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3 months ago

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Grace, Erin, and Tony are on a seesaw. Grace has a mass of 45kg and is seated 0.7m to the left of the fulcrum. Nicole has a mass

of 50kg and is
seated 0.6m to the left of the fulcrum. Tony has a mass of 50kg. How far to the right of the fulcrum must Tony be seated for the seesaw to be in equilibrium? Show work

1 answer:

serg [3.5K]3 months ago

5 0

Utiliza Scoratic, funciona con cualquier tema.

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When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction. What

Softa [3030]

Answer:

Electric field, $E=4\times 10^5\ N/C$

Explanation:

The following values are provided:

Magnitude of charge, $q_o=2\ nC=2\times 10^{-9}\ C$

Force experienced, $F=8\times 10^{-4}\ N$

To determine the electric field at the origin, the formula employed is:

$F=q_o\times E$

$E=\dfrac{F}{q_o}$

$E=\dfrac{8\times 10^{-4}}{2\times 10^{-9}}$

$E=4\times 10^5\ N/C$

Consequently, the electric field at the origin is $4\times 10^5\ N/C$ . Hence, this provides the sought solution.

3 0

2 months ago

The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu

Yuliya22 [3333]

Answer: $0.10233nm$

Explanation:

The mean free path of an atom can be calculated using the following equation:

$\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}$ (1)

Where:

$\lambda=0.2\mu m=0.2(10)^{-6}m$

$R=8.3145J/mol.K$ is referred to as the Universal gas constant

$T=0\°C=273.115K$ represents the absolute standard temperature

$d$ denotes the diameter of helium atoms

$N_{A}=6.0221(10)^{23}/mol$ symbolizes Avogadro's number

$P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3}$ indicates absolute standard pressure

<pFrom this, we can solve for $d$ using (1), aiming to determine the radius $r$ of the helium atom:

$d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}$ (2)

$d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}$ (3)

$d=2.0467(10)^{-10}m$ (4)

If the radius equals half of that diameter:

$r=\frac{d}{2}$ (5)

Eventually:

$r=\frac{2.0467(10)^{-10}m}{2}$ (6)

$r=1.0233(10)^{-10}m$ (7)

Nonetheless, we were tasked with finding this radius in nanometers. Knowing $1nm=(10)^{-9}m$ :

$r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm$ (8)

Ultimately:

$r=0.10233nm$ Represents the radius of the helium atom in nanometers.

5 0

4 months ago