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xenn [34]
2 years ago
9

What is the net force on this object A: 0 newtons b: 200 newtons c:400 newtons D:600 newtons

Physics
1 answer:
Ilya [14]2 years ago
8 0

Answer:

100kg

Explanation:

Total force =400N+600N=1000N

As we have,

F=m×g

1000N=m×10m/s²

1000N÷10m/s²=m

m=100kg

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A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu
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Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = \frac{3Q}{4\pi R^{3} }

So, Q_{1} = \frac{Qr^{3} }{R^{3} }

Applying Gauss law of electrostatics ;

∫ E ds = Q₁/ε₀

Here E is electric field inside the sphere and ds is surface element of sphere of radius r.

Substitute the value of Q₁ in the above equation. Hence,

E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

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3 years ago
Carbon dioxide can be represented by the chemical formula
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Answer:

Your answer is B co2

Explanation:

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A company is interested in buying a new machine to replace their outdated equipment. However, before committing to the purchase,
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So ζ = 1540 / 1600 * 100% = 96.25%
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Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle
Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

v = \frac{2 \pi r}{T}  

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

T^{2} = r^{3}

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

T^{2} = r^{3}

T = \sqrt{(133370 Km)^{3}}

T = \sqrt{(2.372x10^{15} Km)}

T = 4.870x10^{7} Km

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( 1.50x10^{8} Km )

1 AU is defined as the distance between the earth and the sun.

\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU

T = 0.324 AU

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

T = \frac{0.324 AU}{1 AU} . 1 year

T = 0.324 year

That can be expressed in units of days

T = \frac{0.324 year}{1 year} . 365.25 days  

T = 118.60 days

<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>

v = \frac{2 \pi r}{T}

v = \frac{2 \pi (133370 Km)}{(118.60 days)}

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

118.60 days .\frac{86400 s}{1 day} ⇒ 10247040 s

133370 Km .\frac{1000 m}{1 Km} ⇒ 133370000 m

v = \frac{2 \pi (133370000 m)}{(10247040 s)}

v = 81.778 m/s

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

T^{2} = r^{3}

T = \sqrt{(69000 Km)^{3}}

T = \sqrt{(3.285x10^{14} Km)}

T = 1.812x10^{7} Km

\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU

T = 0.120 AU

T = \frac{0.120 AU}{1 AU} . 1 year

T = 0.120 year

T = \frac{0.120 year}{1 year} . 365.25 days  

T = 43.83 days

<em>Circular velocity for the particle in </em><em>D Ring</em><em>:</em>

v = \frac{2 \pi r}{T}

v = \frac{2 \pi (69000 Km)}{(43.83 days)}

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

43.83 days . \frac{86400 s}{1 day} ⇒ 3786912 s

69000 Km . \frac{1000 m}{ 1 Km} ⇒ 69000000 m

v = \frac{2 \pi (69000000 m)}{(3786912 s)}

v = 114.483 m/s

 

\frac{114.483 m/s}{81.778 m/s} = 1.399            

The particle in the D ring is 1399 times faster than the particle in the Encke Division.  

7 0
3 years ago
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