Question 25

Comparing the speed of sound in liquids, gases, and solids, the speed of sound is usually lowest in ( ) and highest in ( ).

nydimaria [60]

Answer:

the answer is the 4th one. the speed of sound is lowest in gas and highest in solid and its because the molecules in solid is closer to each other than the molecules in gas and when the molecules are close to each other, they allow the waves of the sound to travel more quickly.

7 0

3 years ago

As a person pushes a box across a floor, the energy from the person's moving arm is transferred to the box, and the box and the

In-s [12.5K]

Answer:

It is conserved

Explanation:

Converted to heat energy due to the friction caused by the box rubbing on the floor

4 0

3 years ago

A couple of astronauts agree to rendezvous in space after hours. Their plan is to let gravity bring them together. She has a mas

Oduvanchick [21]

Answer:

(a) Acceleration of female astronaut = 9.33*10^-12 m/s^2; Acceleration of male astronaut = 7.56*10^-12 m/s^2

(b) 27.013 days

(c) No, their accelerations would not be constant.

Explanation:

In the given question, we have:

Mass of female astronaut $M_{1}$ = 60.0 kg

Mass of male astronaut $M_{2}$ = 74.0 kg

Distance between them (S) = 23.0 m

(a) The free-body diagram is shown in the attached figure.

Using the equations below, the initial accelerations of the two astronauts can be calculated:

Force of gravity (F) = $\frac{G*M_{1}*M_{2}}{S^{2} }$

G = 6.67*10^-11 $\frac{m^{3} }{kg*s^{2} }$

F = (6.67*10^-11 *60*74)/23^2 = 5.598*10^-10 N

For the female astronaut, her initial acceleration = F/ $M_{1}$ = 5.598*10^-10/60 = 9.33*10^-12 m/s^2

For the male astronaut, his acceleration = F/ $M_{2}$ = 5.598*10^-10/74 = 7.56*10^-12 m/s^2

(b) Since the different between their mass is not much, we can deduce that:

$a_{average} = \frac{a_{1}+a_{2}}{2}$ = (9.33*10^-12 + 7.56*10^-12)/2 = 8.445*10^-12 m/s^2

Using the equation below, we can calculate the the time:

S = ut + 1/2 (at^2) where u = 0

23 = 1/2 (8.445*10^-12)*t^2

t^2 = 5.447*10^12

t = 2333883.044 s = 27.013 days

(c) No, their accelerations will not be constant. It will increase because their radii would be decreasing.

8 0

3 years ago

cycling at 13.0 m/s on your new aero carbon fiber bike, how much times would it take a pro cyclist to ride in 120 km? Give your

sp2606 [1]

We have that the time in seconds, minutes, and hours is

$t=1.083*10^{-4}s$

$T_{min}=1.805*10^{-6}min$

$T_{hours}=3.0083*10^{-8}hours$

From the Question we are told that

Velocity $v=13.0m/s$

Distance $d=120 km$

Generally the equation for the Time is mathematically given as

$t=\frac{13}{120*10^3}\\\\t=1.083*10^{-4}s$

Therefore

$T_{min}=1.083*10^{-4}s/60$

$T_{min}=1.805*10^{-6}min$

And

$T_{hours}=T_{min}/60$

$T_{hours}=3.0083*10^{-8}hours$

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4 0

3 years ago

One mole of a gas is placed in a closed system with a 20 L vessel initially at T = 300 K. The vessel is then isothermally expand

Elden [556K]

Answer:

Given that

P = RT/V + a/V²

We know that

H= U + PV

For T= Constant (ΔU=0)

ΔH= ΔU +Δ( PV)

ΔH= Δ( PV)

P = RT/V + a/V²

P V= RT + a/V

dH/dV = d(RT + a/V)/dV

dH/dV = - a/V²

So the expression of dH/dV

$\dfrac{dH}{dV}=\dfrac{-a}{V^2}$

b)

In isothermal process

$\Delta H=nRT\ln{\dfrac{V_2}{V_1}}$ (ΔU=0)

Now by putting the all values

$\Delta H=nRT\ln{\dfrac{V_2}{V_1}}$

$\Delta H=1\times 0.08206\times 300\ln{\dfrac{40}{20}}$

ΔH = 17.06 L.atm

8 0

3 years ago