If plants do not get watered, they will start to wilt because their central vacuoles are empty. When they get water again, they will stand back up because their vacuoles have filled back up. What process allows their vacuoles to fill back up with water? *
Answers
Answer:
osmosis
Explanation:
The process that allows the vacuoles of plants to fill back up with water after a period of wilting is osmosis.
Osmosis is the movement of water molecules from a region of higher water potential to a region of lower water potential through a semi-permeable membrane. Water enters the vacuoles of plants osmotically once water becomes available in their cells.
The availability of water in the cytosol makes it to have a higher water potential than the cell sap. In this case, the vacuole membrane or the tonoplast acts as the semi-permeable membrane and allows water to diffuse into the vacuole.
Related Questions
what is the formula to calculate Acceleration?
Answers
Answer:
too much :(
Explanation:
\overline{a} = \frac{v - v_0}{t} = \frac{\Delta v}{\Delta t}
\overline{a} = average acceleration
v = final velocity
v_0 = starting velocity
t = elapsed time
Answer:
a = Δv⁄t.
Explanation:
toge
e
6. Refer to the illustration below.
90
www
R₂ 1513
www
R₂ 1013
a. what is the total power in the circuit?
b. What is the total resistance of this
circuit?
Answers
The total resistance of this circuit is 15Ω and the total power of the circuit is 60W.
The power is the ratio of the square of voltage and resistance. The total resistance is obtained from the addition of series and parallel resistance.
From the given,
Total resistance (Requ) = R₁ + R₂
R₁ is a series resistance
R₂ is the parallel resistance
R₂ = 1/15 Ω + 1/10 Ω
= 10×15 / (15+10)
= 150 / 25
= 6Ω
Parallel resistance R₂ = 6Ω
R(equivalent) = R₁ + R₂
= 9 + 6
= 15Ω
Thus, the total resistance is 15 Ω.
The total power, P = E² / R(equivalent)
E represents the voltage
R(equivalent) is the equivalent resistance
P = 30×30 / 15
= 60 watts.
Thus, the total power in the circuit is 60 watts.
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A 1500 kg car is moving to the right with a speed of 20.0 m/s when it collides with a wall and reboubds at a speed of 5.00 m/s.
If the collision lasts for 250 ms, then the magnitude of the average force acring on the car is _____ kN (the answer is 150 but I'm not sure how)
pls help!!
Answers
Answer:
See below.
Explanation:
When the 1500 kg car collides with the wall and rebounds at a speed of 5.00 m/s, we can calculate the change in the car's velocity using the following formula:
Δv = v2 - v1
Where Δv is the change in velocity, v2 is the final velocity, and v1 is the initial velocity. Substituting the given values, we get:
Δv = 5.00 m/s - 20.0 m/s
Δv = -15.0 m/s
The negative sign indicates that the direction of the car's velocity has reversed, or that the car is now moving to the left. To calculate the magnitude of the change in velocity, we take the absolute value:
|Δv| = |-15.0 m/s|
|Δv| = 15.0 m/s
Therefore, the magnitude of the change in velocity is 15.0 m/s.
Now,
To find the magnitude of the average force acting on the car during the collision, we can use the impulse-momentum theorem, which states that:
Impulse = change in momentum
Average force = Impulse / time
The change in momentum of the car is given by:
Δp = mΔv
where Δv is the change in velocity calculated in the previous answer and m is the mass of the car.
Δp = 1500 kg × (-15.0 m/s)
Δp = -22500 kg·m/s
The impulse acting on the car during the collision is equal to the change in momentum:
Impulse = Δp = -22500 kg·m/s
To find the magnitude of the average force acting on the car during the 250 ms collision, we divide the impulse by the duration of the collision:
Average force = Impulse / time
Average force = -22500 kg·m/s / 0.250 s
Average force ≈ -90,000 N
The negative sign indicates that the force is in the opposite direction of the car's motion, or to the left. Therefore, the magnitude of the average force acting on the car during the collision is approximately 90,000 N.
You apply a 325 N force to a heavy crate with a rope that makes a 29.0° angle with the horizontal, if you pull the crate a distance of 5.37 m, how much work was done by you?
Answers
Answer:
W = 1526.43 J
Explanation:
Given that,
Force acting on a heavy crate, F = 325 N
The rope will make 29.0° angle with the horizontal
The distance covered by the crate, d = 5.37 m
We need to find the work done by the crate. The work done by an object is given by :
\(W=Fd\cos\theta\\\\=325 \times 5.37\times \cos29\\\\=1526.43\ J\)
So, the required work done is 1526.43 J.
The parents of a new baby believe they brought the wrong child home from the hospital. Gel electrophoresis was performed using DNA samples from the parents and the child. A section of the gel electrophoresis results is shown below.Which conclusion is valid based on the gel electrophoresis results?
Answers
Can someone help me label this flower??
Answers
Hope this helps!!!!!!!!!!!!!
Frog jumps to the left with average speed of 1.8 m/s for 0.55 s whats the frogs displacement in meters
Answers
Answer: 0.99m left
Explanation:
The formula:
Distance / Time = Rate
Fill in the formula
Distance / 0.55s = 1.8 m/s
Multiply both sides by 0.55s
Distance = 0.99m
Displacement = 0.99m to the left.
Answer:
-0.99 (See attached)
Explanation:
(See attached)
what are the two pieces of evidence that support the theory of continental drift?
Answers
Answer:
the two pieces are:
Evidences of sea floor spreadingGeographical similarities between the opposing coast of Atlantic oceanExplanation:
Evidences of sea floor spreading
It has been reported that from geodetic evidence that Greenland is drifting westward at the rate of 20cm per year.
Geographical similarities between the opposing coast of Atlantic ocean
The outlines of the coast on the other two sides of the Atlantic are such that they can be easily joined together and appear to be a detached portion of one another.
What is the
covering that
surrounds
both plant and
animal cells?
Answers
Answer:
cell membrane ) is the covering that surrounds both plant and animal cells
PLS HELP which ones would be made of cells? and which ones show cell walls?
Cork, Sponge, Wood, Plastic, Tree
Answers
The first question's answer depends on what you mean by "sponge". If you're talking about sea sponges, then all but plastic are made up of cells. Some sponges used for cleaning are also made of plant material but also other, non-organic materials like dyes.
Cell walls are only present in plant cells, so they would be found in cork (derived from a certain tree bark), wood, and trees. Synthetic sponges made with plant material might also contain them, but they wouldn't be made entirely of cells with walls.
From given options or choices all are made up of cells except plastic and among the other four Cork, Sponge, Wood, and tree all show cell walls except Sponge.
A cell is the basic structural unit of all living organisms and contains various cell organelles. On the base of different cell organelles or presence or absence of the certain organelles help in distinct and divide the cell type.
The major types of cells are:
Prokaryotic cellsEukaryotic cellsPlastic is not a living organism as there are no cells and is made up of polymers of hydrocarbon.
The cell wall is one of the major cellular structures that helps in identifying the type of cell organism and protects the organism from the external environment. It classifies the organism on its constituent of the cell wall.
In eukaryotic cells, animal cells have no cell wall, however, fungi cells, plant cells have a cell wall. A sponge is an animal and other cork wood, and trees are plants or plant-based products.
Thus, the sponge does not show the cell wall.
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an object of mass 0.2kg is thrown vertically upwards with a speed of 0.2 m/s. calculate potential energy at the highest point where h=10 and the kinetic energy under the same condition
Answers
Answer:
the potential energy is 19.6J
the kinetic energy is 0.0.004J
Explanation:
using the formulas
P.E = mgh while
K.E =mv²/
A sample of lead is heated up to a temperature of 100°C and then placed in a sample of water with an initial temperature of 5°C. If the mixture is thermally isolated from its surroundings, then it:
exchanges no thermal energy with the environment outside the system as it comes to a final temperature.
gains thermal energy from the environment outside the system as it comes to a final temperature.
both gains and loses thermal energy to the environment outside as it comes to a final temperature.
loses thermal energy to the environment outside as it comes to a final temperature.
None of these choices are correct.
Answers
If the mixture is thermally isolated from its surroundings, then it, loses thermal energy to the environment outside as it comes to a final temperature. The correct answer is d.
When the sample of lead is placed in the water, heat will flow from the lead to the water until they reach a common final temperature. Since the final temperature will be less than the initial temperature of the lead, heat must have flowed out of the lead into the surroundings, causing the lead to lose thermal energy to the environment outside the system.
Since the mixture is thermally isolated from its surroundings, no thermal energy is exchanged between the system (lead and water) and the environment during the process. However, heat can still flow within the system itself until thermal equilibrium is reached. Option d is correct.
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According to the Newton’s first law of motion , which of the following objects will change in motion
A. a biker on a smooth and level road , pedaling at a constant rate .
B. a train pulling on another train , each with the same amount of force .
C. a ball rolling across the floor with friction acting in the opposite direction.
D. A chair sitting on the floor with balanced forces acting in opposite directions.
Answers
A toy car of 1 kg mass is moving at 1 m/s. How much momentum does it have? show the formula substitutions to answer below
Answers
Answer:
momentum = 1 kg m/s
Explanation:
We are told that a toy car has a mass of 1 kg, and a speed of 1 m/s. In order to calculate the car's momentum from this information, we have to use the following formula:
\(\boxed{P = mv}\),
where:
• P = momentum
• m = mass of the object
• v = velocity of the object
Substituting m = 1 kg and v = 1 m/s into the above formula, we can calculate the car's momentum:
P = m × v
= 1 kg × 1 m/s
= 1 kg m/s
Therefore the momentum of the toy car is 1 kg m/s.
see
below
the
radius if Tantalum atom is 142 pm. gow many tantalum atoms would
have to be laid side-by-side to span a distance of 4.20 MM
_____ atoms
Answers
If the radius is 142 pm, approximately 14,788,732 tantalum atoms would need to be laid side-by-side to span a distance of 4.20 MM.
To determine the number of tantalum atoms that would need to be laid side-by-side to span a distance of 4.20 MM, we can use the given radius of a tantalum atom.
First, let's convert the distance of 4.20 MM to picometers (pm) for consistency. Since 1 mm is equal to 1,000,000 pm, 4.20 MM is equal to 4,200,000,000 pm.
Next, we need to calculate the diameter of a tantalum atom. The diameter is simply twice the radius. Therefore, the diameter of a tantalum atom is 2 * 142 pm = 284 pm.
To find the number of tantalum atoms that can fit in the given distance, we divide the distance by the diameter of a tantalum atom. So, 4,200,000,000 pm divided by 284 pm gives us the number of tantalum atoms.
Performing the calculation, we have:
4,200,000,000 pm ÷ 284 pm = 14,788,732.39
Since we cannot have a fraction of an atom, we round down to the nearest whole number. Therefore, approximately 14,788,732 tantalum atoms would need to be laid side-by-side to span a distance of 4.20 MM.
Therefore, the answer is:
Approximately 14,788,732 atoms.
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What is the purpose of setting a deadline for a goal?
Answers
Answer:
To help push yourself to reach the goal, It can also help you pace yourself time wise
Explanation:
what is the best fuse for 1000 w heater?
a)3A
b)4A
c)5A
d)11A
Answers
c)5A
Explanation:
As,
\(I = \frac{P}{V} = \frac{1000}{220} = 5\)
if load is 10 kn, cross sectionl area is 2 square mm, contact area = 5 square mm. what will be bearing stress?
Answers
The bearing stress in this scenario is 2 kN/mm². To calculate the bearing stress, we need to use the formula:
Bearing Stress = Load / Contact Area
Substituting the given values:
Bearing Stress = 10 kn / 5 square mm
Bearing Stress = 2 N/mm^2
It is important to note that bearing stress is a measure of the force per unit area exerted on the contact surface between two components. In this case, the load is distributed over an area of 5 square mm, resulting in a bearing stress of 2 N/mm^2. It is important to ensure that the bearing stress is within the allowable limits to prevent failure or damage to the components.
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In a partially-filled array, the capacity may be less than the array's size.
When inserting a value into a partially-filled array, in ascending order, the insertion position may be the same as capacity.
When inserting elements into a partially-filled array, the array should be declared const.
When comparing two partially-filled arrays for equality, both arrays should not be declared const.
When deleting an element from a partially-filled array, it is an error if the index of the element to be removed is < size.
When inserting a value into a partially-filled array, elements following the insertion position are shifted to the left.
In a partially-filled array, the size represents the allocated size of the array.
In a partially-filled array, the capacity represents the effective size of the array.
In a partially-filled array, all of the elements are not required to contain meaningful values
When inserting an element into a partially-filled array, it is an error if size < capacity.
In a partially-filled array, all of the elements contain meaningful values
When deleting elements from a partially-filled array, the array should be declared const.
In a partially-filled array capacity represents the number of elements that are in use.
When searching for the index of a particular value in a partially-filled array, the array should not be declared const.
When inserting a value into a partially-filled array, in ascending order, the insertion position is the index of the first value smaller than the value.
True or False :
Answers
The statement "When inserting an element into a partially-filled array, it is an error if size < capacity" is true. When inserting an element into a partially-filled array, it is an error if size < capacity.How to insert a value into a partially-filled array?
The array should be traversed starting from the right end, where the last value has been placed, until the position of the insertion value is found. If the value is less than or equal to the value at the current position, move one space to the left. Insert the value in the position to the right of the current position when it is greater than the value at the current position. If the insertion position is the same as the array capacity, the value can be inserted at that location.The insertion of the element into the partially filled array shifts all the elements that come after the insertion position to the right. If the element is to be inserted at index k, and the current elements at positions k to size-1, they will be moved to k+1 to size.If the deletion of an element is to be performed in a partially filled array, it is an error if the index of the element to be removed is greater than or equal to the size of the array. The elements will be shifted to the right to fill the vacant position when an element is deleted.The following are true for a partially-filled array:In a partially-filled array, the capacity represents the effective size of the array.In a partially-filled array, all of the elements are not required to contain meaningful values.In a partially-filled array, the size represents the allocated size of the array.The number of elements that are in use is represented by the capacity in a partially-filled array.
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A vector is defined by its direction and length; its location does not
matter.
A:True
B:False
Answers
False.
a quantity having direction as well as magnitude, especially as determining the position of one point in space relative to another.Vectors are quantities that are fully described by magnitude and direction. Not Length!
Answer:
B.) False
Explanation:
Please put my answer as brainlist answer
I hope it helps you
PLEASE HELP
Three charges are on a line. A positive charge is on the far left labeled q Subscript 1 baseline positive 6 Coulombs. The second charge is 2 m away and is labeled q Subscript 2 baseline negative 4 Coulombs. The third charge is 1 m away and is labeled q Subscript 3 baseline = positive 3 Coulombs.
What is the electrical force between q2 and q3?
Recall that k = 8.99 × 109 N•meters squared over Coulombs squared..
1.0 × 1011 N
–1.1 × 1011 N
–1.6 × 1011 N
1.8 × 1011 N
Answers
Answer:
B) –1.1 × 1011 N
Explanation:
Since the second charge is 2 m away and is labeled q₂ = - 4 Coulombs. The third charge is 1 m away and is labeled q₃ = + 3 Coulombs, the electrical force between q₂ and q₃ is -1.1 × 10¹¹ N
To answer the question, we need to find the electrical force of attraction between two charges and this is given by Coulomb's law
Coulomb's lawThis states that the electrical force of attraction between two charges, F is directly proportional to the product of the charges q and Q and inversely proportional to the square of their distance apart, d.
So, mathematically F = kqQ/d²
Now, given that there are three charges are on a line. A positive charge is on the far left labeled q₁ = + 6 Coulombs. The second charge is 2 m away and is labeled q₂ = - 4 Coulombs. The third charge is 1 m away and is labeled q₃ = + 3 Coulombs.
The electrical force of attraction between q₂ and q₃So, the electrical force of attraction between q₂ and q₃ is F = kq₂q₃/d² where
k = 8.99 × 10⁹ Nm²/C², q₂ = - 4 C, q₃ = + 3 C and d = 1 m (since q₂ and q₃ are 1 m apart)Substituting the values of the variables into the equation, we have
F = kq₂q₃/d²
F = 8.99 × 10⁹ Nm²/C² × (-4 C) × (+ 3 C)/(1 m)²
F = 8.99 × 10⁹ Nm²/C² × (-12 C²)/(1 m)²
F = -107.88 × 10⁹ Nm²/1 m²
F = -1.0788 × 10¹¹ N
F ≅ -1.1 × 10¹¹ N
Since the second charge is 2 m away and is labeled q₂ = - 4 Coulombs. The third charge is 1 m away and is labeled q₃ = + 3 Coulombs, the electrical force between q₂ and q₃ is -1.1 × 10¹¹ N
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The peak production of nox typically occurs when the combustion temperatures are between 2,500 and 2,800 degrees fahrenheit (True or False)
Answers
The peak production of NOx (nitrogen oxides) typically occurs when the combustion temperatures are between 2,500 and 2,800 degrees Fahrenheit is True.
This is because at these temperatures, the nitrogen and oxygen in the air combine to form NOx compounds. This process is more likely to occur in engines that run hot, such as in gas turbines, diesel engines, and boilers. The high temperatures can be caused by factors such as high compression ratios, high air-to-fuel ratios, and high combustion pressures. The production of NOx is undesirable as it contributes to smog and acid rain and can also have adverse effects on human health. Therefore, there are regulations in place to limit the amount of NOx emissions from industrial processes and transportation.For more such question on production curve
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When the sun, the moon, and the Earth line up during a new moon, which of the following is produced?
low tide
high tide
spring tide
neap tide
Answers
Answer:
neap tide i think or high tide
Two runners start at the same point and jog at a constant speed along a straight path. Runner A starts at
time t = 0s, and runner B starts at time t = 2. 5 s. The runners both reach a distance 64 m from the starting
point at time t = 25 s.
If the runners continue at the same speeds, how far from the starting point will each be at time t = 45 s?
options:
Runner A's distance will be 115. 2 m, and runner B's distance will be 120. 7 m.
O Runner A's distance will be 64 m, and runner B's distance will be 71 m.
O Runner A's distance will be 115. 2 m, and runner B's distance will be 133. 65 m.
Runner A's distance will be 64 m, and runner B's distance will be 133. 65 m
Answers
Answer:
Va = 64 / 25 = 2.56 m/sec
Vb = 64 / 22.5 = 2.84 m/s
At t = 45
Sa = 2.56 m/s * 45 s = 115 m
Sb = 2.84 m/s * 42.5 s = 121 m
The first option given is the correct description
A proton, moving north, enters a magnetic field of a certain strength. Because of this field the proton curves downward. What is the direction of the magnetic field
Answers
The direction of the magnetic field is directed downward.
When a charged particle, such as a proton, enters a magnetic field, it experiences a force called the Lorentz force. The direction of this force is perpendicular to both the velocity of the particle and the magnetic field.
According to the right-hand rule, if the proton is moving north, the force on the proton is directed downward. Therefore, the direction of the magnetic field must be perpendicular to the velocity of the proton and downward in order to create the observed downward curvature of the proton's path.
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Use equation 11.27 to calculate the wavelength of the electronic transition in polyenes for n = 6, 8, and 10. comment on the variation of a with l, the length of the molecule.
Answers
Equation 11.27 can be used to calculate the wavelength of electronic transitions in polyenes for different values of n, such as n = 6, 8, and 10. The variation of a with l, the length of the molecule, can be observed and commented upon.
Equation 11.27, which is not provided here, likely relates to the mathematical expression used to calculate the wavelength of electronic transitions in polyenes. By applying this equation for different values of n (such as n = 6, 8, and 10), we can determine the corresponding wavelengths for the electronic transitions in polyenes with varying chain lengths.
By analyzing the results obtained from the calculations, we can comment on the variation of a with l, where a represents the wavelength and l represents the length of the molecule. This analysis will help us understand the relationship between the length of the polyene molecule and the wavelength of its electronic transitions. We may observe a pattern or trend indicating how the wavelength changes as the molecule lengthens.
Further analysis and interpretation of the calculated wavelengths and their relationship to the length of the molecule could provide insights into the behavior of electronic transitions in polyenes. It may help identify any systematic trends or deviations from expected patterns, leading to a better understanding of the structure and properties of polyene systems.
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1 A grandfather clock uses energy stored in raised weights. The weights transfer energy to the clock mechanism as they fall. One clock has a 4.5 kg weight that supplies energy to the chimes (which play a few notes every 15 minutes), and two 3.5 kg weights that power the clock and the mechanism that strikes the hours.
For all questions on this sheet,
use g = 10 N/kg
a Calculate how much energy is stored when all three of these weights are raised by 70 cm. b How far does the 4.5 kg weight have to be lifted to store 45 J of energy?
2 The water tank in a house can hold 200 litres of water. The mass of 1 litre of water is 1 kg. The tank is 2 m above the bathroom taps and 5 m above the kitchen taps. The kitchen taps are 1 m above the floor.
a
Calculate the gravitational potential energy (GPE stored in the water in the tank when it is full. State any assumptions made in your answer.
b Calculate the speed at which the water would come out of the bathroom taps and kitchen taps. You
may assume that no energy is transferred due to friction in the pipes.
3 The Victoria Falls in Africa is one of the world's largest waterfalls. Just over 1000 m° of water pass over the falls every second and fall approximately 100 m. 1 m3 of water has a mass of 1000 kg. a What mass of water goes over the falls every second? Give your answer in standard form.
b
Calculate the GPE of 1 kg of water at the top of the falls.
c If all the GPE stored in 1 kg of water is transferred to kinetic energy, calculate the speed of the water as
it reaches the bottom.
d Suggest why the water will not be falling as fast as your answer to part c suggests. e What is the total energy transferred per second as the GP stored in the water falling in one second is
transferred to other energy stores.
f Suggest the ways in which this energy is finally stored.
4 A post driver is used to drive fence posts into the ground. It is a hollow tube with a closed top, and handles on the side. A person fits the driver over a fence post, then lifts it up and lets it drop.
post driver
50 cm
a A post driver has a mass of 10 kg. Calculate the change in GPE stored when the post driver is lifted by 50 cm above the post, as shown in the diagram.
b
Calculate the speed of the driver when the end hits the post.
C
Explain how much extra energy is stored if the post driver is
fence post
lifted by 1 metre instead of only 50 cm.
d Calculate the speed of the post driver after it falls for 1 m. e A new design of post driver has a mass of 15 kg. Suggest one advantage and one disadvantage of this new design.
Extra challenge
5 F The post driver in question 4a stops in
0.5 seconds when it hits the fence post.
a Calculate the force needed to bring the post driver to a stop. (Hint: use your answer to 4b.)
The momentum of a moving object is the product of its mass and its velocity. The force needed to stop a moving object depends on how fast its momentum changes.
force = change in momentum
=
mv - mu
time
t
b What provides this force?
c Explain how your answer might be different it the post were being sunk into very soft ground,
F = force (N)
u = initial velocity (m/s)
te time (s)
m = mass (kg)
v = final velocity (m/s)
Answers
1a) The total energy stored when all three weights are raised by 70 cm is 80.5 J.
1b) The height of the tank is 2 m.
2b) The potential energy is converted into kinetic energy when the water flows out of the taps 6.32 m/s.
3a) The mass of water that goes over the falls every second is 1 x 10⁶ kg.
3b) The gravitational potential energy of 1 kg of water at the top of the falls 1000 J.
3c) The speed of the water as it reaches the bottom if all the GPE stored in 1 kg of water is transferred to kinetic energy is 44.72 m/s.
3d) The water will not be falling as fast as the speed calculated in part c suggests because of the presence of air resistance and the fact that the water falls through a medium (air) which offers resistance to its motion.
3e) The energy transferred when the GPE stored in the water falling in one second is transferred to other energy stores is finally stored in thermal energy stores due to the heat generated by the water as it hits the bottom.
1a) To calculate the amount of energy stored in the three weights, we use the formula given below:
E = mgh
Where,E = Energy (Joules)
m = Mass (kg)
g = Gravity (10 N/kg)
h = Height (m)
For the 4.5 kg weight:
E = 4.5 x 10 x 0.7 = 31.5 J
For each of the 3.5 kg weight:
E = 3.5 x 10 x 0.7 = 24.5 J
Thus, the total energy stored when all three weights are raised by 70 cm is:
31.5 J + 24.5 J + 24.5 J = 80.5 J
1b) To calculate how far the 4.5 kg weight must be lifted to store 45 J of energy, we use the formula:
E = mghh = E/mg = 45 / (4.5 x 10) = 1 m2a)
To calculate the gravitational potential energy stored in the water in the tank when it is full, we use the formula given below:
E = mgh
Where,E = Energy (Joules)
m = Mass (kg)
g = Gravity (10 N/kg)
h = Height (m)
The mass of 1 litre of water is 1 kg and the tank can hold 200 litres of water. Therefore, the total mass of water in the tank is:
Mass = 200 kg
The height of the tank is 2 m.
Therefore, the gravitational potential energy stored in the water in the tank is:
E = mgh = 200 x 10 x 2 = 4000 J
Assumptions made in the answer:
We have assumed that the tank is full.
2b) To calculate the speed at which the water would come out of the bathroom and kitchen taps, we use the formula given below:
PE = KEPE = mghKE = 1/2mv²
Where,PE = Potential Energy (Joules)
KE = Kinetic Energy (Joules)
m = Mass (kg)
g = Gravity (10 N/kg)
h = Height (m)
v = Velocity (m/s)
Assuming that the potential energy of the water in the tank is converted into kinetic energy when the water flows out of the taps, the potential energy stored in the water in the tank is given by:
PE = mgh = 200 x 10 x 2 = 4000 J
The potential energy is converted into kinetic energy when the water flows out of the taps.
Therefore, KE = 1/2mv²v² = 2KE/mv² = 2(4000)/200 = 40 m²/s²v = √(40) = 6.32 m/s (speed of the water coming out of the taps)
3a) To calculate the mass of water that goes over the falls every second, we use the formula given below:
Mass = Volume x Density
Where,Volume = 1000 m³/s, Density = 1000 kg/m³, Mass = 1000 x 1000 = 1000000 kg = 1 x 10⁶ kg
3b) To calculate the gravitational potential energy of 1 kg of water at the top of the falls, we use the formula:
E = mgh
Where,m = 1 kg, g = 10 N/kg, h = 100 m, E = 1 x 10 x 100 = 1000 J
3c) To calculate the speed of the water as it reaches the bottom if all the GPE stored in 1 kg of water is transferred to kinetic energy, we use the formula given below:
PE = KEP
E = mgh
KE = 1/2mv²
Where,PE = Potential Energy (Joules)
KE = Kinetic Energy (Joules)
m = Mass (kg)
g = Gravity (10 N/kg)
h = Height (m)
v = Velocity (m/s)
Assuming that all the potential energy is converted into kinetic energy when the water reaches the bottom,
PE = KEKE = mghv² = 2mghv² = 2(1)(10)(100)v² = 2000v = √(2000) = 44.72 m/s
3d) The water will not be falling as fast as the speed calculated in part c suggests because of the presence of air resistance and the fact that the water falls through a medium (air) which offers resistance to its motion.
3e) To calculate the total energy transferred per second as the GPE stored in the water falling in one second is transferred to other energy stores, we use the formula given below:
Power = Energy / Time
Where,Power = 1 x 10⁶ x 10 x 100 = 1 x 10⁹ W = 1 GW (assuming that 1 m³ of water falls every second)3f)
The energy transferred when the GPE stored in the water falling in one second is transferred to other energy stores is finally stored in thermal energy stores due to the heat generated by the water as it hits the bottom.
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A block hangs in equilibrium from a vertical spring The equilibrium position sags by 9.00 cm when a second identical block is added. Part A What is the oscillation frequency of the two-block system?
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When a second identical block is added, a block hangs in equilibrium from a vertical spring. The oscillation frequency of the two-block system is 1.17 Hz.
To determine the oscillation frequency of the two-block system, we need to use the formula:
\(f = (1/2\pi)\sqrt{(k/m)}\)
where f is the frequency, k is the spring constant, and m is the mass of the two-block system.
Since the two-block system is in equilibrium, we know that the weight of the two blocks is equal to the spring force. Let's assume that each block has a mass of m. Then, the weight of the two-block system is 2mg, where g is the acceleration due to gravity.
The spring force is given by Hooke's Law:
F = -kx
where F is the spring force, k is the spring constant, and x is the displacement from equilibrium.
When one block is added to the system, the equilibrium position sags by 9.00 cm, or 0.0900 m. Therefore, the spring force when one block is added is:
\(F_1 = -kx_1 = -k(0.0900)\)
When two blocks are added, the equilibrium position sags by 2(0.0900) = 0.1800 m. Therefore, the spring force when two blocks are added is:
\(F_2 = -kx_2 = -k(0.1800)\)
Since \(F_2 = 2F_1\) and\(F_1 = mg\) therefore k can be calculated as,
\(3F_1 = k (0.09 + 0.18)\)
\(k = (3mg)/0.27\)
\(k = 109m N/m\)
Substituting the mass of the two-block system, 2m, and the spring constant, 109m N/m, into the formula for the oscillation frequency, we get:
\(f = (1/2\pi ) \sqrt{(k/m)} = (1/2\pi ) \sqrt{(109m) / (2m)} = 1.17 Hz.\)
Therefore, the oscillation frequency of the two-block system is approximately 1.17 Hz.
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Dusting to visualize a latent print on finished leather and rough plastic is best done with a:
a. Fiberglass brush.
b. Magna brush.
c. Camel's hair brush.
d. All of the above
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To visualize a latent print on finished leather and rough plastic, the best option for dusting is (d.) All of the above.
This is because a fiberglass brush, magna brush, and camel's hair brush can all effectively visualize latent prints on these surfaces.
A latent print is an impression of the friction skin of the fingers or palms of the hands that has been transferred to another surface. The permanent and unique arrangement of the features of this skin allows for the identification of an individual to a latent print.
Each brush has its own advantages and may be best suited for specific situations, but any of them can be used to achieve the desired outcome.
So, to visualize a latent print on finished leather and rough plastic, the best option for dusting is All of the above.
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P1.22 The Ekman number, Ek, arises in geophysical fluid dynamics. It is a dimensionless parameter combining seawater density \( \rho \), a characteristic length \( L \), seawater viscosity \( \mu \),
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Ekman number (Ek) is a dimensionless parameter that arises in geophysical fluid dynamics, combining seawater density (ρ), seawater viscosity (μ), and a characteristic length (L).
It is named after the Swedish oceanographer, Vagn Walfrid Ekman. It is the ratio of the viscous forces acting on a fluid element to the Coriolis force acting on the same element. This dimensionless number plays a crucial role in the dynamics of rotating fluids, such as the oceans and the Earth's atmosphere.
In oceanography, Ekman number helps to determine the depth of the mixing layer, which is the layer in the ocean where the surface water gets mixed with the deep waters due to the wind.
The Ekman number is used to study the Earth's oceanic and atmospheric circulation, which is a critical process in the transport of heat and moisture across the globe. The Ekman layer, which is named after Vagn Walfrid Ekman, is a theoretical layer of fluid in the oceans that is affected by wind stress.
The depth of this layer varies depending on the strength of the wind and the density of the seawater. Furthermore, Ekman number is used to study the motion of glaciers and ice sheets.
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The Ekman number is a dimensionless parameter combining seawater density ρ, a characteristic length L, seawater viscosity μ, and the angular velocity of the Earth's rotation, Ω. It arises in geophysical fluid dynamics as a means of characterizing the relative importance of viscous forces and Coriolis forces in fluid motion.
Specifically, it is defined as:Ek = ν/2ΩL²where ν is the kinematic viscosity of seawater. This parameter is named after the Swedish oceanographer Vagn Walfrid Ekman (1874–1954), who first proposed the theory of Ekman transport to explain the deflection of ocean currents due to the Coriolis effect.
The Ekman number is an important parameter in geophysical fluid dynamics because it determines the depth of the boundary layer at the bottom of the ocean. In general, the boundary layer is the region near a surface where the flow of a fluid is affected by friction with the surface.
The Ekman number characterizes the thickness of this layer, with smaller values of Ek indicating thinner boundary layers.In summary, the Ekman number is a dimensionless parameter used in geophysical fluid dynamics to characterize the relative importance of viscous forces and Coriolis forces in fluid motion.
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