PHYSICS HIGH SCHOOL

A person looks at a gem stone using a converging lens with a focal length of 12.5 cm. The lens forms a virtual image 30.0 cm from the lens. Determine the magnification. Is the image upright or inverted?

Answers

Answer 1

Answer:

THE MAGNIFICATION FROM THE GEMSTONE WILL HAVE 3.4 MAGNIFICATION. THE IMAGE WILL BE PRODUCE UPRIGHT.

Explanation:

WE WILL USE FORMULA ,

$(1)/(f)$ = $(1)/(p) + (1)/(q)$

where q = -30.0 cm

f = 12.5 cm

so, $p =(1)/(f)-(1)/(q)$ )⁻¹

$=(1)/(12.5cm)-(1)/(-30 cm)$ )⁻¹

= 8.82353 cm

m= $(-q)/(p)$

= $-((-30)/(8.82353))$

m= 3.4

so the magnification will be 3.4 which is upright image as magnification is positive.

As a person looks at gem stone using converging lens with focal length of 12.5 cm and virtual image 30 cm from lens the magnification for such will be 3.4 .And the image will be upright.

Answer 2

Answer:

Final answer:

For the case of a gem being observed through a converging lens, creating a virtual image located from the lens, the magnification would be greater than 1 and positive, suggesting an upright and magnified image.

Explanation:

In the given scenario, the gem stone is being observed through a converging lens with a focal length of 12.5 cm, and a virtual image is forming 30.0 cm from the lens. To find the magnification, which is a measure of how much larger or smaller the image is relative to the object, we use the formula for magnification in lens theory: Magnification (m) = - Image Distance (di) / Object Distance (do). Since we're dealing with a virtual image, the image distance is a negative value (-30.0 cm).

However, we are not given the object distance (do). In the context of magnifiers and conjugate rays, and taking into account that our image is virtual, we can assume that the value of the object distance (do) is less than the value of the focal length (12.5 cm) for this so-called 'Case 2' image. Therefore, we can say that the magnification (m) is greater than 1 and positive, which indicates that the image of the gem stone is upright and magnified (larger than the object).

Learn more about Lens Theory here:

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Maria throws an apple vertically upward from a height of 1.3 m with an initial velocity of +2.8 m/s. Will the apple reach a friend in a tree house 5.9 m above the ground? The acceleration due to gravity is 9.81 m/s 21. No, the apple will reach 5.27136 m below the tree house2. Yes, the apple will reach 5.27136 m above the tree house3. No, the apple will reach 1.43117 m below the tree house4. No, the apple will reach 1.5289 m below5. Yes, the apple will reach 1.5289 m above the tree house6. Yes, the apple will reach 1.43117 m above the tree house

Answers

Answer:

No, the apple will reach 4.20041 m below the tree house.

Explanation:

t = Time taken

u = Initial velocity = 2.8 m/s

v = Final velocity = 0

s = Displacement

g = Acceleration due to gravity = -9.81 m/s² = a (negative as it is going up)

Equation of motion

$v^2-u^2=2as\n\Rightarrow s=(v^2-u^2)/(2a)\n\Rightarrow s=(0^2-2.8^2)/(2* -9.81)\n\Rightarrow s=0.39959\ m$

The height to which the apple above the point of release will reach is 0.39959 m

From the ground the distance will be 1.3+0.39959 = 1.69959 m

Distance from the tree house = 5.9-1.69959 = 4.20041 m

No, the apple will reach 4.20041 m below the tree house.

The values in the option do not reflect the answer.

Final answer:

The apple will not reach the friend in the tree house as it will only reach a height of approximately 1.527 m.

Explanation:

To determine whether the apple will reach a friend in a tree house 5.9 m above the ground, we can use the equations of motion. Since the apple is thrown vertically upward, it will experience a negative acceleration due to gravity. Using the equation h = vo*t + (1/2)*a*t^2, where h is the final height, vo is the initial velocity, a is the acceleration, and t is the time, we can calculate the time it takes for the apple to reach a height of 5.9 m. Plugging in the values, we get:

5.9 = 2.8*t + (1/2)*(-9.81)*t^2

Simplifying the equation, we have:

-4.905*t^2 + 2.8*t - 5.9 = 0

Using the quadratic formula, we can solve for t. The quadratic formula is t = (-b ± sqrt(b^2 - 4ac)) / (2a), where a = -4.905, b = 2.8, and c = -5.9.

Plugging in the values, we get:

t = (-2.8 ± sqrt(2.8^2 - 4*(-4.905)*(-5.9))) / (2*(-4.905))

After evaluating the formula, we find that the apple will take approximately 1.527 seconds to reach a height of 5.9 m. Since the apple continues to rise after reaching this height, it will not reach the friend in the tree house.

An unknown substance has a mass of 0.125 kg and an initial temperature of 90.5°C. The substance is then dropped into a calorimeter made of aluminum containing 0.285 kg of water initially at 29.5°C. The mass of the aluminum container is 0.150 kg, and the temperature of the calorimeter increases to a final equilibrium temperature of 32.0°C. Assuming no thermal energy is transferred to the environment, calculate the specific heat of the unknown substance.

Answers

Answer:

The specific heat capacity of the substance = 455.38 J/kgK

Explanation:

Heat lost by the substance = Heat gained by water + heat gained by the aluminum calorimeter

Qs = Qw + Qc.................... equation 1

Where Qs = heat lost by the substance, Qw = heat gain by water, Qc = heat gain by the aluminum calorimeter.

Qs = c₁m₁(T₁-T₃)................ equation 2

Qw = c₂m₂(T₃-T₂)............. equation 3

Qc = c₃m₃(T₃-T₂)............. equation 4

Where c₁ = specific heat capacity of the substance, m₁ = mass of the substance, c₂ = specific heat capacity of water, m₂ = mass of water, c₃ = specific heat capacity of aluminium, m₃ = mass of the aluminum container, T₁ = Initial Temperature of the substance, T₂ = initial temperature of water, T₃ = Final equilibrium temperature.

Substituting equation 2, 3, 4 into equation 1

c₁m₁(T₁-T₃) = c₂m₂(T₃-T₂) + c₃m₃(T₃-T₂)................. equation 5

Making c₁ the subject of equation 5

c₁ = {c₂m₂(T₃-T₂) + c₃m₃(T₃-T₂)}/m₁(T₁-T₃)............... equation 6

Where c₂ = 4200 J/kgK, m₂ = 0.285 kg, m₁ = 0.125 kg, c₃ = 900 J/kgK, m₃= 0.150 kg, T₁ = 90.5°C, T₂ = 29.5°C, T₃ = 32.0°C

Substituting these values into Equation 6,

c₁ = {4200×0.285(32-29.5) + 900×0.150(32-29.5)}/0.125(90.5-32)

c₁ = {1197(2.5) + 135(2.5)}/7.3125

c₁ = {2992.5 + 337.5}/7.3125

c₁ = 3330/7.3125

c₁ = 455.38 J/kgK.

Therefore the specific heat capacity of the substance = 455.38 J/kgK

Jawless fish and ocean reefs were devastated by which extinction? Late Devonian

Ordovician-Silurian

Permian-Triassic

End Triassic

Cretaceous-Tertiary

Answers

Late Devonian (I believe)

The Late Devonian.

If you go back and reach the text, you may find the passage, "Which species did we lose during this extinction (Late Devonian)? About 20% of all animal families and 70-80% of all animal species were lost. Major victims included the following:

....

- Jawless fish"

What is velocity of a particle?

Answers

Velocity of a particle is its speed and the direction in which it is moving.

It can also be described as the instantaneous rate of change of the particle's
distance traveled, and the direction in which the distance is changing.

velocity is generally defined as the distance travelled by a particle in a particular time interval ...........its formula is given as
v=d/t
where v is velocity
d is distance travelled and
t is time interval.....
its units are - kmph , m/s read as kilometers per second and meters per second

Which planet has an AU of 5.20?
A. Uranus
B. Earth
C. Mercury
D. Jupiter

Answers

Jupiter, planet, semi-major axis: 5.20 AU

I agree it is Jupiter.

A mass of gas has a volume of 4m3, a temperature of 290k, and an absolute pressure of 475 kpa. When the gas is allowed to expand to 6.5m3, it's new temperature is 277K. What's the absolute pressure of the gas after expansion?