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sergey [27]
2 years ago
4

A nail driven into a board increases in temperature. If we assume that 60 % of the kinetic energy delivered by a 1.80-kg hammer

with a speed of 7.80 m/s is transformed into heat that flows into the nail and does not flow out, what is the temperature increase of an 8.00-g aluminum nail after it is struck ten times?
Physics
1 answer:
nignag [31]2 years ago
8 0

Answer:

\Delta T=45.63^{\circ}C

Explanation:

Given:

  • mass of hammer, m=1.8\ kg
  • speed of the hammer, v=7.8\ m.s^{-1}
  • part of kinetic energy of hammer getting transformed into heat, F=60\%
  • mass of nail, m'=8\ g=0.008\ kg
  • we've specific heat of aluminium, c=900\ J.kg^{-1}.^{\circ}C^{-1}

<u>Total kinetic energy given to the nail:</u>

KE=10\times \frac{1}{2} \times m.v^2 (Since the aluminium nail is struck 10 times)

<u>The heat that flows to the nail is 0.6 times of the total kinetic energy:</u>

Q=0.6\times 10\times 0.5\times1.8\times 7.8^2

Q=328.536\ J

<u>From the equation of heat:</u>

Q=m'c.\Delta T

328.536=0.008\times 900\times \Delta T

\Delta T=45.63^{\circ}C is the increase in the temperature of the nail.

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Answer:

A. 5.4 * 10^(-4) m

B. 500V

Explanation:

A. Electric potential, V is given as:

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This means that radius, r is

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r = (9 * 10^9 * 30 * 10^(-12))/500

r = (270 * 10^(-3))/500

r = 5.4 * 10^(-4) m

B. Now the radius is doubled and the charge is doubled,

V = (9 * 10^9 * 2 * 30 * 10^(-12))/(2 * 5.4 * 10^(-4) * 2)

V = 500V

7 0
2 years ago
(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500 - μC charge and flies due west at a sp
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(a) 2.64\cdot 10^{-8} N north

We can treat the aircraft as a single point charge moving in a magnetic field. In this case, the magnetic force exerted on the plane is

F=qvB sin \theta

where

q=0.500 \mu C = 0.500\cdot 10^{-6} C is the charge on the plane

v = 660 m/s is the velocity

B=8.00\cdot 10^{-5} T is the magnitude of the magnetic field

\theta=90^{\circ} is the angle between the direction of motion of the jet and of the magnetic field

Substituting,

F=(0.5\cdot 10^{-6})(660)(8.0\cdot 10^{-5})=2.64\cdot 10^{-8} N

The direction can be found by using Fleming's left hand rule. We have:

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- middle finger: velocity of the plane (due west)

- force: thumb --> north

(b) Not negligible

As we can see from part (a), the magnitude of the force is not really big, so the effects are negligible.

For instance, we can compare this force with the weight of a plane. If we take a Boeing 737, its mass is about 80,000 kg, so its weight is

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For a long ideal solenoid having a circular cross-section, the magnetic field strength within the solenoid is given by the equat
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Answer:

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Explanation:

It is given that,

The magnetic field strength within the solenoid is given by the equation,

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The induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, x = 2 m

The electric field due to changing magnetic field is given by :

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x is the distance from the axis of the solenoid

E(2\pi x)=\pi r^2\dfrac{dB}{dt}, r is the radius of the solenoid

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r^2=\dfrac{2\times 2\times 1.1}{(5)}

r = 0.93 meters

So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.

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