Answer:
F = force
f = friction
u = coefficient of friction
R = normal reaction force
a = Acceleration
m = mass of block
g = gravity
f = uR
F = Ma
Say the block is moving to the right.
The 146N force thus acts to the right, and the friction force to the left, since it resists movement.
The 146N force acts to the right, but the horizontal component of it is 146 cos 50 = 93.84: So this is the force to the right.
Since F = uR and we're trying to find u, we need both F and R. R is easy to get since it is just m x g. This is in fact already given as the weight 350N. So R = 350.
The block is moving at a constant speed, so the force to the right must = the force to the left.
F = ma, so 93.84 - f = (350/g) x 0
This means f must be 93.84 also.
so we have f = uR,
93.84 = u x 350
so u = 0.268 or
0.27 to 2dp.
Hope you understand this.
Explanation:
The coefficient of friction is 0.26 if the 46N force is needed to pull a 350 N block across a horizontal surface at a constant speed by a rope making an angle of 50 degrees with the floor.
What is the friction force?It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N). Mathematically it is defined as the product of the coefficient of friction and normal reaction.
We know:
f = uR
Where f is the friction force and u is the coefficient of friction, R is the normal reaction force.
The horizontal component of 146N is:
F' = 146cos50°
F' = 93.84 N
Since F = mass×acceleration
Because the block is traveling at a constant speed, the right-hand force must equal the left-hand force.
F' = f = 93.84 N
93.84 = u x 350 (R = 350N)
u = 0.26
Thus, the coefficient of friction is 0.26 if the 46N force is needed to pull a 350 N block across a horizontal surface at a constant speed by a rope making an angle of 50 degrees with the floor.
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If an ice cube has a mass of 5g and is left in the sun in a beaker, what will happen to its mass?
Answer:
The mass remains the same
Explanation:
An ice cube of mass 5g left in the sun in a beaker will have no mass difference.
For this system, mass is conserved.
The ice cube begins to melt and the state changes. The state change is from solid to liquid. Since no mass is loss, the mass of the liquid formed and the ice will remain the same. In retrospect, this physical change ensures that mass is conserved.Brainliest brainliest help help help mememememememme
Answer:
????????????????????,
Explanation:
I need points sorry
Answer:
honestly this was so long ago can i get brainliest i need 2 more until i am at expert level
Explanation:
2. A 2500 kg car is slowed down uniformly from an initial velocity of 20.0 m/s to
the north by a 6250 N braking force acting opposite the car's motion. Use the
impulse-momentum theorem to answer the following questions:
a. What is the car's velocity after 2.50 s?
b. How far does the car move during 2.50 s?
c. How long does it take the car to come to a complete stop?
Answer:
13.75m/s; 42.2m; 8s
Explanation:
(a) the car's velocity after 2.50 s is 13.75 m/s
(b) The distance traveled by the car is 42.18 m
(c) the time taken for the car to come to complete stop is 8 s.
The given parameters;
mass of the car, m = 2500 kg
initial velocity of the car, u = 20 m/s
breaking applied on the car, f = 6250 N
The acceleration of the car is calculated as follows;
[tex]F = ma \\\\a = \frac{F}{m} = \frac{6250}{2500} = 2.5 \ m/s^2[/tex]
(a) Using impulse-momentum theorem, the car's velocity after 2.5 s is calculated as follows;
[tex]F = \frac{m(u-v)}{t} \\\\m(u-v) = Ft\\\\u-v = \frac{Ft}{m} \\\\v = u - \frac{Ft}{m} \\\\v = 20 - \frac{6250 \times 2.5}{2500} \\\\v = 13.75 \ m/s[/tex]
(b) The distance traveled by the car during the 2.5 s;
[tex]v^2 = u^2 - 2as\\\\2as = u^2 - v^2\\\\s = \frac{u^2 - v^2}{2a} \\\\s = \frac{20^2 - 13.75^2}{2\times 2.5} \\\\s = 42.18 \ m[/tex]
(c) The time taken for the car to come to a complete stop;
when the car stop's the final velocity, v = 0
v = u - at
0 = 20 - 2.5t
2.5t = 20
[tex]t = \frac{20}{2.5} \\\\t = 8 \ s[/tex]
Thus, the time taken for the car to come to complete stop is 8 s.
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A small mass is released from rest at a very great distance from a larger stationary mass. Draw a graphs best represents the gravitational potential energy U of the system of the two masses as a function of time T.
Answer:
attached below
Explanation:
Gravitational potential
energy = [tex]- \frac{GmM}{r}[/tex] ,
V ∝ [tex]- \frac{1}{r}[/tex]
attached below is a graph that represents the gravitational potential energy U
Gravitational potential energy is the energy held in an entity as a result of its vertical position or length. This gravitational pull of a planet on an object causes the information to be stored.
This is connected with gravitational pull since it takes work to lift anything against the gravity of the Earth.
Freshwater in a raised lake or kept behind a dam demonstrates gravitational force.
According to the formula:
Gravitational potential energy [tex]\bold{=-\frac{GMm}{r^2}}[/tex]
Therefore
[tex]\to \bold{V \propto -\frac{1}{r}}[/tex]
Therefore, the answer is "Option d"
Learn more about the Gravitational potential :
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A person runs up 33 stairs of height 0.25 m. The person weighs 599.85 N. How much energy does it take for the person to reach the top of the stairs? I need the answer ASAP
Answer:
149.96J
Explanation:
Given parameters:
Number of stairs = 33
Height = 0.25m
Weight = 599.85N
Unknown:
Energy the person takes to reach the top = ?
Solution:
To solve this problem;
The energy it takes is given as:
Energy = weight x height
Energy = mass x gravity x height
Now,
Insert the given parameters and solve;
Energy = 599.85 x 0.25 = 149.96J
An energy source forces a constant current of 2A to flow through a light bulbfilament for twenty seconds. If 4.6 kJ is given off in the form of light and heatenergy, calculate the voltage drop across the bulb.
Answer:
The voltage drop across the bulb is 115 V
Explanation:
The voltage drop equation is given by:
[tex]V=\frac{\Delta W}{\Delta q}[/tex]
Where:
ΔW is the total work done (4.6kJ)
Δq is the total charge
We need to use the definition of electric current to find Δq
[tex]I=\frac{\Delta q}{\Delta t}[/tex]
Where:
I is the current (2 A)
Δt is the time (20 s)
[tex]2=\frac{\Delta q}{20}[/tex]
[tex]q=40 C[/tex]
Then, we can put this value of charge in the voltage equation.
[tex]V=\frac{4600}{40}=115 V[/tex]
Therefore, the voltage drop across the bulb is 115 V.
I hope it helps you!
The resistance of a wire depends on its length i and on its cross sectional area A the resistance is
Answer:
The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area
Explanation:
In a particular crash test, an automobile of mass 1577 kg collides with a wall and bounces back off the wall. The x components of the initial and final speeds of the automobile are 17 m/s and 1.5 m/s, respectively. If the collision lasts for 0.18 s, find the magnitude of the impulse due to the collision. Answer in units of kg · m/s.
Answer:
Ft=24,443.5 kgm/s
Explanation:
Step one
Given data
Mass of automobile m=1577kg
Initial Velocity u=17m/s
Final Velocity v=1.5m/s
Time t=0.18s
Step two
From the impulse and momentum equation
Ft=mΔv
Substitute
Ft=1577*(17-1.5)
Ft=1577*15.5
Ft=24,443.5 kgm/s
A cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5 m/s2 for 8.0 seconds?
A. 36 m/s
B. 9 m/s
C. 72 m/s
D. 15 m/s
Answer:
V = 15m/s
Explanation:
Given the following data;
Initial velocity = 3m/s
Time = 8secs
Acceleration = 1.5m/s²
To find the final velocity, we would use the first equation of motion;
V = U + at
Substituting into the equation, we have
V = 3 + 1.5*8
V = 3 + 12
V = 15m/s
An electron traverses a vacuum tube with a length of 2 m in 2 X 10- 4
sec. What is the average speed of the
electron during this time?
Answer:
Average speed = 10,000 m/s
Explanation:
Given the following data;
Distance = 2m
Time = 0.0002secs
To find the average speed;
Average speed = distance/time
Average speed = 2/0.0002
Average speed = 10,000 m/s
Therefore, the average speed of the
electron is 10,000 meters per seconds.
A small lead ball, attached to a 1.10-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is 1.3 m above the ground. The ball travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise
Answer:
1.84 m
Explanation:
For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.
So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,
a = g + rω²
= 9.8 m/s² + 1.10 m × (18.85 rad/s)²
= 9.8 m/s² + 390.85 m/s²
= 400.65 m/s²
Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.
v² = u² + 2a(h₂ - h₁)
So, v² - u² = 2a(h₂ - h₁)
h₂ - h₁ = (v² - u²)/2a
h₂ = h₁ + (v² - u²)/2a
substituting the values of the variables into the equation, we have
h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)
h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²
h₂ = 1.3 m + 0.54 m
h₂ = 1.84 m
A motorcycle moving 18.8 m/s has
57800 J of KE. What is its mass?
Answer:
m = 327.07 kg
Explanation:
Given that,
Kinetic energy of a motorcycle, E = 57800 J
Velocity of the motorcycle, v = 18.8 m/s
We need to find the mass of the motorcycle. The kinetic energy of an object is given by :
[tex]E=\dfrac{1}{2}mv^2[/tex]
m is mass
[tex]m=\dfrac{2E}{v^2}\\\\m=\dfrac{2\times 57800 }{(18.8)^2}\\\\m=327.07\ kg[/tex]
So, the mass of the motorcycle is 327.07 kg.
If you travel from Tucson to Argentina, you will see some different constellations in the night sky. true or false
Answer:
its true!!
Explanation: have a nice day !!
A 22.0 kg child is riding a playground merry-go- round that is rotating at 40.0 rev/min. What centripetal force must
Answer:
F = 482.51 N
Explanation:
Given that,
Mass of a child, m = 22 kg
Angular velocity of the merry-go-round, [tex]\omega=40\ rev/min[/tex]
Let the radius of the path, r = 1.25 m
We need to find the centripetal force acting on the child. The formula for the centripetal force is given by :
[tex]F=m\omega^2r\\\\=22\times (4.18879)^2\times 1.25\\\\=482.51\ N[/tex]
So, the required centripetal force is 482.51 N.
All of the following are ways in which sports psychologists help athletes except __________.
A.
staying motivated
B.
managing fear of failure
C.
improving performance
D.
enhancing memory
Please select the best answer from the choices provided
A
B
C
D
Answer:
D-Enhancing memory
Explanation:
What are the three longest wavelengths for standing waves on a270-cm-long string that is fixed at both ends
Answer:
The answer is below
Explanation:
a) What are the three longest wavelengths for standing waves on a 270-cm-long string that is fixed at both ends? b. If the frequency of the second-largest wavelength is 50.0 Hz, what is the frequency of the third-longest wave length?
Solution:
a) The wavelengths (λ) for standing waves is given by the formula:
[tex]\lambda_m=\frac{2*length\ of\ string}{m}\\\\Where\ m=1,2,3,.\ .\ .\\\\Given\ that\ length\ of\ string = 270\ cm=2.7\ m,\ m=1,2,3(three\ longest\ wavelengths)\\\\Hence:\\\\\lambda_1=\frac{2(2.7)}{1}=5.4\ m\\\\\lambda_2=\frac{2(2.7)}{2}=2.7\ m \\\\\lambda_3=\frac{2(2.7)}{3}=1.8\ m[/tex]
b) The frequency (f) and wavelength (λ) is given by:
fλ = constant
Hence:
[tex]f_2\lambda_2=f_3\lambda_3\\\\f_2=50\ Hz\\\\2.7*50=f_3(1.8)\\\\f_3=\frac{2.7*50}{1.8} \\\\f_3=75\ Hz[/tex]
The three longest wavelengths for the standing waves on a 270-cm long string that is fixed at both ends are:
1. 5.4 meters.
2. 2.7 meters.
3. 1.8 meters.
Given the following data:
Length of string = 270 cm to m = [tex]\frac{270}{100} =2.7\;m[/tex]To determine the three (3) longest wavelengths for these standing waves:
Mathematically, the wavelength for standing waves is given by the formula:
[tex]\lambda_n = \frac{2L}{n}[/tex]
Where:
[tex]\lambda_n[/tex] is the wavelength for standing waves.L is the length of string.Note: n = 1, 2, and 3.
When n = 1:
[tex]\lambda_1 = \frac{2\times 2.7}{1} \\\\\lambda_1 = 5.4 \;meters[/tex]
When n = 2:
[tex]\lambda_2 = \frac{2\times 2.7}{2} \\\\\lambda_2 = 2.7 \;meters[/tex]
When n = 3:
[tex]\lambda_3 = \frac{2\times 2.7}{3} \\\\\lambda_3 =\frac{5.4}{3} \\\\\lambda_3 = 1.8 \;meters[/tex]
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Why could it be argued that the respiratory system is most critical to sustaining life?
Explanation:
Energy is the most important ingredient for life. Organisms use energy in diverse ways. Scientifically, energy is defined as the ability to do work. Without this ability, organisms would not exist.
So, the most important process is one that can furnish the body with energy.
The respiratory system happens to be the one that furnishes the body with energy. During respiration, the energy needs of the body is met by series of processes. Oxygen is taken in and use to liberate calories from chemical substances packed with energy. So, without respiration, the bodily energy demands will not be met.What can we conclude from observing an attractive force between a positively charged rod and some object
Answer: that the object is negatively charged.
Explanation:
We know that the force between objects that have the same type of charge is a repulsive force, while for objects with an opposite charge, the force is attractive.
In this case, we know that we have an attractive force between an object and a positively charged rod.
Then the only conclusion we can take in this situation is that the object is negatively charged.
If a man traveled to a different planet
Answer:
He would be in space.
Explanation:
Answer: What would happen if a human traveled to Uranus?
As an ice giant, Uranus doesn't have a true surface. The planet is mostly swirling fluids. While a spacecraft would have nowhere to land on Uranus, it wouldn't be able to fly through its atmosphere unscathed either. The extreme pressures and temperatures would destroy a metal spacecraft.
Explanation:
An object with velocity 141 ft/s has a kinetic energy of 1558.71 ft∙lbf, on a planet whose gravity is 31.5 ft/s2. What is its mass in pounds
Answer:
The mass of the object is 5.045 lbm.
Explanation:
Given;
kinetic energy of the object, K.E = 1558.71 ft.lbf
velocity of the object, V = 141 ft/s
The kinetic energy of the object is calculated as;
[tex]K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \ \times \ \ \ \ 1 \ lbf\ }[/tex]
[tex]m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm[/tex]
Therefore, the mass of the object is 5.045 lbm.
Allen and Jason are chucking a speaker around. On one particular throw, Allen throws the speaker, which is playing a pure tone of frequency f, at a speed of 10 m/s directly towards Jason, but his aim is a bit off. As a result, Jason runs forward towards the speaker at a speed of 6 m/s before catching it. Then, the frequency that Jason hears while running can be written as (m/n)f Hz, where m and n are relatively prime positive integers. Compute m n.
Answer:
Explanation:
We shall apply Doppler's effect of sound .
speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at 6 m/s .
apparent frequency = [tex]f_o\times\frac{V+v_o}{ V-v_s}[/tex]
V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .
Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f
apparent frequency = [tex]f\times \frac{340+6}{340-10}[/tex]
= [tex]f\times \frac{346}{330}[/tex]
So m = 346 , n = 330 .
An inductor is connected to a 120-V, 60-Hz supply. The current in the circuit is 2.4 A. What is the inductive reactance
Answer:
Inductive reactance is 50.00 ohms
Explanation:
Given the following data;
Voltage = 120v
Frequency = 60Hz
Current = 2.4 A
To find the inductive reactance;
Inductive reactance, XL = V/I
Where;
XL represents the inductive reactance. V represents the voltage. I represents the current.Substituting into the equation, we have;
XL = 120/2.4
XL = 50.00 ohms
Scientists create models to better understand Earth. Which evidence has led scientists to conclude that there are different layers within Earth's interior?
A.analysis of seismic wave data
B.measurement of Earth's diameter
C.temperatures taken within each layer
D.rock samples taken from Earth's core
Answer:
it is A or D
Explanation:
Answer:
ANswer:A
Explanation:
A finch rides on the back of a Galapagos tortoise, which walks at the stately pace of 0.060 m>s. After 1.5 minutes the finch tires of
Complete Question:
A finch rides on the back of a Galapagos tortoise, which walks
at the stately pace of 0.060 m/s. After 1.5 minutes the finch tires of
the tortoise’s slow pace, and takes flight in the same direction for
another 1.5 minutes at 11 m/s.
What was the average speed of the finch for this 3.0-minute interval?
Answer:
[tex]Speed = 5.53 m/s[/tex]
Explanation:
Distance is calculated as:
[tex]Distance = Speed * Time[/tex]
First, we calculate the distance for the first 1.5 minutes
For the first 1.5 minutes, we have:
[tex]Speed = 0.060m/s[/tex]
[tex]Time = 1.5\ mins[/tex]
[tex]D_2= 0.060m/s * 1.5\ mins[/tex]
Convert 1.5 mins to seconds
[tex]D_2= 0.060m/s * 1.5 * 60s[/tex]
[tex]D_2= 5.4m[/tex]
Next, we calculate the distance for the next 1.5 minutes
[tex]Speed = 11m/s[/tex]
[tex]Time = 1.5\ mins[/tex]
[tex]D_2= 11m/s * 1.5\ mins[/tex]
Convert 1.5 mins to seconds
[tex]D_2 = 11m/s * 1.5 * 60s[/tex]
[tex]D_2= 990m[/tex]
Total distance is:
[tex]Distance = 990m + 5.4m[/tex]
[tex]Distance = 995.4m[/tex]
The average speed for the 3.0 minute interval is:
[tex]Speed = \frac{Distance}{Time}[/tex]
[tex]Speed = \frac{995.4\ m}{3.0\ mins}[/tex]
Convert 3.0 minutes to seconds
[tex]Speed = \frac{995.4\ m}{3.0 * 60 secs}[/tex]
[tex]Speed = \frac{995.4\ m}{180 secs}[/tex]
[tex]Speed = 5.53 m/s[/tex]
A hammer strikes a nail with a 10 N force for 0.01 seconds. Calculate the impulse of the hammer.
Answer:
0.1Ns
Explanation:
Impulse is the product of Force and time
Impulse = Force * Time
Given
Force = 10N
Time = 0.01s
Substitute into the formula
Impulse = 10 * 0.01
Impulse = 10 * 1/100
Impulse = 10/100
Impulse = 0.1Ns
hence the impulse of the hammer is 0.1Ns
A block weighing 9.3 N requires a force of 3.7 N to push it along at constant velocity. What is the coefficient of friction for the surface
Answer:
0.398
Explanation:
According to friction, the frictional force is directly proportional to the normal reaction
Ff = nR
Ff is the frictional force
n is the coefficient of friction
R is the reaction
Reaction is equal to the weight
R= W = 9.3N
Fm = Ff = 3.7N
Fm is the moving force
Get the coefficient of friction
n = Ff/R
n = 3.7/9.3
n = 0.398
Hence the coefficient of friction for the surface is 0.398
A 9.0 kg test rocket is fired vertically from Cape Canaveral. Its fuel gives it a kinetic energy of 1905 J by the time the rocket engine burns all of the fuel. What additional height will the rocket rise
Answer:
21.6m
Explanation:
Since the rocket engine burns all the fuel hence the kinetic energy will be converted to potential energy
Potential Energy = mass × acceleration due to gravity × height
Given
PE = 1905J
Mass = 9.0kg
Acceleration due to gravity =9.8m/s²
Required
Height h
Substitute into the formula
1905 = 9(9.8)h
1905 = 88.2h
h =1905/88.3
h = 21.6m
Hence the required height is 21.6m
Which is greater, the energy of one photon of orange light or the energy of one quantum ofradiation having a wavelength of 3.36 * 10^-9
The question is incomplete, here is the complete question:
Which is greater, the energy of one photon of orange light or the energy of one quantum of radiation having a wavelength of [tex]3.36\times 10^{-9}m[/tex]
Answer: The energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.
Explanation:
To calculate the energy of one photon, we use the Planck's equation:
[tex]E=\frac{N_Ahc}{\lambda}[/tex]
where,
E = energy of radiation
[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}mol^{-1}[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8 m/s[/tex]
[tex]\lambda}[/tex] = wavelength of radiation
For orange light:For 1 photon, the term [tex]N_A[/tex] does not appear
[tex]\lambda}[/tex] = 620 nm = [tex]620\times 10^{-9}m[/tex] (Conversion factor: [tex]1nm=10^{-9}m[/tex] )
Putting values in above equation, we get:
[tex]E=\frac{6.626\times 10^{-34}Js\times 3\times 10^8m/s}{620\times 10^{-9}m}\\\\E=3.206\times 10^{-19}J[/tex]
For one quantum of radiation:[tex]\lambda}[/tex] = [tex]3.36\times 10^{-9}m[/tex]
Putting values in above equation, we get:
[tex]E=\frac{6.022\times 10^{23}mol^{-1}\times 6.626\times 10^{-34}Js\times 3\times 10^8m/s}{3.36\times 10^{-9}m}\\\\E=3.56\times 10^{7}J/mol[/tex]
Hence, the energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.
The human nervous system can propagate nerve impulses at about 102 m>s. Estimate the time it takes for a nerve impulse to travel 2 m from your toes to your brain.
Answer:
t = 0.196 s
Explanation:
The speed of a pulse is determined by the characteristics of the medium, its density and its resistance to stress, as long as these remain the speed will be constant for which we can use the kinetic expressions of the uniform movement
v = x / t
t = x / v
calculate
t = 2/102
t = 0.196 s
Water runs out of a horizontal drainpipe at the rate of 135 kg/min. It falls 3.1 m to the ground. Assuming the water doesn't splash up, what average force does the water exert on the ground
Answer:
The average force exerted by the water on the ground is 17.53 N.
Explanation:
Given;
mass flow rate of the water, m' = 135 kg/min
height of fall of the water, h = 3.1 m
the time taken for the water to fall to the ground;
[tex]h = ut + \frac{1}{2}gt^2\\\\h = 0 + \frac{1}{2}gt^2\\\\t = \sqrt{\frac{2\times 3.1}{9.8} } \\\\t = 0.795 \ s[/tex]
mass of the water;
[tex]m = m't\\\\m = 135 \ \frac{kg}{min} \ \times \ 0.795 \ s \ \times \ \frac{1 \ \min}{60 \ s} \ = 1.789 \ kg[/tex]
the average force exerted by the water on the ground;
F = mg
F = 1.789 x 9.8
F = 17.53 N
Therefore, the average force exerted by the water on the ground is 17.53 N.