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EleoNora [17]
3 years ago
7

Calculate 1.6 ∙ 103 times 3.0 ∙ 10-6.

Physics
2 answers:
Ray Of Light [21]3 years ago
8 0

Answer:

3,955.2

Explanation:

I think using order of operations PEMDAS this is right

VARVARA [1.3K]3 years ago
6 0
1.6*103*3.0*10-6=4938
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Ivan

Answer:

333.5 MJ

Explanation:

ΔV = m·g·Δh

     = 8500 · 9.81 · (15000-11000)

     = 8500 · 9.81 · 4000

     = 333 540 000

     ≈ 333.5 ·10⁶J = 333.5 MJ

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3 years ago
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scZoUnD [109]
Yes? I think a tutor would help with this!!!!
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3 years ago
One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a s
baherus [9]

Given that,

Central maximum = 1 cm

Distance from the window shade to the wall =4 m

We know that,

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Using formula of average wavelength

\lambda_{avg}=\dfrac{\lambda_{1}+\lambda_{2}}{2}

Put the value into the formula

\lambda_{avg}=\dfrac{400+700}{2}

\lambda_{avg}=550\ nm

(b). We need to calculate the diameter of the pinhole

Using formula for diameter

w=\dfrac{2.44\lambda L}{D}

D=\dfrac{2.44\lambda L}{w}

Put the value into the formula

D=\dfrac{2.44\times550\times10^{-9}\times4}{1\times10^{-2}}

D=0.537\ mm

Hence, (a). The average wavelength 550 nm.

(b). The diameter of the pinhole is 0.537 mm.

7 0
4 years ago
Let’s determine what lens is needed to correct the vision of a myopic eye. Assume that the far point of the myopic eye is 50 cmc
koban [17]

Answer:

Explanation:

Person suffering from myopia is unable to see objects beyond 50 cm  because far point of his vision is 50 cm . He requires a concave lens

( diverging lens ) to correct his vision .

After using the lens ,

ray of light coming from far off place will appear to be coming from 50 cm after refraction through lens so it will be focused on retina .

object distance u = ∝

image distance v =  - 50 cm

focal length f = ?

1 / v - 1 / u = 1 /f

- 1 / 50 - 1 / ∝ = 1 / f

- 1 / 50  = 1 / f

f = - 50 cm

focal length is negative so len is concave.

He requires  concave lens of focal length 50 cm .

7 0
3 years ago
Figure 1 shows the kinetic energy as a function of time for a 2kg object that is released from rest and falls toward Earth’s sur
garri49 [273]

<u><em>Answer:</em></u>

The answer is 1400 J, according to my Physics teacher.

<u><em>Explanation:</em></u>

You need to take into account everything that is listed in the question; it's important to remember that the question is asking about the change in gravitational potential energy of the object-object-Earth system from 0s to 10s, not 0s to 20s. :)

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3 years ago
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