A _____ map provides information about how and when rocks formed in a particular area.A.contourB.geologicC.topographic

which of the glass lenses above, when placed in air, will cause rays of light (parallel to the central axis) to converge?

777dan777 [17]

Convex lenses when placed in the air, will cause rays of light (parallel to the central axis) to converge.

Converging lenses, commonly referred to as convex lenses, have thicker centers and narrower upper and lower margins. The edges are outwardly curled. This lens has the ability to concentrate a beam of parallel light rays coming from the outside onto a spot on the opposite side of the lens.

The image created is referred to be a genuine image when it is inverted relative to the object. On a screen, this kind of image can be recorded. When the object is positioned at a point farther than one focal length from the lens, a converging lens creates a true image.

A virtual image is one that cannot be produced on a screen and is formed when the image is upright in relation to the object. When an item is positioned within one focal length of a converging lens, a virtual image is created. It creates an enlarged image of the object on the same side of the lens as the image. It serves as a magnifier.

Learn more about the convex lens here:

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3 0

6 months ago

In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the sys

nydimaria [60]

Answer : The final energy of the system if the initial energy was 2000 J is, 3500 J

Solution :

(1) The equation used is,

$\Delta U=q+w\\\\U_{final}-U_{initial}=q+w$

where,

$U_{final}$ = final internal energy

$U_{initial}$ = initial internal energy

q = heat energy

w = work done

(2) The known variables are, q, w and $U_{initial}$

initial internal energy = $U_{initial}$ = 2000 J

heat energy = q = 1000 J

work done = w = 500 J

(3) Now plug the numbers into the equation, we get

$U_{final}-(2000J)=(1000J)+(500J)$

(4) By solving the terms, we get

$U_{final}-(2000J)=(1000J)+(500J)$

$U_{final}-(2000J)=1500J$

$U_{final}=2000J+1500J$

$U_{final}=3500J$

(5) Therefore, the final energy of the system if the initial energy was 2000 J is, 3500 J

5 0

2 years ago

A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angl

aliya0001 [1]

Answer:

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i: unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R= 131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

$D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2} }$

$D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2} }$

D₂= 167.21 cm

Direction of the second displacement

$\beta = tan^{-1} \frac{D_{y}}{D_{x} }$

$\beta = tan^{-1} \frac{62.18}{155.22 }$

β= 21.8°

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

6 0

2 years ago

The earth spins on its axis once a day and orbits the sun once a year (365 1/4 days). Determine the average angular velocity (in

lubasha [3.4K]

Answer:

Given that

The earth spins on its axis once a day and orbits the sun once a year (365 1/4 days)

a)

When earth spins on its axis

We know that earth take 1 day to complete one revolution around its own axis.

T= 1 day = 24 hr = 24 x 3600 s

T=86400 s

We know that

T=2π/ω

ω= 2π/T

ω= 2π/86400

ω=7.27 x 10⁻5 rad/s

b)

When earth revolve around earth

T =365 1/4 days = 365.25 days

T= 365.24 x 86400 s

T=31557600

We know that

T=2π/ω

ω= 2π/T

ω= 2π/31557600

ω=1.99 x 10⁻⁷ rad/s

8 0

2 years ago

Read 2 more answers

Submarine a travels horizontally at 11.0 m/s through ocean water. it emits a sonar signal of frequency f 5 5.27 3 103 hz in the

xeze [42]

Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ

6 0

2 years ago