a student standing between two walls shouts once.he hears the first echo after 3 seconds and the next after 5 seconds. calculate the distance between the walls.​

Answers

Answer 1

Explanation:

It took [tex]t_1 =1.5\:\text{s}[/tex] for the sound to reach the 1st wall and at the same time time, the same sound took [tex]t_2 = 2.5\:\text{s}[/tex] to reach the 2nd wall. Assuming that the sound travels at 343 m/s, then let [tex]x_1[/tex] be the distance of the person to the 1st wall and [tex]x_2[/tex] be the distance to the 2nd wall. So the distance between the walls X is

[tex]X = x_1 + x_2 = v_st_1 + v_st_2 = v_s(t_1 + t_2)[/tex]

[tex]\:\:\:\:\:= (343\:\text{m/s})(4.0\:\text{s}) = 1372\:\text{m}[/tex]


Related Questions

What is a benefit of joining the YMCA instead of a private health club?


The YMCA has nicer equipment and buildings.The YMCA has nicer equipment and buildings. , ,

The YMCA is likely to be in an upscale neighborhood.The YMCA is likely to be in an upscale neighborhood. , ,

The YMCA is much more affordable.The YMCA is much more affordable. , ,

The YMCA has restaurants and spa services.

Answers

Answer:

Explanation:

This depends on where you are.

The YMCA where I grew up was affordable. Very. Or my mother would not have let me go.

Help please!!!!!!!!! I will mark brainliest!!!

Answers

Answer:

solving for: velocity

equation: velocity = distance / time

substitution: velocity = 1425 km / 12.5 hrs

answer: 114 km/hr

7) A force of 500N exists between two identical point charges separated by a dis-
tance of 40 cm the magnitude of the two charges is​

Answers

Answer:

q=9.43×10^-5C

Explanation:

F=kq^2/r^2

500= 9×10^9 × q^2/ (0.4)^2

500N×0.16m=9×10^9Nm^2C^2 × q^2

80/(9×10^9)= q^2

√(8.8889×10^9) = q

q= 9.43×10^-5C

since they are identical both charges, q=9.43×10^-5C

The magnitude of the two charges is ​9.43×10^-5 C.

To calculate the magnitude of an electric force, it is necessary to use the following expression:

                                          [tex]F = k \frac{q_{1} \times q_{2} }{d^{2}}[/tex]

Assuming that the constant is:

                                          [tex]k = 9\times 10^{9}[/tex]

We can apply the values ​​in the formula above, obtaining:

                                     [tex]500 = 9 \times 10^{9} \times \frac{q^{2}}{0.4^{2}}[/tex]

                                         [tex]q = \sqrt{0.88\times10^{-8}}[/tex]

                                          [tex]q = 9.43 \times 10^{-5} C[/tex]

So, the magnitude of the two charges is   ​9.43×10^-5 C.

Learn more about electrical force in: brainly.com/question/16888648

The theory of the origin of the universe that is most popular among space scientists suggests that our universe originated how long ago

10-20 million years ago

10-20 trillion years ago,

10-20 billion years ago

10-20 thousand years ago

Answers

Answer:

10 - 20 million years ago

why do we go to hospital​

Answers

Answer:

bcz we want to have fun there lolol

Answer:

for emergency, treatment, medicines,etc.....

A 50.0 ohm and a 30.0 ohm resistor are connected in parallel. What is their equivalent resistance? Unit=Ohms

Answers

R(parallel) = product/ sum

50×30/50+30

1500/80

18,75 ohms

Answer: 18.75

above is right but you need to put a dot after the number 18

What would happen if you changed the position of the screen, but kept the other factors the same?

Answers

Answer:

I wish I could help but iam srry

what Is accuracy ............​

Answers

Answer:

Accuracy is how much the consequence of an estimation adjusts to the right worth or a norm' and basically alludes to how close an estimation is to its concurred esteem

《OAmalaOHopeO》

Answer:

In a set of measurements, accuracy is closeness of the measurements to a specific value, while precision is the closeness of the measurements to each other.

Explanation:

_Hope it helps you_

A 10 n force is applied horizontally on a box to move it 10 m across a frictionless surface. How much work was done to move the box?

Answers

Given from question
Force = 10 N
Displacement = 10 m
Work done = ?
We know that
Work done = force X displacement
So 10 X 10
100
Work done = 100J answer

Answer:

[tex]\boxed {\boxed {\sf 100 \ J}}[/tex]

Explanation:

We are asked to calculate the work done to move a box.

Work is the product of force and distance or displacement.

[tex]W= F*d[/tex]

A 10 Newton force is applied horizontally on the box. Since the surface is frictionless, there is no force of friction, and the net force is 10 Newtons. The force moves the box 10 meters.

F= 10  N d= 10 m

Substitute the values into the formula.

[tex]W= 10 \ N * 10 \ m[/tex]

Multiply.

[tex]W= 100 \ N*m[/tex]

Let's convert the units. 1 Newton meter is equal to 1 Joule, therefore our answer of 100 Newton meters is equal to 100 Joules.

[tex]W= 100 \ J[/tex]

100 Joules of work was done to move the box.

Describe the forces that act on a skydiver before
and after the parachute is opened.
I will give brainliest!!!!

Answers

Answer:

Before the parachute opens: Immediately on leaving the aircraft, the skydiver accelerates downwards due to the force of gravity. There is no air resistance acting in the upwards direction, and there is a resultant force acting downwards. The skydiver accelerates towards the ground.

Once the parachute is opened, the air resistance overwhelms the downward force of gravity. The net force and the acceleration on the falling skydiver is upward. An upward net force on a downward falling object would cause that object to slow down. The skydiver thus slows down.

I HOPE THIS WILL HELP YOU IF NOT THEN SORRY

HAVE A GREAT DAY :)

A liquid is poured into a vessel to a depth of 16cm when viewed from the top, the bottom appears to be raised 4cm. What is the refractive index of the liquid?

Answers

Answer:

Solution

Verified by Toppr

Correct option is

C

3 cm

RI=apparent depthreal depth

Substituting, 34=apparentdepth12

Therefore, apparent depth=412×3=9

The height by which it appears to be raised is 12−9=3cm

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SIMILAR QUESTIONS

A coin is placed at the bottom of a glass tumbler and then water is added. It appeared that the depth of the coin has been reduced because

Medium

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A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

calculate the rate of loss of heat through a glass window of area 200 CM square and thickness 0.5 CM where temperature inside is 35 degree Celsius and outside is -5 degree Celsius conductivity of Glass is 2.2 into 10 to the power 3 cal per s per cm per k .

Answers

Answer:

The inner and outer surfaces of a 0.5-cm thick 2-m by 2-m window glass in winter are 10°C and 3°C, respectively. If the thermal conductivity of the glass

Explanation:

Sound is an example of a:
Select one:
O a. rolling waves
b. longitudinal wave
O c. traverse waves
O d. surface wave

Ez Physics question will mark brainliest.

Answers

Answer:

The answer is B. longitude wave

the treasure map gives the following directions to the buried treasure ​

Answers

Answer:

North

South

East

West

Explanation:

please mark me as brainliest

Current is the rate at which charge is flowing.

a. True
b. Fals

Answers

Answer:

A. True

Explanation:

Our system is a block attached to a horizontal spring on a frictionless table. The spring has a spring constant of 4.0 N/m and a rest length of 1.0 m, and the block has a mass of 0.25 kg.

Compute the PE when the spring is compressed by 0.50 m.

Answers

Answer

E - 1/2 K x^2      potential energy of compressed spring

E = 1/2 * 4 N / m * (.5 m)^2 = 2 * .5^2 N-m = .5 N-m

Which item will be shipped third?

—-

Answers

Answer:

I know it's groceries

Explanation:

electronics ship before clothing

electronics ship after groceries

urgent items are first so

order:

1.) A/Electronics

2.) Clothing/B

3.) Groceries(since groceries aren't urgent)

thing is it's C or D I'm leaning to D since it says it ships last but i dont know so if I'm wrong sorry.

Consider a concave mirror that has a focal length f. In terms of f, determine the object distances that will produce a magnification of

A. -2
B. -3
C. -4

Answers

We have that the magnification of each focal length is given respectively as

A) has [tex]u=3\frac{f}{2}[/tex]

B) has [tex]u=4\frac{f}{3}[/tex]

C) has  [tex]u=5\frac{f}{4}[/tex]

From the question we are told that:

Focal Length F

Generally, the equation for Magnification is mathematically given by

[tex]M=\frac{-v}{u}[/tex]

Therefore

[tex]v=2u[/tex]

For A

[tex]M=-2[/tex]

Therefore

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{2u}[/tex]

Therefore

[tex]u=3\frac{f}{2}[/tex]

For B

[tex]M=-3[/tex]

Therefore

[tex]v=3u[/tex]

Where

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{3u}[/tex]

Therefore

[tex]u=4\frac{f}{3}[/tex]

For C

[tex]M=-4[/tex]

Therefore

[tex]v=4u[/tex]

Therefore

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{4u}[/tex]

Therefore

[tex]u=5\frac{f}{4}[/tex]

Conclusion

From the calculations above we can rightly say that the magnifications of the values above are

A has [tex]u=3\frac{f}{2}[/tex]

B has [tex]u=4\frac{f}{3}[/tex]

C has  [tex]u=5\frac{f}{4}[/tex]

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A student graphs power (p) on the vertical axis and time (t) on the horizontal axis. The graph appears to be a hyperbola.

a) What should the student graph on each axis to test whether the relationship is actually
hyperbolic?

b) If the relationship is actually hyperbolic, what is the general equation for the relationship between power and time?

Answers

Answer: it would be daddy

Explanation:

Because I’m daddy

A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane.

Answers

It will take a shorter amount of time for the cylinder to go down the plane down off the plane Because more pressure is applied one going up then going down there’s no pressure at all it’s the gravity is helping

Good evening everyone Help me i n my hw ,The wall of cinema hall are covered with sound absorbing materials. Why?Answer it ASAP.Good day ​

Answers

what do you mean about it

14. If you are asked to analyze an issue for an essay assignment, what should you do? a. Nb. C. Divide the issue into its main parts and discuss each part. Express your opinion about the issue. Explain what you think the issue means and how you came to that interpretation Describe the main ideas or points about the issue. d.​

Answers

Answer:

Option C

Explanation:

Divide the issue into its main parts and discuss each part.

Cho các máy cắt sử dụng trong công nghiệp có ký hiệu trên nhãn thiết bị: C350; B500. Hãy tính dòng điện bảo vệ ngắn mạch và dòng điện bảo vệ quá tải của từng thiết bị?

Answers

Answer:

ask in the English then I can help you

Explanation:

please mark me as brain list

What is tensor quantity?
Is Inertia a tensor? give reason​

Answers

Answer:

A tensor is a quantity, for example a stress or a strain, which has magnitude, direction, and a plane in which it acts. Stress and strain are both tensor quantities. ... A tensor is a quantity, for example a stress or a strain, which has magnitude, direction, and a plane in which it acts.

Inertia Tensor. where I = the inertia tensor. The angular momentum of a rigid body rotating about an axis passing through the origin of the local reference frame is in fact the product of the inertia tensor of the object and the angular velocity. ... As shown in [7], the inertia tensor is symmetric.

Explanation:

Hope dis help

A plastic box with objects has a mass of 4 kg and is on a shelf at a height of 2.4 m. What will it's potential gravitational energy?

Answers

Answer:

potential energy=mgh

4×9.8×2.4

Explanation:

may be hope this will help you

The pressure of sea water increases by 1.0atm for each 10m increase in the depth, by what percentage is the density of water increased in the deepest ocean of water of 12km. Compressibility is 5.0×10^-5 atm

Answers

The percentage by which the water density increased is 4.1[tex]\mathbf{\overline 6}[/tex] %

The known values are;

The increase in pressure per 10 meter increase in depth = 1.0 atm

The depth of the deepest ocean = 12 km = 12,000 m

The compressibility of the ocean = 5.0 × 10⁻⁵ 1/atm

The unknown

The percentage the density of water increased in the deepest ocean

Strategy;

Find the pressure at the deepest point of the deepest ocean and apply the compressibility

We have;

[tex]\mathbf{Compressibility = \dfrac{1}{V} \times \dfrac{\partial V}{\partial p}}[/tex]

The change in pressure, [tex]\partial p[/tex] = (12,000 m/(10 m)) × 1.0 atm = 1,200 atm

Therefore, we have for one cubic meter of water

[tex]\mathbf{5.0 \times 10^{-5} \ atm^{-1} = \dfrac{1}{1 \, m^3} \times \dfrac{\partial V}{1,200 \, atm}}[/tex]

Therefore;

[tex]\mathbf{\partial}[/tex]V = 5.0 × 10⁻⁵ atm⁻¹ × 1 m³ × 1,200 atm = 0.06 m³

The new volume = V - [tex]\mathbf{\partial}[/tex]V

∴ The new volume = 1 m³ - 0.06 m³ = 0.94 m³

The initial density = mass/(1 m³)

The new density = mass/(0.96 m³)

The percentage increase in density, [tex]\partial[/tex]ρ%, is given as follows;

[tex]\mathbf{\partial p \% = \dfrac{ \dfrac{Mass}{0.96 \ m^3} - \dfrac{Mass}{1 \ m^3} }{ \dfrac{Mass}{1 \ m^3}} \times 100 = \dfrac{25}{6} \% = 4.1 \overline 6 \%}[/tex]

∴  [tex]\mathbf{\partial}[/tex]ρ% =  4.1[tex]\mathbf {\overline 6}[/tex] %

The percentage by which the water density increased, [tex]\partial[/tex]ρ% = 4.1[tex]\mathbf{\overline 6}[/tex] %

Learn more about compressibility here;

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A stone that is dropped freely from rest traveled half the total height in the last second. with what velocity will it strike the ground

Answers

Answer:

hellooooo :) ur ans is 33.5 m/s

At time t, the displacement is h/2:

Δy = v₀ t + ½ at²

h/2 = 0 + ½ gt²

h = gt²

At time t+1, the displacement is h.

Δy = v₀ t + ½ at²

h = 0 + ½ g (t + 1)²

h = ½ g (t + 1)²

Set equal and solve for t:

gt² = ½ g (t + 1)²

2t² = (t + 1)²

2t² = t² + 2t + 1

t² − 2t = 1

t² − 2t + 1 = 2

(t − 1)² = 2

t − 1 = ±√2

t = 1 ± √2

Since t > 0, t = 1 + √2.  So t+1 = 2 + √2.

At that time, the speed is:

v = at + v₀

v = g (2 + √2) + 0

v = g (2 + √2)

If g = 9.8 m/s², v = 33.5 m/s.

A train starts from rest (at position zero) and moves with constant acceleration. On the first observation, its velocity is 20m/s and 80seconds later the velocity became 60m/s. At 80s calculate the position, average velocity, and the constant acceleration over the interval.(7-points)

Answers

The value of the acceleration is a = 0.5 m/s². The position at 80 s is x = 3200 m and finally the average velocity is v = 40 m/s.

Acceleration:

We can use the fallowing kinematic equation to get the acceleration at 80 s.

[tex]a=\frac{v_{f}-v_{i}}{t}[/tex]            

Where:

v(i) is the initial velocity (20 m/s)v(f) is the final velocity (60 m/s)t is the interval (80 s)

The, we have:

[tex]\vec{a}=\frac{60-20}{80}[/tex]

[tex]\vec{a}=0.5\: m/s^{2}[/tex]

Position:

Knowing the acceleration we can find the position using the falling equation.

[tex]\vec{x}=v_{i}t+0.5at^{2}[/tex]

[tex]\vec{x}=20*80+0.5*0.5*80^{2}[/tex]

[tex]\vec{x}=3200 m[/tex]

Average velocity:

The definition of the average velocity is:

[tex]\vec{v}=\frac{\Delta x}{t}[/tex]

[tex]\vec{v}=\frac{x_{f}-x_{i}}{t}[/tex]

[tex]\vec{v}=\frac{3200-0}{80}[/tex]

[tex]\vec{v}=40\: m/s[/tex]

Learn more about the kinematic equations here:

https://brainly.com/question/13143668

I hope it helps you!

can you guys pls also solve for average speed.

Answers

Answer:

d_t = 3.05km

v_a = 4.3km/h

Explanation:

42mins*(2/3) = 28mins

42mins-28mins = 14mins

d = v*t

d_1 = (4km/h)*(1h/60mins)*(28mins)

d_1 = 1.87km

d_2 = (5km/h)*(1h/60mins)*(14mins)

d_2 = 1.17km

d_t = d_1+d_2

d_t = 1.87km+1.17km

d_t = 3.05km

v_a = (v_1+v_2)/2

v_a = [(2*4km/h)+5km/h)]/3

v_a = 4.3km/h

Earth’s Moon has a diameter of 3,474 km and orbits at an average distance of 384,000 km. At that distance it subtends and angle just slightly larger than half a degree in Earth’s sky. Pluto’s moon Charon has a diameter of 1,186 km and orbits at a distance of 19,600 km from the dwarf planet. Compare the appearance of Charon in Pluto’s skies with the Moon in Earth’s skies. Describe where in the sky Charon would appear as seen from various locations on Pluto.

Answers

The result of the comparison of the appearance of Charon on Pluto and   times the Moon from Earth is that; Charon as seen from Pluto appears approximately 7 times larger than the Moon

Charon is directly overhead from the side of Pluto locked to the side of Charon

Charon appears at the horizon from the poles of the axis of rotation of Jupiter around Charon

The reason for arriving at the above solutions is as follows:

The given dimensions and distance from the Earth of the Moon are;

The diameter of the Moon, d = 3,474 km

The average distance of the Moon from the Earth, R = 384,000 km

Required:

The comparison between Charon's appearance in Pluto and the Moon's appearance on Earth Earth

Solution:

The distance of the Moon's travels in an orbit, C = 2·π·R

∴ C = 2 × π × 384,000 km

The angle subtended by the Moon, θ = d/C × 360°

∴ θ = 3,474/(2 × π × 384,000) × 360° ≈ 0.518°

Pluto's moon Charon, has the following parameters;

The diameter of the Charon, d₂ = 1,186 km

The average distance of the Charon from Pluto, R₂ = 19,600 km

Therefore, the distance of the Moon's travels in an orbit, C₂ = 2·π·R₂

∴ C₂ = 2 × π × 19,600 km

The angle subtended by the Moon, θ₂ = d₂/C₂ × 360°

∴ θ₂ = 1,186/(2 × π × 19,900) × 360° ≈ 3.415°

The angle subtended by Charon in Pluto's sky ≈ 3.415°

Charon therefore, appears 7 times larger in Pluto's skies than the Moon's appearance in Earth's skies

Required:

The appearance of Charon as seen from different locations on Pluto

Solution:  

Charon is gravitationally locked to Pluto, therefore, the same side of Pluto is faced with the same side of Charon

Therefore;

Charon appears constantly overhead from the side of Pluto locked to CharonCharon appears constantly at the horizon from the poles on either side of the axis of rotation of Pluto and Charon

Learn more about Pluto's moon Charon here:

https://brainly.com/question/3920772

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