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3 years ago

Which of the following statements is false?

Physics

2 answers:

Allushta [10]3 years ago

7 0

Answer:

Explanation:

The weight of an object is 1/6 of the earth's weight in the moon

For example if your weight is 54 kg on earth,

Weight on moon=1/6 of 54=1/6 x 54 = 9kg

olasank [31]3 years ago

7 0

Answer:

C. The weight of an object is the same on the Earth and the moon

Explanation:

The mass of an object is ⅙ of earth's weight on moon. So, option (C) is the false statement.

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Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th

Mariulka [41]

Answer:

a) 0.0288 grams

b) $2.6*10^{-10} J/kg$

Explanation:

Given that:

A typical human body contains about 3.0 grams of Potassium per kilogram of body mass

The abundance for the three isotopes are:

Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.

Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.

However, the amount of potassium that is present in such person is :

0.012% × 240 grams

= 0.012/100 × 240 grams

= 0.0288 grams

the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:

First the Dose in (Gy) = $\frac{energy \ absorbed }{mass \ of \ the \ body}$

= $\frac{1.10*10^6*1.6*10^{-14}}{80}$

= $2.2*10^{-10} \ J/kg$

Effective dose (Sv) = RBE × Dose in Gy

Effective dose (Sv) = $1.2 *2.2*10^{-10} \ J/kg$

Effective dose (Sv) = $2.6*10^{-10} J/kg$

5 0

4 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csf%7B%20What%20%5C%3A%20is%20%5C%3A%20a%20%5C%3A%20parallel%20%5C%3A%20circuit%3F%20%7D"

crimeas [40]

Answer:

parallel circuit is a kind of circuits in which all components are connected between the same two sets of electrically common points, creating multiple paths for the current to flow from one end of the battery to the other:

8 0

3 years ago

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4. show your understanding of electric and magnetic forces<br> below.

Vitek1552 [10]

Answer:

Explanation:

Electric forces exist among stationary electric charges; both electric and magnetic forces exist among moving electric charges. ... The magnetic force between two moving charges may be described as the effect exerted upon either charge by a magnetic field created by the other.

8 0

3 years ago

Please help! 34 points!

solmaris [256]

I believe the answer is Compression C

8 0

3 years ago

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Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =

Troyanec [42]

Answer:

The magnitude of the electrostatic force on $\mathrm{+32\,\mu C}$ is: $\mathbf{12\;N}$

Explanation:

Consider

$q_1=+32\;\mu\text{C}$ is at position $x_1=0\;\text{m}$
$q_2=+20\;\mu\text{C}$ is at position $x_2=40\;\text{cm}$
$q_3=-60\;\mu\text{C}$ is at position $x_3=60\;\text{cm}$

The sum of all horizontal forces on $q_1$ is given as

$\sum F_x=F_{12}+F_{13}$

where

$F_{12}$ is the force exerted by $q_2$
$F_{13}$ is the force exerted by $q_3$

The force on $q_1$ exerted by $q_2$ is repulsive, so the direction of the force $F_{12}$ is to the left (negative direction). Thus

$F_{12}=-\frac{K\,|q_1|\,|q_2|}{r_{12}^2}\\F_{12}=-\frac{K\,|q_1|\,|q_2|}{(x_2\;-\;x_1)^2}\\F_{12}=\mathrm{-\frac{(8.988\times10^9\;\frac{Nm^2}{C^2})\,|+32\;\mu C|\,|+20\;\mu C|}{(40\;cm\;-\;0\;cm)^2}}\\F_{12}=\mathrm{-\frac{(8.988\times10^9\;\frac{Nm^2}{C^2})\,(32\;\mu C)\,(20\;\mu C)}{(40\;cm)^2}}\\F_{12}=\mathrm{-\frac{(8.988\times10^9\;\frac{Nm^2}{C^2})\,(32\times10^{-6}\;C)\,(20\times10^{-6}\;C)}{(0.40\;m)^2}}\\F_{12}=\mathrm{-36\;N}$

The force on $q_1$ exerted by $q_3$ is attractive, so the direction of the force $F_{13}$ is to the right (positive direction). Thus

$F_{13}=+\frac{K\,|q_1|\,|q_3|}{r_{13}^2}\\F_{13}=+\frac{K\,|q_1|\,|q_3|}{(x_3\;-\;x_1)^2}\\F_{13}=\mathrm{+\frac{(9\times10^9\;\frac{Nm^2}{C^2})\,|+32\;\mu C|\,|-60\;\mu C|}{(60\;cm\;-\;0\;cm)^2}}\\F_{13}=\mathrm{+\frac{(9\times10^9\;\frac{Nm^2}{C^2})\,(32\;\mu C)\,(60\;\mu C)}{(60\;cm)^2}}\\F_{13}=\mathrm{+\frac{(9\times10^9\;\frac{Nm^2}{C^2})\,(32\times10^{-6}\;C)\,(60\times10^{-6}\;C)}{(0.60\;m)^2}}\\F_{13}=\mathrm{+48\;N}$

Therefore

$\sum F_x=\mathrm{-36\;N+48\;N} \;\;\;\;\;\Rightarrow\;\;\;\;\; \mathbf{\sum F_x=+12\;N}$

the electrostatic force on $q_1$ is to the right and has a magnitude of $\mathbf{12\;N}$

8 0

3 years ago