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julia-pushkina [17]
3 years ago
8

How far does light year travel in one year

Physics
1 answer:
SOVA2 [1]3 years ago
6 0
Light travels approximately 186,000 miles per second in empty space.

Then, if you do the math, it goes 5.88 Trillion miles per year.
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A 20.0 μf capacitor is charged to a potential difference of 850 v. the terminals of the charged capacitor are then connected to
liberstina [14]
  • (a) Q = 1.70\times 10^{-2}\;\text{C};
  • (b) V_\text{final} = 5.31\times 10^{2}\;\text{V};
  • (c) E_\text{final} = 4.52\;\text{J};
  • (d) \Delta E = 2.82\;\text{J}.

All four values are in 3 sig. fig.

<h3>Explanation</h3>

(a)

Q = C\cdot V = 20.0\times 10^{-6} \times 850\;\text{V} = 1.70\times 10^{-2}\;\text{J}.

(b)

Sum of the final charge on the two capacitors should be the same as the sum of the initial charge. Voltage of the two capacitors should be the same. That is:

C_1\cdot V_\text{final} +C_2 \cdot V_\text{final} = C_1\cdot V_\text{initial};

(C_1+C_2)\cdot V_\text{final} = C_1\cdot V_\text{initial};

\displaystyle V_\text{final} = \frac{C_1}{C_1+C_2}\cdot V_\text{initial}\\\phantom{V_\text{final}} = \frac{20.0\;\mu\text{F}}{20.0\;\mu\text{F} + 12.0\;\mu\text{F}} \times 850\;\text{V}\\\phantom{V_\text{final}} =531\;\text{V}.

(c)

\displaystyle E = \frac{1}{2}\cdot C\cdot V^{2}.

\displaystyle E_\text{final} = \frac{1}{2} (C_1 + C_2) \cdot {V_\text{final}}^{2} \\\phantom{E_\text{final}} = \frac{1}{2} \times (20.0\times 10^{-6} + 12.0\times 10^{-6}) \times 531.25\\\phantom{E_\text{final}} = 4.52\;\text{J}.

(d)

Initial energy of the system, which is the same as the initial energy in the 20.0\;\mu\text{F} capacitor:

\displaystyle E_\text{initial} = \frac{1}{2} \times 20.0\times 10^{-6} \times 850^{2} = 7.225\;\text{J}.

Change in energy:

\Delta E = 7.225\;\text{J} - 4.516\;\text{J} = 2.70\;\text{J}.

4 0
3 years ago
I need an answer ASAP .....
Harrizon [31]

Answer:

Hope this helped :

Explanation:

When light passes from a medium with one index of refraction (m1) to another medium with a lower index of refraction (m2), it bends or refracts away from an imaginary line perpendicular to the surface (normal line). As the angle of the beam through m1 becomes greater with respect to the normal line, the refracted light through m2 bends further away from the line.

At one particular angle (critical angle), the refracted light will not go into m2, but instead will travel along the surface between the two media (sine [critical angle] = n2/n1 where n1 and n2 are the indices of refraction [n1 is greater than n2]). If the beam through m1 is greater than the critical angle, then the refracted beam will be reflected entirely back into m1 (total internal reflection), even though m2 may be transparent!

4 0
3 years ago
Read 2 more answers
A cube of metal has a mass of 11 grams and a volume of 1 cm . When fully submerged in water this metal cube hanging from an accu
Anarel [89]

Answer:

0.098 N

Explanation:

From the question,

Spring scale reading = W-U............... Equation 1

Where W = weight of the cube, U = upthrust.

W = mg

Where m =  mass of the cube, g = acceleration due to gravity.

Given: m = 11 g = 0.011 kg, g = 9.8 m/s².

W = 0.011(9.8)

W = 0.1078 N.

From Archimedes principle,

Upthrust = weight of water displaced.

U = (Density of water×volume of metal cube)×acceleration due to gravity.

U = (D×V)g

Given: D = 1000 kg/m², V = 1 cm³ = (1/1000000) = 1×10⁻⁶ m³, g - 9.8 m/s²

U = 1000(9.8)(10⁻⁶)

U = 0.0098 N.

Substitute the value of W and U into equation 1

Reading of the spring scale = 0.1078-0.0098

Reading of the spring scale = 0.098 N

7 0
3 years ago
A+20 N force acts on a car and at the same time, a -30 N force acts on the
katovenus [111]

Answer:

-10 N, not balanced

Explanation:

Force is a vector quantity, so in order to find the net force on an object, we must use vector addition rule.

This means that if two forces acting on an object along the same line, we have to choose one direction as positive, and write the two forces with the correct sign.

In this problem, we have two forces acting along a line on the car:

- A first force of

F_1 = +20 N

- A second force of

F_2=-30 N

Therefore, the net force on the car is:

\sum F=F_1+F_2=+20 +(-30)=-10 N

Moreover, the net force on an object is said to be "balanced" if it is zero: in this case, it is not zero, so it is not balanced.

5 0
3 years ago
A ball is thrown upward. As it passes 5.0 m height it is traveling at 4.0 m/s up. What was its initial upward velocity? (a) 7.0
sveticcg [70]

Answer:

c) 10.7m/s

Explanation:

From the exercise we know that at 5m the ball  is traveling at 4m/s

To calculate its initial velocity we need to solve the following equation:

v_{y}^{2}=v_{oy}^{2}+2g(y-y_{o})

Since the initial height is 0

Solving for v_{o}

v_{oy}=\sqrt{v_{y}^{2}-2gy}=\sqrt{(4m/s)^2-2(-9.8m/s^2)(5m)}=10.7m/s

5 0
3 years ago
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