = .8964m ,v = .1634m

m = .1634 / .8964

=.182

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GETTING TIMED PLS HELP! Do the runners in the picture above represent kinetic or potential energy? you need to explain why.
prohojiy [21]

Answer:

<em>They represent kinetic energy</em>

Explanation:

<u>Kinetic Energy </u>

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\displaystyle K=\frac{mv^2}{2}

Both runners are moving in a horizontal path, thus they have kinetic energy, given by the above equation. If they could jump below ground level, then they will also have potential energy

8 0
2 years ago
A golfer hits a 42 g ball, which comes down on a tree root and bounces straight up with an initial speed of 15.6 m/s. Determine
nevsk [136]

Answer:

12.2 m

Explanation:

Given:

v₀ = 15.6 m/s

v = 0 m/s

a = -10 m/s²

Find: Δy

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5 0
3 years ago
Read 2 more answers
A proton moving in the positive x direction with a speed of 9.9 105 m/s experiences zero magnetic force. When it moves in the po
Alex

Answer:

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

Explanation:

In order to calculate the magnitude and direction of the magnetic field, you take into account the following equation for the magnetic force on the proton:

\vec{F_B}=q\vec{v}\ X\ \vec{B}       (1)

v: speed of the proton = 9.9*10^5 m/s

q: charge of the proton = 1.6*10^-19C

B: magnetic field = ?

FB: magnetic force on the proton = 1.6*10^-13N

When the proton travels in the positive y direction (^j), you have that the proton experiences a force in the positive z direction (+^k). To obtain this direction of the magnetic force on the proton, it is necessary that the magnetic field points in the negative x direction, in fact, you have:

^j X (-^i) = -(-^k)=^k

To obtain the magnitude of the magnetic field you use:

F_B=qvBsin90\°=qvB\\\\B=\frac{F_B}{qv}=\frac{1.6*10^{-13}N}{(1.6*10^{-19}C)(9.9*10^5m/s)}\\\\B=1.01T

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

8 0
3 years ago
A key falls from a bridge that is 44 m above the water. It falls directly into a model boat, moving with constant velocity, that
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The time t it takes for the key to fall 44 m is

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The boat, moving at a presumably constant speed, then has 3.0 s to travel 19 m to the point of impact, which means its speed must be

v=\dfrac{19\,\mathrm m}{3.0\,\mathrm s}=6.3\,\dfrac{\mathrm m}{\mathrm s}

6 0
3 years ago
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