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jekas [21]
3 years ago
12

What problem do refractor telescopes have that reflectors don't? Group of answer choices bad seeing light loss from secondary el

ements diffraction limited resolution spherical aberration chromatic aberration
Physics
1 answer:
Paul [167]3 years ago
8 0

chromatic aberration problem do refractor telescopes have that reflectors don't

<u>Explanation:</u>

Chromatic aberration is a phenom in which light rays crossing through a lens focus at various points, depending on their wavelength. Chromatic aberration is a dilemma in which lens or refracting, telescopes undergo from. The various image distances for the respective colors affect various image sizes for them.

This involves the creation of disturbing color fringes in the image. Chromatic aberration can be pretty well adjusted by the use of an achromatic doublet. Here, a positive biconvex lens is coupled with a negative lens placed backward with greater dispersion. Thus partly compensates for the chromatic aberration.

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Help me with the question in the image I provided​
inna [77]

Answer: 10 s, 30 m/s , 150 m

Explanation:

Given

The speed of motorcyclist is  u_1=15\ m/s

The initial speed of a police motorcycle is u_2=0\ m/s

acceleration of police motorcycle is a=3\ m/s^2

Police will catch the motorcyclist when they traveled equal distances

distance traveled by motorcyclist in time t is

\Rightarrow s_1=15\times t

Distance traveled by Police in time t is

\Rightarrow s_2=u_2t+\frac{1}{2}at^2\\\Rightarrow s_2=0+0.5\times 3\times t^2

put s_1=s_2

\Rightarrow 15t=0.5\times 3\times t^2\\\\\Rightarrow t(1.5t-15)=0\\\\\Rightarrow t=\dfrac{15}{1.5}=10\ s

Police officer's speed at that time is

\Rightarrow v_2=u_2+at\\\Rightarrow v_2=0+3\times 10=30\ m/s

Distance traveled by each vehicle is

\Rightarrow s_1=s_2=15\times t=15\times 10=150\ m

8 0
3 years ago
What name is used for a high-power variable resistor?
const2013 [10]
The answer would be C: Rheostat.  :)
3 0
3 years ago
Read 2 more answers
A 34.0 %-efficient electric power plant produces 800 MW of electric power and discharges waste heat into 20∘C ocean water. Suppo
mojhsa [17]

Answer:

77647

Explanation:

\eta = Efficiency = 34%

Power used in 1 home = 0.02 MW

Total power is

P=\dfrac{800}{\eta}\\\Rightarrow P=\dfrac{800}{0.34}\\\Rightarrow P=2352.94117\ MW

Waste of power

2352.94117-800=1552.94117\ W

Number of homes would be given by

n=\dfrac{1552.94117}{0.02}=77647.0585\ homes

The number of homes that could be heated with the waste heat of this one power plant is 77647

8 0
3 years ago
(15pts) A hungry 12.0 kg fish is coasting from west to east at 75 cm/s when it suddenly swallows a 1 kg fish swimming towards it
faust18 [17]

Answer:

The speed of the big fish after swallowing the small fish is 0.38 m/s.

Explanation:

Consider west to east direction as positive and the opposite direction as negative.

Given:

Mass of big fish (m₁) = 12.0 kg

Initial velocity of big fish (u₁) = 75 cm/s = 0.75 m/s

Mass of small fish (m₂) = 1 kg

Initial velocity of small fish (u₂) = -4 m/s (Direction is opposite to u₁)

After swallowing the small fish, both the fishes move together with same velocity. Let the velocity be 'v'.

So, as there are no effects of drag or any other forces, the given scenario can be considered as a case of inelastic collision where the objects move together with same velocity after collision.

The momentum is conserved in inelastic collision. Therefore,

Initial momentum of the fishes = Final momentum of the fishes

m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}

Now, plug in the given values and solve for 'v'. This gives,

v=\frac{12.0\times 0.75+1\times (-4)}{12.0+1}\\\\v=\frac{9-4}{13}\\\\v=\frac{5}{13}=0.38\ m/s

Therefore, the speed of the big fish after swallowing the small fish is 0.38 m/s

3 0
3 years ago
A major artery with a cross sectional area of 1.00cm^2 branches into 18 smaller arteries, each with an average cross sectional a
dybincka [34]

Here we can say that rate of flow must be constant

so here we will have

A_1v_1 = 18 A_2v_2

now we know that

A_1 = 1 cm^2

A_2 = 0.4 cm^2

now from above equation

1 cm^2 v_1 = 18(0.400 cm^2)v_2

\frac{v_2}{v_1} = \frac{1}{18\times 0.4}

\frac{v_2}{v_1} = 0.14

so velocity will reduce by factor 0.14

3 0
4 years ago
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