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2 years ago

The two major parts of the optical telescope

1 answer:

5 0

The "objective" (lens or mirror) is the major major major part of
the optical telescope. It's really the only part you need in order
to make a telescope (besides something to hold the objective).
You can put a piece of film or a CCD right at the focal point of
the objective lens or mirror and capture 'images' (pictures) there.

If you want to use the telescope for looking through and seeing stuff
with your eye, then you need the other major part ... the eyepiece lens.

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suppose 384g of steam originally at 100C is quickly cooled to produce liquid water at 31C. How much heat must be removed from th

dlinn [17]

Answer:

$Q=977216.256\ J=977.216\ kJ$

Explanation:

Given:

mass of steam, $m=384\ g$

temperature of steam, $T_{is}=100^{\circ}C$

temperature of resultant water, $T_{fw}=31^{\circ}C$

We have,

latent heat of vapourization of water, $L=2256\ J.g^{-1}$

specific heat capacity of water, $c=4.186\ J.g^{-1}$

When we cool the steam of 100°C then firstly it loses its latent heat to convert into water of 100°C and the further cools the water.

Now the heat removed from steam to achieve the final state of water:

$\rm Q=latent\ heat\ of\ vapourization+sensible\ heat\ of\ water$

$Q=m(L+c.\Delta T)$

$Q=384(2256+4.186\times (100-31))$

$Q=977216.256\ J=977.216\ kJ$

3 0

3 years ago

A compound whose molecules contain one boron atom and three fluorine atoms would be named monoboron fluoride. Please select the

Slav-nsk [51]

I know it is false...............

5 0

2 years ago

If we add 50 Joules of thermal energy to a heat engine, and that heat engine does 30 Joules of work, how much thermal energy is

Natalka [10]

Answer:

The correct answer should be

A. 20 Joules

Explanation:

I'm taking the K12 Unit Test: Energy - Part 1 right now

7 0

2 years ago

A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa

Lesechka [4]

At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
$U_i= \frac{1}{2} ka^2$
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
$K_i =0$
And the total energy of the system is
$E_i = U_i+K= \frac{1}{2}ka^2$

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
$U_f = 0$
while the mass is moving at speed v, and therefore the kinetic energy is
$K_f = \frac{1}{2} mv^2$
And the total energy is
$E_f = U_f + K_f = \frac{1}{2} mv^2$

For the law of conservation of energy, the total energy must be conserved, therefore $E_i = E_f$ . So we can write
$\frac{1}{2} ka^2 = \frac{1}{2}mv^2$
that we can solve to find an expression for v:
$v= \sqrt{ \frac{ka^2}{m} }$

6 0

3 years ago

In a lab, four balls have the same velocities but different masses.

olya-2409 [2.1K]

Answer:

New Momentum of Ball B $=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$