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Sati [7]
3 years ago
9

N Rutherford's gold foil experiment the alpha particles pass through which part of the atom?

Physics
1 answer:
OleMash [197]3 years ago
3 0
It passes through C the Electron Cloud
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Electromagnetic
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3 years ago
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When a 25000-kg fighter airplane lands on the deck of the aircraft carrier, the carrier sinks 0.23cm deeper into the water.
Genrish500 [490]

Answer:

10604 square meters

Explanation:

0.23 cm = 0.0023m

Assume the carrier has a shape of a rectangular box. When the carrier sinks 0.0023m deeper into water, the extra volume submerged into water is the same as the extra water volume being replaced. This extra volume would add an additional buoyant force to counter balance the extra weight created by the 25000 kg fighter.

With g being constant, the mass of the extra water displaced is the same as the mass of the fighter airplane m_f. And mass of the water displaced is its volume V times its density \rho

V\rho = m_f

1025V = 25000

V = 25000/1025 = 24.39 m^3

We assume the carrier has a shape of a rectangular box, so this extra displaced volume is the extra depth d = 0.0023 m times cross-section area A

V = dA

24.39 = 0.0023A

A = 24.39 / 0.0023 = 10604 m^2

4 0
4 years ago
Which exoplanet has the most Earth-like orbit?​
Digiron [165]

Answer: Kepler-452b

I think

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4 0
3 years ago
Two motorcycles travel along a straight road heading due north. At t = 0 motorcycle 1 is at x = 50 m and moves with a constant s
Oduvanchick [21]

Answer:

Vf = 23 m/s

Explanation:

First we need to find the distance covered by the motorcycle 2 when it passes motorcycle 1. Using the uniform speed equation for motorcycle 1:

s₁ = v₁t₁

where,

s₁ = distance covered by motorcycle 1 = ?

v₁ = speed of motorcycle 1 = 6.5 m/s

t₁ = time = 10 s

Therefore,

s₁ = (6.5 m/s)(10 s)

s₁ = 65 m

Now, for the distance covered by motorcycle 2 at the meeting point. Since, the motorcycle started 50 m ahead of motorcycle 2. Therefore,

s₂ = s₁ + 50 m

s₂ = 65 m + 50 m

s₂ = 115 m

Now, using second equation of motion for motorcycle 2:

s₂ = Vi t + (1/2)at²

where,

Vi = initial velocity of motorcycle 2 = 0 m/s

Therefore,

115 m = (0 m/s)(10 s) + (1/2)(a)(10 s)²

a = 230 m/100 s²

a = 2.3 m/s²

Now, using 1st equation of motion:

Vf = Vi + at

Vf = 0 m/s + (2.3 m/s²)(10 s)

Vf = 23 m/s

3 0
3 years ago
A body accelerates by 25m/s 2 when it applied by 40n forces.what would be acceleration if it is applied by 80n forces
il63 [147K]

Answer:

50m/s²

Explanation:

cross muliplyy write down the values acceleration and force

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