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-BARSIC- [3]
3 years ago
14

Which of these is an experiment generally regarded as being first carried out by James Joule?

Physics
2 answers:
Sholpan [36]3 years ago
8 0
1. The correct answer among the choices provided is the third option. Measuring the temperature increase of water from doing work by stirring it is an experiment generally regarded as being first carried out by James Joule.
2.  Joule's experiment directly shows that heat is a form of energy. He wanted to make a different way of measuring energy.
marusya05 [52]3 years ago
4 0

Answer 1: The correct answer is: 'measuring the temperature increase of water from doing work stirring it'.

Answer 2: The correct answer is :'Heat is a form of energy'.

Explanation:

In his experiment he took three fluids : water, oil, mercury in an insulating container with stirrer agitated in the fluid.

He measured the amount of work done by the rotating stirrer in the fluid and the temperature rise in the fluid due to the rotation of the stirrer was also observed.

From his observation he concluded that the heat possessed by the fluid and work done on the fluid is directly linked to each other. Which he put down is simple words "heat is a form of energy"

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These days magnets are made artificially in various shapes and sizes depending on their use. One of the most common magnets - the bar magnet - is a long, rectangular bar of uniform cross-section that attracts pieces of ferrous objects. The magnetic compass needle is also commonly used. The compass needle is a tiny magnet which is free to move horizontally on a pivot. One end of the compass needle points in the North direction and the other end points in the South direction.

The end of a freely pivoted magnet will always point in the North-South direction. The end that points in the North is called the North Pole of the magnet and the end that points South is called the South Pole of the magnet. It has been proven by experiments that like magnetic poles repel each other whereas unlike poles attract each other.

3 0
4 years ago
If R = 12 cm, M = 520 g, and m = 20 g (below), find the speed of the block after it has descended 50 cm starting from rest. Solv
timofeeve [1]

Answer:

v = 0.84 m/s

Explanation:

given,

R = 12 cm

M (mass of pulley )= 520 g

m  (mass of block)=  20 g

s = 50 cm = 0.5 m

using conservation of energy

Potential energy = Kinetic energy

 m g h = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2

   I_{disk}= \dfrac{1}{2}MR^2  and v = r ω

 m g h = \dfrac{1}{2}mv^2 + \dfrac{1}{2}(\dfrac{1}{2}MR^2)(\dfrac{v}{R})^2

 m g h = \dfrac{1}{2}mv^2 +\dfrac{1}{4}Mv^2

 m g h = \dfrac{1}{2}v^2(m +\dfrac{1}{2}M)

 v=\sqrt{\dfrac{2mgh}{m + 0.5 M}}

 v=\sqrt{\dfrac{2\times 0.020 \times 9.8 \times 0.5}{0.02 + 0.5\times 0.52}}

      v = √0.7

      v = 0.84 m/s

5 0
3 years ago
Particle A of charge 3.06 10-4 C is at the origin, particle B of charge -5.70 10-4 C is at (4.00 m, 0), and particle C of charge
Alex

Answer:

F_net = 26.512 N

Explanation:

Given:

Q_a = 3.06 * 10^(-4 ) C

Q_b = -5.7 * 10^(-4 ) C

Q_c = 1.08 * 10^(-4 ) C

R_ac = 3 m

R_bc = sqrt (3^2 + 4^2) = 5m

k = 8.99 * 10^9

Coulomb's Law:

F_i = k * Q_i * Q_j / R_ij^2

Compute F_ac and F_bc :

F_ac = k * Q_a * Q_c / R^2_ac

F_ac =  8.99 * 10^9* ( 3.06 * 10^(-4 ))* (1.08 * 10^(-4 )) / 3^2

F_ac = 33.01128 N

F_bc = k * Q_b * Q_c / R^2_bc

F_bc =  8.99 * 10^9* ( 5.7 * 10^(-4 ))* (1.08 * 10^(-4 )) / 5^2  

F_bc = - 22.137 N

Angle a is subtended between F_bc and y axis @ C

cos(a) = 3 / 5

sin (a) = 4 / 5

Compute F_net:

F_net = sqrt (F_x ^2 + F_y ^2)

F_x = sum of forces in x direction:

F_x = F_bc*sin(a) = 22.137*(4/5) = 17.71 N

F_y = sum of forces in y direction:

F_y = - F_bc*cos(a) + F_ac = - 22.137*(3/5) + 33.01128 = 19.72908 N

F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N

Answer: F_net = 26.512 N

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4 years ago
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Answer:

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Explanation:

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