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ozzi
3 years ago
15

Only conservative forces perform work on an object over a span of time. During this time, the total mechanical energy of the obj

ect ______.
decreases

increases continuously

remains constant

increases to a certain maximum
Physics
1 answer:
siniylev [52]3 years ago
4 0

During this time, the total mechanical energy of the object remains constant.

Answer: Option C

<u>Explanation:</u>

The sum total of potential energy and the kinetic energy presented in the system is called mechanical energy. The total mechanical energy in the system, which represents the combined potential and kinetic energies, remains constant as long as the only force work at conservative forces, and mechanical energy is maintained on this principle.

For example, a gravity box in which we throw the ball straights up, and then leave the hand with a specific amounts of kinetic energy. In the first half of the track, there is no kinetic energy, but it has potential energy similar to kinetic energy that it had when that left our hand. When we catch that again, it has the same kinetic energy as when that left our hand. That is why gravity belongs to the category of conservative forces.

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A horizontal beam of light of intensity 25 W/m2 is sent through two polarizing sheets. The polarizing direction of the first mak
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Answer:

option (B)

Explanation:

Intensity of unpolarised light, I = 25 W/m^2

When it passes from first polarisr, the intensity of light becomes

I'=\frac{I_{0}}{2}=\frac{25}{2}=12.5 W/m^{2}

Let the intensity of light as it passes from second polariser is I''.

According to the law of Malus

I'' = I' Cos^{2}\theta

Where, θ be the angle between the axis first polariser and the second polariser.

I'' = 12.5\times Cos^{2}15

I'' = 11.66 W/m^2

I'' = 11.7 W/m^2

7 0
3 years ago
A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

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3 years ago
Does this diagram illustrate the second law of thermodynamics? Why or why not
Savatey [412]
<h2>Answer:</h2>

The diagram is not showing the second law of thermodynamics. It is the demonstration of 1st law of thermodynamics.

<h3>Explanation:</h3>

Second law of thermodynamics describes the entropy of the system increase with time, it does not decrease with time. It is constant for ideal systems.

While in first law of thermodynamics, it is stated that the energy of a system can not be lost but it is transferred from one form to other form.

And in this picture, it is shown that the energy released from heat source to cold sink is used in doing work.

Work and heat are forms of energy.

5 0
4 years ago
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