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bagirrra123 [75]
2 years ago
11

What is the approximate initial height of the counterweight?

Physics
1 answer:
BabaBlast [244]2 years ago
5 0

Answer: I don't know this one but I'm just came here for points

Explanation:

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A cat falls from a table of height 1.3m. What is the impact speed of the cat
Lorico [155]

Answer: 12.753

Explanation: Multiply 1.3 and 9.81 (the speed that everything drops at.)

Also, i feel bad for the cat

it might be wrong sorry

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Which of these is a benefit of social networking ?
Alisiya [41]

Answer:

Staying connected to friends

Explanation:

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2 years ago
Its mass is 20 grams, and its density is 7.87 g/cm3. What’s the larger cube’s volume?
Alik [6]
Volume = mass / density
Volume = 20 / 7.87
Volume = 2.54 (2 s.f)
3 0
3 years ago
A heavy boy and a lightweight girl are balanced on a massless seesaw. The boy moves backward, increasing his distance from the p
victus00 [196]

Answer:

The side the boy is sitting on will tilt downward.

Explanation:

According to the law of moments when the same force is applied at a greater distance from the pivot then the effect of moment is greater about that point.

<u>Mathematically momentum is given as:</u>

M=F\times r

where:

F is the applied force at a distance 'r' acting in a direction perpendicular to the line joining the point of application and the hinge.

  • Moment is the rotational effect of the applied force on the body.

<em>When the boy of a heavier mass than the girl was sitting on a balanced see-saw then it is certain that he was closer to the hinge than the girl to balance the turning effect (in case of an unbiased see-saw). When the body moves farther his weight is same but the radial distance from the hinge increases which increases his moment of weight.</em>

6 0
2 years ago
Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,
Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:  

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F = \frac{k*q_1*q_2}{d^2} Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)  

d: distance between the charges in meters (m)

Equivalence  

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

F_{13} = \frac{k*q_1*q_3}{d_1^2}

F_{23} = \frac{k*q_2*q_3}{d_2^2}

F₁₃ = F₂₃

\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2} We eliminate k and q₃ on both sides

\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}

\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}

\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2} We eliminate 10⁻⁶ on both sides

(1-x)^2 = \frac{7}{5} x^2

1-2x+x^2=\frac{7}{5} x^2

5-10x+5x^2=7 x^2

2x^2+10x-5=0

We solve the quadratic equation:

x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m

x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in  opposite way .

x = 0.458m = 45.8cm

8 0
3 years ago
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