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3 years ago

12

YALLL I NEED HELP ASAP... THIS DUE AT 11:59 AND ITS 10.42 RIGHT NOW

1 answer:

Stells [14]3 years ago

7 0

Yeah it’s probably A

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Which objects switch places in a double-replacement reaction?

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A. compounds because the reactants of a double replacement reaction do not have charges

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3 years ago

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at the moment a truck traveling with a constant velocity of 13.6 m/s passes a dog at rest the dog begins moving with a constant

goblinko [34]

I don’t know how to help you hope you figure it out have a good day

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2 years ago

Compute the ratio of the rate of heat loss through a single-pane window with area 0.15 m2 to that for a double-pane window with

Rudik [331]

Answer:

2.80321285141

Explanation:

$L_g$ = Thickness of glass = 4.5 mm

$k_g$ = Thermal conductivity of glass = 0.8 W/mK

$R_0$ = Combined thermal resistance = $0.15\times m^2K/W$

$L_a$ = Thickness of air = 6.6 mm

$k_a$ = Thermal conductivity of air = 0.024 W/mK

The required ratio is the inverse of total thermal resistance

$\dfrac{2(L_g/k_g)+R_0+(L_a/k_a)}{(L_g/k_g)+R_0}\\ =\dfrac{2(4.5\times 10^{-3}/0.8)+0.15+(6.6\times 10^{-3}/0.024)}{(4.5\times 10^{-3}/0.8)+0.15}\\ =2.80321285141$

The ratio is 2.80321285141

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3 years ago

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Kilo- - 1,000<br>deka- - 10<br><br>How much larger is a kilo- than a deka-?

Ilya [14]

$lets's \ take \ an \ example \ one \ gram: \\ \\ 0,001kg \\ \\ 0,01hg \\ \\ 0,1dag \\ \\ \boxed{1g} \\ \\ 10dg \\ \\ 100cg \\ \\ 1000mg \\ \\ the \ units \ are \ rising \ and \ decreasing \ from \ 10 \ to \ 10 \ results \ \\ \\ a \ kilogram \ is \ 100 \ times \ larger \ than \ a \ dekagram$

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3 years ago

Particle-X has a speed of 0.720 c and a momentum of 4.350x1019 kgm/s. What is the mass of the particle? 2.0206 10-27 kg Hints: T

kirill115 [55]

Explanation:

Given that,

Speed of particle = 0.720 c

Momentum = 4.350\times10^{-19}\ kgm/s[/tex]

(I). We need to calculate the mass of the particle

Using formula of momentum

$P=mv$

$m =\dfrac{P}{v}$

$m=\dfrac{4.350\times10^{-19}}{ 0.720\times3\times10^{8}}$

$m=2.013\times10^{-27}\ Kg$

We need to calculate the rest mass of particle

Using formula of rest mass

$m=\dfrac{m_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}$

Where, $m_{0}$ = rest mass

Put the value into the formula

$m_{0}=2.013\times10^{-27}\times\sqrt{1-(\dfrac{0.720 c}{c})^2}$

$m_{0}=2.013\times10^{-27}\times\sqrt{1-(0.720)^2}$

$m_{0}=1.4\times10^{-27}\ kg$

(b). We need to calculate the rest energy of the particle

Using formula of energy

$E_{0}=m_{0}c^2$

Put the value into the formula

$E_{0}=1.4\times10^{-27}\times(3\times10^{8})^2$

$E_{0}=1.26\times10^{-10}\ J$

(c). We need to calculate the kinetic energy of the particle

Using formula of kinetic energy

$K.E=mc^2-m_{0}c^2$

$K.E=(m-m_{0})\timesc^2$

$K.E=(2.013\times10^{-27}-1.4\times10^{-27})\times3\times10^{8}$

$K.E=1.84\times10^{-19}\ J$

(d). We need to calculate the total energy of the particle

Using formula of energy

$E=mc^2$

Put the value into the formula

$E=2.013\times10^{-27}\times(3\times10^{8})^2$

$E=1.812\times10^{-10}\ J$

Hence, This is the required solution.

8 0

3 years ago