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tatyana61 [14]
3 years ago
8

Why is it that the farther away from earth you go the bigger the planets get?

Physics
1 answer:
Tanzania [10]3 years ago
3 0
Because you're getting closer to other planets?
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After a check up, a person now has a far point of 100 cm, but with good near point vision. He needs to wear a new pair of correc
lakkis [162]

Answer:

so his far point according to this pair of glass is 200 cm

Explanation:

power of old pair of corrective glasses is given as

P = -0.5 dioptre

now we have

f = \frac{1}{P}

f = -2 m

f = -200 cm

now we know that for normal vision the maximum distance of vision is for infinite distance

so by lens formula we have

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

\frac{1}{d_i} + 0 = \frac{1}{200}

d_i = 200 cm

so his far point according to this pair of glass is 200 cm

7 0
3 years ago
1. What is evaporation and how does it affect weather?
alina1380 [7]

Answer:

Explanation:don’t know

8 0
3 years ago
Behavioral Adaptations: Behavior that animals begin life with that helps them meet their needs is called ________.
Arturiano [62]
I believe the blank would simply be behaviour adaptations. Behavioural adaptations are behaviours that organisms demonstrate to help them better survive and reproduce in a habitat. Hope that helps!!
6 0
3 years ago
Read 2 more answers
A 6.00-mH solenoid is connected in series with a 5.0-μF capacitor and an AC source. The solenoid has internal resistance 3.0 Ω w
son4ous [18]

Answer:

5773.50269 Hz

23 A

Explanation:

L = Inductance = 6 mH

C = Capacitance = 5 μF

R = Resistance = 3 Ω

\epsilon = Maximum emf = 69 V

Resonant angular frequency is given by

\omega=\dfrac{1}{\sqrt{LC}}\\\Rightarrow \omega=\dfrac{1}{\sqrt{6\times 10^{-3}\times 5\times 10^{-6}}}\\\Rightarrow \omega=5773.50269\ Hz

The resonant angular frequency is 5773.50269 Hz

Current is given by

I=\dfrac{\epsilon}{R}\\\Rightarrow I=\dfrac{69}{3}\\\Rightarrow I=23\ A

The current amplitude at the resonant angular frequency is 23 A

7 0
3 years ago
If the average velocity during the athlete's walk back
goblinko [34]

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

From the question we are told

If the average velocity during the athlete's walk back  to the starting line in Guided Example 2.5 is – 1.50 m/s,

Generally the equation Time spent  is mathematically given as

T=\frac{d}{v}

Therefore

T=\frac{d}{1.50}

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

For more information on this visit

brainly.com/question/22271063

4 0
3 years ago
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