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oksano4ka [1.4K]
3 years ago
5

A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves

. What does this demonstrate about the propagation of waves through a medium?
A) Waves transmit energy but not matter as they progress through a medium.
B) Waves transmit matter but not energy as they progress through a medium.
C) Waves do not transmit matter or energy as they progress through a medium.
D) Waves transmit energy as well as matter as they progress through a medium.
Physics
1 answer:
KIM [24]3 years ago
4 0

Answer:A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?

A) Waves transmit energy but not matter as they progress through a medium.

B) Waves transmit matter but not energy as they progress through a medium.

C) Waves do not transmit matter or energy as they progress through a medium.

D) Waves transmit energy as well as matter as they progress through a medium.

Explanation:

A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?

A) Waves transmit energy but not matter as they progress through a medium.

B) Waves transmit matter but not energy as they progress through a medium.

C) Waves do not transmit matter or energy as they progress through a medium.

D) Waves transmit energy as well as matter as they progress through a medium.

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Heat transfers energy from a hot object to a cold object. Both objects are isolated from their surroundings. The change in entro
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To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as

S = \frac{Q}{T}

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Q = Total Heat

T = Temperature

The total change of entropy from a cold object to a hot object is given by the relationship,

\Delta S = \frac{Q}{T_{cold}}-\frac{Q}{T_{hot}}

From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'

Change in entropy \Delta S_{hot} is smaller than \Delta S_{cold}

Therefore the correct answer is C. Will always have a smaller magnitude than the change in entropy of the cold object

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3 years ago
You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c
Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

  • Since the block slides across the floor at constant speed, this means that it's not accelerated.
  • According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
  • This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

       F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N  (1)

  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

      F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N  (2)

⇒    169 N + Fn = Fg = 216 N  (3)

  • This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
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The nucleus of an atom contains a. only neutrons b protons and neutrons
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B. protons and neutrons
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3 years ago
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An athlete whirls a 7.00 kg hammer 1.8 m from the axis of rotation in a horizontal circle, as
iogann1982 [59]

Answer:

A-500 N

Explanation:

The computation of the tension in the chain is shown below

As we know that

F = ma

where

F denotes force

m denotes mass = 7

And, a denotes acceleration

Now for the acceleration we have to do the following calculations

The speed (v) of the hammer is

v = Angular speed × radius

where,

Angular seed = 2 × π ÷ Time Period

So, v = 2 × π × r ÷ P

v = 2 × 3.14 × 1.8 ÷ 1

= 11.304 m/s

Now

a = v^2 ÷ r

= 70.98912 m/s^2

Now the tension is  

T = F = m × a

= 7 × 70.98912

= 496.92384 N

= 500 N

5 0
3 years ago
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