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cluponka [151]
3 years ago
15

How much energy ( in joules ) is released when 0.06 kilograms of mercury is condensed to a liquid at the same temperature ?

Physics
2 answers:
aleksley [76]3 years ago
5 0

Answer:

B. 17,705.1 J

Explanation:

The hear released when the mercury condenses into a liquid is given by:

Q=m \lambda_v

where

m = 0.06 kg is the mass of the mercury

\lambda_v is the latent heat of vaporization

For mercury, the latent heat of vaporization is \lambda_v = 296 kJ/kg, so the heat released during the process is:

Q=(0.06 kg)(296 kJ/kg)=17.76 kJ = 17,760 J

So, the closest option is

B. 17,705.1 J

valina [46]3 years ago
5 0

The energy ( in joules ) released when 0.06 kilograms of mercury is condensed to a liquid at the same temperature is about 697.08 J. in other to solve this problem, should know the latent heat of fusion of mercury which is equal to about 11.4 kJ/kg and multiplying by mass of mercury.

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If you know that the period of a pendulum is 1.87 seconds, what is the length of that pendulum? (assume that we are on earth and
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The length of the pendulum is 0.087 m. Option d is correct.

<h3>What is Simple harmonic motion?</h3>

Simple harmonic motion is periodic motion caused by a restoring force that is proportionate to the deviation from equilibrium.

Simple harmonic motion is periodic motion but many other conditions are dependent.

The time period of the pendulum is found as;

\rm T= 2 \pi \sqrt{\frac{L}{g} } \\\\ \rm 1.87 \ sec= 2 \times 3.14  \sqrt{\frac{L}{9.81 m/s^2} } \\\\ L=0.087 \ m

The length of the pendulum is 0.087 m

Hence, option d is correct.

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2 years ago
Determine the number of ways the faces of a regular tetrahedron can be colored with three colors
GREYUIT [131]
Tetrahedron is a geometrical figure or solid that is formed in similar manner as a pyramid. The faces of the formed pyramid are triangles. Also is the base. 

By the statement above, it can be deduced that the figure has four sides. The number of ways in which the sides can be colored by any of the three (3) colors is,
             3 x 3 x 3 x 3 = 81

Since the sides can still have the same colors, we deduct this value by 3 in order to eliminate the uniform colors of the four sides.
             = 81 - 3 = 78

Also, we can eliminate the concern on only two colors can mix,
              = 3 x 2 = 6
 Subtracting,
                      78 - 6 = 72

Hence, the answer is 72. 
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3 years ago
Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw
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Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of  the distance between them, acting along  the  line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the  factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

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3 years ago
Move the chemistry book across the physics book again. study the thermometer to determine whether there’s a change in the physic
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Heat is always transferred through the movement of matter in the convection mode of heat transmission.

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6 0
2 years ago
A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130
konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

W=Fd      (1)

F: applied force = 130N

d: distance = 5.0m

W=(130N)(5.0m)=650J

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

W_f=F_fd=\mu Mgd       (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

W = 0J

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

\Delta K=W_T         (3)

You calculate the total work:

W_T=Fd-F_fd=(F-F_f)d     (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N

Next, you replace the values of all parameters in the equation (4):

W_T=(130N-105.84N)(5.00m)=120.80J

\Delta K=120.80J

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

W_T=\frac{1}{2}M(v_2^2-v_1^2)       (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}

The final speed of the box is 2.58m/s

5 0
3 years ago
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