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
Brums [2.3K]
3 years ago
9

Sound travels about 750 miles per hour. if you stand in a canyon and sound a horn, you will hear an echo. suppose it takes about

3.5 seconds to hear the echo. how far away is the canyon wall, in feet
Physics
2 answers:
diamong [38]3 years ago
5 0
The answer to this question would be:3850ft

To answer this question, you need to convert the speed velocity from miles/hour into feet/second. The equation would be: 750 miles/hour x 5280 foot/mile x 1 hour/3600second = 1100 ft/s
Then multiply the time with the velocity= 3.5 second x 1100 ft/s= 3850ft
umka21 [38]3 years ago
3 0

The canyon is 3.8 ×10³ feet if the echo is heard after 3.5 seconds.

Further Explanation

Speed is how fast an object moves or how far an object travels per unit time. If the distance traveled and the total time of travel are known, the speed can be calculated using the formula:

speed = \frac{distance}{time}

In the problem, we are given:

  • speed = 750 mi/h
  • time = 3.5 seconds

We are looking for the distance traveled by sound which can be represented by the variable <em>d.</em>

We can solve for d by manipulating the speed formula to get the equation:

distance \ (d) \ = speed \times time

But before we can plug in the given into the formula, the speed must be converted in terms of seconds since the given time is also in seconds:

speed \ = \frac{750 \ mi}{1 \ h} \times \frac{1 \ h}{60 min} \times \frac{1 \ min}{60 sec}\\\\\boxed {speed = 0.2083 \ \frac{mi}{sec}}

Now that the speed in miles per second is known, the distance can now be calculated using the formula above:

distance \ = speed \times \ time\\distance \ = 0.2083 \frac{mi}{sec} \times 3.5 sec\\\boxed {distance = 0.729 \ mi/tex]Now, since the problem requires the distance in feet, the conversion factor is used: 1 mi = 5280 ft[tex]0.729 mi \times \frac{5280 \ ft}{1 \ mi} = \boxed {3,849.12 \ ft}\\

Since the given has two significant digits, the answer must also have two. Therefore, the final answer must be 3.8 × 10³ feet.

<h3>LEARN MORE</h3>
  • Learn more about Velocity brainly.com/question/862972
  • Learn more about Acceleration brainly.com/question/4134594
  • Learn more about Distance - Time Graphs brainly.com/question/1378025

Keywords: speed, distsance, time

You might be interested in
A person with mass mp = 76 kg stands on a spinning platform disk with a radius of R = 1.98 m and mass md = 191 kg. The disk is i
nalin [4]
<span>1.7 rad/s The key thing here is conservation of angular momentum. The system as a whole will retain the same angular momentum. The initial velocity is 1.7 rad/s. As the person walks closer to the center of the spinning disk, the speed will increase. But I'm not going to bother calculating by how much. Just remember the speed will increase. And then as the person walks back out to the rim to the same distance that the person originally started, the speed will decrease. But during the entire walk, the total angular momentum remained constant. And since the initial mass distribution matches the final mass distribution, the final angular speed will match the initial angular speed.</span>
3 0
4 years ago
What is the momentum of a golf ball with a mass of 62 g moving at 73 m/s?
Vikentia [17]

Explanation:

Momentum = mass × Velocity

p = 62×73

p =4526

4 0
2 years ago
What is anything that has mass and volume?
Reika [66]
Matter, substance. Material howya call it.
3 0
3 years ago
After today i might not be here no more
Norma-Jean [14]

Answer:

aw why? are you deleting the app for school?

6 0
2 years ago
Read 2 more answers
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0
3 years ago
Other questions:
  • How many significant figures from 0,020170 kg ?<br> a. 3<br> b. 4<br> c. 5<br> d. 6<br> e. 7
    5·1 answer
  • Two movers are pushing a large crate with a force of 60.0 n each. one pushes north, the other east. what is the equilibrant forc
    15·1 answer
  • 3. True or false: Evaporation is the change of state from a gas to a liquid. *
    12·2 answers
  • Suggest why it might be a good idea to reduce the pressure of the air in car tyres if the car is to be driven over soft sand or
    6·1 answer
  • The magnetic field 0.02 m from a wire is 0.1 t. what is the magnitude of the magnetic field 0.01 m from the same wire?
    14·2 answers
  • Douglas has a segment with endpoints I(5, 2) and J(9, 10) that is divided by a point K such that IK and KJ form a 2:3 ratio. He
    14·1 answer
  • A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 degrees with the horizontal. The rope moves
    11·1 answer
  • What will be the mass of a body at the center of the earth as compared to other places on
    11·2 answers
  • LPG is a better domestic fuel than wood?​
    13·1 answer
  • Can someone help me :)
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!