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3 years ago

Radon is the heaviest naturally radioactive ________________ gas. A) noble B) halogen C) group 1 D) diatomic

Physics

2 answers:

Sever21 [200]3 years ago

6 0

(A) noble gases / group 18

Effectus [21]3 years ago

6 0

Answer: A) noble

Explanation:

Noble gases are group 18 elements which have completely filled valence shell. thus they do not lose or gain electrons and are called as nonreactive or noble.

Electron configuration of radon is:

$Rn:[Xe]4f^{14}5d^{10}6s^26p^6$

Halogens are group 17 elements and have a valency of 1 as they can gain one electron to attain stable configuration.

Group 1 elements are alkali metals which have a valency of 1 as they can lose their outermost electron to attain stable configuration.

Diatomic means the molecule contains two atoms in one molecule.

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4 years ago

On a vacation flight, you look out the window of the jet and wonder about the forces exerted on the window. Suppose the air outs

user100 [1]

Answer:

A) $\Delta P = 14512.5 Pa = 14.512 kPa$

B) F = 1632.65 N

Explanation:

Given details

outside air speed is given as $v_2 = 150 m/s$

since inside air is atmospheric , $v_1 = 0 m/s$

a) By using bernoulli equation between outside and inside of flight

$P_1 + \frac{1}{2} \rho v_1^2 + \rho gh = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh$

$\Delta P = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2$

$\Delta P = \frac{1}{2} \rho[ v_2^2 -v_1^2]$

$\Delta P = \frac{1}{2} 1.29 [ 150^2 - 0^2]$

$\Delta P = 14512.5 Pa = 14.512 kPa$

b) force exerted on window

Area of window $= 25\times 45 = 1125 cm^2 = 0.1125 m^2$

We know that force is given as

$F = P\times A$

$F = 14512.5 \times 0.1125$

F = 1632.65 N

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4 years ago

A 30.0 kg child, initially at rest, slides down a 2.0 m tall slide. The child reaches the bottom of the slide with a speed of 6

HACTEHA [7]

Answer:

Explanation:

Total energy is constant

E = mgh + ½mv² + Fd

At the top of the slide, all energy is potential

E = mgh + 0 + 0

At the bottom of the slide, all potential energy has converted to kinetic and work of friction.

mgh = ½mv² + W

W = mgh - ½mv²

W = 30.0[(9.81)(2.0) - ½6²]

W = 48.6 J

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2 years ago