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anzhelika [568]
3 years ago
12

How much more light can the James Webb Space Telescope (with its 6-m diameter mirror) gather than the Hubble Space Telescope (wi

th a diameter of 2.4 m)?
Physics
1 answer:
OverLord2011 [107]3 years ago
4 0

Answer:

James Webb Space Telescope has 6.25 times the light collecting area than the mirror of Hubble Space Telescope.

Explanation:

Radius of Hubble Space Telescope mirror = 2.4/2 = 1.2 m

Radius of James Webb Space Telescope mirror = 6/2 = 3 m

Area of mirror of Hubble Space Telescope= πr² = π × 1.2² = 4.52 m²

Area of mirror of James Webb Space Telescope = π × 3² = 28.27 m²

Ratio of mirror area

[tex]\frac{A_J}{A_H}=\frac{28.27}{4.52}\\\Rightarrow A_J=6.25A_H[\tex]

mirror of James Webb Space Telescope has 6.25 times the area of mirror of Hubble Space Telescope.

Which means the James Webb Space Telescope has 6.25 times the light collecting area than the mirror of Hubble Space Telescope.

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To verify the identity, we can make use of the basic trigonometric identities:
cot θ = cos θ / sin θ 
sec θ = 1 / cos <span>θ
csc </span>θ = 1 / sin θ<span>

Using these identities:
</span>cot θ ∙ sec θ = (cos θ / sin θ ) (<span> 1 / cos </span><span>θ)
</span>
We can cancel out cos <span>θ, leaving us with
</span>cot θ ∙ sec θ = 1 / sin θ
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6 0
4 years ago
Ví dụ 1 : Hai vật xuất phát từ A và B cách nhau 400m chuyển động cùng chiều theo hướng từ A đến B. Vật thứ nhất chuyển động đều
Mashcka [7]

Answer:

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3 years ago
1. How much power does a light bulb contain if it does 600J of work in 5 seconds?
dimaraw [331]

Answer:

120 watts

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#1: 120 watts

#2: 667 watts

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6 0
4 years ago
Read 2 more answers
a nonrelativistic proton is confined to a length of 2.0 pm on the x-axis. what is the kinetic energy of the proton if its speed
Elina [12.6K]

Answer:

K = 1.29eV

Explanation:

In order to calculate the kinetic energy of the proton you first take into account the uncertainty principle, which is given by:

\Delta x \Delta p\geq \frac{h}{4\pi}      (1)

Δx : uncertainty of position = 2.0pm = 2.0*10^-12m

Δp: uncertainty of momentum = ?

h: Planck's constant = 6.626*10^-34 J.s

You calculate the minimum possible value of Δp from the equation (1):

\Delta p=\frac{h}{4\pi \Delta x}=\frac{6.626*10^{-34}J.s}{4\pi(2.0*10^{-12}m)}\\\\\Delta p=2.63*10^{-23}kg.\frac{m}{s}

The minimum kinetic energy is calculated by using the following formula:

k=\frac{(\Delta p)^2}{2m}       (2)

m: mass of the proton = 1.67*10^{-27}kg

k=\frac{(2.63*10^{-23}kgm/s)^2}{2(1.67*10^{-27}kg)}=2.08*10^{-19}J

in eV you have:

2.08*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.29eV

The kinetic energy of the proton is 1.29eV

7 0
3 years ago
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