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77julia77 [94]
3 years ago
6

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:
pantera1 [17]3 years ago
6 0

Answer:

It's A

Explanation:

As the waves progress through the medium, the particles they are made of move perpendicular to the direction in which the waves move. The particles do not move with the wave. So waves transmit energy but not matter as they progress through a medium.

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What coefficients would balance the following equation? __C2H6 + __O2 → __CO2 + __H2O A. 1C2H6 + 5O2 → 2CO2 + 3H2O B. 2C2H6 + 5O
sladkih [1.3K]
It's C, with the 2/7/4/6 in front of each reactant and product.
4 0
3 years ago
A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 50.9 J and a maximum d
Llana [10]

(a) 2446 N/m

When the spring is at its maximum displacement, the elastic potential energy of the system is equal to the total mechanical energy:

E=U=\frac{1}{2}kA^2

where

U is the elastic potential energy

k is the spring constant

A is the maximum displacement (the amplitude)

Here we have

U = E = 50.9 J

A = 0.204 m

Substituting and solving the formula for k,

k=\frac{2E}{A^2}=\frac{2(50.9)}{(0.204)^2}=2446 N/m

(b) 50.9 J

The total mechanical energy of the system at any time during the motion is given by:

E = K + U

where

K is the kinetic energy

U is the elastic potential energy

We know that the total mechanical energy is constant: E = 50.9 J

We also know that at the equilibrium point, the elastic potential energy is zero:

U=\frac{1}{2}kx^2=0 because x (the displacement) is zero

Therefore the kinetic energy at the equilibrium point is simply equal to the total mechanical energy:

K=E=50.9 J

(c) 8.55 kg

The maximum speed of the block is v = 3.45 m/s, and it occurs when the kinetic energy is maximum, so when

K = 50.9 J (at the equilibrium position)

Kinetic energy can be written as

K=\frac{1}{2}mv^2

where m is the mass

Solving the equation for m, we find the mass:

m=\frac{2K}{v^2}=\frac{2(50.9)}{(3.45)^2}=8.55 kg

(d) 2.14 m/s

When the displacement is

x = 0.160 m

The elastic potential energy is

U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J

So the kinetic energy is

K=E-U=50.9 J-31.3 J=19.6 J

And so we can find the speed through the formula of the kinetic energy:

K=\frac{1}{2}mv^2 \rightarrow v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(19.6)}{8.55}}=2.14 m/s

(e) 19.6 J

The elastic potential energy when the displacement is x = 0.160 m is given by

U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J

And since the total mechanical energy E is constant:

E = 50.9 J

the kinetic energy of the block at this point is

K=E-U=50.9 J-31.3 J=19.6 J

(f) 31.3 J

The elastic potential energy stored in the spring at any time is

U=\frac{1}{2}kx^2

where

k = 2446 N/m is the spring constant

x is the displacement

Substituting

x = 0.160 m

we find the elastic potential energy:

U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J

(g) x = 0

The postion at that instant is x = 0, since it is given that at that instant  the system passes the equilibrium position, which is zero.

4 0
3 years ago
This equation shows the reaction that occurs when calcium carbonate
yarga [219]

Answer:

a. endothermic

Explanation:

The reaction shown is an endothermic reaction. In such reaction, they require an input of energy.

  • Endothermic changes involves absorption of heat from the surrounding.
  • The surrounding becomes colder at the end of the reaction.
  • When heat is on the reactant side, it suggests a reaction that requires a considerable input of energy.
3 0
3 years ago
Read 2 more answers
A pitcher claims he can throw a 0.148-kg baseball with as much momentum as a 2.00-g bullet moving with a speed of 1.50 ✕ 103 m/s
Svetlanka [38]
(a) The mass of the bullet is m_b = 2.00g=0.002 kg and its speed is v_b = 1.5 \cdot 10^3 m/s, so its momentum is
p=m_b v_b = (0.002 kg)(1.5 \cdot 10^3 m/s)=3 kg m/s

The pitcher claims that he can throw the ball with the same momentum p of the ball. Since the mass of the ball is m=0.148 kg, this means that the velocity of the ball must be:
v= \frac{p}{m}= \frac{3 kg m/s}{0.148 kg}=20.3 m/s

(b) The kinetic energy of the bullet is:
K_b =  \frac{1}{2} m_b v_b^2= \frac{1}{2}(0.002 kg)(1.5 \cdot 10^3 m/s)^2=2250 J

while the kinetic energy of the ball is:
K= \frac{1}{2}mv^2= \frac{1}{2}(0.148 kg)(20.3 m/s)^2=30.5 J

So, the bullet has greater kinetic energy than the ball.
4 0
3 years ago
A cloud mass moving across the ocean at an altitude of 2000 m encounters a coastal mountain range. As it rises to a height of 35
GREYUIT [131]

Answer:

snow

Explanation:

Since the process undergoes adiabatic expansion, hence q = 0 and ΔU = w.

We can sole this problem using the following derivation:

ln(\frac{T_2}{T_1} )=-(\gamma -1)ln(\frac{V_f}{V_i} )=-(\gamma -1)ln(\frac{T_2}{T_1}\frac{P_i}{P_f}  )\\=-(\gamma -1)ln(\frac{T_2}{T_1})-(\gamma -1)ln(\frac{P_i}{P_f})\\=-(\frac{\gamma -1}{\gamma})ln(\frac{P_i}{P_f})\\=-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\ln(\frac{T_2}{T_1} )==-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\Substituting\ values:\\\\

ln(\frac{T_2}{T_1} )=-(\frac{\frac{28.86}{28.86-8.314} -1}{\frac{28.86}{28.86-8.314}})ln(\frac{0.802\ atm}{0.602\ atm})=-0.0826\\\\ln(\frac{T_2}{T_1} )=-0.0826\\\\Taking\ exponential\ of\ both \ sides:\\\\\frac{T_2}{T_1} =e^{-0.0826}\\\\T_2=0.9207T_1\\\\T_2=0.9207*288\\\\T_2=265\ K\\

Since T2 = 265 K, we should expect a snow

4 0
3 years ago
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