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3 years ago

12

How much does a 200 kg car weight?

1 answer:

marin [14]3 years ago

6 0

Answer:

22 kgs

Explanation:

440.925 pounds if thats what you mean but kgs is the weight system for everyone but america

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The nonmetal family of the periodic table that wants to gain, lose, or share four electrons is the __________ family.

vivado [14]

<span>d. halogen is the answer</span>

6 0

3 years ago

Read 2 more answers

Please please help

dsp73

Answer:

true

Explanation:

The law of conservation of charge states that whenever electrons are transferred between objects, the total charge remains the same.

3 0

4 years ago

Tony charges a capacitor and then discharges it through a resistor. He notices that, after two time constants, the voltage acros

grigory [225]

Answer:

Half of its value prior to the initiation of the discharge

Explanation:

7 0

3 years ago

7. A double-slit experiment uses coherent light of wavelength 633 nm with a slit separation of 0.100 mm and a screen placed 2.0

Illusion [34]

Answer:

The distance between first-order and second-order bright fringes is 12.66mm.

Explanation:

The physicist Thomas Young establishes through its double slit experiment a relationship between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.

$\Lambda x = L\frac{\lambda}{d}$ (1)

Where $\Lambda x$ is the distance between two adjacent maxima, L is the distance of the screen from the slits, $\lambda$ is the wavelength and d is the separation between the slits.

The values for this particular case are:

$L = 2.0m$

$\lambda = 633nm$

$d = 0.100mm$

Notice that is necessary to express L and $\lambda$ in units of milimeters.

$L = 2.0m \cdot \frac{1000mm}{1m}$ ⇒ $2000mm$

$\lambda = 633nm \cdot \frac{1mm}{1x10^{6}nm}$ ⇒ $6.33x10^{-4}mm$

Finally, equation 1 can be used:

$\Lambda x = (2000mm)\frac{(6.33x10^{-4}mm)}{(0.100mm)}$

$\Lambda x = 12.66mm$

Hence, the distance between first-order and second-order bright fringes is 12.66mm.

8 0

3 years ago

Radon-222 ( 222/86 Rn) is a radioactive gas with a half-life of 3.82 days. A gas sample contains 4.1 e 8 radon atoms initially.

kow [346]

Answer :

(a) The number of radon atoms will remain after 12 days is, $4.67\times 10^7$

(b) The number of radon nuclei have decayed by this time will be, $3.6\times 10^8$

Explanation :

<u>For part (a) :</u>

Half-life = 3.82 days

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{3.82\text{ days}}$

$k=1.81\times 10^{-1}\text{ days}^{-1}$

Now we have to calculate the number of radon atoms will remain after 12 days.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $1.81\times 10^{-1}\text{ days}^{-1}$

t = time passed by the sample = 12 days

a = initially number of radon atoms = $4.1\times 10^8$

a - x = number of radon atoms left = ?

Now put all the given values in above equation, we get

$12=\frac{2.303}{1.81\times 10^{-1}}\log\frac{4.1\times 10^8}{a-x}$

$a-x=4.67\times 10^7$

Thus, the number of radon atoms will remain after 12 days is, $4.67\times 10^7$

<u>For part (b) :</u>

Now we have to calculate the number of radon nuclei will have decayed by this time.

The number of radon nuclei have decayed = Initial number of radon atoms - Number of radon atoms left

The number of radon nuclei have decayed = $(4.1\times 10^8)-(4.67\times 10^7)$

The number of radon nuclei have decayed = $3.6\times 10^8$

Thus, the number of radon nuclei have decayed by this time will be, $3.6\times 10^8$

5 0

3 years ago