Answer:
Object could only be moving with increasing speed.
Explanation:
Let us consider the general formula of acceleration:
a = (Vf - Vi)/t
Vf = Vi + at -------- equation 1
where,
Vf = Final Velocity
Vi = Initial Velocity
a = acceleration
t = time
FOR POSITIVE ACCELERATION:
Vf = Vi + at
since, both acceleration and time are positive quantities. Hence, it means that the final velocity of the object shall be greater than the initial velocity of the object.
Vf > Vi
It clearly shows that if an object moves with positive acceleration. It could only be moving with increasing speed.
Solving the same equation for negative acceleration shows that the final velocity will be less than initial velocity and object will be moving with decreasing speed.
And for the constant velocity final and initial velocities are equal and thus, acceleration will be zero.
Assume the angular momentum of a diatomic molecule is quantized according to the relation . What are the allowed rotational kinetic energies
Answer:
The answer to this question can be defined as follows:
Explanation:
In the given question, an equation is missing which can be defined as follows:
[tex]I \omega =\sqrt{J(J+1)}h[/tex]
solution:
Angular momentum:
[tex]L=I \omega =\sqrt{J(J+1)}h[/tex]
Convert Angular momentum in terms of kinetic energy:
[tex]K = \frac{L^2}{2I}[/tex]
[tex]= \frac{h^2(J(J+1))}{2I}[/tex]
A boat develops a leak and, after its passengers are rescued, eventually sinks to the bottom of a lake. When the boat is at the bottom, what is the force of the lake bottom on the boat?
Answer:
Explanation:
The force of the lake bottom on the boat = force the boat exerts on the bottom
force the boat exerts on the bottom = weight of the boat - buoyant force on the boat
so
The force of the lake bottom on the boat = weight of the boat - buoyant force on the boat
So
The force of the lake bottom on the boat will be less than its weight .
5. Sandor fills a bucket with water and whirls it in a vertical circle to demonstrate that the
water will not spill from the bucket at the top of the loop. If the length of the rope from his
hand to the centre of the bucket is 1.24 m, what is the minimum tension in the rope (at the
top of the swing)? How slow can he swing the bucket? Explain your answer.
Given that,
radius = 1.24 m
According to question,
The rope cannot push outwards. It must always have some slight tension or the bucket will fall.
We need to calculate the tension in the rope
At the top the force of gravity is
[tex]F=mg[/tex]
The force needed to move the bucket in a circle is centripetal force.
So, if mg is ever greater than centripetal force then the bucket and the contents will start to fall.
The rope have a tension of less than zero.
We need to calculate the velocity of swing bucket
Using centripetal force
[tex]F=\dfrac{mv^2}{r}[/tex]
[tex]mg=\dfrac{mv^2}{r}[/tex]
[tex]g=\dfrac{v^2}{r}[/tex]
[tex]v^2=gr[/tex]
[tex]v=\sqrt{gr}[/tex]
Put the value into the formula
[tex]v=\sqrt{9.8\times1.24}[/tex]
[tex]v=3.49\ m/s[/tex]
Hence, The minimum tension in the rope is less than zero .
The bucket swings with the velocity of 3.49 m/s.
Which kind of image can never be projected and forms where light rays appear to originate?
virtual
real
inverted
enlarned
Answer:
a) virtual
Explanation:
This is a lens problem, in this case the equation that describes the process is the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, weights the distance to the object and qq the distance to the image
In a projection system the image that we see is q, which can have several characteristics
a) virtual It can never be projected, since this image is formed by the extensions of the light rays, therefore it is not real but a construction of the brain that interprets where the rays must come from.
b) Real. This image can be projected since light rays pass through the image
c) Inverted. Inverted images are real so they can be projected, so rays pass through the image
d) expanded. In this case the image is greater than the object, this occurs when the object the distance to the image is greater than the distance to the object, therefore the distance q is negative, therefore this image is straight and is formed by the extensions from the rays and you can't project
Answer:
option A) virtual
Explanation:
hope this helps :)
When moving to a new apartment, you rent a truck and create a ramp with a 244 cm long piece of plywood. The top of the moving ramp lies on the edge of the truck bed at a height of 115 cm. You load your textbooks into a wooden box at the bottom of the ramp (the coefficient of kinetic friction between the box and ramp is = 0.2). Then you and a few friends give the box a quick push and it starts to slide up the ramp. A) What angle is made by the ramp and the ground?B) Unfortunately, after letting go, the box only tables 80cm up the ramp before it starts coming back down! What speed was the box initially traveling with just after you stopped pushing it?
Answer:
A) θ = 28.1º , B) v = 2.47 m / s
Explanation:
A) The angle of the ramp can be found using trigonometry
sin θ = y / L
Φ = sin⁻¹ y / L
θ = sin⁻¹ (115/244)
θ = 28.1º
B) For this pate we can use the relationship between work and kinetic energy
W =ΔK
where the work is
W = -fr x
the negative sign is due to the fact that the friction force closes against the movement
Lavariacion of energy cineta is
ΔEm = ½ m v² - mgh
-fr x = ½ m v² - m gh
the friction force has the equation
fr = very N
at the highest part there is no speed and we take the origin from the lowest part of the ramp
To find the friction force we use Newton's second law. Where one axis is parallel to the ramp and the other is perpendicular
Axis y . perpendicular
N- Wy = 0
cos tea = Wy / W
Wy = W cos treaa
N = mg cos tea
we substitute
- (very mg cos tea) x = ½ m v²2 - mgh
v2 = m (gh- very g cos tea x)
let's calculate
v = Ra (9.8 0.80 - 0.2 9.8 0.0 cos 28.1)
v = RA (7.84 -1.729)
v = 2.47 m / s
In a ballistic pendulum experiment, a small marble is fired into a cup attached to the end of a pendulum. If the mass of the marble is 0.0215 kg and the mass of the pendulum is 0.250 kg, how high h will the pendulum swing if the marble has an initial speed of 5.15 m/s? Assume that the mass of the pendulum is concentrated at its end so that linear momentum is conserved during this collision.
Answer:
h = 8.48*10^-3m
Explanation:
In order to calculate the height reached by the pendulum with the marble, you first take into account the momentum conservation law, to calculate the speed of both pendulum and marble just after the collision.
The total momentum of the system before the collision is equal to the total momentum after:
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex] (1)
Here you used the fact that the pendulum has its total mass concentrated at the end of the pendulum.
m1: mass of the marble = 0.0215kg
m2: mass of the pendulum concentrated at its end = 0.250kg
v1: horizontal speed of the arble before the collision = 5.15m/s
v2: horizontal speed of the pendulum before the collision = 0m/s
v: horizontal speed of both marble and pendulum after the collision = ?
You solve the equation (1) for v, and replace the values of the other parameters:
[tex]v=\frac{m_1v_1+m_2v_2}{m_1+m_2}\\\\v=\frac{(0.0215kg)(5.15m/s)+(0.250kg)(0m/s)}{0.0215kg+0.250kg}=0.40\frac{m}{s}[/tex]
Next, you use the energy conservation law. In this case the kinetic energy of both marble and pendulum (just after the collision) is equal to the potential energy of the system when both marble and pendulum reache a height h:
[tex]U=K\\\\(m_1+m_2)gh=\frac{1}{2}(m_1+m_2)v^2\\\\h=\frac{v^2}{2g}[/tex]
v = 0.40m/s
g: gravitational acceleration = 9,8m/s^2
[tex]h=\frac{(0.40m/s)^2}{2(9.8m/s^2)}=8.48*10^{-3}m[/tex]
Then, the height reached by marble and pendulum is 8.48*10^-3m
gg The sound source of a ship’s sonar system operates at a frequency of 22.0 kHzkHz . The speed of sound in water (assumed to be at a uniform 20∘C∘C) is 1482 m/sm/s . What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling straight toward the ship at 4.95 m/sm/s ? Assume that the ship is at rest in the water.
Answer:
Δf = 73.72Hz
Explanation:
In order to calculate the difference in frequency between the direct waves and the reflected waves, you first take into account the Doppler's effect for an observer getting closer to the source:
[tex]f'=f(\frac{v+v_o}{v-v_s})[/tex] (1)
You can assume that the reflected waves come from a source "the whale". Then you have:
f': frequency of the reflected waves = ?
f: frequency of the source = 22.0*kHz = 22.0*10^3 Hz
v: speed of the sound in water = 1482m/s
vs: speed of the source = 4.95m/s
vo: speed of the observer = 0m/s
You replace the values of the parameters in the equation (1):
[tex]f'=(22.0*10^3Hz)(\frac{1482m/s}{1482m/s-4.95m/s})=22073.72Hz[/tex]
Then, the difference in frequency is:
[tex]\Delta f = f'-f=22000Hz-22073.72Hz=73.72Hz[/tex]
A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.60 cm distant from the first, in a time interval of 3.20×10−6s3.20×10 −6 s. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.
Answer:
E = 326.17 N/C
Explanation:
(a) In order to calculate the magnitude of the electric field between the parallel plates you first calculate the acceleration of the proton. You use the following formula:
[tex]x=v_ot+\frac{1}{2}at^2[/tex] (1)
vo: initial speed of the proton = 0m/s
t: time that the proton takes to cross the space between the plates = 3.20*10^-6 s
a: acceleration of the proton = ?
x: distance traveled by the proton = 1.60cm = 0.016m
You solve the equation (1) for a, and replace the values of all parameters:
[tex]a=\frac{2x}{t^2}=\frac{2(0.016m)}{(3.20*10^{-6}s)^2}=3.125*10^{10}\frac{m}{s^2}[/tex]
Next, you use the Newton second law for the electric force, to find the magnitude of the electric field:
[tex]F_e=qE=ma[/tex] (2)
q: charge of the proton = 1.6*10^-19C
m: mass of the proton = 1.77*10^-27kg
You solve the equation (2) for E:
[tex]E=\frac{ma}{q}=\frac{(1.67*10^{-27}kg)(3.125*10^{10}m/s^2)}{1.6*10^{-19}C}\\\\E=326.17\frac{N}{C}[/tex]
The magnitude of the electric field in between the parallel plates is 326.17N/C
For the last part of the lab, you should have found the mass of the meter stick. So if a mass of 85 g was placed at the 2 cm MARK and the pivot point was moved to the 38.6 cm MARK, what would have been the mass of the meter stick
Answer:
272.89g
Explanation:
Find the diagram to the question in the attachment below;.
Using the principle of moment to solve the question which states that the sum of clockwise moment is equal to the sum of anticlockwise moment.
Moment = Force * Perpendicular distance
Taking the moment of force about the pivot.
Anticlockwise moment:
The 85g mass will move in the anticlockwise
Moment of 85g mass = 85×36.6
= 3111gcm
Clockwise moment.
The mass of the metre stick M situated at the centre (50cm from each end) will move in the clockwise direction towards the pivot.
CW moment = 11.4×M = 11.4M
Equating CW moment to the ACW moment we will have;
11.4M = 3111
M = 3111/11.4
M = 272.89g
The mass of the metre stick is 272.89g
When an old LP turntable was revolving at 3313 rpm, it was shut off and uniformly slowed down and stopped in 5.5 seconds. What was the magnitude of its angular acceleration (in rad/s2) as it slowed down?
Answer:
-0.63 rad/s²
Explanation:
Given that
Initial angular velocity of the turntable, w(i) = 33 1/3 rpm
Final angular velocity of the turntable, w(f) = 0 rpm
Time taken to slow down, t = 5.5 s
The calculation is attached in the photo below
what is thermodynamic?
Answer:
Thermodynamics is a branch of physics which deals with the energy and work of a system. It was born in the 19th century as scientists were first discovering how to build and operate steam engines. Thermodynamics deals only with the large scale response of a system which we can observe and measure in experiment.
Answer:
thermodynamics is the branch of physics which deals with the study of heat and other forms energy and their mutual relationship(relation ship between them)
Explanation:
i hope this will help you :)
A ballast is dropped from a stationary hot-air balloon that is at an altitude of 576 ft. Find (a) an expression for the altitude of the ballast after t seconds, (b) the time when it strikes the ground, and (c) its velocity when it strikes the ground. (Disregard air resistance and take ft/sec2.)
Answer:
a) [tex]S = \frac{1}{2}gt^2\\[/tex]b) 6secsc) 192ftExplanation:
If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;
[tex]S = ut + \frac{1}{2}at^{2}[/tex]
S is the altitude of the ballest
u is the initial velocity
a is the acceleration of the body
t is the time taken to strike the ground
Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s
Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g
Substituting this values into the equation of the motion;
[tex]S = 0 + \frac{1}{2}gt^2\\ S = \frac{1}{2}gt^2\\[/tex]
a) An expression for the altitude of the ballast after t seconds is therefore
[tex]S = \frac{1}{2}gt^2\\[/tex]
b) Given S = 576ft and g = 32ft/s², substituting this into the formula in (a);
[tex]576 = \frac{1}{2}(32)t^2\\\\\\576*2 = 32t^2\\1152 = 32t^2\\t^2 = \frac{1152}{32} \\t^2 = 36\\t = \sqrt{36}\\ t = 6.0secs[/tex]
This means that the ballast strikes the ground after 6secs
c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.
v = 0 + 32(6)
v = 192ft
C.
(11) in parallel
A potentiometer circuit consists of a
battery of e.m.f. 5 V and internal
resistance 1.0 12 connected in series with a
3.0 12 resistor and a potentiometer wire
AB of length 1.0 m and resistance 2.0 12.
Calculate:
(i) The total resistance of the circuit
The current flowing in the circuit
(iii) The lost volt from the internal
resistance of battery across the
battery terminals
(iv) The p.d. across the wire AB
(v) The e.m.f. of a dry cell which can be
balanced across 60 cm of the wire
AB.
Assume the wire has a uniform cross-
sectional area.
Answer:
fggdfddvdghyhhhhggghh
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Complete question:
A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Answer:
The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.
Explanation:
Given;
magnitude of applied force, F = 1.5 N
Apply Newton's second law of motion;
F = ma
[tex]F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]
The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;
[tex]\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N[/tex]
Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.
An ac circuit consist of a pure resistance of 10ohms is connected across an ae supply
230V 50Hz Calculate the:
(i)Current flowing in the circuit.
(ii)Power dissipated
Plz check attachment for answer.
Hope it's helpful
what tools use cut wood
Answer:
hand saws
power saws
Circular Saw
Explanation:
that is all that i know
A particle of charge = 50 µC moves in a region where the only force on it is an electric force. As the particle moves 25 cm, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference acting on the partice
Answer:
nvbnncbmkghbbbvvvvvvbvbhgggghhhhb
What is the relationship between the magnitudes of the collision forces of two vehicles, if one of them travels at a higher speed?
Explanation:
The collision forces are equal and opposite. Therefore, the magnitudes are equal.
5) what is the weight of a
if its weight
is 5N in moon?
body in the earth,
Answer:
Weight of object on moon is 5N ,as we know. Weight of object on moon is 1/3 the of object on earth,so
let weight of object on earth = X
5N= X/3
X = 3×5 = 15N
Hence the weight of the object on earth will
be 15N
The resonance tube used in this experiment produced only one resonance tone. What length of tube would be required to produce a second tone under the same experimental conditions? Explain your answer.
Answer:
the length that would produce a sound tone under the same experimental contditions must be increased by Δl = [tex]\frac{v}{2f}[/tex]
Explanation:
Recall
V = f ×λ
where λ is ⁴/₃l₂ for second resonance
f = [tex]\frac{3v}{4l_{2} }[/tex]
l₂ = [tex]\frac{3v}{4f}[/tex]
where λ is 4l₁ for 1st resonance
f = [tex]\frac{v}{4l_{1} }[/tex]
l₁ = [tex]\frac{v}{4f}[/tex]
∴ Δl = l₂ - l₁ = [tex]\frac{3v}{4f}[/tex] ⁻ [tex]\frac{v}{4f}[/tex]
Δl= [tex]\frac{2v}{4f}[/tex]
Δl = [tex]\frac{v}{2f}[/tex]
Therefore, the length should increase by [tex]\frac{v}{2f}[/tex]
In an experiment to measure the acceleration due to gravity g, two independent equally reliable measurements gave 9.67 m/s2 and 9.88 m/s2. Determine (i) the percent difference of the measurements (ii) the percent error of their mean. [Take the theoretical value of g to be 9.81 m/s2]
Answer and Explanation:
a. The computation of the percent difference between the measurements is shown below:-
The first value of g is 9.67 and the second value is 9.88
So, difference = 9.88 - 9.67
= 0.21
Percentage difference in measurement is
[tex]= \frac{0.21}{9.88}\times100[/tex]
= +/-2.13
Percent difference with 9.88
Difference = 9.88 - 9.81
= 0.07
[tex]= \frac{0.07}{9.81}\times100[/tex]
= +/-0.71%
b. The Computation of percent error of their mean is shown below:-
Mean of two values is
= [tex]\frac{9.67 + 9.88}{2}[/tex]
= 9.775
Difference = 9.81 - 9.775
= 0.035
Percentage difference is
[tex]= \frac{0.035}{9.81}\times 100[/tex]
= +/- 0.36%
Which of the following biotic organisms makes its own energy from inorganic substances?
producers
consumers
decomposers
minerals
Answer:
producers make its own energy frominorganic substances.
A uniform disk, a uniform hoop, and a uniform solid sphere are released at the same time at the top of an inclined ramp. They all roll without slipping. In what order do they reach the bottom of the ramp
Answer:
sphere, disk, hoop
Explanation:
See attached file
A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that is 0.60 m from the center is closest to
Complete Question
The complete question is shown on the first uploaded image
Answer:
The electric field at that point is [tex]E = 7500 \ N/C[/tex]
Explanation:
From the question we are told that
The radius of the inner circle is [tex]r_i = 0.80 \ m[/tex]
The radius of the outer circle is [tex]r_o = 1.20 \ m[/tex]
The charge on the spherical shell [tex]q_n = -500nC = -500*10^{-9} \ C[/tex]
The magnitude of the point charge at the center is [tex]q_c = + 300 nC = + 300 * 10^{-9} \ C[/tex]
The position we are considering is x = 0.60 m from the center
Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as
[tex]E = \frac{k * q_c }{x^2}[/tex]
substituting values
[tex]E = \frac{k * q_c }{x^2}[/tex]
where k is the coulomb constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
substituting values
[tex]E = \frac{9*10^9 * 300 *10^{-9}}{0.6^2}[/tex]
[tex]E = 7500 \ N/C[/tex]
In a fixed-target experiment positrons are fired at a target of electrons at rest. What positron energy is required to produce a Z (mZ = 91.188 GeV)?
Answer:
Explanation:
electron rest mass = 0.511MeV/C²
postion rest mass = 0.511MeV/C²
boson rest mass = 91.188GeV/C²
= 91188 MeV/C²
A passenger jet flies from one airport to another 1,233 miles away in 2.4 h. Find its average speed. = ____ m/s
Speed = (distance) / (time)
Speed = (1,233 mile) / (2.4 hour)
Speed = 513.75 mile/hour
Speed = (513.75 mi/hr) x (1609.344 meter/mi) x (1 hr / 3600 sec)
Speed = (513.75 x 1609.344 / 3600) (mile-meter-hour/hour-mile-second)
Speed = 229.7 meter/second
a 1000 kg car accelerates from 0 to 15 m/s in 5.0 s with negligible friction and air resistance. what is the average power delivered by the engine
Answer:
30.16 hp
Explanation:
Given :
m= 1000 kg
v=15 m/s
t=5.0 s
The average power delivered by the engine can be determined by using the given formula
[tex]Average\ power\ =\ \frac{0.5*m*v^2\ }{t}[/tex]
where m=,mass, v=velocity and t=time
Now putting the value of m,v and t in the previous equation we get
[tex]Average \ power\ =\ \frac{0.5*1000*15*15}{5} \\Average \ power\ =\ 22,500\ w\\Average \ power\ =\frac{22,500}{746} \\Average \ power\ =30.16\ hp[/tex]
To convert w to hp we divide by 746
Therefore 30.16 hp is the answer.
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 12.9 rad/s in 2.98 s.
(a) Find the magnitude of the angular acceleration of the wheel.
(b) Find the angle in radians through which it rotates in this time interval.
Explanation:
(a) Find the magnitude of the angular acceleration of the wheel.
angular acceleration = angular speed /timeangular acceleration = 12.9/2.98 = 4.329rad/s²(b) Find the angle in radians through which it rotates in this time interval.
angular speed = 2x3.14xf12.9rad = 2 x3.14rad = 6.28/12.9rad = 0.487Now we convert rad to angle
1 rad = 57.296°0.487 = unknown angleunknown angle =57.296 x 0.487 = 27.9°The angle in radians = 27.9°
You illuminate a slit with a width of 77.7 μm with a light of wavelength 721 nm and observe the resulting diffraction pattern on a screen that is situated 2.83 m from the slit. What is the width, in centimeters, of the pattern's central maximum
Answer:
The width is [tex]Z = 0.0424 \ m[/tex]
Explanation:
From the question we are told that
The width of the slit is [tex]d = 77.7 \mu m = 77.7 *10^{-6} \ m[/tex]
The wavelength of the light is [tex]\lambda = 721 \ nm[/tex]
The position of the screen is [tex]D = 2.83 \ m[/tex]
Generally angle at which the first minimum of the interference pattern the light occurs is mathematically represented as
[tex]\theta = sin ^{-1}[\frac{m \lambda}{d} ][/tex]
Where m which is the order of the interference is 1
substituting values
[tex]\theta = sin ^{-1}[\frac{1 *721*10^{-9}}{ 77.7*10^{-6}} ][/tex]
[tex]\theta = 0.5317 ^o[/tex]
Now the width of first minimum of the interference pattern is mathematically evaluated as
[tex]Y = D sin \theta[/tex]
substituting values
[tex]Y = 2.283 * sin (0.5317)[/tex]
[tex]Y = 0.02 12 \ m[/tex]
Now the width of the pattern's central maximum is mathematically evaluated as
[tex]Z = 2 * Y[/tex]
substituting values
[tex]Z = 2 * 0.0212[/tex]
[tex]Z = 0.0424 \ m[/tex]
The Huka Falls on the Waikato River is one of New Zealand's most visited natural tourist attractions. On average, the river has a flow rate of about 300,000 L/s. At the gorge, the river narrows to 20-m wide and averages 20-m deep.
(a) What is the average speed of the river in the gorge?
(b) What is the average speed of the water in the river downstream of the falls when it widens to 60 m and its depth increases to an average of 40 m?
Answer:
(a) V = 0.75 m/s
(b) V = 0.125 m/s
Explanation:
The speed of the flow of the river can be given by following formula:
V = Q/A
V = Q/w d
where,
V = Speed of Flow of River
Q = Volume Flow Rate of River
w = width of river
d = depth of river
A = Area of Cross-Section of River = w d
(a)
Here,
Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s
w = 20 m
d = 20 m
Therefore,
V = (300 m³/s)/(20 m)(20 m)
V = 0.75 m/s
(b)
Here,
Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s
w = 60 m
d = 40 m
Therefore,
V = (300 m³/s)/(60 m)(40 m)
V = 0.125 m/s