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Ksju [112]
2 years ago
5

Determine how many times per second it would move back and forth across a 6.0-m-long room on the average, assuming it made very

few collisions with other molecules. Express your answer using two significant figures.
Physics
1 answer:
timofeeve [1]2 years ago
4 0

Answer:

The right solution is "24.39 per sec".

Explanation:

According to the question,

⇒ v=\frac{502.1}{\sqrt{3} }

      =289.9 \ m/s

The time will be:

⇒ t=\frac{d}{v}

      =\frac{2\times 6}{289.9}

      =\frac{12}{289.9}

      =0.041 \ sec

hence,

⇒ N=\frac{1}{t}

        =\frac{1}{0.041}

        =24.39 \ per \ sec

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Consider two identical objects of mass m = 0.250 kg and charge q = 4.00 μC. The first charge is held in place at the origin of a
Gnom [1K]

Answer:

a = 640 m/s²

Explanation:

From work-kinetic energy principles,

The net force acting on the second object is the gravitational force and the electric force due to the first object.

So, the gravitational force on the mass is F₁ = Gm₁m₂/r² since m₁ = m₂ = m, U = -Gm²/r²

Also, the electric force on the charge is F₂ = kq₁q₂/r² since q₁ = q₂ = q, U = kq²/r²

The net Force F = ma

So, -F₁ + F₂ = F     (F₁ is negative since it is an attractive force in the negative x -direction and F₂ is positive since it is a repulsive force in the positive x- direction)

-Gm²/r² + kq²/r² = ma

ma = -Gm²/r² + kq²/r²

a = (-Gm²/r² + kq²/r²)/m

a = (-G + kq²/m²)m/r²

Since m = 0.250 kg, q = 4.00 μC = 4.00 × 10⁻⁶ C, r = 3.00 cm = 3.00 × 10⁻² m, G = 6.67 × 10⁻¹¹ Nm²/kg², k = 9 × 10⁹ Nm²/C² and a = acceleration of second mass.

Substituting the variables into the equation, we have

a = (m/r²)(-G + k(q/m)²)]

a = (0.250 kg/{3.00 × 10⁻² m}²)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(4.00 × 10⁻⁶ C/0.250 kg)²)

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(16 × 10⁻⁶ C/kg)²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(256 × 10⁻¹² C²/kg²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 2304 × 10⁻³ Nm²/kg²  ]

a = (0.250 kg/9.00 × 10⁻⁴ m)(2.304 Nm²/kg²)

a = 0.576 Nm²/kg /9.00 × 10⁻⁴ m²

a = 0.064 × 10⁴N/kg

a = 64 × 10 N/kg)

a = 640 m/s²

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3 years ago
If a car is speeding up,its initial speed is ______than its final speed
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Initial speed is less than final speed
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3 protons should be your answer
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3. When two things are the same temperature:
JulijaS [17]
The answer is definitely C.) their molecules move at the same average speed
3 0
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You are traveling at 14 m/s for 20 seconds. What is your displacement?
enyata [817]

Velocity = 14 m/s

Time = 20 s

Displacement = Velocity×Time = (14×20) m = 280 m

The displacement is 280 m towards the direction of motion.

5 0
2 years ago
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