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natka813 [3]
3 years ago
9

A piece of transparent meterial which has one or two spherical surfaces is called​

Physics
1 answer:
Iteru [2.4K]3 years ago
4 0

Answer:

Lens

Explanation:

Lenses fit this description.

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It is about 100oC at a pressure of 1.1 atmosphere. Hope this helps.
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To understand the interplay of observations and models you must first be able to distinguish between things that we observe and
Svetllana [295]

Answer: a) Observation

b) observation

c) observation

d) observation

e) inference

f) inference

g) inference

h) inference

Explanation:

observation: The photosphere is made mostly of hydrogen and helium,The photosphere emits mostly visible light,The corona is hotter than the photosphere, The Sun emits neurtrinos

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3 years ago
What happens to the current in a circuit if the resistance triples? If the voltage triples?
Sauron [17]
(Example 1 ) 
<span>If the Voltage that furnishes the current is an ideal (no internal resistance) Voltage source. Then; </span>

<span>V/R = i </span>
<span>V/2R = i/2 If external resistance doubles, current reduced to 1/2 of original value </span>
<span>V/3R = i/3 If external resistance triples, current reduced to 1/3 of original value </span>

<span>(Example 2) </span>
<span>But if the Voltage that furnishes the current is a practical [contains an internal resistance (Ri)] Voltage source. Then the current is a function of the Voltage source`s internal resistance, which does not double nor triple, plus the external resistance which is being doubled and tripled. </span>

<span>V/(R + Ri) = i </span>
<span>V/(2R + Ri) = greater than i/2 but less than I. </span>
<span>V/(3R + Ri) = greater than i/3 but less than i/2</span>
7 0
3 years ago
An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m
Aneli [31]

Answer:

\displaystyle y_m=3.65m

Explanation:

<u>Motion in The Plane</u>

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of v_o and \theta\\ as the initial speed and angle, then we have

\displaystyle v_x=v_o\ cos\theta

\displaystyle v_y=v_o\ sin\theta-gt

\displaystyle x=v_o\ cos\theta\ t

\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}

If we want to know the maximum height reached by the object, we find the value of t when v_y becomes zero, because the object stops going up and starts going down

\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt

Solving for t

\displaystyle t=\frac{v_o\ sin\theta }{g}

Then we replace that value into y, to find the maximum height

\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2

Operating and simplifying

\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}

We have

\displaystyle v_o=20\ m/s,\ \theta=25^o

The maximum height is

\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}

\displaystyle y_m=3.65m

7 0
3 years ago
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