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harkovskaia [24]

1 year ago

13

For the meter stick shown in figure 10-4, the force F1 10.0 N acts at 10.0 cm. What is the magnitude of torque due to F1 about a

n axis through point A perpendicular to the page? Is it clockwise, or is it counterclockwise? Show your work and give correct units.

1 answer:

Phantasy [73]1 year ago

8 0

Torque is equal position vector times (r) times force vector (F). Since F= 10 N and r = 0.1 m, so the torque is equal to (10 N) x ( 0.1 m) = 1Nm. The direction of the torque would be into the screen, clockwise rotation.

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Suggest reasons why poaching for subsistence is likely to be less damaging to the biodiversity of an area than poaching for prof

dlinn [17]

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5 0

1 year ago

A metal disk weighing 1 N is resting on an index card that is balanced on top of a glass. When the index card is quickly pulled

Burka [1]

Answer: D

Step by step explanation:

8 0

2 years ago

Read 2 more answers

he first excited state of the helium atom lies at an energy 19.82 eV above the ground state. If this excited state is three-fold

bekas [8.4K]

Answer:

Relative population is 2.94 x 10⁻¹⁰.

Explanation:

Let N₁ and N₂ be the number of atoms at ground and first excited state of helium respectively and E₁ and E₂ be the ground and first excited state energy of helium respectively.

The ratio of population of atoms as a function of energy and temperature is known as Boltzmann Equation. The equation is:

$\frac{N_{1} }{N_{2} }$ = $\frac{g_{1}e^{\frac{-E_{1} }{KT} } }{g_{2}e^{\frac{-E_{2} }{KT} }}$

$\frac{N_{1} }{N_{2} }$ = $\frac{g_{1}e^{\frac{-(E_{1}-E_{2}) }{KT} } }{g_{2}}$

Here g₁ and g₂ be the degeneracy at two levels, K is Boltzmann constant and T is equilibrium temperature.

Put 1 for g₁, 3 for g₂, -19.82 ev for (E₁ - E₂) and 8.6x10⁵ ev/K for K and 10000 k for T in the above equation.

$\frac{N_{1} }{N_{2} }$ = $\frac{1\times e^{\frac{-(-19.82)}{8.6\times 10^{-5}\times 10000} } }{3}$

$\frac{N_{1} }{N_{2} }$ = 3.4 x 10⁹

$\frac{N_{2} }{N_{1} }$ = 2.94 x 10⁻¹⁰

5 0

2 years ago

A cube with edges exactly 2 cm long is made of material with a bulk modulus of 3.5 x 109 n/m2. when it is subjected to a pressur

Igoryamba

As we know by the formula of bulk modulus

$B = \frac{\Delta P}{-\Delta V/V}$

now we can rearrange it as

$\Delta V/V = -\frac{\Delta P}{B}$

$V_f - V = -V\frac{\Delta P}{B}$

now the final volume after pressure is applied is given as

$V_f = V(1 - \frac{\Delta P}{B})$

now we know that

$\Delta P = 3 \times 10^5 Pa$

$V = 2^3 cm^3 = 8 cm^3$

$B = 3.5 \times 10^9 N/m^2$

now plug in all data

$V_f = 8(1 - \frac{3 \times 10^5}{3.5 \times 10^9})$

$V_f = 7.999 cm^3$

so volume is above after pressure is applied over it

5 0

2 years ago

Suppose I have an infinite plane of charge surrounded by air. What is the maximum charge density that can be placed on the surfa

gregori [183]

Answer:

$53.1\mu C/m^2$

Explanation:

We are given that

Electric field,E= $3\times 10^6V/m$

We have to find the value of maximum charge density that can be placed on the surface of the plane before dielectric breakdown of the surrounding air occurs.

We know that

$E=\frac{\sigma}{2\epsilon_0}$

Where $\epsilon_0=8.85\times 10^{-12}$

Using the formula

$3\times 10^6=\frac{\sigma}{2\times 8.85\times 10^{-12}}$

$\sigma=3\times 10^6\times 2\times 8.85\times 10^{-12}$

$\sigma=5.31\times 10^{-5}C/m^2$

$\sigma=53.1\times 10^{-6}C/m^2=53.1\mu C/m^2$

$1\mu C=10^{-6} C$

3 0

1 year ago