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Novay_Z [31]
3 years ago
14

Two copper wires have the same length, but one has twice the diameter of the other. Compared to the one that has the smaller dia

meter, the one that has the larger diameter has a resistance that is
Physics
1 answer:
enot [183]3 years ago
6 0

Answer:

Ratio of resistance of wire with larger diameter to smaller diameter is 4.

Explanation:

Resistance is defined as the property of the material/wire to oppose the flow of current through it.

Resistance (R) of the wire is determine by the relation:

R = (рL)/A

Here р is resistivity of the wire and it depends upon the material of the wire, L is length of the wire and A is the area of the wire.

Area of wire, A = π (d/2)²

Here d is diameter of the wire.

According to the question, let R₁ and R₂ be the resistance of the two copper wires, d₁ and d₂ be there diameters respectively and L and р be the length and resistivity of the two wire respectively. Since, L and р are same foe two wires.

Hence, there ratio of resistance is given by:

\frac{R_{1} }{R_{2} }  = \frac{A_{2} }{A_{1} }  

\frac{R_{1} }{R_{2} } = \frac{d_{2}^{2}  }{d_{1}^{2}  }

We know that d₂ = 2d₁

\frac{R_{1} }{R_{2} } =  \frac{(2d_{1})^{2}  }{d_{1}^{2}  } = 4

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A 3.5 kW drill transfers 5 000 kJ of kinetic energy during 15 seconds of use. What is the percentage efficiency of the drill?
Marizza181 [45]

Answer:

1.04%

Explanation:

Given that,

The power of drill = 3.5 kW = 3500 W

Transferred kinetic energy = 5000 kJ during 15 seconds of use.

We need to find the percentage efficiency of the drill. It can be given by :

\eta=\dfrac{P_o}{P_i}\times 100

Where

Po and Pi are output and input powers.

P_o=\dfrac{5000\times 10^3}{15}\\\\=3.34\times 10^5\ W

So,

\eta=\dfrac{3500}{3.34\times10^{5}}\times100\\\\=1.04\%

So, the percentage efficiency of the drill is 1.04%.

4 0
3 years ago
It is well known that runners run more slowly around a curved track than a straight one. One hypothesis to explain this is that
barxatty [35]

Answer:

percentage = 6.27 %

Explanation:

As we know that when runner is moving on straight track then the net force on his feet is given as

F_s = mg

while when runner is moving on circular track then we have

F_c = \sqrt{(mg)^2 + (\frac{mv^2}{R})^2}

F_c = m\sqrt{g^2 + \frac{v^4}{R^2}}

F_c = m\sqrt{9.8^2 + \frac{8^4}{18^2}}

F_c = m(10.42)

now percentage change is given as

percentage = \frac{F_c - F_s}{F_s} \times 100

percentage = \frac{m(10.42) - m(9.81)}{m(9.81)}\times 100

percentage = 6.27 %

6 0
3 years ago
Convert the Metric measurements below.<br> 76km =<br> m
Novosadov [1.4K]

Answer:

76,000 meters

Explanation:

4 0
3 years ago
Read 2 more answers
A 1.1 kg hammer strikes a nail. Before the impact, the hammer is moving at 4.5 m/s; after the impact it is moving at 1.5 m/s in
Talja [164]

Answer:

132 N

Explanation:

Given that a  1.1 kg hammer strikes a nail. Before the impact, the hammer is moving at 4.5 m/s; after the impact it is moving at 1.5 m/s in the opposite direction. If the hammer is in contact with the nail for 0.025 s, what is the magnitude of the average force exerted by the hammer on the nail

From Newton 2nd law of motion,

Change in momentum = impulse.

Change in momentum = m( V - U )

Substitute all the parameters into the formula

Change in momentum = 1.1 ( 4.5 - 1.5 )

Change in momentum = 1.1 × 3

Change in momentum = 3.3 kgm/s

Impulse = Ft

That is,

Ft = 3.3

Substitute time t into the formula above

F × 0.025 = 3.3

F = 3.3 / 0.025

F = 132 N

Therefore, the magnitude of the average force exerted by the hammer on the nail is 132 N.

3 0
4 years ago
PLEASE HELP
Aleksandr [31]
B and C because an earthquake does not directly affect gravity, and precipitation doesn’t fall, it is water surfacing
5 0
3 years ago
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