Answer:
D
Explanation:
right edge 2022
As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock. As the waves recede, they carry the sediment away. In this scenario, which process represents weathering, and which process represents erosion?
Answer:
WEATHERING is represented by the scenario (As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock).
Erosion is represented by the scenario (As the waves recede, they carry the sediment away).
Explanation:
A wave is a disturbance which travels through a medium and transfers energy from one point to another. When wind blows over a water body like the ocean, ocean waves are formed. As the generated energy from the wind is transported through the water by the waves, the can hit against rocks on the shores leading to its break down with time. WEATHERING occurs when tiny bit of sediments from rocks are loosened due to the impact of ocean waves.
Erosion can be described as the wearing away of the earth's surface due to the impact of wind, rainfall ( water) or waves. There are different types of erosion which is classified according it's cause of formation.
Wave erosion occurs when sediments such as sand, shell and other materials are carried to the shoreline by ocean waves. This erodes the shore over time as the sediments act like sandpapers.
A student is driving through a mountainous region where the road is at some times flat, at some times inclined upward, and at some time inclined downward. The student maintains a speed of 20 m/s on the roadway, but is required to make an emergency stop on the three sepearte occasions. On levels roadway, it takes 25 m to stop. On a downward-sloping roadway, it takes 40 m to stop. On an upward-sloping roadway, it takes 18 m to stop. Explain why the stopping distances are different. (Focus answer using work and energy, other concepts may be used as well but be sure work and energy are included.)
Answer:
Explanation:
It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .
At levelled road , for stoppage
Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .
At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .
At upward inclined road , for stoppage
Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force = (friction force + gravitational force ) x displacement .
Hence displacement is less .
At downward slopping road , friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle also .
At downward inclined road , for stoppage
Kinetic energy of vehicle + work done by gravitational force = Work done by frictional force = friction force x displacement .
Hence displacement is more .
Hence displacement is more in the downward slopping.
What is Displacement?Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.
It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .
At levelled road , for stoppage
Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .
At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .
At upward inclined road , for stoppage
Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force = (friction force + gravitational force ) x displacement .
Hence displacement is less .
At downward slopping road , friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle also .
At downward inclined road , for stoppage
Kinetic energy of vehicle + work done by gravitational force = Work done by frictional force = friction force x displacement .
Hence displacement is more in the downward slopping.
To learn more about displacement refer to the link:
brainly.com/question/11934397
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How much work does a 50.0kg person do in walking up one flight of stairs, equivalent to 3.0m?
Answer:
1470J
Explanation:
Given parameters:
Mass of the person = 50kg
height = 3m
Unknown:
Work done = ?
Solution:
To solve this problem, we use the expression below:
Work done = mass x acceleration due gravity x height
So;
Work done = 50 x 9.8 x 3 = 1470J
please help!!!!!! 26 points
A Boeing 787 is initially moving down the runway at 6.0 m/s preparing for takeoff. The pilot pulls on the throttle so that the engines give the plane a constant acceleration of 1.9 m/s2. The plane then travels a distance of 1700 m down the runway before lifting off. How long does it take from the application of the acceleration until the plane lifts off, becoming airborne
Answer:
39.26 s
Explanation:
From the question given above, the following data were obtainedb
Initial velocity (u) = 6 m/s
Acceleration (a) = 1.9 m/s²
Distance travelled (s) = 1700 m
Time (t) =?
Next, we shall determine the final velocity of the plane. This can be obtained as follow:
Initial velocity (u) = 6 m/s
Acceleration (a) = 1.9 m/s²
Distance travelled (s) = 1700 m
Final velocity (v) =?
v² = u² + 2as
v² = 6² + (2 × 1.9 × 1700)
v² = 36 + 6460
v² = 6496
Take the square root of both side
v = √6496
v = 80.6 m/s
Finally, we shall determine the time taken before the plane lifts off. This can be obtained as follow:
Initial velocity (u) = 6 m/s
Acceleration (a) = 1.9 m/s²
Final velocity (v) = 80.6 m/s
Time (t) =?
v = u + at
80.6 = 6 + 1.9t
Collect like terms
80.6 – 6 = 1.9t
74.6 = 1.9t
Divide both side by 1.9
t = 74.6 / 1.9
t = 39.26 s
This, it will take 39.26 s before the plane lifts off.
Why does the sky change colors at sunset?
The atmosphere reflects colors.
The atmosphere bends light.
The sun changes colors.
The sky changes colors at sunset because the atmosphere ABSORBS some colors of light more than other colors.
The second choice ("The atmosphere bends light") is a correct statement, but it's not the reason that the sky changes colors at sunset.
Answer:
The atmosphere bends light.
Explanation:
I got the test, and passed! =)
What is the purpose of a tractor?
To plow fields.
To harvest wheat.
To haul loads of manure, hay etc.
To power other machines.
to plow fields beacuse it's much easier to plow on tractor than on foot
Two identical cylinders with a movable piston contain 0.7 mol of helium gas at a temperature of 300 K. The temperature of the gas in the first cylinder is increased to 564 K at constant volume by doing work W1 and transferring energy Q1 by heat. The temperature of the gas in the second cylinder is increased to 564 K at constant pressure by doing work W2 while transferring energy Q2 by heat.
A. Find ΔEint, 1, Q1, and W1 for the process at constant volume.
B. Find ΔEint, 2, Q2, and W2 for the process at constant pressure.
Answer:
Explanation:
A ) At constant volume :
ΔEint = n Cv x ΔT , n is no of moles , Cv is specific heat at constant volume , ΔT is increase in temperature .
For helium Cv = 3/2 R = 1.5 x 8.3 J = 12.45 J
ΔEint = .7 x 12.45 x ( 564 - 300 )
= 2300.76 J .
W₁ = 0 because volume is constant so work done by gas is zero .
Q₁ = ΔEint = 2300.76 J
B )
At constant pressure
Q₂ = n Cp Δ T , Cp is specific heat at constant pressure .
For monoatomic gas ,
Cp = 5/2 R = 2.5 x 8.3 J = 20.75 J
Q₂ = .7 x 20.75 x 264 J
= 3834.6 J
W₂ = work done by gas
= PΔV = nRΔT
= .8 x 8.3 x 264
= 1752.96 J
ΔEint = Q₂ - W₂
= 3834.6 - 1752.96
= 2081.64 J.
ΔEint, 1, Q1, and W1 for the process at constant volume. and ΔEint, 2, Q2, and W2 for the process at constant pressure is mathematically given as
a)
dE1= 2300.76 J .
W1=0 as Volume is constant
Q1= 2300.76 J as Q= dE1
b)
Q2= 3834.6 J
W2= 1752.96 J
dE2= 2081.64 J.
What is ΔEint, Q1 and W1 for the process at constant volume and pressure?a)
Generally, the equation for the Constant Volume is mathematically given as
dE = n Cv x dT
Where
Cv = 3/2
R = 1.5 x 8.3 J
R= 12.45 J
Therefore
dE = 0.7 x 12.45 x ( 564 - 300 )
dE1= 2300.76 J .
W1=0 as Volume is constant
Q1= 2300.76 J as Q= dE1
b)
Generally
Q2 = n Cp dT
Where
Cp = 5/2
R = 2.5 x 8.3 J
R= 20.75 J
Hemce
Q2 = 0.7 x 20.75 x 264 J
Q2= 3834.6 J
For Work done
W=PdV
W= nRdT
Therefore
W= 0.8 x 8.3 x 264
W2= 1752.96 J
Hence
dE = Q₂ - W₂
dE= 3834.6 - 1752.96
dE2= 2081.64 J.
For more information on Pressure
https://brainly.com/question/25688500
When a rattlesnake strikes, its head accelerates from rest to a speed of 22 m/s in 0.48 seconds. Assume for simplicity that the only moving part of the snake is its head of mass 170 g. How much (average) power does the rattlesnake need to accelerate its head that fast? Answer in units of W.
Answer:
P = 85.72 W
Explanation:
Given that,
Initial speed, u = 0
Final speed, v = 22 m/s
Time, t = 0.48 s
Mass, m = 170 g = 0.17 kg
Let a be the acceleration of the rattlesnake.
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{22-0}{0.48}\\\\a=45.84\ m/s^2[/tex]
Let x is the displacement of a rattlesnake. It can be given by :
[tex]x=ut+\dfrac{1}{2}at^2\\\\x=0+\dfrac{1}{2}\times 45.84\times (0.48)^2\\\\x=5.28\ m[/tex]
The power of the rattlesnake is given by :
[tex]P=\dfrac{W}{t}\\\\P=\dfrac{m\times a\times x}{t}\\\\P=\dfrac{0.17\times 45.84\times 5.28}{0.48}\\\\P=85.72\ W[/tex]
So, the power of the rattlesnake is 85.72 W.
Which has the most mass?
O The Moon
O A Pencil
O Your teacher.
O Earth
Answer:
Earth
lol... ....
How much work is done in pushing a 10kg box, 2.0m to the right
with a force of 20.0N?
Answer:
40J
Explanation:
Given parameters:
Mass of box = 10kg
distance moved = 2m
Force applied = 20N
Unknown:
Work done in pushing the box = ?
Solution:
The work done in moving a body is a function of the force applied to move it through a particular distance.
Work done = Force x distance
So;
Work done = 20 x 2 = 40J
a car accelerates from 2 m/s to 28m/s at a constant rate of 3 m/s^2. How far does it travel while accelerating?
Answer:
Distance, S = 130m
Explanation:
Given the following data;
Initial velocity = 2m/s
Final velocity = 28m/s
Acceleration = 3m/s²
To find the distance, we would use the third equation of motion.
V² = U² + 2aS
Substituting into the equation, we have;
28² = 2² + 2*3*S
784 = 4 + 6S
6S = 784 - 4
6S = 780
S = 780/6
Distance, S = 130m
a 14n force is applied for 0.33 seconds, calculate the impulse
Answer:
4.62 N-s
Explanation:
recall that the formula for impulse is given by
Impulse = Force x change in time
in our case, we are given
Force = 14 N
change in time = 0.33s
Simply substituting the above into the equation for impulse, we get
Impulse = Force x change in time
Impulse = 14 x 0.33
= 4.62 N-s
[tex]\\ \sf\longmapsto Impulse=Force(Time)[/tex]
[tex]\\ \sf\longmapsto Impulse=14(0.33)[/tex]
[tex]\\ \sf\longmapsto Impulse=4.62Ns[/tex]
What is the 59th element
Answer:
The 59th element is Praseodymium it's symbol is Pr.
An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the center of the fluid surface is depressed. At what angular velocity
Answer:
Explanation:
To find the angular velocity of the tank at which the bottom of the tank is exposed
From the information given:
At rest, the initial volume of the tank is:
[tex]V_i = \pi R^2 h_i --- (1)[/tex]
where;
height h which is the height for the free surface in a rotating tank is expressed as:
[tex]h = \dfrac{\omega^2 r^2}{2g} + C[/tex]
at the bottom surface of the tank;
r = 0, h = 0
∴
[tex]h = \dfrac{\omega^2 r^2}{2g} + C[/tex]
0 = 0 + C
C = 0
Thus; the free surface height in a rotating tank is:
[tex]h=\dfrac{\omega^2 r^2}{2g} --- (2)[/tex]
Now; the volume of the water when the tank is rotating is:
dV = 2π × r × h × dr
Taking the integral on both sides;
[tex]\int \limits ^{V_f}_{0} \ dV = \int \limits ^R_0 \times 2 \pi \times r \times h \ dr[/tex]
replacing the value of h in equation (2); we have:
[tex]V_f} = \int \limits ^R_0 \times 2 \pi \times r \times ( \dfrac{\omega ^2 r^2}{2g} ) \ dr[/tex]
[tex]V_f = \dfrac{ \pi \omega ^2}{g} \int \limits ^R_0 \ r^3 \ dr[/tex]
[tex]V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{r^4}{4} \Big]^R_0[/tex]
[tex]V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{R^4}{4} \Big] --- (3)[/tex]
Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.
Then [tex]V_f = V_i[/tex]
Replacing equation (1) and (3)
[tex]\dfrac{\pi \omega^2}{g}( \dfrac{R^4}{4}) = \pi R^2 h_i[/tex]
[tex]\omega^2 = \dfrac{4g \times h_i }{R^2}[/tex]
[tex]\omega =\sqrt{ \dfrac{4g \times h_i }{R^2}}[/tex]
[tex]\omega = \sqrt{\dfrac{4 \times 9.81 \ m/s^2 \times 0.7 \ m}{(0.5)^2} }[/tex]
[tex]\omega = \sqrt{109.87 }[/tex]
[tex]\mathbf{\omega = 10.48 \ rad/s}[/tex]
Finally, the angular velocity of the tank at which the bottom of the tank is exposed = 10.48 rad/s
Your __________ environment includes the people you spend time with.
Answer:
social
Explanation:
Your social environment includes the people you spend time with – your family, your friends, classmates and other people in your community. Your social environment is healthier when you choose friends who show concern for their own health and yours.
Describe the position of the sun, moon, and earth during a new moon and a full moon.
During a new moon, (erase this and insert your answer describing how the earth, moon, and sun are aligned).
During a full moon (erase this and insert your answer describing how the earth, moon, and sun are aligned).
Why do we experience the different phases of the moon?
We experience the different phases of the moon because
Answer:
* he new moon phase when the position is Sun - Moon - Earth,
* have of the Full Moon when the position is Sun - Earth - Moon,
*All the phases of the moon are governed by the movement of the Moon around the Earth.
Explanation:
In the solar system, the planets revolve around the sun, which is much more massive, in the case of the Earth it is more massive than its satellite, therefore the Moon revolves around the Earth in a period of approximately 28 days.
It is said that the moon is in the new moon phase when the position is Sun - Moon - Earth, so the moon cannot be seen
It is in the phase of the Full Moon when the position is
Sun - Earth - Moon, in this case the moon can be observed by the light reflected from it.
All the phases of the moon are governed by the movement of the Moon around the Earth.
What do I know about dance movements?
Answer:
Hi lol
Explanation:
follow me lol
sarivigakarthi this is my account...
what is energy? list the three mechanical forms of energy and their associated equations.
Answer:
mm hope that helps
Explanation:
What are the 3 forms of energy?
Forms of energy
Potential energy.
Kinetic energy.
Which statements correctly describe the formula or name of a compound? Select all that apply.
A. The formula of nitrogen trifluoride is NF 3
OB. The formula of ammonia is NH3
OC. The name of AlF3 is trialuminum fluoride.
D. The formula of calcium chloride is cach,
DE The name of Lise is lithium selenate.
F. The formula of dinitrogen monoxide is NO
G. The formula of sulfur trioxide is so
H. The formula of magnesium hydroxide is Mg(OH)2
Answer:
The formula of nitrogen trifluoride is NF 3
The formula of ammonia is NH3
The formula of magnesium hydroxide is Mg(OH)2
Explanation:
The formula of a compound refers to the correct representation of the compound using appropriate chemical symbols.
The formula of a compound is derived from the symbols of its constituent elements. The formula of the compound must correctly show the number of atoms of each element in one formula unit of the compound.
In the answer section, the correct names or formulas of some compounds have been shown
Answer:
A. The formula of nitrogen trifluoride is NF3.
B. The formula of ammonia is NH3.
D. The formula of calcium chloride is CaCI2.
H. The formula of magnesium hydroxide is Mg(OH)2.
Explanation:
On March 27, 2004, the United States successfully tested the hypersonic X-43A scramjet, which flew at Mach 7.0 (seven times the speed of sound) for 11 seconds. (A scramjet gets its oxygen directly from the air, rather than from fuel.) For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Swim competition. Part A At this rate, how many minutes would it take such a scramjet to carry passengers the approximately 5000 kmkm from San Francisco to New York? (Use 331 m/sm/s for the speed of soun
Answer:
Explanation:
Speed of sound = 331 m /s
speed of jet = 7 .00 Mach = 7 times speed of sound
= 7 x 331 = 2317 m /s
distance to be covered = 5000 x 1000 = 5 x 10⁶ m
Time taken = distance / speed of jet
= 5 x 10⁶ / 2317
= 2.158 x 10³ s
= 35.96 minutes .
Why is deforestation a serious global environment problem
Answer:
we all know that deforestation is when tress are cut down but this may cause cardio dioxide to go up into the atmosphere and this may cause the rise in sea level and temperates tend to fluctuate
Explanation:
I'm just that smart yah dig
what is energy? list the three mechanical forms of energy and their associated equations.
Answer:
See the explanation below.
Explanation:
Energy is the ability of bodies to perform work and produce changes in themselves or other bodies.
There are several types of energy, but let's talk specifically about mechanical energy.
Mechanical energy is associated or subdivided into kinetic, potential, and elastic energies.
Kinetic energy
[tex]E_{kin}=\frac{1}{2}*m*v^{2}[/tex]
Potential energy
[tex]E_{pot}=m*g*h[/tex]
Elastic Energy
[tex]E_{elas}=\frac{1}{2} *k*x^{2}[/tex]
What energy store is in the hammer
BEFORE it hits the nail?
Answer:
mechanical energy
Explanation:
when raised up has potential energy
Answer:
potential energy
Explanation:
potential energy
According to Newton's law of universal gravitation, which statements are true?
As we move to higher altitudes, the force of gravity on us decreases.
O As we move to higher altitudes, the force of gravity on us increases,
O As we gain mass, the force of gravity on us decreases.
O Aswe gain mass, the force of gravity on us increases.
DAs we move faster, the force of gravity on us increases.
Suppose you drop a superball of massMand a marble of mass m(both treated as point masses) from a heighthwith the marble just on top of the superball. A superbowl has essentially elastic collisions with hard objects. Thesuperball hits the floor, rebounds and then collides with the marble. How highdoes the marble go if all the motion is vertical
Answer:
h '= [ ( \frac{ M-m}{M +m } )+ 2 (\frac{M}{M+m})]² h
Explanation:
Let's analyze this problem, first the two bodies travel together, second the superball bounces, third it collides with the marble and fourth the marble rises to a height h ’
let's start by finding the velocity of the two bodies just before the collision, let's use the concepts of energy
starting point. Starting point
Em₀ = U = m g h
final point. Just before the crash
Em_f = K = ½ m v²
as there is no friction the mechanical energy is conserved
Em₀ = Em_f
mg h = ½ m v²
v = √2gh
this speed is the same for the two bodies.
Second point. The superball collides with the ground, this process is very fast, so we will assume that the marble has not collided, let's use the concept of conservation of moment
initial instant. Just when the superball starts contacting the ground
p₀ = M v
this velocity is negative because it points down
final instant. Just as the superball comes up from the floor
p_f = M v '
the other body does not move
p₀ = p_f
- m v = M v '
v ’= -v
Therefore, the speed of the asuperbola is the same speed with which it arrived, but in the opposite direction, that is, upwards.
Let's use the subscript 1 for the marble and the subscript 2 for the superball
Third part. The superball and the marble collide
the system is formed by the two bodies, so that the forces during the collision are internal and the moment is conserved
initial instant. Moment of shock
p₀ = M [tex]v_{1'}[/tex]+ m v_2
final instant. When the marble shoots out.
P_f = Mv_{1f'}+ m v_{2f}
p₀ = p_
M v_{1'}+ m v_2 = M v_{1f'} + m v_{2f}
M (v_{1'} - v_{1f'}) = -m (v_2 - v_{2f})
in this expression we look for the exit velocity of the marble (v2f), as they indicate that the collision is elastic the kinetic nerve is also conserved
K₀ = K_f
½ M v_{1'}² + m v₂² = M v_{1f'}² + ½ m v_{2f}²
M (v_{1'}² - v_{1f'}²) = - m (v₂² - v_{2f}²)
Let's set the relation (a + b) (a-b) = a² - b²
M (v_{1'} + v_{1f'}) (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})
let's write our two equations
M ( v_{1'} - v_{1f'}) = -m (v₂ - v_{2f}) (1)
M (v_{1'} + v_{1f'}) (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})
if we divide these two expressions
(v_{1'}+ v_{1f'}) = (v₂ + v_{2f} )
we substitute this result in equation 1 and solve
v_{1f'}= (v₂ + v_{2f}) - v_{1'}
M (v_{1'} - [(v₂ + v_{2f}) - v_{1'}] = -m (v₂ - v_{2f})
-M v₂ - M v_{2f1'} + 2M v_{1'} = m v₂ - m v_{2f}
-M v_{2f} -m v_{2f} = m v₂ -M v₂ + 2M v_{1'}
v_{2f} (M + m) = - v₂ (M-m) + 2 M v_{1'}
v_{2f} = - [tex]( \frac{ M-m}{M +m } )[/tex]) v₂ + 2 [tex](\frac{M}{M+m})[/tex] v_{1'}
now we can substitute the velocity values found in the first two parts
[tex]v_{2f}[/tex] = - ( \frac{ M-m}{M +m } ) √2gh + 2(\frac{M}{M+m}) √2gh
we simplify
v_{2f} = [( \frac{ M-m}{M +m } ) + 2 (\frac{M}{M+m})] [tex]\sqrt{2gh}[/tex]
let's call the quantity in brackets that only depends on the masses
A = ( \frac{ M-m}{M +m } )+ 2 (\frac{M}{M+m})]
v_{2f}= A \sqrt{2gh}
in general, the marble is much lighter than the superball, so its speed is much higher than the speed of the superball
finally with the conservation of energy we find the height that the marble reaches
Starting point
Emo = K = ½ mv_{2f}²
Final point
Emf = U = m g h'
Em₀ = Em_f
½ m v_{2f}² = m g h ’
h ’= ½ v_{2f}² / g
h ’= ½ (A \sqrt{2gh})² / g
h ’= A² h
h '= [ ( \frac{ M-m}{M +m } )+ 2 (\frac{M}{M+m})]² h
. A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 192 m>s and a frequency of 240 Hz. The amplitude of the standing wave at an antinode is 0.400 cm. (a) Calculate the amplitude at points on the string a distance of (i) 40.0 cm; (ii) 20.0 cm; and (iii) 10.0 cm from the left end of the string. (b) At each point in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement
Answer:
Explanation:
The formula holding the relationship between wavelength and frequency of a wave can be expressed as:
[tex]\lambda = \dfrac{v}{f}[/tex]
where;
[tex]\lambda =[/tex] wavelength
v = speed
f = frequency
Given that:
v = 192 m/s and f = 240Hz
Then;
[tex]\lambda = \dfrac{192 \ m/s}{240 \ Hz}[/tex]
[tex]\lambda = 0.800 \ m[/tex]
Now, to estimate the respective amplitude of the string, we need to approach it by using the concept of wave equation which is:
y = A sin kx
here;
A = amplitude of the standing wave
k = wave number
x = maximum displacement
y = distance from center
recall that:
[tex]k = \dfrac{2 \pi}{\lambda}[/tex]
∴
[tex]y = A sin \dfrac{2 \pi}{\lambda }x[/tex]
Now;
for A = 0.400 cm ; [tex]\lambda[/tex] = 0.800 m ; k = 40 cm
Then;
[tex]y =(0.400 \ cm ) \ sin \dfrac{2 \pi}{0.800 m }\times (40 \ cm \times \dfrac{10^{-2} \ m}{1 \ cm } )[/tex]
y = 0.400 sin π
y = 0 cm
At distance 40 cm; the amplitude = 0 cm
Thus, it is a node.
For k = 20cm
Then:
[tex]y =(0.400 \ cm ) \ sin \dfrac{2 \pi}{0.800 m }\times (20 \ cm \times \dfrac{10^{-2} \ m}{1 \ cm } )[/tex]
y = 0.400 sin π/2
y = 0.400 cm
At distance 20 cm; the amplitude = 0.400 cm
Thus, it is antinode.
For k = 10cm
Then:
[tex]y =(0.400 \ cm ) \ sin \dfrac{2 \pi}{0.800 m }\times (10 \ cm \times \dfrac{10^{-2} \ m}{1 \ cm } )[/tex]
4y = 0.400 sin π/2
y = 0.283 cm
At distance 10 cm; the amplitude = 0.283 cm
b)
The required time taken to go through the displacement( i.e. from largest upward to downward) is the time required to cover half of the wavelength.
This is expressed as:
[tex]T = \dfrac{1}{2} \times \dfrac{1}{f}[/tex]
[tex]T= \dfrac{1}{2} \times \dfrac{1}{240 \ Hz}[/tex]
T = 0.00208
T = 2.08 × 10⁻³ s
Your friend, a world-class long jumper, is trapped on the roof of a burning building. His only escape route is to jump to the roof of the next building. Fortunately for him, he is in telephone contact with you, a Physics 161 student, for advice on how to proceed. He has two options. He can jump to the next building by using the long-jump technique where he jumps at 45o to the horizontal. Or, he can take his chances by staying where he is in the hopes that the fire department will rescue him. You learn from the building engineers that the next building is 10 m away horizontally and the roof is 3 m below the roof of the burning building. You also know that his best long-jump distance is 7.9 m. What do you advise him to do
Answer:
y = -2.69 m
the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.
Explanation:
his problem must be solved with the missile launch equations.
Let's start by looking for the jumper's initial velocity
R = v₀² sin 2θ / g
for the long jump the angle used is tea = 45º, in the exercise they indicate that the best record is R = 7.9m
v₀² = R g / sin 2te
v₀ = [tex]\sqrt{ \frac{7.9 \ 9.8}{1 }[/tex]
v₀ = 8.80 m / s
Now suppose you jump with this speed to get to the other building, let's use trigonometry for the components of the speed
sin 45 = [tex]v_{oy}[/tex] /v₀
cos 45 = v₀ₓ / v₀
v_{oy} = v₀ sin 45
v₀ₓ = v₀ cos 45
v_{oy} = 8.8 sin 45 = 6.22 m / s
v₀ₓ = 8.8 cos 45 = 6.22 m / s
now let's calculate the sato with these speeds
x = [tex]v_{ox}[/tex] t
the minimum jump is x = 10 m
t = x / v₀ₓ
t = 10 / 6.22
t = 1.61 s
let's find the vertical distance for this time
y = v_{oy} t - ½ g t²
where zero is placed on the jump building
y = 6.22 1.61 - ½ 9.8 1.61²
y = -2.69 m
Let's analyze this result, the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.
Please help. I'm stuck!
What is the mass of a catamaran moving at 7.65 m/s that has a momentum of 530145 kg x m/s?
A cars mass is 950kg and it travels at a speed of 35 m/s when it rounds a flat curve of radius 215 m.
a. Determine the value of the frictional force exerted on the car.
b. Determine the value of the coefficient of friction between the tires and the road.
(a) It's the force of (static) friction that keeps the car on the road and prevents it from skidding, and this friction is directed toward the center of the curve.
Recall that centripetal acceleration has a magnitude a of
a = v ² / R
where
v = tangential speed
R = radius of the curve
so that
a = (35 m/s)² / (215 m) ≈ 5.69767 m/s² ≈ 5.7 m/s²
Parallel to the road, the only force acting on the car is friction. So by Newton's second law, we have
∑ F = Fs = m a
where
Fs = magnitude of static friction
m = mass of the car
Then
Fs = (950 kg) (5.7 m/s²) ≈ 5412.79 N ≈ 5400 N
(b) Perpendicular to the road, the car is in equilbrium, so its weight and the normal force of the road on the car are equal in magnitude. By Newton's second law,
N - W = 0
where
N = magnitude of normal force
W = weight
so that
N = W = m g = (950 kg) (9.8 m/s²) = 9310 N
Friction is proportional to the normal force by a factor of µ, the coefficient of static friction:
Fs = µ N
Assuming 35 m/s is the maximum speed the car can travel without skidding, we find
µ = Fs / N = (5400 N) / (9310 N) ≈ 0.581395 ≈ 0.58