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anastassius
2 days ago
5

Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and

Physics
1 answer:
ValentinkaMS [3K]2 days ago
7 0

Answer:

20 cm

Explanation:

The electric potential energy U is calculated with the formula U = kq₁q₂/r, where q₁ = 5 nC (5 × 10⁻⁹ C) and q₂ = -2 nC (-2 × 10⁻⁹ C) and r is determined as √(x - 2)² + (0 - 0)² + (0 - 0)² = x - 2. This leads to U = -0.5 µJ (-0.5 × 10⁻⁶ J), where k = 9 × 10⁹ Nm²/C².

Thus, solving for r gives us r = kq₁q₂/U

which leads to x - 2 = kq₁q₂/U

Then, rearranging gives x = 0.02 + kq₁q₂/U m

So, x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

Resulting in x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

This simplifies to x = 0.02 + 0.18 = 0.2 m, or 20 cm

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serg [3222]

Answer:

The response to your inquiry is: 15 m/s²

Explanation:

Equation    x = at³ - bt² + ct

a = 4.1 m/s³

b = 2.2 m/s²

c = 1.7 m/s

First we calculate x at t = 4.1 s

x = 4.1(4.1)³ - 2.2(4.1)² + 1.7(4.1)

x = 4.1(68.921) - 2.2(16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we calculate speed

v = x/t = 252.58/ 4.1 = 61.6 m/s

Finally

acceleration = v/t = 61.6/4.1 = 15 m/s²

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If you double the mass of an object while leaving the net force unchanged what is the result
kicyunya [2911]

Answer:

If the net force acting on an object doubles, its acceleration also doubles. Conversely, if the mass is doubled, the acceleration will be halved. If both the net force and mass are doubled, the acceleration remains the same.

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[[TAG_9]][[TAG_10]]
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3 days ago
a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio
kicyunya [2911]

Answer:

a) factor b=\sqrt{2}

b) factor b=\frac{1}{2}

c) factor b=1

d) factor b=1

Explanation:

For an oscillating spring-mass system, the time period is expressed as:

T=\frac{1}{f}

T={2\pi} \sqrt{\frac{m}{k} }

where:

f= represents the frequency of oscillation

m= signifies the mass linked to the spring

k= is the spring's stiffness constant

a) If the mass is doubled:

  • New mass, m'=2m

Thus, the new time period:

T'=2\pi\sqrt{\frac{m'}{k} }

T'=2\pi\sqrt{\frac{2m}{k} }

T'=\sqrt{2}\times 2\pi\sqrt{\frac{m}{k} } }

T'=\sqrt{2} \times T

this leads to factor b=\sqrt{2} as per the question.

b) When the stiffness constant is quadrupled, holding other factors constant:

New stiffness constant, k'=4k

Thus, the new time period:

T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T

this results in factor b=\frac{1}{2} as required.

c) When both mass and stiffness constant are quadrupled:

New stiffness, k'=4k

New mass, m'=4m

Thus, the new time period:

T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T

which leads to factor b=1 as stated in the question.

d) If amplitude is quadrupled, the time period remains unaffected because T does not depend on amplitude as demonstrated by the equation.

Thus, factor b=1

7 0
27 days ago
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