answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Allisa [31]
2 years ago
12

An electric buzzer is activated, then sealed inside a glass chamber. When all of the air is pumped out of the chamber, how is th

e sound from the buzzer affected?
Physics
1 answer:
Aleksandr-060686 [28]2 years ago
4 0
The sound is increased because sound waves are in fact mech. waves which means the  that they can't travel through empty space and thus need a medium to travel through
You might be interested in
A roller coaster car drops a maximum vertical distance of 35.4 m. Determine the maximum speed of the car at the bottom of that d
marissa [1.9K]

Answer:

The maximum speed of the car at the bottom of that drop is 26.34 m/s.

Explanation:

Given that,

The maximum vertical distance covered by the roller coaster, h = 35.4 m

We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :

mgh=\dfrac{1}{2}mv^2

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 35.4}

v = 26.34 m/s

So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.

8 0
2 years ago
A boat traveled 280 miles downstream and back. The trip downstream took 7 hours. Trip back took 14 hours. What is the speed of t
avanturin [10]
Assuming 280 miles is the total distance travelled:
Let b = boat speed in still water
Let c = current speed.
For the downstream trip the speed is b + c. In 7 hours at the speed of (b + c) mph the boat travels 140 miles.
7(b + c) = 140 .............(1)
For the upstream trip the speed is b - c. In 14 hours at the speed of (b - c) mph the boat travels 140 miles.
14(b - c) = 140 ............(2)
The left hand sides of equations (1) and (2) are equal. Therefore we can write
7b + 7c = 14b - 14c ...........(3)
Rearranging equation (3) we get
21c = 7b
c = b/3 .......................(4)
The value for c obtained in equation (4) should now be substituted into equation (1) which can then be solved to find the value of b.
7 0
2 years ago
Read 2 more answers
During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s
Ray Of Light [21]

Answer:

part a : <em>The dry unit weight is 0.0616  </em>lb/in^3<em />

part b : <em>The void ratio is 0.77</em>

part c :  <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}

Here

  • \gamma_d is the dry unit weight which is to be calculated
  • γ is the bulk unit weight given as

                                              \gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3

  • w is the moisture content in percentage, given as 12%

Substituting values

                                              \gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3

<em>The dry unit weight is 0.0616  </em>lb/in^3<em />

Part b

Void Ratio

The void ratio is given as

                                                e=\frac{G_s \gamma_w}{\gamma_d} -1

Here

  • e is the void ratio which is to be calculated
  • \gamma_d is the dry unit weight which is calculated in part a
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72

Substituting values

                                              e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

S=\frac{G w}{e}

Here

  • e is the void ratio which is calculated in part b
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

                                      S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}

Here

  • \gamma_{zav} is  the zero air unit weight which is to be calculated
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

                                      \gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\

Now as the volume is known, the the overall weight is given as

weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0
2 years ago
A 60-μC charge is held fixed at the origin and a −20-μC charge is held fixed on the x axis at a point x = 1.0 m. If a 10-μC char
Aleksandr [31]

Answer:

Ek =  8,79 [J]

Explanation:

We are going to solve this problem, using  the energy conservation principle

State 1 or initial state (charges at rest t=0)

E₁  = Ek  + U₁

As charge are at rest Ek = 0

And  U₁ has two components

U₁₂   = K * Q₁*Q₂ / 0,4          and    U₃₂  = K*Q₃*Q₂ / 0,6

U₁₂  = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,4  ⇒ U₁₂ = 9*60*10*10⁻³/0,4

U₃₂ =  - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,6  ⇒ U₃₂ = - 9*20*10*10⁻³/0,6

U₁₂ = 540*10⁻2/0,4 [J]   ⇒13,5 [J]

U₃₂ = - 180*10⁻² /0,6 [J] ⇒ - 3 [J]

Then   E₁ = E₁₂ + E₃₂    

E₁ = 10,5 [J]

At  the moment of Q₂ passing x = 40 cm  or 0,4 m

E₂ = Ek + U₂

We can calculate the components of U₂ in this new configuration

U₂  =  U₁₂  + U₃₂

U₁₂  = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,7   ⇒  U₁₂ = 9*60*10*10⁻³/0,7

U₁₂ = 540*10⁻²/0,7       U₁₂ = 7,71 [J]

U₃₂ =  - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,3  ⇒ U₃₂ = -  9*20*10*10⁻³/0,3

U₃₂ = -  9*20*10⁻²/0,3  

U₃₂ = - 6

U₂ = 7,71 -6

U₂ = 1,71 [J]

Then as  

E₂  = Ek + U₂  and  E₂ = E₁

Then

Ek + U₂ = E₁

Ek =  10,5 - U₂    

Ek  = 10,5 - 1,71

Ek =  8,79 [J]

5 0
1 year ago
A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of
ivann1987 [24]

Answer:

a rock of 50kg should be placed =drock=0.5m from the pivot point of see saw

Explanation:

τchild=τrock  

Use the equation for torque in this equation.

(F)child(d)child)=(F)rock(d)rock)

The force of each object will be equal to the force of gravity.

(m)childg(d)child)=(m)rockg(d)rock)

Gravity can be canceled from each side of the equation. for simplicity.

 (m)child(d)child)=(m)rock(d)rock)  

Now we can use the mass of the rock and the mass of the child. The total length of the seesaw is two meters, and the child sits at one end. The child's distance from the center of the seesaw will be one meter.

(25kg)(1m)=(50kg)drock

Solve for the distance between the rock and the center of the seesaw.

drock=25kg⋅m50kg

drock=0.5m

6 0
2 years ago
Other questions:
  • a bullet of mass 4g when fired with a velocity of 50m/s can enter a wall upto a depth of 10cm how much will be the average resis
    6·1 answer
  • Which of the following solid fuels has the highest heating value?
    13·2 answers
  • Which of the following statements is/are true? Check all that apply. Check all that apply. The work done by a nonconservative fo
    13·1 answer
  • A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc
    5·1 answer
  • There are many well-documented cases of people surviving falls from heights greater than 20.0 m. In one such case, a 55.0 kg wom
    7·1 answer
  • A ceiling fan has five blades, each with a mass of 0.34 kg and a length of 0.66 m. The fan is operating in its "low" setting at
    6·1 answer
  • An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass
    8·1 answer
  • The drawing shows three identical springs hanging from the ceiling. Nothing is attached to the first spring, whereas a 4.50-N bl
    6·1 answer
  • A ball bearing of radius of 1.5 mm made of iron of density
    11·1 answer
  • When 999mm is added to 100m ______ is the result​
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!