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aleksandr82 [10.1K]

3 years ago

11

Example 2: Write the slope-intercept form of an equation for the line that passes through (-3.-2) and is perpendicular to y = 4x

+ 12.

1 answer:

Brilliant_brown [7]3 years ago

8 0

Answer:

bubu kayo mga tanga gago Mali ang sa got dito

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The burning of 1.0 g of coal produces 8.4 kcal. How many kilojoules are produced?

Andre45 [30]

The conversations need to solve this problem:

1 cal = 4.184 joules
1 Kcal= 1000 calories
1 kj= 1000 joules

or a more direct approach---->> 1 Kcal = 4.184 Kjoules

8.4 kcal (1000 calories/ 1 Kcal) x (4.184 joules/ 1 cal) x (1 Kj/ 1000 joules)= 35.1 Kj

or 8.4 kcal (4.184 Kj/ 1 kcal)= 35.1 Kj (same answer)

7 0

3 years ago

Which of the following equations demonstrates the law of conservation of mass?

VLD [36.1K]

Answer:

Mg + O2 ---> MgO

Explanation: V2O5 + CaS → CaO + V2S5. 5V2O5 + 5CaS → 10CaO + 5V2S5. 3V2O5 + 3CaS → 3CaO + 3V2S5.

7 0

3 years ago

In DNA, what base pairs with cytosine?

adell [148]

The answer is C guanine and thymine goes with adenine these pairs will always be the same a=t g=c

8 0

3 years ago

Read 2 more answers

A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second l

madreJ [45]

Explanation:

From first source, kinetic energy ( $K.E_{1}$ ) ejected is 1 eV and wavelength of light is $\lambda$ .

From second source, kinetic energy ( $K.E_{2}$ ) ejected is 4 eV and wavelength of light is $\frac{\lambda}{2}$ .

Relation between work function, wavelength, and kinetic energy is as follows.

K.E = $\frac{hc}{\lambda} - \phi$

where, h = Plank's constant = $6.63 \times 10^{-34}$ J.s

c = speed of light = $3 \times 10^{8}$ m/s

Also, it is known that 1 eV = $1.6 \times 10^{-19}$ J

Therefore, substituting the values in the above formula as follows.

From first source,

$K.E_{1}$ = $\frac{hc}{\lambda} - \phi$

1 eV = $1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi$

$1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (1)

From second source,

$K.E_{2}$ = $\frac{hc}{\lambda} - \phi$

$4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi$

$6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (2)

Now, divide equation (2) by 2. Therefore, it will become

${6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}$

$3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}$ ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

$1.6 \times 10^{-19} = \frac{\phi}{2}$

$\phi$ = $3.2 \times 10^{-19}$

= 2 eV

Thus, we can conclude that work function of the metal is 2 eV.

3 0

4 years ago

The theory of plate tectonics states that the earths surface is made up of moving ________

andriy [413]

Answer:

Plates

Explanation:

for example when 2 plates rub together that creates an earthquake a tsunami or even a mountain

7 0

3 years ago