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3 years ago

Match these terms with their definitions.

Physics

2 answers:

Tomtit [17]3 years ago

8 0

1. Entropy

2. Thermodynamics

3. First law of thermodynamics

4. Second law of thermodynamics

5. Maximum entropy

6. The kinetic and potential energy of molecules of a substance

adoni [48]3 years ago

7 0

The correct matches are as follows:

1. measure of disorder
entropy

2. the study of energy conversions
thermodynamics

3. conservation of energy when heat is added
second law of thermodynamics

4. disorder increasing
the kinetic and potential energy of the molecules of a substance

5. reaching a state of total disorder
maximum entropy 

6. internal energy
first law of thermodynamics

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Answer:

Explanation:

let the charge is q. velocity, v = 24.7 m/s

magnetic force, F = 2.38 x 10^-4 N

Let the magnetic field is B.

Velocity, v' = 5.64 m/s

angle, θ = 21.2°

The force experienced by a charged particle placed in a magnetic field is given by

F = q x v x B x Sinθ

in first case

2.38 x 10^-4 = q x 24.7 x B x Sin 90 .... (1)

in second case

F = q x 5.64 x B x Sin 21.2° .... (2)

Divide equation (2) by equation (1), we get

$\frac{F}{2.38\times 10^{-4}}=\frac{5.64\times Sin 21.2}{24.7\times Sin 90}$

F = 1.97 x 10^-5 N

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4 years ago

A hollow plastic sphere is held below the surface of a fresh-water lake by a cord anchored to the bottom of the lake. The sphere

Naya [18.7K]

Answer: a) B = 6811N

b) m = 603.2kg

c) 86.8%

Explanation: Buoyant force is a force a fluid exerts on a submerged object.

It can be calculated as:

$B=\rho_{fluid}.V_{obj}.g$

where:

$\rho_{fluid}$ is density of the fluid the object is in;

$V_{obj}$ is volume of the object;

g is acceleration due to gravity, is constant and equals 9.8m/s²

a) For the hollow plastic sphere, density of water is 1000kg/m³:

$B=10^{3}.0.695.9.8$

B = 6811N

b) Anchored to the bottom, the forces acting on the sphere are Buoyant, Tension and Force due to gravity:

B = T + $F_{g}$

B = T + mg

mg = B - T

$m=\frac{B - T}{g}$

Calculating:

$m=\frac{6811 - 900}{9.8}$

m = 603.2kg

c) When the shpere comes to rest on the surface of the water, there are only buoyant and gravity acting on it:

B = m.g

$\rho_{w}.V_{sub}.g=m.g$

$V_{sub}=\frac{m}{\rho_{w}}$

$V_{sub}=\frac{603.2}{1000}$

$V_{sub}$ = 0.6032m³

Fraction of the submerged volume is:

$\frac{V_{sub}}{V_{obj}}$ = $\frac{0.6932}{0.695}$ = 0.868

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Answer:

Explanation:

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