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1 month ago

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

Physics

1 answer:

ValentinkaMS [3.4K]1 month 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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A large box of mass m sits on a horizontal floor. You attach a lightweight rope to this box, hold the rope at an angle θ above t

inna [3103]

Answer:

The answer to the specified question will be " $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$ ".

Explanation:

Referring to the question,

$\sum F_{x}$

⇒ $TCos \theta-F_{s}=0$

⇒ $T_{m}Cos \theta =F_{s}$ ...(equation 1)

$\sum F_{y}$

⇒ $TSin \theta+F_{N}=m_{g}$

⇒ $M_{g}-TSin \theta=F_{N}$ ...(equation 2)

Now,

From equation 1 and equation 2, we conclude

⇒ $T_{m} Cos \theta = \mu_{s}F_{N}$

By substituting the value of $F_{N}$ , we derive

⇒ $T_{m} Cos\theta = \mu_{s}(M_{g}-T_{m}Sin \theta)$

⇒ $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$

4 0

2 months ago

The position of an object is given by x = at3 - bt2 + ct,where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI un

serg [3582]

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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2 months ago

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Yuliya22 [3333]

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1 month ago