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3 years ago

Match the following moves to the correct total distance and displacement of the moves.

1 answer:

5 0

A. Move 2 m east and then 12 m east; displacement is 14 m east and the distance is 14 m
B. Move 10 m east and then 12 m west, the displacement is 2 m west and the distance is 22 m.
C. Move 8 m west and then 16 m east; the displacement is 8 m east and the distance is 24 m
D. Move 12 m west and then 8 m east; the displacement is 4 m and the distance is 20 m

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Because of surface tension, it is possible, with care, to support an object heavier than water on the water surface. The maximum

boyakko [2]

Answer:

$\pi 1 = \frac{h}{l}$

$\pi 2 =$ б / $l^2gp$

$\frac{h}{l} =$ Ф ( б / $l^2gp$ )

Explanation:

<u>Develop a suitable set of dimensionless parameters for this problem</u>

The set of dimensionless parameters for this problem is :

$\pi 1 = \frac{h}{l}$

$\pi 2 =$ б / $l^2gp$

$\frac{h}{l} =$ Ф ( б / $l^2gp$ )

and they are using the pi theorem, MLT systems

attached below is a detailed solution

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3 years ago

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Answer:

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2 years ago

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tester [92]

Answer:

node

Explanation:

on the graph node is higher than antinode

so it can get or hear loud sounds faster

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3 years ago

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3 years ago

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Through what voltage must an αα-particle, with its charge of +2e+2e, be accelerated so that it has just enough energy to reach a

mezya [45]

Answer:5101.35v

Explanation:

Radius of gold nucleus=7.3×10-15m and a charge of +79e

Q= 79e

e=1.6×10^-19

q= +2e

The nucleus is considered as the point charge where the potential energy between the charges are

U = 1/(4×3.142×Eo)×(qQ)/r

Where r is distance between the charges and the nucleus

r=R+d

V=U/q

U= 1/(4×3.142×Eo)×Q/r

V= 1/(4×3.142×Eo)×Q/(r+d)

9.0×10^9 ×(79×10^-19)/(7.3×10^-15)+(1.5×10^-14)

V= 9.0×10^9 ×(1.264×10^-17)/(2.23×10^-14)

V= 9×10^9×(5.67×10^-14)= 5101.35v

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3 years ago