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FinnZ [79.3K]
3 years ago
15

A square metal block with mass (m) is suspended by five wires with the same length. One of these wires is made of iron at the ce

nter of the block and has a radius (1.2 mm) while the other four wires have the same cross-sectional area and they are made of copper at the block corners. If the tension in the iron wire equals (0.3 mg), find the radius of a copper wire. (Take Yiron = 2×1011, Ycopper =1. 2×1011 N/m2 )
Physics
1 answer:
blsea [12.9K]3 years ago
5 0

Answer:

1.18 mm

Explanation:

The mass of the block is m, so the weight of the block is mg.  If the tension in the iron wire is 0.3 mg, then the total tension in the 4 copper wires is 0.7 mg.  So the tension in each copper wire is 0.175 mg.

For each wire:

ΔL/L = F / (EA)

where ΔL is the change in length,

L is the initial length,

F is the force,

E is Young's modulus (modulus of elasticity),

and A is the cross sectional area.

The five wires have the same initial length, and stretch by the same amount.

ΔL/L = ΔL/L

F / (EA) = F / (EA)

(0.3 mg) / (2×10¹¹ Pa × π(1.2 mm)²) = (0.175 mg) / (1.2×10¹¹ Pa × πr²)

0.3 / (2 (1.2 mm)²) = 0.175 / (1.2 r²)

0.36 r² = 0.504 mm²

r = 1.18 mm

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3 years ago
A car with a momentum of 3,200 kg-m/s moves foward at a rate of 2 m/s. What is the mass of the vehicle?
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1600 kg

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5 0
2 years ago
Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the sp
kotegsom [21]

Answer:

m2  = 83.3 g

Explanation:

by conservation of momentum principle we have

m_1v_{i1} + m_2v_{i2} = m_2v_{f2}

as both sphere has same speed so v_{i2} = v_{i1}

m_2 = \frac{m_1}[\frac{v_f2}{v_{f1}}+1}

from conservation of kinetic energy principle we have

\frac{1}{2}m_1v^{2}_{i1} + \frac{1}{2}m_2v^{2}_{i2} = \frac{1}{2}m_2v^{2}_f2

v_{f1} = \sqrt {\frac{(m_1+m_2) v^2_i1}{m_2}

v_{f1} =  v_{i2}\sqrt {\frac{(m_1+m_2)}{m_2}

\frac{v_{f1}}{v_{i2}} =\sqrt {\frac{(m_1+m_2)}{m_2}

substituting this value in above equation to get m2 value

m_2 = \frac{m_1}{\sqrt {\frac{(m_1+m_2)}{m_2}+1}}

solving for m2 we  get

m2 = \frac{m_1}{3}

m_1 = 250 g

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7 0
3 years ago
What total energy (in J) is stored in the capacitors in the figure below (C1 = 0.900 µF, C2 = 16.0 µF) if 1.80 10-4 J is stored
Musya8 [376]

The total energy  stored in the capacitors is determined as  2.41 x 10⁻⁴ J.

<h3>What is the potential difference of the circuit?</h3>

The potential difference of the circuit is calculated as follows;

U = ¹/₂CV²

where;

  • C is capacitance of the capacitor
  • V is the potential difference

For a parallel circuit the voltage in the circuit is always the same.

The energy stored in 2.5 μf capacitor is known, hence the potential difference of the circuit is calculated as follows;

U = ¹/₂CV²

2U = CV²

V = √2U/C

V = √(2 x 1.8 x 10⁻⁴ / 2.5 x 10⁻⁶)

V = 12 V

The equivalent capacitance of C1 and C2 is calculated as follows;

1/C = 1/C₁ + 1/C₂

1/C = (1)/(0.9 x 10⁻⁶)  +  (1)/(16 x 10⁻⁶)

1/C = 1,173,611.11

C = 1/1,173,611.11

C = 8.52 x 10⁻⁷ C

The total capacitance of the circuit is calculated as follows;

Ct = 8.52 x 10⁻⁷ C   +   2.5 x 10⁻⁶ C

Ct = 3.35 x 10⁻⁶ C

The total energy of the circuit is calculated as follows;

U =  ¹/₂CtV²

U =  ¹/₂(3.35 x 10⁻⁶ )(12)²

U = 2.41 x 10⁻⁴ J

Learn more about energy stored in a capacitor here: brainly.com/question/14811408

#SPJ1

7 0
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