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Gnesinka [82]
3 years ago
6

How can scientists study the inner structure of an atom?

Physics
1 answer:
Arturiano [62]3 years ago
8 0
A microscope is one way
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A skier starts from rest at the top of a hill that is inclined 10.5° with respect to the horizontal. The hillside is 200 m long,
Svetlanka [38]

Answer:

d) 289.31 m

Explanation:

Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m

Energy used by friction = μmgcos 10.5 x 200 = .075 x m x 9.8 x cos 10.5 x200 = 144.54 m .

Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d

To equate

357.18 m -144.54 m = .735 m d

d = 289.31 m .

4 0
3 years ago
A generator with �# ' = 300 V and Zg = 50 Ω is connected to a load ZL = 75 Ω through a 50-Ω lossless line of length l = 0.15λ. (
ki77a [65]

Answer:

a. Zin = 41.25 - j 16.35 Ω

b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.  Pin = 216 w

d. PL = Pin = 216 w

e. Pg = 478.4 w , Pzg = 262.4 w

Explanation:

a.

Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]  

βl = 2π / λ * 0.15 λ = 54 °

Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]

Zin = 41.25 - j 16.35 Ω

b.

I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶

V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)

V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.

Pin = ¹/₂ * Re * [V₁ * I₁]

Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)

Pin = 216 w

d.

The power PL and Pin are the same as the line is lossless input to the line ends up in the load so

PL = Pin

PL = 216 w

e.

Pg Generator

Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)

Pg = 478.4 w

Pzg dissipated

Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50

Pzg = 262.4 w

4 0
3 years ago
The table shows the diameters of the planets in our solar system. Assume that a basketball whose diameter is 25
tino4ka555 [31]

Answer:

Mercury:

.85

pea

Venus:

2.1

gumball

Earth:

2.2

gumball

Mars:

1.2

marble

Uranus:

9

grapefruit

Neptune:

8.6

softball

Explanation:

I have no clue if I'm right but hopefully, I am

3 0
2 years ago
The muzzle velocity of a rifle bullet is 709 m s−1along the direction of motion. If the bullet weighs 35 g, and the uncertainty
nydimaria [60]

Answer:

Uncertainty in position of the bullet is \Delta x=1.07\times 10^{-33}\ m

Explanation:

It is given that,

Mass of the bullet, m = 35 g = 0.035 kg

Velocity of bullet, v = 709 m/s

The uncertainty in momentum is 0.20%. The momentum of the bullet is given by :

p=mv

p=0.035\times 709=24.81\ kg-m/s

Uncertainty in momentum is,

\Delta p=0.2\%\ of\ 24.81

\Delta p=0.049

We need to find the uncertainty in position. It can be calculated using Heisenberg uncertainty principal as :

\Delta p.\Delta x\geq \dfrac{h}{4\pi}

\Delta x=\dfrac{h}{4\pi \Delta p}

\Delta x=\dfrac{6.62\times 10^{-34}}{4\pi \times 0.049}

\Delta x=1.07\times 10^{-33}\ m

Hence, this is the required solution.

7 0
3 years ago
Why is it that, in a spoon, you can see yourself upside down? Bamboozled me for years.
Delvig [45]
Here is your answer.

6 0
3 years ago
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