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klio [65]
3 years ago
6

What happens to the particles in a substance when temperature rises

Physics
1 answer:
kolbaska11 [484]3 years ago
7 0

Answer:

They gain kinetic energy

Explanation:

When the temperature of the particles in a substance rise, its internal energy increases. This internal energy is thus translated to kinetic energy of the particles in the substance. If the substance is a solid, as the kinetic energy increases, the vibrations of the particles of the substance increases causing it to undergo a change of state to liquid when the temperature reaches its melting point.

This change of state also occurs from the liquid state to the gaseous state as the internal energy of the substance, and thus the kinetic energy of the particles increase when the temperature reaches the substance's boiling point.

<u>So, as the temperature of a substance rises, the particles of the substance gain kinetic energy.</u>

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The magnetic field inside a superconducting solenoid is 4.00 T. The solenoid has an inner diameter of 6.20 cm and a length of 26
Delvig [45]

Answer:

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(b) The energy stored in the magnetic field within the solenoid is 5 kJ

Explanation:

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length of solenoid, L = 26 cm = 0.26 m

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u _B = \frac{B^2}{2\mu_o} \\\\u _B = \frac{(4)^2}{2(4\pi*10^{-7})}\\\\u_B = 6.366*10^6 \ J/m^3

(b) The energy stored in the magnetic field within the solenoid

U_B = u_B V\\\\U_B = u_B AL

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3 years ago
A bullet of mass 11 g strikes a ballistic pendulum of mass 1.9 kg. The center of mass of the pendulum rises a vertical distance
Ymorist [56]

Answer:

The bullet's initial speed is 243.21 m/s.

Explanation:

Given that,

Mass of the bullet, m_b=11\ g=0.011\ kg

Mass of the pendulum, m_p=19\ kg

The center of mass of the pendulum rises a vertical distance of 10 cm.

We need to find the bullet's initial speed if it is assumed that the bullet remains embedded in the pendulum. Let it is v. In this case, the energy of the system remains conserved. The kinetic energy of the bullet gets converted to potential energy for the whole system. So,

\dfrac{1}{2}(m_b+m_p)V^2 =(m_b+m_p)gh\\\\V=\sqrt{2gh} \ .................(1)

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Now using conservation of momentum as :

m_bv=(m_p+m_b)V

Put the value of V from equation (1) in above equation as :

v=\dfrac{(m_p+m_b)}{m_b}\sqrt{2gh} \\\\v=\dfrac{(1.9+0.011)}{0.011}\sqrt{2\times 9.8\times 0.1}\\\\v=243.21\ m/s

So, the bullet's initial speed is 243.21 m/s.

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