Answer:
If an object is moving with a constant velocity, then by definition it has zero acceleration. So there is no net force acting on the object. The total work done on the object is thus 0 (that's not to say that there isn't work done by individual forces on the object, but the sum is 0 ).
Explanation:
In the middle, when the object was changing position at a constant velocity, the acceleration was 0. This is because the object is no longer changing its velocity and is moving at a constant rate.
1. A 63 kg driver gets into an empty taptap to start the day's work. The springs compress 1.5x10-2 m. What is the effective spring constant of the spring system in the taptap?
2. After driving a portion of the route, the taptap is fully loaded with a total of 24 people including the driver, with an average mass of 68 kg per person. In addition, there are three 15-kg goats, five 3-kg chickens, and a total of 25 kg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much are the springs compressed?
(1) When the driver is at rest, the restoring force exerted by spring is equal in magnitude to the driver's weight, so that
∑ F = s - mg = 0 ==> s = mg = 617.4 N
If the spring is compressed 0.015 m, then the spring constant k is such that
617.4 N = k (0.015 m) ==> k = 41,160 N/m ≈ 41 kN/m
(2) The total mass of the passengers is
24 (68 kg) + 3 (15 kg) + 5 (3 kg) + 25 kg = 1717 kg
so that if everyone is at rest, the spring is compressed a distance x such that
kx = (1717 kg) g ==> x ≈ 0.41 m
A cylinder that is 18 cm tall is filled with water. If a hole is made in the side of the cylinder, 5.0 cm below the top level. Assume that the cylinder is large enough so that the level of the water in the cylinder does not drop significantly. How far will the stream land from the base of the cylinder?
Answer:
The distance is 22.45 cm.
Explanation:
Height of cylinder, H = 14 cm
depth of hole, h = 5 cm
The distance of landing of stream from the base of cylinder is
[tex]R = 2\sqrt{H(H-h)}\\\\R = 2\sqrt{14(14-5)}\\\\R = 22.45 cm[/tex]
If you dive underwater, you notice an uncomfortable pressure on your eardrums due to the increased pressure. The human eardrum has an area of about 70 mm217 * 10-5 m22, and it can sustain a force of about 7 N without rupturing. If your body had no means of balancing the extra pressure (which, in reality, it does), what would be the maximum depth you could dive without rupturing your eardrum
Answer:
[tex]h=10m[/tex]
Explanation:
From the question we are told that:
Area [tex]a=70 x 10^{-6}[/tex]
Force [tex]F=7N[/tex]
Generally the equation for Pressure is mathematically given by
Pressure = Force/Area
[tex]P=\frac{F}{A}[/tex]
[tex]P=\frac{ 7}{(70 * 10^{-6})}[/tex]
[tex]P= 1*10^{5} Pa[/tex]
Generally the equation for Pressure is also mathematically given by
[tex]P=hpg[/tex]
Therefore
[tex]h=\frac{P}{hg}[/tex]
[tex]h=\frac{10000}{1000*9.8}[/tex]
[tex]h=10m[/tex]
In a double-slit experiment, the slit separation is 1.75 mm, and two coherent wavelengths of light, 425 nm and 510 nm, illuminate the slits. At what angle from the centerline on either side of the central maximum will a bright fringe from one pattern first coincide with a bright fringe from the other pattern
Answer:
the required angle is 0.0834879⁰
Explanation:
Given the data in the question;
slit separation; d = 1.75 mm = 1.75 × 10⁻³ m
wavelength λ₁ = 425 nm = 425 × 10⁻⁹ m
wavelength λ₂ 510 nm = 510 × 10⁻⁹ m
Now, we know that, the angle at which a particular bright fringe occurs on either side of the central bright fringe will be;
tanθ = [tex]y_m[/tex] / D = mλ/d
since they both coincides;
tanθ₁ = tanθ₂
m₁λ₁/d = m₂λ₂/d
multiply both sides by d
so,
m₁/m₂ = λ₂/λ₁
we substitute
m₁/m₂ = 510 nm / 425 nm
m₁/m₂ = 510 nm / 425 nm
divide through by 85
m₁/m₂ = 6 / 5
hence m₁ and m₂ are 6 and 5
so, from the previous formula
tanθ₂ = m₂λ₂/d
we substitute
tanθ₂ = [ 5 × ( 510 × 10⁻⁹ m ) ] / 1.75 × 10⁻³ m
tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m
tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m
tanθ₂ = 0.00145714
θ₂ = tan⁻¹( 0.00145714 )
θ₂ = 0.0834879⁰
Therefore, the required angle is 0.0834879⁰
what aspect of the US justice system has its roots in Jewish scripture?
The aspect of the US justice system that has its roots in Jewish scripture is:
the idea that all people are subject to the same rules and laws.
It is the doctrine of "equality before the law." Equality before the law means that every individual is equal in the eyes of the law, whether the individual is a lawmaker, a judge, a law enforcement officer, etc. Equality before the law is also known as equality under the law, equality in the eyes of the law, legal equality, or legal egalitarianism. It is a legal principle that treats each independent being equally and subjects each to the same laws of justice and due process.
Answer:
answer is C
the idea that all people are subject to the same rules and laws
Explanation:
hope this helps!
Convert the unit of 0.00023 kilograms into grams. (Answer in scientific notation)
Answer:
2.3 × [tex]10^{-1}[/tex]
Explanation:
1 kg = 1000 g.
0.00023 kg x 1000 g = 0.23 grams
Answer:
0.23×10⁴
Explanation:
kilogram to gram ÷ 1000
0.00023kg ÷ 1000
=0.23g
scientific notation=0.23×10⁴
The coefficients of friction between a race cars tyres and the track surface are
the question is about tyres of a race car, which are made of rubber and will be in contact with a race track, which is generally made from asphalt, the static coefficient of friction is in the range of (0.5–0.8), in dry conditions (Source: Friction and Friction Coefficients ).
Explanation:
please mark me as a brainlieast
train starts from rest and accelerates at 1m/ s²
for 10 seconds how far does it move
Answer:
s=50m
Explanation:
you can use the formula
s=ut+1/2at²
s=0t+1/2(1)10²
=1/2(100)
=50
I hope this helps
I NEEED HELP IN PHYSICS PLEASE!
Answer:
in which topic you need help
Assume that the car on the left makes a quick turn to the left. According to inertia, your body will resist a change and still want to go in the original direction. In which direction with the passenger slide?
Answer:
to the right
Explanation:
if the car turns to the lift, the body forces energy to the left side, so according to the first law of Newton, the body will move to the right side to resist the sudden motion.
What is the Ah rating of a battery that can provide 0.8 A for 76 h?
Answer:
6.08
Explanation:
Given that,
Current, I = 0.8 A
Time, t = 76 h
We need to find the Ah rating of a battery. It can be calculated by taking the product of current and time. So,
Ah = (0.8)(76)
= 6.08 Ah
So, the Ah rating of the battery is 6.08.
A friend lends you the eyepiece of his microscope to use on your own microscope. He claims that since his eyepiece has the same diameter as yours but twice the focal length, the resolving power of your microscope will be doubled. Is his claim valid? Explain.
Answer:
The resolving power remains same.
Explanation:
The resolving power of the lens is directly proportional to the diameter of the lens not on the focal length.
As the diameter is same but the focal length is doubled so the resolving power remains same.
A long string is moved up and down with simple harmonic motion with a frequency of 46 Hz. The string is 579 m long and has a total mass of 46.3 kg. The string is under a tension of 3423 and is fixed at both ends. Determine the velocity of the wave on the string. What length of the string, fixed at both ends, would create a third harmonic standing wave
Answer:
a) [tex]v=206.896m/s[/tex]
b) [tex]L=6.749m[/tex]
Explanation:
From the question we are told that:
Frequency [tex]F=46Hz[/tex]
Length [tex]l=579m[/tex]
Total Mass [tex]T=4.3kg[/tex]
Tension [tex]T=3423[/tex]
a)
Generally the equation for velocity is mathematically given by
[tex]v=\sqrt{\frac{T}{\rho}}[/tex]
Where
[tex]\pho=m*l\\\\\pho=46*579\\\\\pho=0.0799kg/m[/tex]
Therefore
[tex]v=\sqrt{\frac{3423}{0.0799}}[/tex]
[tex]v=206.896m/s[/tex]
b)
Generally the equation for length of string is mathematically given by
[tex]L=\frac{3\lambda}{2}[/tex]
Where
[tex]\lambda=\frac{v}{f}[/tex]
[tex]\lambda=\frac{206.89}{46}[/tex]
[tex]\lambda=4.498[/tex]
Therefore
[tex]L=\frac{3*4.498}{2}[/tex]
[tex]L=6.749m[/tex]
A redox reaction is always a single-displacement reaction, but a single-
displacement reaction isn't always a redox reaction.
A. True
B. False
SUBMIT
These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor
Answer:
Following are the solution to the given question:
Explanation:
For charging plates that are connected in a similar manner:
Calculating the total charge:
[tex]\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C[/tex]
Calculating the common potential:
[tex]\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\[/tex]
Calculating the charge after redistribution:
[tex]When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2[/tex]
[tex]\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\ \\ \to q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C[/tex]
The area around a charged object that can exert a force on other charged objects is an electric ___
The exponent of the exponential function contains RC for the given circuit, which is called the time constant. Use the units of R and C to find units of RC. Write ohms in terms of volts and amps and write farads in terms of volts and coulombs. Simplify until you get something simple. Show your work below.
Answer:
The unit of the time constant RC is the second
Explanation:
The unit of resistance, R is the Ohm, Ω and resistance, R = V/I where V = voltage and I = current. The unit of voltage is the volt, V while the unit of current is the ampere. A.
Since R = V/I
Unit of R = unit of V/unit of I
Unit of R = V/A
Ω = V/A
Also, The unit of capacitance, C is the Farad, F and capacitance, F = Q/V where Q = charge and V = voltage. the unit of charge is the coulomb, C while the unit of voltage is the volt, V
Since C = Q/V
Unit of C = unit of Q/unit of V
Unit of C = C/V
F = C/V
Now the time constant equals RC.
So, the unit of the time constant = unit of R × unit of C = Ω × F = V/A × C/V = C/A
Also. we know that the 1 Ampere = 1 Coulomb per second
1 A = 1 C/s
So, substituting 1 A in the denominator, we have
unit of RC = C/A = C ÷ C/s = s
So, the unit of RC = s = second
So, the unit of the time constant RC is the second
A 69.0-kg astronaut is floating in space, luckily he has his trusty 28.0-kg physics book. He throws his physics book and accelerates at 0.0130 m/s2 in the opposite direction. What is the magnitude of the acceleration of the physics book?
Answer:
0.032 [tex]m/s^2[/tex]
Explanation:
Given :
Weight of the astronaut = 69 kg
Weight of the physics book = 28 kg
Acceleration of the astronaut = 0.0130 [tex]m/s^2[/tex]
The force that is applied on the astronaut :
[tex]F=ma[/tex]
[tex]$=69 \times 0.013$[/tex]
= 0.897 N
Therefore, by Newton's 3rd law, we know that the force applied on the physics book is also F = 0.897 N
Therefore, the acceleration of the physics book is given by :
[tex]$a = \frac{\text{Force on physics book}}{\text{mass of physics book}}$[/tex]
[tex]$a = \frac{0.897}{28}$[/tex]
a = 0.032 [tex]m/s^2[/tex]
Hence, the acceleration of the physics book is 0.032 [tex]m/s^2[/tex].
Answer:
The acceleration of astronaut is 5.27 x 10^-3 m/s^2.
Explanation:
mass of astronaut, M = 69 kg
Mass of book, m = 28 kg
acceleration of book, a = 0.013 m/s^2
Let the acceleration of astronaut is A.
According to the Newton's third law, for every action there is an equal and opposite reaction.
So, the force acting on the book is same as the force acting on the astronaut but the direction is opposite to each other.
M A = m a
69 x A = 28 x 0.013
A = 5.27 x 10^-3 m/s^2
A parallel plate capacitor is constructed using two square metal sheets, each of side L = 10 cm. The plates are separated by a distance d = 2 mm and a voltage applied between the plates. The electric field strength within the plates is E = 4000 V/m. The energy stored in the capacitor is
Answer:
The energy stored is 1.4 x 10^-9 J.
Explanation:
Side of square, L = 10 cm = 0.1 m
Distance, d = 2 mm = 0.002 m
Electric field, E = 4000 V/m
The energy stored in the capacitor is
[tex]U = 0.5 C V^2[/tex]
The capacitance is given by
[tex]C = \frac{\varepsilon o A}{d}\\\\So \\\\U = 0.5\frac{\varepsilon o A}{d}\times E^2 d^2\\\\U = 0.5\times 8.85\times 10^{-12}\times 0.1\times 0.1\times 4000\times 4000\times 0.002\\\\U = 1.4\times10^{-9} J[/tex]
Two motors in a factory are running at slightly different rates. One runs at 825.0 rpm and the other at 786.0 rpm. You hear the sound intensity increase and then decrease periodically due to wave interference. How long does it take between successive instances of the sound intensity increasing
Answer:
[tex]T=1.54s[/tex]
Explanation:
From the question we are told that:
Speed of Motor 1 [tex]\omega_1=825rpm=>2 \pi 13.75[/tex]
Speed of Motor 2 [tex]\omega_1=786rpm=>2 \pi 13.1[/tex]
Therefore
Frequency of Motor 1 [tex]f_1=13.75[/tex]
Frequency of Motor 2 [tex]f_2= 13.1[/tex]
Generally the equation for Time Elapsed is mathematically given by
[tex]T=\frac{1}{df}[/tex]
Where
[tex]df=f_1-f_2[/tex]
[tex]df=13.75-13.1[/tex]
[tex]df=0.65Hz[/tex]
Therefore
[tex]T=\frac{1}{65}[/tex]
[tex]T=1.54s[/tex]
A magnetic field is passing through a loop of wire whose area is 0.015 m2. The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.20 T/s. (a) Determine the magnitude of the emf induced in the loop. (b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B
This question is incomplete, the complete question is;
A magnetic field is passing through a loop of wire whose area is 0.015 m2. The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.20 T/s.
(a) Determine the magnitude of the emf induced in the loop.
(b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B = 1.5 T, if the induced emf is to be zero? (Give the magnitude of the rate of change of the area.) (m2/s)
Answer:
a) the magnitude of the emf induced in the loop is 0.003 V
b) dA/dt = 0.002 m²/s
Explanation:
Area of the loop wire A = 0.015 m²
magnitude of the field is increasing dB/dt = 0.20 T/s
a)
Determine the magnitude of the emf induced in the loop.
V = A( dB/dt )
we substitute
V = 0.015 m² × 0.20 T/s
V = 0.003 V
Therefore, the magnitude of the emf induced in the loop is 0.003 V
b) the induced emf is;
V = B( dA/dt ) + A( dB/dt )
given that; induced emf is 0, B = 1.5
so we substitute
0 = [ 1.5T × ( dA/dt ) ] + [ 0.015 m² × 0.20 T/s ]
-[ 1.5T × ( dA/dt )] = 0.003 m²T/s
dA/dt = -[ 0.003 m²T/s / 1.5T ]
dA/dt = -0.002 m²/s
the negative shows that the area is decreasing
hence, dA/dt = 0.002 m²/s
A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand.
(a) At what angle was the ball thrown if its initial speed was 12.0 m/ s, assuming that the smaller of the two possible angles was used?
(b) What other angle gives the same range, and why would it not be used?
(c) How long did this pass take?
Answer:
a) θ = 14.23º, b) θ₂ = 75.77, c) t = 0.6019 s
Explanation:
This is a missile throwing exercise.
a) the reach of the ball is the distance traveled for the same departure height
R = [tex]\frac{v_o^2 \ sin 2 \theta }{g}[/tex]
sin 2θ = [tex]\frac{Rg}{v_o^2}[/tex]
sin 2θ = 7.00 9.8 / 12.0²
2θ = sin⁻¹ (0.476389) = 28.45º
θ = 14.23º
the complementary angle that gives the same range is the angle after 45 that the same value is missing to reach 90º
θ ’= 90 -14.23
θ’= 75.77º
b) the two angles that give the same range are
θ₁ = 14.23
θ₂ = 75.77
the greater angle has a much greater height so the time of the movement is greater and has a greater chance of being intercepted by the other team.
C) the time of the pass can be calculated with the expression
x = v₀ₓ t
t = x / v₀ₓ
t = 7 / 11.63
t = 0.6019 s
You have 150 W/m^2 hitting your roof each day. You can convert 13% of it into
usable energy, and you need 3.5 kW to run your house for a day. Show the MATH,
answer and units, to determine the size solar panel you will need to succeed.
Answer:
Energy = .13 W / m^2 energy of incident energy
N = 3500 Watts / day power needed
N = 3500 Watts (3600 * 24 sec) = .0405 Watts/sec
The problem must mean that one needs 3.5 Kw-days
3.5 Kw-days = 3500 watts * 86400 sec = 3.02E8 joules
150 J/sec-m^2 * .13 = 19.5 J / sec-m^2 usable energy
In one day 19.5 J/sec-m^2 = 1.68E6 J/m^2 usable energy received
Area = 3.028E8 J / 1.68E6 J/m2 = 180 m^2
One would need 180 m^2 of solar panels
That's quite a lot of energy
A 1100 watt microwave oven uses 1.1 kW while running so 3.5 kW for 24 hours seems to be quite a lot.
a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its kinetic energy when it's 2.00 m above the ground
Answer:
KE_2 = 3.48J
Explanation:
Conservation of Energy
E_1 = E_2
PE_1+KE_1 = PE_2+KE_2
m*g*h+(1/2)m*v² = m*g*h+(1/2)m*v²
(0.0780kg)*(9.81m/s²)*(5.36m)+(.5)*(0.0780kg)*(4.84m/s)² = (0.0780kg)*(9.81m/s²)*(2m)+KE_2
4.10J+0.914J = 1.53J + KE_2
5.01J = 1.53J + KE_2
KE_2 = 3.48J
A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.680 Hz. The pendulum has a mass of 2.00 kg, and the pivot is located 0.340 m from the center of mass. Determine the moment of inertia of the pendulum about the pivot point.
Answer:
Therefore, the moment of inertia is:
[tex]I=0.37 \: kgm^{2} [/tex]
Explanation:
The period of an oscillation equation of a solid pendulum is given by:
[tex]T=2\pi \sqrt{\frac{I}{Mgd}}[/tex] (1)
Where:
I is the moment of inertiaM is the mass of the pendulumd is the distance from the center of mass to the pivotg is the gravityLet's solve the equation (1) for I
[tex]T=2\pi \sqrt{\frac{I}{Mgd}}[/tex]
[tex]I=Mgd(\frac{T}{2\pi})^{2}[/tex]
Before find I, we need to remember that
[tex]T = \frac{1}{f}=\frac{1}{0.680}=1.47\: s[/tex]
Now, the moment of inertia will be:
[tex]I=2*9.81*0.340(\frac{1.47}{2\pi})^{2}[/tex]
Therefore, the moment of inertia is:
[tex]I=0.37 \: kgm^{2} [/tex]
I hope it helps you!
Each rarefraction on a longitudinal wave correspond to what point on a transverse wave?
In a rolling race, two objects are released from the top of two identical ramps. They then roll without slipping to the bottom of the ramp. If the two objects are 2 hoops of the same radius but different masses, which reaches the bottom first?
a. The lighter one reaches the bottom first
b. The heavier one reaches the bottom first
c. We don’t have enough information
d. They reach the bottom at the same time
Answer:
b. The heavier one reaches the bottom first.
Answer:
B
Explanation:
The answer is B the heavier item has more g force pushing it making it roll faster reaching the bottom of the ramp first.
HELP MEEEEEEE PLEASEEEEEEEEE
Answer:
D) Q = 80 C
Explanation:
Given;
current flowing in the light bulb, I = 2A
time of current flow, t = 40,000 ms = 40,000 x 10⁻³ s = 40 s
The quantity of the charge is calculated as;
Q = It
where;
Q is the quantity of the charge (Coulombs)
Q = (2 ) x (40)
Q = 80 C
Therefore, the quantity of charge flowing in the circuit is 80 C
D) Q = 80 C
Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in J) is stored in this inductor when 21.0 A of current flows through it? J (c) How fast (in s) can it be turned off if the induced emf cannot exceed 3.00 V? s
Answer:
(a) The self inductance, L = 21.95 mH
(b) The energy stored, E = 4.84 J
(c) the time, t = 0.154 s
Explanation:
(a) Self inductance is calculated as;
[tex]L = \frac{N^2 \mu_0 A}{l}[/tex]
where;
N is the number of turns = 1000 loops
μ is the permeability of free space = 4π x 10⁻⁷ H/m
l is the length of the inductor, = 45 cm = 0.45 m
A is the area of the inductor (given diameter = 10 cm = 0.1 m)
[tex]A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2[/tex]
[tex]L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH[/tex]
(b) The energy stored in the inductor when 21 A current ;
[tex]E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J[/tex]
(c) time it can be turned off if the induced emf cannot exceed 3.0 V;
[tex]emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s[/tex]
Identify each action as a wave erosion war wind erosion
Answer:Lesson Objectives
Describe how the action of waves produces different shoreline features.
Discuss how areas of quiet water produce deposits of sand and sediment.
Discuss some of the structures humans build to help defend against wave erosion.
Vocabulary
arch
barrier island
beach
breakwater
groin
refraction
sea stack
sea wall
spit
wave-cut cliff
wave-cut platform
Introduction
Waves are important for building up and breaking down shorelines. Waves transport sand onto and off of beaches. They transport sand along beaches. Waves carve structures at the shore.
Wave Action and Erosion
All waves are energy traveling through some type of material, such as water (Figure below). Ocean waves form from wind blowing over the water.
Ocean waves are energy traveling through water.
The largest waves form when the wind is very strong, blows steadily for a long time, and blows over a long distance.
The wind could be strong, but if it gusts for just a short time, large waves won’t form. Wave energy does the work of erosion at the shore. Waves approach the shore at some angle so the inshore part of the wave reaches shallow water sooner than the part that is further out. The shallow part of the wave ‘feels’ the bottom first. This slows down the inshore part of the wave and makes the wave ‘bend.’ This bending is called refraction.
Wave refraction either concentrates wave energy or disperses it. In quiet water areas, such as bays, wave energy is dispersed, so sand is deposited. Areas that stick out into the water are eroded by the strong wave energy that concentrates its power on the wave-cut cliff (Figure below).
The wave erodes the bottom of the cliff, eventually causing the cliff to collapse.
Other features of wave erosion are pictured and named in Figure below. A wave-cut platform is the level area formed by wave erosion as the waves undercut a cliff. An arch is produced when waves erode through a cliff. When a sea arch collapses, the isolated towers of rocks that remain are known as sea stacks.
(a) The high ground is a large wave-cut platform formed from years of wave erosion. (b) A cliff eroded from two sides produces an arch. (c) The top of an arch erodes away, leaving behind a tall sea stack.
Wave Deposition
Rivers carry sediments from the land to the sea. If wave action is high, a delta will not form. Waves will spread the sediments along the coastline to create a beach (Figure below). Waves also erode sediments from cliffs and shorelines and transport them onto beaches.
Sand deposits in quiet areas along a shoreline to form a beach.
Beaches can be made of mineral grains, like quartz, rock fragments, and also pieces of shell or coral (Figure below).
Quartz, rock fragments, and shell make up the sand along a beach.
Waves continually move sand along the shore. Waves also move sand from the beaches on shore to bars of sand offshore as the seasons change. In the summer, waves have lower energy so they bring sand up onto the beach. In the winter, higher energy waves bring the sand back offshore.
Some of the features formed by wave-deposited sand are in Figure below. These features include barrier islands and spits. A spit is sand connected to land and extending into the water. A spit may hook to form a tombolo.
Examples of features formed by wave-deposited sand.
Shores that are relatively flat and gently sloping may be lined with long narrow barrier islands (Figure below). Most barrier islands are a few kilometers wide and tens of kilometers long.
(a) Barrier islands off of Alabama. A lagoon lies on the inland side. (b) Barrier islands, such as Padre Island off the coast of Texas, are made entirely of sand. (c) Barrier islands are some of the most urbanized areas of our coastlines, such as Miami Beach.
In its natural state, a barrier island acts as the first line of defense against storms such as hurricanes. When barrier islands are urbanized (Figure above), hurricanes damage houses and businesses rather than vegetated sandy areas in which sand can move. A large hurricane brings massive problems to the urbanized area.
Protecting Shorelines
Intact shore areas protect inland areas from storms that come off the ocean (Figure below).
Dunes and mangroves along Baja California protect the villages that are found inland.
Explanation:
Answer: Below
Explanation: Correct on Edmentum
