Choose the box as the system of interest. What objects in the surroundings exert significant forces on this system

Answers

Answer 1

Answer:

Hello your question is incomplete attached below is the complete question

Your are lifting a heavy box we will consider this process for different choices of system and surroundings

Choose the box as the system of interest. What objects in the surroundings exert significant forces on this system

The floor the box you nonethe earth

answer : The earth and YOU

Explanation:

The objects that will exert significant force on this system would be the force exerted by the earth and force exerted by you

This is because the force exerted by the earth is an internal force and it can be calculated  as : m*g ( mass of the object * acceleration due to gravity ) also the force exerted by you are force due to normal reaction when lifting a load


Related Questions

Any body moving with simple harmonic motion is being acted on by a force that is:__________.
a) proportional to a sine or cosine function of the displacement
b) directly proportional to the displacement
c) proportional to the inverse square of the displacement
d) constant

Answers

Answer:

B

Explanation:

Because this oscillations occur when the restoring force is directly proportional to displacement, given as

F=-kx

Where k= force constant

X= displacement

A point source emits electromagnetic energy at a rate of 100 W. The intensity 10 m from the source is

Answers

Answer:

[tex]I=0.0795\ W/m^2[/tex]

Explanation:

Given that,

Power of electromagnetic energy, P = 100 W

We need to find the intensity at a distance of 10 m from the source. Intensity is equal to the power per unit area. So,

[tex]I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{100}{4\pi (10)^2}\\\\I=0.0795\ W/m^2[/tex]

So, the intensity at a distance of 10 m is [tex]0.0795\ W/m^2[/tex]

A worker is pushing a crate of tools up a ramp into a truck. The crate has a mass of 120 kg and is accelerating at a rate of 1.05 m/s^2. Once he finishes pushing that crate up, he then takes another crate that is 3 times as massive and pushes it up the ramp with an acceleration of 0.71 m/s^2. What is the ratio of the force the worker used on the on the 120 kg crate and the more massive crate?

Answers

Answer:

B) 0.493

Explanation:

First Crate

Formula: F=ma

120kg*1.05m/s^2= 126N

Second Crate

Formula: F=ma

3*120kg=360kg

360kg*0.71m/s^2=255.6

Find the ratio by dividing

120kg/255.6kg=0.4929

Round 0.4929=0.493

Hope this Helps!!

A creature moves at a speed of 7.86 furlongs/fortnight (not a very common unit of speed). If 1.0 furlong = 220 yards 1 fortnight = 14 days 1 yard = 0.9144 meter and 1 day= 86400 sec find the speed of the creature (which is probably a snail).
1. 2.28745 × 109 m/s
2. 0.00130719 m/s
3. 1.16706 × 107 m/s
4. 47261.3 m/s
5. 0.25621 m/s
6. 39516.4 m/s
7. 5.29359 × 10−6 m/s

Answers

Answer:

2 0.00130719 m/s

Explanation:

Now we have

7.86 furlongs/fortnight

7.86 x 220 yards / fortnight

1729.2yards / fortnight

1647.8 yards / 14 days

123.5yards / day

So

123.5 x 0.9144 meter / day

112.meters / day

=112.9meters / 86400s

= 0.001307m/s

An agent has just listed a power plant. This is an example of what type of real estate?

Answers

Answer: industrial properties

Explanation:

The power plant listed by the agent falls under industrial property. An industria property is a property which is either used in manufacturing, rentage, warehousing, processing e.t.c.

in this case the plant Would be used in the generation of power which would be sold. so yes the property falls under the category

How many valence electrons does each Cholrine atom have?

Answers

Answer:

7 valence electrons is right answer

Explanation:

2 elections in first shell and 5 electrons in second shell which total makes seven electrons.

hope it helped you:)

What is the largest wavelength that will give constructive interference at an observation point 141 m from one source and 355 m from the other source

Answers

Answer:

Explanation:

For constructive interference of two coherent light waves at a point , the condition is as follows

path difference = n x wavelength where n is any integer like  1 , 2 , 3 ,  .... etc .

Given in the question , path difference of light from the two sources

= 355 - 141 = 214 m

So applying the formula above ,

214 = n x wavelength

wavelength = 214 / n

For largest value of wavelength , n should be smallest . The smallest value of n is 1 .

so ,

largest wavelength possible for constructive interference

= 214 / 1

= 214 m .

When a piano tuner strikes both the A on the piano and a 440 Hz tuning fork, he hears a beat every 2 seconds. The frequency of the piano's A is

Answers

Answer:

The  frequency is  [tex]f = 439.5 \ Hz[/tex]        

Explanation:

From the question we are told that

   The frequency of the  tuning fork is  [tex]f_t = 440 \ Hz[/tex]

   The beat period is  [tex]T = 2 \ s[/tex]

Generally the beat frequency is mathematically represented as

       [tex]f_b = \frac{1}{T}[/tex]

       [tex]f_b = \frac{1}{2}[/tex]

      [tex]f_b = 0.5 \ Hz[/tex]

The  beat frequency is also represented mathematically as

     [tex]f_b = f_t - f[/tex]

Where  [tex]f[/tex] is the frequency of the piano

 So

       [tex]f = 440 - 0.5[/tex]  

       [tex]f = 439.5 \ Hz[/tex]        

Mass and Weight: What is the mass of an object that experiences a gravitational force of 685 N near Earth's surface where g

Answers

Answer:

m = 69.9 kg

Explanation:

The mass and the weight of an object are two different quantities. Mass is basically the amount of matter that is present in a body. It remains same everywhere in the universe and measured in kilograms.

Weight is basically a force. It is the force by which earth attracts everything towards itself. The weight of an object changes from planet to planet, with the change in value of the gravitational acceleration (g).

Therefore, the relation between mass and weight of an object is given by the following formula:

W = mg

m = W/g

where,

m = mass = ?

W = Weight = 685 N

g = 9.8 m/s²

Therefore,

m = (685 N)/(9.8 m/s²)

m = 69.9 kg

The mass of an object that experience a gravitational force of 685N near the earth is approximately 69.90 kg

The mass and weight are 2 different quantities.

The mass of a substance is the amount of matter present in it. Mass is measured in kilograms

The Weight is the measurement of gravitational pull of planet towards the object. Therefore, weight is a force and measured in Newton.

Mathematically, the weight can be express as follows

W = mg

where

m = mass

g = acceleration due to gravity = 9.8 m/s²

Therefore,

685 = 9.8 × m

m = 685 / 9.8

m = 69.8979591837

m = 69.90 kg

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Which explanation is based on scientific evidence?
A. Observations made using microscopes reveal that all known
organisms are composed of tiny structures called cells.
B. Magnets have hidden powers that can have health effects.
C. Performing good deeds for other people will make a person happy.
D. illnesses are caused by bad smells in the air.

Answers

A.Observations made using microscopes revel that all known.

All known species are made up of small structures called cells, according to observations obtained using a microscope; this theory is supported by empirical data..Option A is corect.

What is the investigation?

An investigation is defined as a thorough search or inspection. The investigative process is a series of tasks or procedures that include obtaining evidence, analyzing the material, and developing and validating theories.

Respectfully recognize any information or points of view that contradict your contention. You have to demonstrate to the reader in the refutation why your position is superior to the opposing notion.

Observations made using microscopes reveal that all known organisms are composed of tiny structures called cells the explanation is based on scientific evidence.

Hence option A is corect.

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For small angles, does the pendulum's period of oscillation depend on initial angular displacement from equilibrium? Explain.

Answers

Answer:

No, the pendulum's period of oscillation does not depend on initial angular displacement.

Explanation:

Given that,

For small angle, the pendulum's period of oscillation depend on initial angular displacement from equilibrium.

We know that,

The time period of pendulum is defined as

[tex]T=2\pi\sqrt{\dfrac{l}{g}}[/tex]

Where, l = length of pendulum

g = acceleration due to gravity

So, The time period of pendulum depends on the length of pendulum and acceleration due to gravity.

It does not depend on the initial angular displacement.

Hence, No, the pendulum's period of oscillation does not depend on initial angular displacement.

What will be the force if the particle's charge is tripled and the electric field strength is halved? Give your answer in terms of F

Answers

Answer:

F' = (3/2)F

Explanation:

the formula for the electric field strength is given as follows:

E = F/q

where,

E = Electric Field Strength

F = Force due to the electric field

q = magnitude of charge experiencing the force

Therefore,

F = E q   ---------------- equation (1)

Now, if we half the electric field strength and make the magnitude of charge triple its initial value. Then the force will become:

F' = (E/2)(3 q)

F' = (3/2)(E q)

using equation (1)

F' = (3/2)F

When the charge of the particle is tripled, and magnetic field strength is halved, the force increases by factor of 1.5 (1.5 F)

The force experienced by a particle placed in magnetic field is given as;

F = qvB

where;

q is the charge of the particlev is the velocity of the particleB is the magnetic field strength

When the charge of the particle is tripled, q₂ = 3q and magnetic field strength is halved, B₂ = 0.5B.

The magnetic force of the particle is determined as;

[tex]F = qvB\\\\ v = \frac{F}{qB} \\\\ \frac{F_1}{q_1B}_1 = \frac{F_2}{q_2B_2}\\\\ F_2 = \frac{F_1 q_2B_2}{q_1 B_1} \\\\ F_2 = \frac{F \times 3q_1\times 0.5B_1}{q_1B_1} \\\\ F_2 = 1.5 F[/tex]

Thus, when the charge of the particle is tripled, and magnetic field strength is halved, the force increases by factor of 1.5 (1.5 F).

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You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 14.8m above the ground, it is traveling at a speed of 25.0 m/s upward.
A) Use the work-energy theorem to find its speed just as it left the ground. What is it?
B) Use the work-energy theorem to find its maximum height. What is it?

Answers

Answer:

(A) The speed just as it left the ground is 30.25 m/s

(B) The maximum height of the rock is 46.69 m

Explanation:

Given;

weight of rock, w = mg  = 20 N

speed of the rock at 14.8 m, u = 25 m/s

(a) Apply work energy theorem to find its speed just as it left the ground

work = Δ kinetic energy

F x d = ¹/₂mv² - ¹/₂mu²

mg x d = ¹/₂m(v² - u²)

g x d = ¹/₂(v² - u²)

gd = ¹/₂(v² - u²)

2gd = v² - u²

v² = 2gd  + u²

v² = 2(9.8)(14.8) + (25)²

v² = 915.05

v = √915.05

v = 30.25 m/s

B) Use the work-energy theorem to find its maximum height

the initial velocity of the rock = 30.25 m/s

at maximum height, the final velocity = 0

- mg x H = ¹/₂mv² - ¹/₂mu²

- mg x H = ¹/₂m(0) - ¹/₂mu²

- mg x H = - ¹/₂mu²

2g x H = u²

H = u² / 2g

H = (30.25)² / 2(9.8)

H = 46.69 m

Does anybody know to solve this?

Answers

Answer:

25600 cm

2.56 x 10⁴ cm

Explanation:

Step 1: Find conversion

There are 100 cm in 1 m.

Step 2: Set up dimensional analysis

[tex]256m(\frac{100cm}{1 m} )[/tex]

Step 3: Multiply and cancel out units

25600 cm

2.56 x 10⁴ cm

The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Determine the location ӯ of the center of mass G of the pendulum, then calculate the mass moment of inertia of the pendulum about z axis passing through G. b) Calculate the mass moment of inertia about z axis passing the rotation center O.

Answers

Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

[tex]y=\frac{0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m}{(\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m} \\\\y=0.88\ m[/tex]

b)  the mass moment of inertia about z axis passing the rotation center O is:

[tex]I_G=\frac{1}{12}*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\\0.888)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+\frac{1}{12}*3(1.5)(1.5)^2+\\3(1.5)(0.888-0.75)^2\\\\I_G=13.4\ kgm^2[/tex]

c) The mass moment of inertia about z axis passing the rotation center O is:

[tex]I_o=\frac{1}{12}*3(0.8)(0.8)^2+ \frac{1}{3}* 3(1.5)(1.5)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\\\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\\\\I_o=13.4\ kgm^2[/tex]

Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm until the height of the alcohol column is 20 cm. The two liquids do not mix. What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3and the density of alcohol is 790 kg/m3.

Answers

Answer:

Difference in height = 7.5 cm

Explanation:

We are given;.

Height of ethyl alcohol;h2 = 20 cm = 0.2 m

Density of glycerin: ρ1 = 1260 kg/m³

Density of ethyl alcohol; ρ2 = 790 kg/m³

To get the difference in height, the pressure at the top of the open end must be equal to the pressure at the point where the liquids do not mix since both points will be at different levels after the pouring.

Thus;

P1 = P2

Formula for pressure is; P = ρgh

Thus;

ρ1 × g × h1 = ρ2 × g × h2

g will cancel out to give;

ρ1 × h1 = ρ2× h2

Making h1 the subject, we have;

h1 = (ρ2× h2)/ρ1

h1 = (790 × 0.2)/1260

h1 = 0.125 m

Difference in height will be;

Δh = h2 - h1

Δh = 0.2 - 0.125

Δh = 0.075 m = 7.5 cm

The difference in height between the top surface of the glycerin and the top surface of the alcohol as per Pascal's law is 7.5 cm.

What is Pascal's law?

As per Pascal's law, "The pressure at any point of the vessel filled with incompressible liquid is same".

Given data -

The height of both section is, h2 = 20 cm = 0.20 m.

The density of glycerine is, [tex]\rho_{1} = 1260 \;\rm kg/m^{3}[/tex].

The density of alcohol is, [tex]\rho_{2} = 790 \;\rm kg/m^{3}[/tex].

The pressure at the top of the open end must be equal to the pressure at the point where the liquids do not mix since both points will be at different levels after the pouring. Therefore,

P1 = P2

The formula for pressure is;

P = ρgh

Thus;

ρ1 × g × h1 = ρ2 × g × h2

ρ1 × h1 = ρ2× h2

h1 = (ρ2× h2)/ρ1

h1 = (790 × 0.2)/1260

h1 = 0.125 m

Difference in height will be;

Δh = h2 - h1

Δh = 0.2 - 0.125

Δh = 0.075 m

Δh = 7.5 cm

Thus, we can conclude that the difference in height between the top surface of the glycerin and the top surface of the alcohol is 7.5 cm.

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Unit 1 Horizontal Kinematics Homework #3 Acceleration 1. A bowling ball starts from rest and moves 300 m down the gutter in 22.4 seconds. a) What is its speed at the end of the track? b) What is its acceleration?

Answers

Answer:

Explanation:

since the object start its motion from rest and ends its motion at 300 m so

ΔS=S2-S1=300m-0m=300 m

given time is 22.4 sec therefore

v=ΔS/t

v=300/22.4=14.7 m/s

so the speed of the ball at the end of the track is 14.7 m/s

now to find the acceleration we know that

a=v/t

a=14.7/2.4

a=0.65m/sec²

A security guard walks at a steady pace, traveling 190 m in one trip around the perimeter of a building. It takes him 260 s to make this trip.what is his speed. express answer in 2 significant figures

Answers

Answer:

0.7 m/s

Explanation:

The following data were obtained from the question:

Distance travalled (d) = 190 m

Time (t) = 260 secs

Speed (S) =..?

Speed is simply defined as the distance travelled per unit time. Mathematically, it is expressed as:

Speed = Distance (d) /time(t)

S = d/t

With the above formula, we can calculate the speed of the security guard as follow:

Distance travalled (d) = 190 m

Time (t) = 260 secs

Speed (S) =..?

S = d/t

S = 190/260

S = 0.7 m/s

Therefore, the speed of the security guard is 0.7 m/s

A security guard walks at a steady pace, traveling 190 m in one trip around the perimeter of a building. The speed of the security guard is approximately 0.73 m/s.

Given:

Distance (d) = 190 m

Time (t) = 260 s

To calculate the speed of the security guard, we can use the formula:

Speed = Distance / Time

Speed = 190  / 260 = 0.73 m/s

Therefore, A security guard walks at a steady pace, traveling 190 m in one trip around the perimeter of a building. The speed of the security guard is approximately 0.73 m/s (rounded to two significant figures).

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A 21.3 A current flows in a long, straight wire. Find the strength of the resulting magnetic field at a distance of 45.7 cm from the wire.

Answers

Answer:

The magnetic field strength due to current flowing in the wire is9.322 x 10⁻ T.

Explanation:

Given;

electric current, I = 21.3 A

distance of the magnetic field from the wire, R = 45.7 cm = 0.457 m

The strength of the resulting magnetic field at the given distance is calculated as;

[tex]B = \frac{\mu_o I}{2\pi R}[/tex]

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/A

[tex]B = \frac{\mu_o I}{2\pi R}\\\\B = \frac{4\pi*10^{-7} *21.3}{2\pi(0.457)}\\\\B = 9.322 *10^{-6} \ T[/tex]

Therefore, the magnetic field strength due to current flowing in the wire is 9.322 x 10⁻ T.

"For the lowest harmonic of pipe with two open ends, how much of a wavelength fits into the pipe’s length?"

Answers

Answer:

0.5 lambda(wavelength)

Explanation:

We know that

The first harmonic for both side open ended pipe is

L= 1/2lambda

So L = 0.5*wavelength

The aqueous humor in a person's eye is exerting a force of 0.242 N on the 1.21 cm2 area of the cornea. What pressure is this in mm Hg?

Answers

Answer:

15.00124mmHg

Explanation:

Pressure is defined as the ratio of force applied to an object to its area.

Pressure = Force/Area

Given parameters

Force = 0.242N

Area = 1.21cm²

Required parameters

Pressure = 0.242/1.21

Pressure = 0.2N/cm²

Using the conversion to convert the pressure to mmHg

1N/cm² = 75.0062mmHg

0.2N/cm² = y

y = 0.2 * 75.0062

y = 15.00124mmHg

Hence the pressure in mmHg is 15.00124mmHg

Consider position [x] = L, time [t] = T, velocity [v] = L/T and acceleration [a] = L/T 2 . Find the exponent A in the equation v = a^2 t^ A /x

Answers

Answer:

The exponent A in the equation is 3.

Explanation:

v = a^2 t^ A /x

[tex]v = \frac{a^2t^A}{x} \\\\vx = a^2t^A\\\\(\frac{L}{T})(L) = (\frac{L}{T^2})^2(T)^A\\\\ \frac{L^2}{T}= (\frac{L^2}{T^4})(T)^A\\\\ \frac{L^2}{T} *\frac{T^4}{L^2} = (T)^A\\\\T^3 = T^A\\\\\frac{T^3}{T^3} = \frac{T^A}{T^3}\\\\T^{3-3} = T^{A-3}\\\\3-3 = A-3\\\\0 = A-3\\\\A = 3[/tex]

Therefore, the exponent A in the equation is 3.

The volume of a ball was measured at 500.0 cm3, and then mass was measured on this triple beam balance. What would be the most accurate reading for the mass of the ball?Using this mass, what would be the most accurate calculation of density for the ball? Round to the nearest ten thousandths.

Answers

Answer:

the answer is 0.8084

Explanation:

i just did it and got it correct

An ion has a mass of 6.79 × 10-27 kg and a charge of +4e. What is the magnitude of the electric field that will balance the gravitational force on the particle?

Answers

Answer:

Explanation:

200.34

In the park, there are 25 pigeons, 15 squirrels, 5 rabbits, and 5 stray cats. Why
can't you draw a line graph using only this information?

Answers

Answer: Line graph should be used to show how one variable changes over time not to show multiple categories or variables are at one specific point in time.

Explanation:

In maths, statistics, and related fields, graphs are used to visually display variables and their values. In the case of line graphs, these are mainly used to display evolution or change of a variable over time. For example, a line graph can show how the number of divorces changed from 1920 to 2010.

In this context, the number of different animals in the park cannot be represented through a line graph because this situation does not imply a variable changing over time. Moreover, this situation includes multiple variables or categories of animals and the data shows only one specific point in time, which can be better represented through a bar graph.

A ball is thrown straight up in the air. For which situations are both the instantaneous velocity and the acceleration zero? (Select all that apply.)

Answers

Answer:

Therefore, the situation in which both the instantaneous velocity and acceleration become zero, is the situation when the ball reaches the highest point of its motion.

Explanation:

When a ball is thrown upward under the free fall action of gravity, it starts to loose its Kinetic Energy as it moves upward. As the ball moves in upward direction, its kinetic energy gradually converts into its potential energy. As a result the speed of the ball starts to decrease as it moves up. Therefore, at the highest point during its motion, the velocity of ball becomes zero and it stops at the highest point for a moment, and then it starts to fall back down, under the influence of gravitational force.

Therefore, the situation in which both the instantaneous velocity and acceleration become zero, is the situation when the ball reaches the highest point of its motion.

Calculate the location xcm of the center of mass of the Earth-Moon system. Use a coordinate system in which the center of the Earth is at x

Answers

Answer:

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

Explanation:

Let suppose that planet and satellite can be treated as particles. The masses of Earth and Moon ([tex]m_{E}[/tex], [tex]m_{M}[/tex]) are [tex]5.972\times 10^{24}\,kg[/tex] and [tex]7.349\times 10^{22}\,kg[/tex], respectively. The distance between centers is 384,403 kilometers. The location of the center of mass can be found by using weighted averages:

[tex]\bar x = \frac{x_{E}\cdot m_{E}+x_{M}\cdot m_{M}}{m_{E}+m_{M}}[/tex]

If [tex]x_{E} = 0\,km[/tex] and [tex]x_{M} = 384,403\,km[/tex], then:

[tex]\bar x = \frac{(0\,km)\cdot (5.972\times 10^{24}\,kg)+(384,403\,km)\cdot (7.349\times 10^{22}\,kg)}{5.972\times 10^{24}\,kg+7.349\times 10^{22}\,kg}[/tex]

[tex]\bar x = 4.673\,km[/tex]

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

Four students measured the same line with a ruler like the one shown below. The results were as follows: 5.52 cm, 6.63 cm, 5.5, and 5.93. Even though you cannot see the line they actually measured, which of the recorded measurements are possible valid measurements for this instrument, according to its precision? Select all that apply. 1. 5.52 2. 6.63 3. 5.5 4. 5.93

Answers

Answer:

correct answer is 1 and 3

Explanation:

In direct measurement with an instrument, the precision or absolute error of the instrument is given by its appreciation, in this case we see that the measurements have two decimal places, so the appreciation of the instrument must be 0.01 cm

Based on this appreciation, the valid measurements are 5.52 and 5.5.

the other two measurements have errors much higher than the assessment of the instrument, for which there must have been some errors in the measurement.

The correct answer is 1 and 3

Alice drops a rock to a well. She hears the splash of the rock 4.1s later. The speed of sound is 340m/s. How deep is the well

Answers

Answer:

697 m

Explanation:

From the question above,

Note: Alice hears the splash of the rock in the well through the help of echo.

Applying,

v = 2x/t........................ Equation 1

Where v = speed of sound, x = depth of the well, t = time.

make x the subject of the formula in equation 1

x = vt/2................... Equation 2.

Given: v = 340 m/s, t = 4.1 s

Substitute these values into equation 2

x = 340(4.1)/2

x = 697 m.

two horses are pulling a 325 kg wagon, initially at rest. The horses exert 250 N and 178 N forward forces, respectively. Ignoring friction, how fast is the wagon moving 3.50 s later?

Answers

Answer:

AFter 3.5 s, the wagon is moving at:   [tex]4.62\,\,\frac{m}{s}[/tex]

Explanation:

Let's start by finding first the net force on the wagon, and from there the wagon's acceleration (using Newton's 2nd Law):

Net force = 250 N + 178 N = 428 N

Therefore, the acceleration from Newton's 2nd Law is:

[tex]F=m\,*\,a\\a = \frac{F}{m} \\a= \frac{428}{325}\, \frac{m}{s^2} \\a\approx 1.32 \,\,\frac{m}{s^2}[/tex]

So now we apply this acceleration to the kinematic expression for velocity in an object moving under constant acceleration:

[tex]v_f=v_i+a\,*\,t\\v_f=0+1.32\,*\,3.5\,\,\frac{m}{s} \\v_f=4.62\,\,\frac{m}{s}[/tex]

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