1answer.
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
Nataly_w [17]
3 years ago
11

If the student hit a drum in his bedroom how would the sound wave behave differently than if he hit it in a swimming pool? Pleas

e respond in full sentences. Make sure you use the words faster, slower, medium, air, water in your answer.
Physics
1 answer:
zlopas [31]3 years ago
4 0
The sound wave would behave differently in a swimming pool than in his bedroom because sound waves travel faster in more dense mediums; such as water. The wave will travel faster in water, and slower in air.
You might be interested in
Which statement describes the speed of electromagnetic waves?
goldfiish [28.3K]

Answer:

Their speed in a vacuum is a constant value.

Explanation:

Electromagnetic waves consits of oscillations of electric field and magnetic field. The oscillations of these fields occur in a direction perpendicular to the direction of propagation of the waves, so they are transverse wave. Electromagnetic waves, contrary to mechanical waves, do not need a medium to propagate, so they can also travel through a vacuum. In a vacuum, their speed is constant and has always the same value, the speed of light:

c=3\cdot 10^8 m/s

7 0
3 years ago
Read 2 more answers
How to read a micrometer on a clark cm-100 vickers hardness tester
allsm [11]

Answer:

Explanation:

Equipment manufactured by LECO(8 Corporation, St. Joseph, Michigan is warranted free from defect in material

and workmanship for a period of six months from the date of purchase. Equipment not manufactured by LECO is

covered to the extent of warranty provided by the original manufacturer and this warranty does not cover any

equipment, new or used, purchased from anyone other than the LECO Corporation. All replacement parts shall

be covered under warranty for a period of thirt days from date of purchase. LECO MAKES NO OTHER

REPRESENTATION OR WARRANTY OF ANY OTHER KIND, EXPRESSED OR IMPLIED, WITH RESPECT TO

THE GOODS SOLD HEREUNDER, WHETHER AS TO MERCHANTABILITY, FITNESS FOR PURPOSE, OR

OTHERWISE.

Expendable items such as crucibles, combustion tubes, chemicals and items of like nature are not covered by

this warranty.

LECO's sole obligation under this warranty shall be to repair or replace any part or parts which, to our

satisfaction, prove to be defective upon return prepaid to LECO Corporation, St. Joseph, Michigan. This

obligation does not include labor to install replacement parts, nor does it cover any failure due to accident, abuse,

neglect, or use in disregard of instructions furnished by LECO. In no event shall damages for defective goods

exceed the purchase price of the goods, and LECO SHALL NOT BE LIABLE FOR INCIDENTAL OR

CONSEQUENTIAL DAMAGES WHATSOEVER.

All claims in regard to the parts or equipment must be made within ten (10) days after Purchaser learns of the

facts upon which the claim is based. Authorization must be obtained from LECO prior to returning any other

parts. This warranty is voided by failure to comply with these notice requirements.

NOTICE

The warranty on LECO equipment remains valid only when genuine LECO replacernent parts are employed.

Since LECO has no control over the quality or purity of consumable products not manufactured by LECO, the

specifications for accuracy of results using LECO instruments are not guaranteed unless genuine LECO

consumables are employed in conjunction with LECO instruments. If purchaser defaults in making payment for

any parts or equipment, this warranty shall be void and shall not apply to such parts and equipment. No late

payment or cure of default in payment shall extend the warranty period provided herein.

LECO Corporation is not responsible for damage to any associated instruments, equipment or apparatus nor wil

LECO be held liable for loss of profit or other special damages resulting from abuse, neglect, or use in disregard

of instructions. The Buyer, their employees, agents and successors in interest assume all risks and liabilities for

the operation, use and/or misuse of the product(s) described herein and agree to indemnify, hold harmless and

defend the seller from any and all claims and actions arising from any cause whatsoever, including seller's

negligence for personal injury incurred in connection with the use of said product(s) and any and all damages

proximately resulting therefrom.

CAUTION

The instrument should be operated only by technically qualified individuals who have fully read and understand

these instructions. The instrument should be operated only in accordance with these instructions.

The operator should follow all ,of the warnings and cautions set forth in the manual and the operator should follow

and employ all applicable standard laboratory safety procedures.

LECO'" is a registered trademark of the LECO Corporatio

5 0
3 years ago
Can somebody tell me what time is it
kodGreya [7K]

Answer:

8:21am

Explanation:

I hate stupid school

5 0
3 years ago
Read 2 more answers
A rubber bullet of mass m=0.025\,\mathrm{kg}m=0.025kg traveling at velocity v_0 = 50\,\mathrm{m/s}v 0 ​ =50m/s hits an iron bloc
Pavel [41]

Answer: 0.001 m

Explanation:

In order to get the máximum height reached by the iron block, we can use the energy conservation principle, as all the kinetic energy impressed upon the iron blck by the rubber bullet during the collision, becomes gravitational potential energy, as follows:

½ m v2 = m. g. h (1)

We don´t know the value of v, but if we look to the collision, and we asume no external forces act during it, the total momentum must be conserved.

The initial momentum, as the block is at rest, is just the one due to the rubber bullet:

P1 = mb . vb = 0.025 kg. 50 m/s = 1.25 kg. m/s

The final momemtum, is just the sum of the one due to the bullet (after being bounced back) and the one for the iron block:

P2 = mb . vfb  + mib . vib= 0.025 Kg. (-35 m/s) + 15 Kg.vib

As we have already said, P1 = P2, so we can write the following equation:

0.025 Kg. (-35 m/s) + 15 Kg. vib = 1.25 Kg. m/s.

Solving for vib, we have:

vib = 0.14 m/s

Now, we can replace this value in the equation (1) above:

½ . 15 Kg. (0.14)2  (m/s)2 = 15 Kg. 9.8 m/s2. H

Solving for H, we have:

H = 0.001 m

3 0
3 years ago
A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back
castortr0y [4]

Answer:

a) 5 m/s downwards

b) 17.86 m/s

c) 24.82 m/s

d) 0.228

Explanation:

We can set the frame of reference with the origin on the top of the building and the X axis pointing down.

The rock will be subject to the acceleration of gravity. We can use the equation for position under constant acceleration and speed under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

V(t) = V0 + a * t

In this case

X0 = 0

V0 = -5 m/s

a = 9.81 m/s^2

To know the speed it will have when it falls back past the original point we need to know when it will do it. When it does X will be 0.

0 = -5 * t + 1/2 * 9.81 * t^2

0 = t * (-5 + 4.9 * t)

One of the solutions is t = 0, but this is when the rock was thrown.

0 = -5 + 4.69 * t

4.9 * t = 5

t = 5 / 4.9

t = 1.02 s

Replacing this in the speed equation:

V(1.02) = -5 + 9.81 * 1.02 = 5 m/s (this is speed downwards because the X axis points down)

When the rock is at 15 m above the street it is 15 m under the top of the building.

15 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 15 = 0

Solving electronically:

t = 2.33 s

At that time the speed will be:

V(2.33) = -5 + 9.81 * 2.33 = 17.86 m/s

When the rock is about to reach the ground it is at 30 m under the top of the building:

30 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 30 = 0

Solving electronically:

t = 3.04 s

At this time it has a speed of:

V(3.04) = -5 + 9.81 * 3.04 = 24.82 m/s

---------------------

Power is work done per unit of time.

The work in this case is:

L = Ff * d

With Ff being the friction force, this is related to weight

Ff = μ * m * g

μ: is the coefficient of friction

L = μ * m * g * d

P = L/Δt

P = (μ * m * g * d)/Δt

Rearranging:

μ = (P * Δt) / (m * g * d)

1 horsepower is 746 W

20 minutes is 1200 s

μ = (746 * 1200) / (100 * 9.81 * 4000) = 0.228

8 0
3 years ago
Other questions:
  • Will an object with a density of 1.05 float or sink? Explain.
    8·2 answers
  • What is the magnitude of the acceleration of an electron at a point where the electric field has magnitude 6377 n/c and is direc
    15·1 answer
  • In a Young’s double-slit experiment, the angle that locates the second dark fringe on either side of the central bright fringe i
    10·1 answer
  • How do you measure potential and kinetic energy?
    8·1 answer
  • The temperature of a 50 g sample of aluminum is raised from 20°c to 60°c when 440 cal of heat are added. the specific heat capac
    9·1 answer
  • What is the weight of a 75kg. Mass?
    8·2 answers
  • Adjacent antinodes of a standing wave on a string are 15.0 cm apart. A particle at an antinode oscillates in simple harmonic mot
    5·1 answer
  • How can you write a sentence using gland and hormone in the same sentence?
    12·1 answer
  • A veterinarian thinks that a dog has swallowed a key ring. Which types of electromagnetic waves is the doctor most likely to use
    12·2 answers
  • The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!