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2 years ago

13

During which of the following activities may in inflatable personal flotation device to be used

1 answer:

Alla [95]2 years ago

5 0

They may be used whenever you are not experienced with swimming and you are near the water such as swimming, canoeing, etc

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What conventions are used in SI to indicate units

klemol [59]

Answer:

<u>Conventions used in SI to indicate units are as follows:</u>

Only singular form of units are used. for example: use kg and not kgs.
Do not use full stop after the abbreviations of any unit. for example: do not use kg. or cm.
Use one space between last numeric digit and SI unit. for example: 10 cm, 9 km.
Symbols and words should not be mixed. for example: use Kilogram per cubic and not kilogram/m3.
While writing numerals, only the symbols of the units should be written. for example: use 10 cm and not Ten cm.
Units named after a scientist should be written in small letters. for example: newton, henry.
Degree sign should not be used when the kelvin unit is used. for exmaple: use 37° and not 37°k

5 0

2 years ago

6. The hole on a level, elevated golf green is a horizontal distance of 150 m from the tee and at an elevation of 12.4 m above t

Georgia [21]

Answer:

u = 104.68 m/s

Explanation:

given,

horizontal distance = 150 m

elevation of 12.4 m

angle = 8.6°

horizontal motion = x = u cos θ. t .............(1)

vertical motion =

$y = u sin \theta - \dfrac{1}{2}gt^2$ ................(2)

from equation(1) and (2)

$y = x tan \theta - \dfrac{gx^2}{2u^2cos^2\theta}$ ..........{3}

$12.4 = 150\times tan (8.6) - \dfrac{9.8\times 150^2}{2u^2cos^2(8.6)}$

$\dfrac{9.8\times 150^2}{2u^2cos^2(8.6)} = 10.29$

$\dfrac{9.8\times 150^2}{2\times 10.29\times cos^2(8.6)} = u^2$

$u = \sqrt{10959.34}$

u = 104.68 m/s

The initial speed of the ball is u = 104.68 m/s

8 0

2 years ago

An electron has an initial velocity of (19.0 j + 18.0 k) km/s and a constant acceleration of (3.00 ✕ 1012 m/s2)i in a region in

asambeis [7]

Answer:

$E=(-17.08 i +7.2 j -7.6 k )N/C$

Explanation:

$v= (19.0j+18.k)km/s$

$a=3.0x10^{13}m/s^2$

$\beta= 400x10{-6}T$

Electron information needed to solve the question:

$m_e=9.11x10^{-31}kg$

$q=-1.6x10{-19}C$

$F=F_E+F_B=q*(E+Vx \beta)$

$F=m*a$

$m*a=q*(E+Vx\beta)$

$E=\frac{m*a}{q}-(Vx\beta )$

$E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]$

$E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C$

$E=(-17.08 i +7.2 j -7.6 k )N/C$

6 0

2 years ago

Nikitich [7]

Answer:

Protons and Neutrons are found in the nucleus and the electrons are found on the outside of the nucleus

Explanation:

Hope this helps

7 0

2 years ago

Which of the following elements is in the same period as phosphorus?

Arte-miy333 [17]

The answer is B. magnesium I am pretty sure

5 0

2 years ago