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3 years ago

What can one say about the image produced by a thin lens that produces a positive magnification?

Physics

1 answer:

Dafna11 [192]3 years ago

5 0

Answer:

The image is virtual and upright

Explanation:

The magnification of a lens can be written as follows:

$M=\frac{y'}{y}=-\frac{q}{p}$

where

y' is the size of the image

y is the size of the object

q is the location of the image with respect to the lens

p is the location of the object with respect to the lens

In this situation, the magnification is positive. This means that:

- y' (the image) has same sign as y (the object) --> the image is upright (same orientation as the object)

- q has opposite sign to p --> this means that the image is located on the same side as the object, so it is a virtual image

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A projectile is fired with an initial velocity of 450 feet per second at an angle of 70° with the horizontal.

pav-90 [236]

<h2>After 26.28 seconds projectile returns 26.28 seconds.</h2>

Explanation:

Initial velocity = 450 ft/s = 137.16 m/s

Angle, θ = 70°

Consider the vertical motion of projectile,

When the projectile return to the ground we have

Displacement, s = 0 m

Acceleration, a = -9.81 m/s²

Initial velocity, u = 137.16 x sin70 = 128.89 m/s

Substituting in s = ut + 0.5 at²

s = ut + 0.5 at²

0 = 128.89 x t + 0.5 x (-9.81) x t²

t² - 26.28 t = 0

t ( t- 26.28) = 0

t = 0 s or t = 26.28 s

After 26.28 seconds projectile returns 26.28 seconds.

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3 years ago

A speed skater moving across frictionless ice at 9.2 m/s hits a 5.0 m wide patch of rough ice. She slows steadily, then continue

enot [183]

Answer:

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

Explanation:

The distance travelled on the rough ice is equal to the width of the rough ice.

distance d = 5.0 m

Initial speed u = 9.2 m/s

Final speed v = 5.8 m/s

The time taken to move through the rough ice can be calculated using the equation of motion;

d = 0.5(u+v)t

time t = 2d/(u+v)

Substituting the given values;

t = 2(5)/(9.2+5.8)

t = 2/3 = 0.66667 second

The acceleration is the change in velocity per unit time;

acceleration a = ∆v/t

a = (v-u)/t

Substituting the values;

a = (5.8-9.2)/0.66667

a = -5.099974500127

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

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3 years ago

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Ugo [173]

Answer:

$+1 m/s^2$

Explanation:

Acceleration is given by

$a=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in velocity

$\Delta t$ is the time interval in which the change in velocity occurs

To find the acceleration at 1 second, we can take the data at t = 1 s and t = 2. We find:

$\Delta t = -3 -(-4) = 1 s$

$\Delta v = 2 - (1) = +1 m/s$

So, the acceleration is

$a=\frac{+1}{1}=+1 m/s^2$

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leonid [27]

The change in velocity is 5m/s which added to the initial 3m/s makes the final velocity 8m/s

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3 years ago

How is refraction different from reflection?

olga_2 [115]

<span>The angle of refraction is not necessarily equal to the angle
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4 years ago