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Diano4ka-milaya [45]
3 years ago
11

The second law of thermodynamics dictates that ____. 1 point low-quality energy becomes static some high-quality energy is alway

s degraded the amount of energy in the universe is constantly changing low-quality energy is required to get high-quality energy
Physics
1 answer:
zvonat [6]3 years ago
6 0
The correct option for this completion exercise is: "Some high-quality energy is always degraded". Therefore, the complete text is shown below:
<span> 
 The second law of thermodynamics dictates that "</span>S<span>ome high-quality energy is always degraded.
 
 It is important to know that the "Thermodynamics" is defined as a branch of physics that studies the m</span><span>echanical action of heat </span><span>and the others forms of energy.</span>
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An airplane flies 2500 miles east in 245 seconds what is the velocity of the plane?
forsale [732]

Speed = (distance) / (time)

Speed = (

Velocity = speed, and its direction

The velocity of the plane is 10.2 miles per second East.

(about 48 times the speed of sound)

8 0
3 years ago
A man is walking away from a lamppost with a light source h = 6 m above the ground. the man is m = 2 m tall. how long is the man
Llana [10]

Answer;

= 4 m is the length of the man's shadow.

Explanation;

2/x=6/(8+x) cross multiply.

6x=2(8+x)

6x=16+2x

6x-2x=16

4x=16

x=2=16/4

x=4 m. is the length of the man's shadow.

3 0
3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
2 years ago
Imani stirs a cup of hot sencha tea with a cool silver spoon. She notices that the silver spoon becomes warmer. What energy chan
Lina20 [59]

Answer:

D. The tea loses heat to the spoon causing the spoon to become warmer

Explanation:

When the silver spoon at a lower temperature than the tea, is added to the tea, it makes thermal contact. Hence, the heat transfer starts between the two until the equilibrium is reached. We know that the heat transfer takes place from the body with a higher temperature to a body with a lower temperature. As a result, the body with higher temperature loses heat and its temperature lowers down. While the body with a lower temperature gains heat and its temperature rises.

Therefore, the correct option is:

<u>D. The tea loses heat to the spoon causing the spoon to become warmer</u>

7 0
2 years ago
Which of the following examples illustrates static friction?
vivado [14]

Answer:

A box sits stationary  on a ramp

Explanation:

Static friction is a force which keeps an object at rest as it is in the case of the box. It has to be overcome for the object to be set into motion.

Static force of friction is calculated as follows:

F= μη

F is static force of friction.

μ is the coefficient of static friction.

η is the normal force.

6 0
3 years ago
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