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
1. A body moving with uniform acceleration of 10 m/s2 covers a distance of 320 m. if its initial velocity was 60 m/s. Calculate its final velocity.
Answer:
v² = u² + 2as
v² = 3600 + 6400
v² = 10000
v = 100
Explanation:
final velocity is 100 m/s
According to third equation of kinematics
[tex]\boxed{\sf v^2-u^2=2as}[/tex]
[tex]\\ \sf\longmapsto v^2=u^2+2as[/tex]
[tex]\\ \sf\longmapsto v^2=(60)^2+2(10)(320)[/tex]
[tex]\\ \sf\longmapsto v^2=3600+3400[/tex]
[tex]\\ \sf\longmapsto v^2=10000[/tex]
[tex]\\ \sf\longmapsto v=\sqrt{10000}[/tex]
[tex]\\ \sf\longmapsto v=100m/s[/tex]
The double bond between two oxygen atoms (a molecule of oxygen air) has
two characteristics. What are they?
A. Four valence electrons are shared.
B. A metallic bond is formed.
C. Valence electrons are shared between oxygen atoms.
D. An ionic bond is formed.
Answer:
valance electrons are shared between oxygen atoms.. making them have eight in the outer most shells.
I hope this helps
A charged particle having mass 6.64 x 10-27 kg (that of a helium atom) moving at 8.70 x 105 m/s perpendicular to a 1.30-T magnetic field travels in a circular path of radius 18.0 mm. What is the charge of the particle
Answer:
the charge of the particle is 2.47 x 10⁻¹⁹ C
Explanation:
Given;
mass of the particle, m = 6.64 x 10⁻²⁷ kg
velocity of the particle, v = 8.7 x 10⁵ m/s
strength of the magnetic field, B = 1.3 T
radius of the circle, r = 18 mm = 1.8 x 10⁻³ m
The magnetic force experienced by the charge is calculated as;
F = ma = qvB
where;
q is the charge of the particle
a is the acceleration of the charge in the circular path
[tex]a = \frac{v^2}{r} \\\\ma = qvB\\\\q = \frac{ma}{vB} \\\\q = \frac{mv^2}{rvB} = \frac{mv}{rB} \\\\q = \frac{(6.64\times 10^{-27} ) \times (8.7\times 10^5)}{(1.8\times 10^{-2}) \times (1.3)} \\\\q = 2.47 \ \times 10^{-19} \ C[/tex]
Therefore, the charge of the particle is 2.47 x 10⁻¹⁹ C
A nylon string on a tennis racket is under a tension of 285 N . If its diameter is 1.10 mm , by how much is it lengthened from its untensioned length of 29.0 cm ? Use ENylon=5.00×109N/m2.
Answer:
1.74×10⁻³ m
Explanation:
Applying,
ε = Stress/strain............. Equation 1
Where ε = Young's modulus
But,
Stress = F/A.............. Equation 2
Where F = Force, A = Area
Strain = e/L.............. Equation 3
e = extension, L = Length.
Substitute equation 2 and 3 into equation 1
ε = (F/A)/(e/L) = FL/eA............. Equation 4
From the question,
Given: F = 285 N, L = 29 cm = 0.29 m, ε = 5.00×10⁹ N/m²,
A = πd²/4 = 3.14(0.0011²)/4 = 9.4985×10⁻⁶ m²
Substitute these values into equation 4
5.00×10⁹ = (285×0.29)/(9.4985×10⁻⁶×e)
Solve for e
e = (285×0.29)/(5.00×10⁹×9.4985×10⁻⁶)
e = 82.65/4.74925×10⁴
e = 1.74×10⁻³ m
If 56.5 m3 of a gas are collected at a pressure of 455 mm Hg, what volume will the gas occupy if the pressure is changed to 632 mm Hg? *
Assuming ideal conditions, Boyle's law says that
P₁ V₁ = P₂ V₂
where P₁ and V₁ are the initial pressure and temperature, respectively, and P₂ and V₂ are the final pressure and temperature.
So you have
(455 mm Hg) (56.5 m³) = (632 mm Hg) V₂
==> V₂ = (455 mm Hg) (56.5 m³) / (632 mm Hg) ≈ 40.7 m³
Một vật được ném lên trên theo phương thẳng đứng. Người quan sát
thấy vật đó đi qua vị trí có độ cao h hai lần và khoảng thời gian giữa hai lần đó là
t. Tìm vận tốc ban đầu và thời gian chuyển động của vật từ lúc ném đến khi vật
rơi về vị trí ban đầu.
Answer:
Language -English plz I cant understand
3 of 3 : please help got an extra day for a test and i don’t get this (must show work) points and brainliest!
Explanation:
[tex]qV = \frac{1}{2}mv^2[/tex]
Multiply both sides by 2 and then divide by m to get
[tex]\dfrac{2qV}{m} = v^2[/tex]
Take the square root of both sides to get
[tex]v = \sqrt{\dfrac{2qV}{m}}[/tex]
two bodies A and B with some asses 20 kg and 30 kg respectively above the ground which have greater potential
Answer:
B has greater potential
Explanation:
We know;
Potential Energy (PE) = mgh
where, m=mass of body
g=acceleration due to gravity
h=height of body
From the formula,
PE is directly proportional to the mass of the body
so the body with greater mass has greater potential.
You are a detective investigating why someone was hit on the head by a falling flowerpot. One piece of evidence is a smartphone video taken in a 4th-floor apartment, which happened to capture the flowerpot as it fell past a window. In a span of 8 frames (captured at 30 frames per second), the flowerpot falls 0.84 of the height of the window. You visit the apartment and measure the window to be 1.27 m tall.
Required:
Assume the flowerpot was dropped from rest. How high above the window was the flowerpot when it was dropped?
Answer:
0.37 m
Explanation:
Given :
Window height, [tex]h_1[/tex] = 1.27 m
The flowerpot falls 0.84 m off the window height, i.e.
[tex]h_2[/tex] = (1.27 x 0.84 ) m in a time span of [tex]$t=\frac{8}{30}$[/tex] seconds.
Assuming that the speed of the pot just above the window is v then,
[tex]h_2=ut+\frac{1}{2}gt^2[/tex]
[tex]$(1.27 \times 0.84) = v \times \left( \frac{8}{30} \right) + \frac{1}{2} \times 9.81 \times \left( \frac{8}{30} \right)^2$[/tex]
[tex]$v=\left(\frac{30}{8}\right) \left[ (1.27 \times 0.84) - \left( \frac{1}{2} \times 9.81 \times \left( \frac{8}{30 \right)^2 \right) \right]}$[/tex]
[tex]$v= 2.69$[/tex] m/s
Initially the pot was dropped from rest. So, u = 0.
If it has fallen from a height of h above the window then,
[tex]$h = \frac{v^2}{2g}$[/tex]
[tex]$h = \frac{(2.69)^2}{2 \times 9.81}$[/tex]
h = 0.37 m
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
Cho các máy cắt sử dụng trong công nghiệp có ký hiệu trên nhãn thiết bị: C350; B500. Hãy tính dòng điện bảo vệ ngắn mạch và dòng điện bảo vệ quá tải của từng thiết bị?
Answer:
ask in the English then I can help you
Explanation:
please mark me as brain list
Describe the change in motion and kinetic energy of the particles as thermal energy is added to a liquid. Which change of state might happen?
please ill put brainliest!!!
Answer:
If a liquid is heated the particles are given more energy and move faster and faster expanding the liquid. The most energetic particles at the surface escape from the surface of the liquid as a vapour as it gets warmer. Liquids evaporate faster as they heat up and more particles have enough energy to break away
A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?
Answer:
A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?
Explanation:
write any two physical hazard occuring in the late choldhood
Answer:
Hazards during late childhood
Health Problems: Chronic health ailments like T.B., Pneumonia etc will hinder the child's motor abilities.Accidents: School age children are more adventurous in nature, they run fast, play hard, ride bicycles and scooters and engage in a variety of sports.how many rings does saturn have
Answer:
From far away, Saturn looks like it has seven large rings. Each large ring is named for a letter of the alphabet. The rings were named in the order they were discovered.
Find the ratio of the diameter of aluminium to copper wire, if they have the same
resistance per unit length. Take the resistivity values of aluminium and copper to
be 2.65× 10−8 Ω m and 1.72 × 10−8 Ω m respectively
Answer:
1.24
Explanation:
The resistivity of copper[tex]\rho_1=2.65\times 10^{-8}\ \Omega-m[/tex]
The resistivity of Aluminum,[tex]\rho_2=1.72\times 10^{-8}\ \Omega-m[/tex]
The wires have same resistance per unit length.
The resistance of a wire is given by :
[tex]R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}[/tex]
According to given condition,
[tex]\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24[/tex]
So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.
The force of gravity is an inverse square law. This means that, if you double the distance between two large masses, the gravitational force between them Group of answer choices weakens by a factor of 4. strengthens by a factor of 4. weakens by a factor of 2. also doubles. is unaffected.
Answer:
the force decreases by a factor of 4
Explanation:
The expression for the law of universal gravitation is
F = [tex]G \frac{m_1m_2}{r^2}[/tex]
let's call the force Fo for the distance r
F₀ = [tex]G \frac{m_1m_2}{r^2}[/tex]
They indicate that the distance doubles
r ’= 2 r
we substitute
F = [tex]G \frac{m_1m_2}{(r')^2}[/tex]
F = [tex]G \frac{m_1m_2}{r^2} \ \frac{1}{4}[/tex]
F = ¼ F₀
consequently the correct answer is that the force decreases by a factor of 4
If the distance between two large masses are doubled, the gravitational force between them weakens by a factor of 4.
Let the initial force be F
Let the initial distance apart be r
Thus, we can obtain the final force as follow:
Initial force (F₁) = F
Initial distance apart (r₁) = r
Final distance apart (r₂) = 2r
Final force (F₂) =?F = GM₁M₂ / r²
Fr² = GM₁M₂ (constant)
Thus,
F₁r₁² = F₂r₂²
Fr² = F₂(2r)²
Fr² = F₂4r²
Divide both side by 4r²
F₂ = Fr² /4r²
F₂ = F / 4From the illustration above, we can see that when the distance (r) is doubled, the force (F) is decreased (i.e weakens) by a factor of 4
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A proton moves perpendicular to a uniform magnetic field at a speed of 1.75 107 m/s and experiences an acceleration of 2.25 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.
Answer:
B = 0.013(-j) T
Explanation:
Given that,
The speed of a proton, [tex]v=1.75\times 10^7\ m/s[/tex]
Acceleration experienced by the proton,[tex]a=2.25\times 10^3\ m/s[/tex]
We need to find the magnitude and the direction of the magnetic field. At equilibrium,
[tex]ma=qvB\\\\B=\dfrac{ma}{qv}\\\\B=\dfrac{1.67\times 10^{-27}\times 2.25\times 10^{13}}{1.6\times 10^{-19}\times 1.75\times 10^{7}}\\\\B=0.013\ T[/tex]
The velocity is in +z direction, force in +x direction, then the field must be in -y direction.
A seesaw has an irregularly distributed mass of 30 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced when a 60-kg person sits on one end and a 78-kg person sits on the other end.
Required:
Find a displacement of the center of mass of the system relatively to the seesaw's midpoint.
Answer:
x = 0.9 m
Explanation:
For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive
∑ τ = 0
60 1.5 - 78 1.5 + 30 x = 0
where x is measured from the left side of the fulcrum
90 - 117 + 30 x = 0
x = 27/30
x = 0.9 m
In summary the center of mass is on the side of the lightest weight x = 0.9 m
Which circuit has the larger equivalent resistance: a circuit with two 10 ohm resistors connected in parallel or a circuit with two 10 ohm resistors connected in series?
Answer:
A circuit with two 10 ohm resistors connected in series.
Explanation:
The formula for the equivalent resistance for resistors in parallel is
[tex]\frac{1}{Rt} = \frac{1}{R1} + \frac{1}{R2}[/tex] So if R1=R2= 10 [tex]\frac{1}{Rt} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} <=> Rt =\frac{10}{2} =5 ohm[/tex]
The formula for the equivalent resistance for resistors in series is
Rt = R1 + R2 So Rt= 10 + 10 = 20
26. A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?
Answer:
Explanation:
The formula for determining the Emf induced in a loop is:
[tex]\varepsilon = \dfrac{d \phi}{dt}[/tex]
[tex]\varepsilon = \dfrac{d (B*A)}{dt}[/tex]
[tex]\varepsilon = A \times \dfrac{dB}{dt}[/tex]
[tex]\varepsilon = (side (l))^2 \times \dfrac{dB}{dt}[/tex]
where;
square area A = ( l²)
l² = 6.0 cm = 6.0 × 10⁻²
∴
[tex]\varepsilon = ( 6.0 \times 10^{-2})^2 \times 5.0 \times 10^{-3} \ T/S[/tex]
[tex]\varepsilon =18 \times 10^{6} \ V[/tex]
Recall that:
The resistivity of copper = [tex]1.68 \times 10^{-8}[/tex] ohm m
We can as well say that the length of the copper wire = perimeter of the square loop;
The perimeter of the square loop = 4L
Thus, the length of the copper wire = 4 (6.0 × 10⁻² )m
= 24× 10⁻² m
Finally, the current in the loop is determined from the formula:
V = IR
where,
V = voltage
I = current and R = resistance of the wire
Making "I" the subject:
I = V/R
where;
[tex]R = \dfrac{\rho \times l}{A}[/tex]
[tex]R = \dfrac{\rho \times l}{\pi * r^2}[/tex]
[tex]R = \dfrac{1.68 *10^{-8} \times 24*10^{-2}}{\pi * (1*10^{-3})^2}[/tex]
[tex]R = 0.001283 \ ohms[/tex]
∴
[tex]I = \dfrac{18*10^{-6}}{0.001283}[/tex]
I = 14.029 mA
Tres personas, A, B y C jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección horizontal y la persona B aplica a su vez 5 en dirección horizontal, ¿Cuál es el valor de la fuerza que debe ejercer la persona C, para que la caja esté en equilibrio físico?
Answer:
Un objeto se encuentra en equilibrio físico si la fuerza neta que se le aplica es igual a 0.
En este caso solo se aplican fuerzas en el eje horizontal, por lo que las podremos sumar directamente.
La persona A aplica una fuerza:
Fa = -3N
La persona B aplica una fuerza:
Fb = 5N
La persona C aplica una fuerza Fc, la cual aún no conocemos.
Pero sabemos que la caja está en equilibrio físico, por lo que:
Fa + Fb + Fc = 0N
reemplazando los valores que conocemos, obtenemos:
-3N + 5N + Fc = 0N
Ahora podemos resolver esto para Fc, la fuerza que aplica la persona C.
Fc = 0N + 3N - 5N
Fc = -2N
Podemos concluir que la persona C aplica una fuerza horizontal de -2N
A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?
PLEASE EXPLAIN HOW YOU GOT THE ANSWER THANK YOU SO MUCH
Answer:
0
Explanation:
The speed of the ball when it reaches the floor is 0 because when an object is at rest or in uniform motion, it has no speed/velocity
The final speed of the ball when it reaches the floor is 7.10 m/s.
What is the conservation of energy?The conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but only converted from one form to another or transferred from one system to another. In other words, the total amount of energy in a closed system remains constant over time, even though it may be converted from one form to another.
This principle is based on the first law of thermodynamics, which states that the total energy of a closed system is always conserved, and can only be changed by the transfer of heat, work, or matter into or out of the system. The conservation of energy has important applications in various fields of physics, including mechanics, thermodynamics, and electromagnetism, and is a fundamental principle in the understanding of the natural world.
Here in the Question,
We can use the conservation of energy to solve this problem. Initially, the ball has kinetic energy due to its motion on the tabletop, but no potential energy since it is at a constant height. When the ball rolls off the edge of the table, it loses some kinetic energy due to friction but gains potential energy as it moves upward. When it reaches the floor, it has gained potential energy but lost kinetic energy due to friction. We can assume that the energy lost due to friction is converted to thermal energy, so the total energy of the system is conserved.
Let's start by calculating the potential energy gained by the ball as it moves from the edge of the table to the floor:
ΔPE = mgh
where ΔPE is the change in potential energy, m is the mass of the ball, g is the acceleration due to gravity, and h is the vertical distance traveled by the ball.
ΔPE = (0.50 kg)(9.81 m/s^2)(1.0 m) = 4.905 J
Now we can use the conservation of energy to find the final kinetic energy of the ball, which will allow us to calculate its final speed:
KEi + ΔPEi = KEf + ΔPEf
where KEi and ΔPEi are the initial kinetic and potential energies of the ball, respectively, and KEf and ΔPEf are the final kinetic and potential energies of the ball, respectively.
Since the ball is not bouncing, we can assume that its initial and final potential energies are zero. Therefore:
KEi = KEf + ΔKE
where ΔKE is the change in kinetic energy due to friction.
We can assume that the coefficient of kinetic friction between the ball and the incline is constant, and use the work-energy principle to find ΔKE:
Wfric = ΔKE
where Wfric is the work done by friction.
The work done by friction can be expressed as:
Wfric = ffricd
where ffric is the force of friction and d is the distance traveled by the ball on the incline.
The force of friction can be expressed as:
ffric = μmg
where μ is the coefficient of kinetic friction, and m and g have their usual meanings.
Putting it all together, we get:
KEi = KEf + ffricd
KEi = KEf + μmgd
(1/2)mv^2 = (1/2)mu^2 + μmgd
v^2 = u^2 + 2gd
where u is the initial speed of the ball on the tabletop, and v is the final speed of the ball on the floor.
Plugging in the given values, we get:
v^2 = (5.0 m/s)^2 + 2(9.81 m/s^2)(1.0 m)
v^2 = 50.405
v = 7.10 m/s
Therefore, the final speed of the ball when it reaches the floor is 7.10 m/s.
To learn more about the Law of Conservation of Momentum click:
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You drive 7.5 km in a straight line in a direction east of north.
a. Find the distances you would have to drive straight east and then straight north to arrive at the same point.
b. Show that you still arrive at the same point if the east and north legs are reversed in order.
Answer:
a) a = 5.3 km, b) sum fulfills the commutative property
Explanation:
This is a vector exercise, If you drive east from north, we can find the vector using the Pythagorean theorem
R² = a² + b²
where R is the resultant vector R = 7.5 km and the others are the legs
If we assume that the two legs are equal to = be
R² = 2 a²
r = √2 a
a = r /√2
we calculate
a = 7.5 /√2
a = 5.3 km
therefore, you must drive 5.3 km east and then 5.3 km north and you will reach the same point
b) As the sum fulfills the commutative property, the order of the elements does not alter the result
a + b = b + a
therefore, it does not matter in what order the path is carried out, it always reaches the same end point
Để sử dụng nguồn điện xoay chiều 220V/50Hz thắp sáng bóng đèn 12V/3W, ta chọn điện trở giảm áp có giá trị:
Explanation:
Hi Linda,
How's it going?
Sorry I haven't been in touch for such a long time but I've had exams so I've been studying every free minute. Anyway, I'd love to hear all your news and I'm hoping we can get together soon to catch up. We just moved to a bigger flat so maybe you can come and visit one weekend?
How's the new job?
Looking forward to hearing from you!
Helga
A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r < R) from the center of the sphere the electric field has a value E. If the same charge Q were distributed uniformly throughout a sphere of radius 2R the magnitude of the electric field at a radius r would be equal to:__________
Answer:
Hence the answer is E inside [tex]= KQr_{1} /R^{3}[/tex].
Explanation:
E inside [tex]= KQr_{1} /R^{3}[/tex]
so if r1 will be the same then
E [tex]\begin{bmatrix}Blank Equation\end{bmatrix}[/tex] proportional to 1/R3
so if R become 2R
E becomes 1/8 of the initial electric field.
Answer:
The electric field is E/8.
Explanation:
The electric field due to a solid sphere of uniform charge density inside it is given by
[tex]E =\frac{\rho r}{3}[/tex]
where, [tex]\rho[/tex] is the volume charge density and r is the distance from the center.
For case I:
[tex]\rho = \frac{Q}{\frac{4}{3}\pi R^3}[/tex]
So, electric field at a distance r is
[tex]E = \frac { 3 Q r}{3\times 4\pi R^3}\\\\E = \frac{Q r}{4\pi R^3}[/tex]
Case II:
[tex]\rho = \frac{Q}{\frac{4}{3}\pi 8R^3}[/tex]
So, the electric field at a distance r is
[tex]E' = \frac { 3 Q r}{3\times 32\pi R^3}\\\\E' = \frac{Q r}{8\times 4\pi R^3}\\\\E' = \frac{E}{8}[/tex]
A car whose tire have radii 50cm travels at 20km/h. what is the angular velocity of the tires?
Convert to m/s
[tex]\\ \sf\longmapsto v=20\times 5/18=5.5m/s[/tex]
We know
[tex]\boxed{\sf \omega=\dfrac{rv}{|r|^2}}[/tex]
[tex]\\ \sf\longmapsto \omega=\dfrac{0.5(5.5)}{|0.5|^2}[/tex]
[tex]\\ \sf\longmapsto \omega=\dfrac{2.75}{0.25}[/tex]
[tex]\\ \sf\longmapsto \omega=11rad/s[/tex]
How to calculate voltage U1 ?
Please help!
Answer:
he is a baby art and design
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!
In first case a mass M is split into two parts with one part being 1/6.334 th of the original mass. In second case M is split into two equal parts. In both the cases the two parts are separated by same distance. What ratio of the magnitude of the gravitational force in first case to the magnitude of the gravitational force in the second case
Answer:
[tex]F_r=0.132:0.25[/tex]
Explanation:
From the question we are told that:
[tex]M_1=M*\frac{1}{6.334}[/tex]
Therefore
[tex]M_2=M-M*\frac_{1}{6.334}[/tex]
[tex]M_2=M*\frac{5.334}{6.334}[/tex]
Generally the equation for Gravitational force of attraction is mathematically given by
For Unequal split
[tex]F=\frac{GM_1M_2}{d^2}[/tex]
[tex]F=\frac{G(M*\frac_{1}{6.334})(M*\frac{5.334}{6.334})}{d^2}[/tex]
[tex]F=\frac{GM^2}{d^2}*(0.132)[/tex]
For equal split
[tex]F=\frac{GM_1M_2}{d^2}[/tex]
[tex]F=\frac{G(\frac{M}{2})((\frac{M}{2}}{d^2}[/tex]
[tex]F=0.25 \frac{GM^2}{d^2}[/tex]
Therefore the ratio of the gravitational force is
[tex]F_r=0.132:0.25[/tex]