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2 years ago

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If the only force exerted on a star far from the center of the Galaxy (r = 7.40 ✕ 1020 m) is the gravitational force exerted by

the ordinary matter (Mord = 1.90 ✕ 1041 kg), find the speed of the star. Assume a circular orbit and assume all the Galaxy's matter is concentrated at the center.

1 answer:

lina2011 [118]2 years ago

8 0

Answer:

The value is $v = 1.309*10^{5}\ m/s$

Explanation:

The radius is $r = 7.40 *10^{20} \ m$

The mass of the ordinary matter is $M_{rod} = 1.90 *10^{41}\ kg$

Generally the speed of the star is mathematically represented as

$v = \sqrt{\frac{G * M}{r} }$

Here G is the gravitational constant with a value

$G = 6.67384 * 10^{-11}$

So

$v = \sqrt{\frac{6.67384 * 10^{-11} * 1.90 *10^{41}}{7.40 *10^{20}} }$

=> $v = 1.309*10^{5}\ m/s$

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6 The energy stored in the system increases.

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10 The energy stored increases and then decreases.

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What do radio waves and gamma rays have in common?

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Answer:

A - They are both electromagnetic waves.

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Time t = 6 sec

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2 years ago

A charge of 63.0 nC is located at a distance of 3.40 cm from a charge of -47.0 nC. What are the x- and y-components of the elect

Serjik [45]

Answer:

$Ep_x = 288.97*10^3\frac{N}{C}$

$Ep_y = 2770.6*10^3\frac{N}{C}$

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/r²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

r: distance from charge q to point P in meters (m)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Graphic attached

The attached graph shows the field due to the charges:

Ep₁: Total field at point P due to charge q₁. As the charge is positive ,the field leaves the charge.

Ep₂: Total field at point P due to charge q₂. As the charge is negative, the field enters the charge.

Known data

q₁ = 63 nC = 63×10⁻⁹ C

q₂ = -47 nC = -47×10⁻⁹ C

k = 8.99*10⁹ N×m²/C²

d₁ = 1.4cm = 1.4×10⁻² m

d₂ = 3.4cm = 3.4×10⁻² m

Calculation of r and β

$r=\sqrt{d_1^2 + d_2^2} = \sqrt{(1.4*10^{-2})^2 + (3.4*10^{-2})^2} = 3.677*10^{-2}m$

$\beta = tan^{-1}(\frac{d_1}{d_2}) = tan^{-1}(\frac{1.4}{3.4}) = 22.38^o$

Problem development

Ep: Total field at point P due to charges q₁ and q₂.

$Ep = Ep_x i + Ep_y j$

Ep₁ₓ = 0

$Ep_{2x}=\frac{-k*q_2*Cos\beta}{r^2}=\frac{8.99*10^9*47*10^{-9}*Cos(22.38)}{(3.677*10^{-2})^2}=288.97*10^3\frac{N}{C}$

$Ep_{1y}=\frac{-k*q_1}{d_1^2}=\frac{8.99*10^9*63*10^{-9}}{(1.4*10^{-2})^2}=2889.6*10^3\frac{N}{C}$

$Ep_{2y}=\frac{-k*q_2*Sen\beta}{r^2}=\frac{-8.99*10^9*47*10^{-9}*Sen(22.38)}{(3.677*10^{-2})^2}=-119*10^3\frac{N}{C}$

Calculation of the electric field components at point P

$Ep_x = Ep_{1x} + Ep_{2x} = 0 + 288.97*10^3 = 288.97*10^3\frac{N}{C}$

$Ep_y = Ep_{1y} + Ep_{2y} = 2889.6*10^3 - 119*10^3 = 2770.6*10^3\frac{N}{C}$

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2 years ago