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Alexandra [31]
4 years ago
7

A particular wire has a resistivity of 6.47×10-8 Ωm and a cross-sectional area of 2.32 mm2. A length of this wire is to be used

as a resistor that will develop 130 W of power when connected to a 9.00 V battery. What length of wire is required?
Physics
1 answer:
PolarNik [594]4 years ago
6 0

Using  the formula for power:

P = V^2 / R  

130 W = (9.00 V)^2 / R  

Solve for r:

R = 81/130

R = 0.623 ohms

Now solve for the length of wire:

R = rho L / A  

A must be in m^2 - 2.32 mm^2 * 1 m^2/10^6 mm^2 = 2.132x10^-6 m^2  

Now you have:

0.623 = (6.47x10^-8) L / (2.32x10^-6)  

L = 22.34 m (Round answer as needed.)

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maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
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v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
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3 0
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vovikov84 [41]

Answer:

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creativ13 [48]

Answer:

v = 12.4 [m/s]

Explanation:

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V=2.88*4.523*10^{-4} \\V=1.302*10^{-3} [m^{3}/s ]

Despite the fact that you cover the inlet with the finger, the volumetric flow rate is the same.

v=V/A\\v=1.302*10^{-3} /1.05*10^{-4} \\v=12.4[m/s]

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Answer:

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