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
g100num [7]
1 year ago
13

The air temperature is 20°C. You are swimming underwater when you hear a boat noise. Then, 3.5 s later, you hear a crash. If the

speed of sound in water is 1450 m/s, how long after you hear the crash does your friend on the dock beside you hear the crash?
Physics
1 answer:
GaryK [48]1 year ago
8 0

To answer this question we first need to know the distance from the crash to the swimmer and the dock.

The distance is given by:

d=vt

This means that the distance from the crash to the swimmer is:

d=(1450)(3.5)=5075

Therefore the crash happened at 5075 meters from the swimmer.

Now, to determine the time it takes the sound to reach the deck we need to determine the speed of sound on air at that temperature, this is given by:

v=331\sqrt[]{1+\frac{T}{273}}

then if the temperature is 20°C we have:

v=331\sqrt[]{1+\frac{20}{273}}=342.91

Then it takes the sound to reach the deck:

t=\frac{5075}{342.91}=14.8

Finally to determine the time it takes after you hear it we subtract the time it takes for you to hear it, then:

14.8-3.5=11.3

Therefore your friend hear the crash 11.3 seconds after you do.

You might be interested in
The potential energy of a negative charge moved from point A to point B will increase.A negative charge moved from point A to po
AysviL [449]

Answer:

<em>The K.E from A to B won't increase...</em>

Explanation:

That's because the P.E from A to B is increasing. The K.E will increase if charge moves from a higher potential to a lower potential i.e., from B to A.

That is the reason there is no effect on net K.E when moving from a potential to same potential over and over (A to C).

4 0
3 years ago
you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k
alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}

→ v_{p}=150 km/h ,  v_{w}=50 km/h

→ v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11 km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is tan^{-1}\frac{50}{150}=18.4°

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0
3 years ago
What mechanism of energy is transferred by mass motion of fluid from one region of space to another?​
lora16 [44]
Convection, because it is the process of heat transfer from one location to the next by the movement of fluids. The moving fluid carries energy within it.
5 0
3 years ago
Why must lslamic believers travel to Mecca once during their lives?
SSSSS [86.1K]

Answer: The Islamic believers traveled to Mecca once during their lives because they would perform Hajj ( this is piligrimage) and it is the main part of the islam

Explanation:

Hajj is very important to islam people so in order to do this they had to go to Mecca. Hajj is rituals and rites that need to be fulfilled. Its not the only pillar for islam but it is one out of five.

4 0
3 years ago
A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on
SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

\sum F = ma

the resultant of the forces must be zero: \sum F = 0 (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

F, the push applied by the worker

F_f=-\mu mg, the force of friction, with \mu=0.25 being the coefficient of friction, m=30.0 kg being the mass of the crate, and g=9.8 m/s^2. The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

W=Fd cos \theta

where

F = 73.5 N is the force

d = 4.5 m is the displacement

\theta=0^{\circ} is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

W=F_f d cos \theta

where

F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N is the magnitude of the force of friction

d = 4.5 m is the displacement

\theta=180^{\circ} is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

W=F_f d cos \theta

We notice that both gravity and normal force are perpendicular to the displacement: therefore, \theta=90^{circ}, and so

cos \theta=0

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0
3 years ago
Other questions:
  • For your senior project, you are designing a Geiger tube for detecting radiation in the nuclear physics laboratory. This instrum
    13·1 answer
  • Match the term to its definition.
    15·2 answers
  • A particle moving along a straight line is subjected to a deceleration ???? = (−2???? 3 ) m/???? 2 , where v is in m/s. If it ha
    9·1 answer
  • Which of the following best describes a consumer?
    6·1 answer
  • An air-conditioning system's automatic controller might directly control the A. conversion of pneumatic energy to hydraulic sign
    13·1 answer
  • 1. Transverse waves _____.
    7·2 answers
  • If used with godly wisdom, the earth has sufficient resources to sustain its human population.
    11·1 answer
  • ball A is dropped from a hot air balloon rising at a costant velocity of 14,7 m.s'1 at a height of 19,7 m above the ground.the b
    9·1 answer
  • WILL MARK BRAINLIEST IF CORRECT ANSWERS - Is the reaction below a balanced chemical reaction? How do you know?
    13·2 answers
  • a___of water a. of bread a. of soap a. of juice a. salt. a. sand a. of glass a. of corn fill in the blanks I follow him but righ
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!