Final answer:
The speed at which the water leaves the hole is 17.7 m/s, and the diameter of the hole is approximately 4.8 mm using Torricelli's theorem and the flow rate equation.
Explanation:
The scenario described involves principles of fluid dynamics within the field of physics, specifically Bernoulli's equation and the equation of continuity. To calculate the speed at which the water leaves the hole (a) and the diameter of the hole (b), we employ the Torricelli's theorem and the flow rate equation.
Using Torricelli's theorem, the speed (v) of water exiting the hole can be determined by the formula v = √(2gh), where g is the acceleration due to gravity (9.8 m/s²) and h is the height of the water column above the hole (16.0 m).
v = √(2 * 9.8 m/s² * 16.0 m) = √(313.6 m²/s²) = 17.7 m/s
To find the diameter of the hole, we use the flow rate Q = A * v, where Q is the flow rate (2.50 × 10³ m³/min), A is the area of the hole, and v is the speed of water leaving the hole. First, we convert the flow rate to m³/s by dividing by 60 (since there are 60 seconds in a minute):
Q (in m³/s) = (2.50 × 10³ m³/min) / 60 = 4.17 × 10µ m³/s
To find the area: A = Q / v, then to find the diameter (d), we use the area of a circle A = π * (d/2)². Rearranging and solving for d gives us:
d = 2 * √(Q / (π * v))
d = 2 * √((4.17 × 10µ m³/s) / (π * 17.7 m/s)) = 4.8 mm
(a) The speed at which the water leaves the hole is approximately 17.7m/s.
(b) The diameter of the hole is approximately 1.74mm.
Given:
- The depth of the hole below the water level, [tex]\( h = 16.0 \)[/tex] m
- The rate of flow, [tex]\( Q = 2.50 \times 10^{-3} \) m\(^3\)/[/tex]min
Part (a): Speed of Water Leaving the Hole
We can use Torricelli's Law to determine the speed at which the water leaves the hole:
[tex]\[ v = \sqrt{2gh} \][/tex]
where:
- [tex]\( g \)[/tex] is the acceleration due to gravity[tex](\( 9.8 \) m/s\(^2\))[/tex]
- [tex]\( h \)[/tex] is the height of the water above the hole
Plugging in the values:
[tex]\[ v = \sqrt{2 \times 9.8 \times 16.0} \][/tex]
[tex]\[ v = \sqrt{2 \times 9.8 \times 16.0} \][/tex]
[tex]\[ v = \sqrt{313.6} \][/tex]
[tex]\[ v \approx 17.7 \text{ m/s} \][/tex]
Part (b): Diameter of the Hole
First, convert the flow rate to cubic meters per second:
[tex]\[ Q = 2.50 \times 10^{-3} \text{ m}^3/\text{min} \][/tex]
[tex]\[ Q = \frac{2.50 \times 10^{-3}}{60} \text{ m}^3/\text{s} \][/tex]
[tex]\[ Q = 4.17 \times 10^{-5} \text{ m}^3/\text{s} \][/tex]
The flow rate ( Q ) can also be expressed as:
[tex]\[ Q = A \cdot v \][/tex]
where:
- ( A ) is the cross-sectional area of the hole
- ( v ) is the speed of the water leaving the hole
Rearranging to solve for ( A ):
[tex]\[ A = \frac{Q}{v} \][/tex]
[tex]\[ A = \frac{4.17 \times 10^{-5}}{17.7} \][/tex]
[tex]\[ A \approx 2.36 \times 10^{-6} \text{ m}^2 \][/tex]
The area ( A) of a circular hole is given by:
[tex]\[ A = \pi \left( \frac{d}{2} \right)^2 \][/tex]
Solving for the diameter[tex]\( d \)[/tex]:
[tex]\[ d = 2 \sqrt{\frac{A}{\pi}} \][/tex]
[tex]\[ d = 2 \sqrt{\frac{2.36 \times 10^{-6}}{\pi}} \][/tex]
[tex]\[ d = 2 \sqrt{7.51 \times 10^{-7}} \][/tex]
[tex]\[ d \approx 1.74 \times 10^{-3} \text{ m} \][/tex]
[tex]\[ d \approx 1.74 \text{ mm} \][/tex]
Tech A says that most wheels have a drop center or deep well that is used in installing a tire on the wheel. Tech B says that the drop center or deep well is used to prevent the tire from coming off the wheel in the case of low tire pressure. Who is correct?
Final answer:
Tech A is correct in stating that the drop center is used for tire installation, while Tech B is incorrect; the drop center's primary purpose isn't to prevent the tire from detaching at low pressure.
Explanation:
Tech A says that most wheels have a drop center or deep well that is used in installing a tire on the wheel. Tech B says that the drop center or deep well is used to prevent the tire from coming off the wheel in the case of low tire pressure. The answer here is that Tech A is correct. The drop center or deep well feature on a wheel is indeed primarily designed to help in the installation and removal of the tire. It allows the bead of the tire to drop into a recessed area, giving enough slack to work the tire over the rim. On the other hand, Tech B’s statement is not accurate. While the drop center might incidentally help keep the tire on the rim in the case of low pressure, its primary purpose is not to prevent the tire from coming off the wheel.
During any process involving a heat transfer, the entropy of a system must always increase, it can never decrease.
True or False?
Answer:
True
Explanation:
Entropy is a thermodynamic variable that always increases or remains constant, but does not decrease.
This indicates that energy is always transferred from a higher temperature system to a lower temperature system and not the other way around.
Entropy can also be thought of as what drives a system into equilibrium, and systems reach equilibrium through heat transmission.
The fact that the entropy of a closed system never decreases is a statement of the second law of thermodynamics.
Which of the following is an example of Newton's second law of motion?
a. Catie's skateboard hits a curb and she falls forward onto the grass. b. A tissue and a rock are dropped from a ladder.
c. The rock exerts a greater force when it hits the ground.
d. A person stands on a chair and pushes down, and the chair pushes upwards on the person.
e. Tires from a bike push against the road and the road push back.
Answer:
B
Explanation:
Newton’s Second Law of Motion
Newton’s Second Law of Motion states that ‘when an object is acted on by an outside force, the mass of the object equals the strength of the force times the resulting acceleration’.
This can be demonstrated dropping a rock or and tissue at the same time from a ladder. They fall at an equal rate—their acceleration is constant due to the force of gravity acting on them.
The rock's impact will be a much greater force when it hits the ground, because of its greater mass. If you drop the two objects into a dish of water, you can see how different the force of impact for each object was, based on the splash made in the water by each one.
Answer: B
Explanation:
Let θ (in radians) be an acute angle in a right triangle and let x and y, respectively, be the lengths of the sides adjacent to and opposite θ. Suppose also that x and y vary with time. At a certain instant x=9 units and is increasing at 9 unit/s, while y=5 and is decreasing at 19 units/s. How fast is θ changing at that instant?
Answer:
Explanation:
According to question
tan θ = y / x
Differentiate with respect to t on both the sides
[tex]Sec^{2}\theta \times \frac{d\theta }{dt}=\frac{x\times dy/dt-y\times dx/dt}{x^{2}}[/tex]
[tex]\frac{d\theta }{dt}=\frac{x\times dy/dt-y\times dx/dt}{x^{2}\times Sec^{2}\theta}[/tex] .... (1)
According to question,
tan θ = 5 / 9
So, Sec θ = 10.3 / 9 = 1.14
dx/dt = 9 units/s
dy/dt = 19 units/s
Substitute the values in equation (1), we get
[tex]\frac{d\theta }{dt}=\frac{9\times 19-5\times 9}{81\times 1.14^{2}}[/tex]
dθ/dt = 1.2 units/s
For all turning vehicles, the rear wheels follow a ______ than the front wheels.
a. slower pathb. faster pathc. longer pathd. shorter path
Answer:
Shorter path
Explanation:
For all turning vehicles, the rear wheels follow Shorter path than the front wheels.
Any turning vehicle, the rear(the back part of something, especially a vehicle.) wheels follow a shorter path than the front wheels. The longer the vehicle is, the greater the difference will be in path. Trucks initially swing out before making a turn
A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position yi such that the spring is at its rest length. The object is then released from yi and oscillates up and down, with its lowest position being 10 cm below yi.
(a) What is the frequency of the oscillation?
(b) What is the speed of the object when it is 8.0 cm below the initial position?
(c) An object of mass 300 g is attached to the first object, after which the system oscillates with half the original frequency. What is the mass of the first object?
(d) How far below yi is the new equilibrium (rest) position with both objects attached to the spring?
While certain parts of your question cannot be answered without specific data such as the spring constant, we can deduce from the half frequency that the mass of the initial object must also be 300g. More mass on the spring would pull the equilibrium further down but we can't quantify without more details.
Explanation:This question involves simple harmonic motion and Hooke's Law. Hooke's Law is F = -kx where F is the force, k is the spring constant, and x is the displacement. Simple Harmonic Motion deals with oscillations like that of a spring. However, without values for the spring constant (k) or initial speed of the object, we cannot calculate the frequency of the oscillation or speed of the object at any given point.
As for part (c), the frequency of the oscillation halves when a second mass is introduced. This implies that the system's mass has doubled as frequency is inversely proportional to the square root of the mass. Therefore, the initial object's mass would be 300g as well.
For part (d), without any specific figures such as the spring constant, the new equilibrium position below yi cannot be determined. Commonly, the force produced by the weight of the objects will be counteracted by the spring force when the system is in equilibrium. More mass would pull the spring further down until this equilibrium is achieved.
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The frequency of oscillation can be determined using the formula f = 1 / T. The speed of the object when it is 8 cm below the initial position can be calculated using the formula v = A * ω. The mass of the first object can be found by setting up two equations and solving for M. The new equilibrium position is slightly lower than the initial position yi.
Explanation:(a) The frequency of an oscillation can be determined using the formula:
frequency (f) = 1 / period (T)
Since the object oscillates up and down, it completes one full cycle (2π radians) every time it goes from the highest point to the lowest point and back. The period (T) is therefore the time it takes for one complete cycle. The object reaches its lowest position when it is 10 cm below yi, so the total distance traveled is 10 cm. Since the object moves up and down symmetrically, we can calculate the period by dividing the total distance traveled by 2 times the amplitude (A):
T = 2π * sqrt(m / k)
where m is the mass of the object and k is the spring constant. Given that the spring is massless, the mass of the object does not affect the period. Therefore, we can calculate the period as:
T = 2π * sqrt(k)
From the period, we can compute the frequency using the formula provided above.
(b) The speed of the object when it is 8.0 cm below the initial position can be determined using the formula for the speed of an object in simple harmonic motion:
speed (v) = amplitude (A) * angular frequency (ω)
The angular frequency is related to the period by the formula:
angular frequency(ω) = 2π / period (T)
Using the given values, we can calculate the angular frequency and then substitute it into the equation for speed.
(c) When an object of mass 300 g is attached to the first object, the system oscillates with half the original frequency. Let's denote the mass of the first object as M and the spring constant as K. We can set up two equations to solve for the mass of the first object:
Equation 1: f = (1 / (2π)) * sqrt(K / M) (original frequency)
Equation 2: (f/2) = (1 / (2π)) * sqrt(K / (M + 0.3)) (new frequency)
By solving these equations simultaneously, we can find the mass of the first object.
(d) The new equilibrium position, with both objects attached to the spring, is determined by the combined mass of the two objects. Since the first object is now heavier due to the addition of the second object, the equilibrium position will be slightly lower than the initial position yi. The exact displacement can be calculated using the formulas for equilibrium position in simple harmonic motion.
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Under normal conditions, you are just barely able to lift a mass of 74 kg. Your friend drops a box of volume 2.4 m3 into a lake. If you are just able to lift it to the surface (so that it is still completely submerged), what is the mass of the box
Answer:
mass=2326kg
Explanation:
Archimedes' principle states that when an object is submerged into a liquid, it appears lighter in weight due to the buoyant force applied by the liquid in upward direction. Buoyant force is equal to the weight of the liquid displaced by the object. An object floats on the surface of liquid or floats partially submerged when weight of the displaced liquid is greater than weight of the object.
the mass f he box=74kg
densty=mass /vlme
1000=mass/2.4
mass =2400kg
apparent mass wen completely submerged will be
2400-74kg
mass=2326kg
What force is described as the attraction between a sample of matter and all other matter in the universe?
Answer:
Gravitational Force
Explanation:
Gravitational force also called gravity or gravitation is an attractive force that keeps two objects in space. Gravitational force is an attractive force that tends to pull matters together. Every objects in the universe experience gravitational pull. Planets, stars, galaxies, are held together by gravity. It is a weak force. The weight of an object is the product of gravitational force acting on its mass.
Newton's Law of Universal Gravitation states the force of attraction between two masses m₁ and m₂ in the universe is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.
[tex]F = G\frac{m_{1}m_{2}}{r^2}[/tex]
Where;
F is the gravitational force,
G is the gravitational constant = 6.67 × 10¹¹ m³/kg/s,
m1 and m2 are the masses of the objects,
r is the distance between the centers of the masses
Let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t ≥ 0 . After her parachute opens, her velocity satisfies the differential equation dvdt=−2v−32 , with initial condition v(0) = −50 . It is safe to land when her speed is 20 feet per second. At what time t does she reach this speed?
Answer:
t = 1.07 seg
Explanation:
First we are going to solve the differential equation for the velocity:
[tex]\frac{dv}{dt} = -2v-32[/tex]
This is a differential equation of separable variables
[tex]\frac{dv}{-2v-32} = dt[/tex]
Multiplying by -1 to both sides of the equation
[tex]\frac{dv}{2v+32} = -dt[/tex]
We integrate the left side with respect to the velocity and the right side with respect to time
[tex]\frac{ln(2v+32)}{2} = -t +k[/tex]
where k is a integration constant
ln(2v+32) = -2t + k
[tex]2v +32 = e^{-2t+k}[/tex]
[tex]2v + 32 = ce^{-2t}[/tex]
[tex]v = ce^{-2t} -16[/tex]
We determine the constant c with the initial condition v(0) = -50
[tex]-50 = ce^{-2(0)} -16[/tex]
-50 + 16 = c
c = -34
Then
[tex]v(t) = -34e^{-27}-16[/tex]
When the velocity is -20 ft/s the time is:
[tex]-20 = -34e^{-2t}-16[/tex]
[tex]\frac{-4}{-34} = e^{-2t}[/tex]
[tex]ln(\frac{4}{34} ) = -2t[/tex]
t = 1.07 seg
To determine the time at which the skydiver's velocity reaches 20 feet per second, we integrate the given differential equation dv/dt = -2v - 32 with initial conditions and solve for the time t when the velocity function v(t) equals -20.
Explanation:The student's question involves solving a first-order linear differential equation to find the time at which a skydiver's velocity reaches a certain value after her parachute opens. The differential equation dv/dt = -2v - 32 represents the velocity of the skydiver, and we must use initial conditions to solve it. Given that the skydiver's initial velocity is -50 feet per second and the safe landing speed is 20 feet per second, we need to integrate the differential equation to find the function v(t) and then solve for t when v(t) equals -20 feet per second (since velocity is negative when descending).
How does a rotating coil inside a magnetic field generate electricity?
Answer:
When an electrical current passes through a wire, a magnetic field is generated around it. Likewise, if the magnetic field around a wire is changed ( for example by rotating a coil inside a stationary manger), electricity will move through the wire.
If you are over-driving your headlights and you see an object ahead, you will_____. be given the right-of-way from other vehicles not be able to stop in time to miss the object need to execute a high-speed U-turn be able to stop, but may graze the object.
Over-driving your headlights implies a situation where you are driving so fast that you won't be able to stop within the area illuminated by your headlights. Therefore, if you see an object in your path under such circumstances, you are likely not going to be able to stop in time to avoid hitting it. Headlights' range is usually 350 feet, and driving at a speed that requires a stopping distance greater than 350 feet is considered over-driving your headlights.
Explanation:The phrase 'over-driving your headlights' refers to a situation where a driver is traveling at such a speed that their stopping distance is further than the distance illuminated by their headlights. Thus, if an object is within your path, you won't have enough time to stop your vehicle before hitting it, especially if you're speeding.
On most roads, the farthest your headlights can help you see ahead is around 350 feet. If you're driving faster than a speed that permits you to stop within these 350 feet, you're said to be 'over-driving' your headlights. If you're over-driving your headlights, and you see an object ahead then you will most likely not be able to stop in time to miss the object. This is because your vehicle's stopping distance will be greater than your visual distance, which is dependent on the capabilities of your headlights.
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Solve for the tension in the left rope, TL, in the special case that x=0. Be sure the result checks with your intuition. Express your answer in terms of W and the dimensions L and x. Not all of these variables may show up in the solution.
The tension in the left rope is expressed by [tex]T_{L} = W\cdot \left(1-\frac{x}{L} \right)[/tex]. For [tex]x = 0[/tex], [tex]T_{L} = W[/tex].
How to apply moment formulas to determine the tension in the left ropeA representation of the system is presented in the image attached below. Let suppose that the entire system represents a rigid body, that is, a body whose geometry cannot be neglected and does not experiment deformation due to the application of external loads.
The tension in the left rope ([tex]T_{L}[/tex]) can be found by applying Newton's laws and D'Alembert's Principle applied on the rightmost point of the bar:
[tex]\Sigma F = W\cdot (L-x)-T_{L}\cdot L = 0[/tex]
[tex]T_{L} = \frac{W\cdot (L-x)}{L}[/tex]
[tex]T_{L} = W\cdot \left(1-\frac{x}{L} \right)[/tex] (1)
The tension in the left rope is expressed by [tex]T_{L} = W\cdot \left(1-\frac{x}{L} \right)[/tex]. For [tex]x = 0[/tex], [tex]T_{L} = W[/tex]. [tex]\blacksquare[/tex]
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Considering a situation where there is no horizontal acceleration, the tension in the left rope (TL) equals the weight of the person when x equals zero (x=0). Since the only forces on the static system are weight (W) and the tensions in the two ropes, and they are in equilibrium, the tension in either rope will be the same as the weight.
Explanation:The question requires us to find the tension in the left rope (TL), given that x equals zero. Using the equilibrium equation under the condition of no horizontal acceleration and no external forces except the weight (W) and the two tensions, TL and TR, we can conclude the magnitudes of the tensions TL and TR must be equal.
Given the scenario, the only forces acting on the system are the weight of the walker (W) and the tensions in the ropes. If the system is static, the net external force is zero, which means the sum of the forces should also equal zero according to Newton's second law. Hence, expressing the tension TL in terms of W becomes simple given that the system is in equilibrium. Under these conditions, the tension in the rope should equal the weight of the supported mass, making TL equal to W.
Therefore, since x is zero and we are neglecting other variables like rope weight and the angle of the ropes, the tension TL in the left rope becomes equivalent to the weight of the person. Hence the tension (TL) equals W.
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While attempting to tune the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string. (a) What are the possible frequencies of the string? (b) When she tightens the string slightly, she hears 3.00 beats/s. What is the frequency of the string now? (c) By what percentage should the piano tuner now change the tension in the string to bring it into tune?
Answer
given,
frequency = 523 Hz
piano tuner hears = 2.00 beats/s
a) possible frequency = 523 ± 2
525 Hz and 521 Hz
b) On tightening piano wire, tension will be increased which will cause an increase of frequency of piano wire. Since no. of beats heard is also increased, the frequency of piano wire must be greater than the frequency of oscillator.
f p' = 5+ f₀
f p' = 3 + 523 Hz = 526 Hz.
c) f = k x √T
Let the tension at 523 Hz be T₀ and at 526 Hz it be T₁
523 = k x √T₀
T₀ = 523²/ k²
T₁ = 526²/ k²
% change in tension
= [tex]\dfrac{T_0- T_1}{T_1}[/tex]
= [tex]\dfrac{523^2- 526^2}{526^2}[/tex]
= 1.13 %
An object is pulled northward by a force on 10 N and at the same time another force of 15 N pulls it southward. The magnitude of the resultant force on the object is _____
Answer:
The magnitude on the resultant force on the object = 18.03 N
Explanation:
Resultant Force: The resultant force of two or more forces acting on a body in a given direction is the single force obtained when the two forces are combined, which produces the same effect as the two forces acting together.
Since both forces makes an angle of 90° with each other,
Fr = √(F₁² + F₂²)......................... Equation 1
Where Fr = resultant force on the object F₁ = The force pulling the body northward, F₂ = The force pulling the body southward.
Given: F₁ = 10 N, F₂ = 15 N
Substituting these values into equation 1
Fr = √(10² + 15²)
Fr = √(100 + 225)
Fr = √325
Fr = 18.03 N
Therefore the magnitude on the resultant force on the object = 18.03 N
The resultant force when an object is pulled by 10 N northward and 15 N southward is 5 N towards the south, found by subtracting the smaller force from the larger one.
Explanation:In physics, when multiple forces act on an object, they combine to create a resultant force. If two forces of equal magnitude are pulling in opposite directions, they cancel out and the resultant force is zero. However, in this case, the forces acting on the object are different: 10 N northward and 15 N southward.
To find the resultant force, we simply subtract the smaller force from the larger one, considering the opposing directions. So, here 15 N (south) - 10 N (north) gives us 5 N towards the south.
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What exposure indicator number would be generated if 1 mR exposed a Carestream imaging plate?
Answer:
2000
Explanation:
Exposure indicator can be obtained using Carestream's formula for imagine pate.
Exposure indicator number = 2000 + [1000 x (log of exposure in mR)].
Given; log of exposure in mR = 1 mR
Exposure indicator number = 2000 + [1000 x (log 1)]
Exposure indicator number = 2000 + [1000 x (0)]
Exposure indicator number = 2000 + [0]
Exposure indicator number = 2000
What type of intermolecular force happens with a polar covalent molecule when dipoles are created due to un-equal sharing of electrons?
Answer:
Dipole-dipole interaction force
Explanation:
When one of the constituent atom of the covalent bonding is at least 1.5 times more electronegative than the other atom sharing the electron in the covalent bond then the shared pair of electrons are shifted towards the more electronegative atom developing a partial negative charge on it and similarly develops an equal partial positive charge on the other atom involved in the covalent bond.
This happens in water molecules and the resulting dipole is the cause of hydrogen bonding between two molecules of water. Hydrogen bond also exists in (HF) hydrogen fluoride molecules.When determining the net force given a force of 12 N and a force of 7 N, what would these forces be called?
Answer:Resultant force
Explanation:
The net given force is known as the resultant force. The resultant force is the single force that acts in place of other forces combined together.
The force of gravity on a 1 kg object on the Earth's surface is approximately 9.8 N. For the same object in low-earth orbit around the earth (at the height of the International Space Station), which is the closest value to the force of gravity on the object?
Answer:
g = 8.61 m/s²
Explanation:
distance of the International Space Station form earth is 200 Km
mass of the object = 1 Kg
acceleration due to gravity on earth = 9.8 m/s²
mass of earth = 5.972 x 10²⁴ Kg
acceleration due to gravity = ?
r = 6400 + 200 = 6800 Km = 6.8 x 10⁶ n
using formula
[tex]g = \dfrac{GM}{r^2}[/tex]
[tex]g = \dfrac{6.67\times 10^{-11}\times 5.972\times 10^24}{(6.8\times 10^6)^2}[/tex]
g = 8.61 m/s²
Final answer:
The force of gravity on a 1 kg object in low-earth orbit around the Earth depends on the height of the orbit and is smaller than on the Earth's surface.
Explanation:
The force of gravity on a 1 kg object on the Earth's surface is approximately 9.8 N. In low-earth orbit around the Earth, the force of gravity on the object is much smaller. This is because the force of gravity decreases as you move further away from the Earth's surface. However, the exact value of the force of gravity in low-earth orbit depends on the height of the orbit. For example, at the height of the International Space Station, the force of gravity on the object would be about 88% of the force on the Earth's surface, which is approximately 8.6 N.
Lemon juice and coffee are both acidic solutions. Lemon juice has a pH of 2, and coffee has a pH of 5. Which of these solutions is the stronger acid and why?
Answer:
Lemon Juice
Explanation:
Acidic levels rise as pH levels decrease. The solution with the lowest pH level will always be more acidic. In this case, lemon juice is the solution with a lower pH level, therefore lemon juice is more acidic than coffee.
Hope this helps!
Final answer:
Lemon juice is the stronger acid compared to coffee, with a pH of 2 versus a pH of 5, meaning it is 1000 times more acidic.
Explanation:
The acidity of a solution is determined by its pH level, which measures the concentration of hydrogen ions (H+) in the solution. The pH scale ranges from 0 to 14, with lower numbers indicating more acidic solutions. In comparing lemon juice with a pH of 2 and coffee with a pH of 5, lemon juice is the stronger acid. Since pH is measured on a logarithmic scale, each pH unit represents a tenfold difference in hydrogen ion concentration. Thus, the difference between pH 2 and pH 5 is 103 or 1000 times. Therefore, lemon juice is 1000 times more acidic than coffee.
A military base a few miles from a medium- sized city is being decommissioned. The base is a large one, sited along a coastline, and includes some small hills, pristine woodlands and wetlands, streams, a river, and a small lake. There is a railroad spur that connects the base with the nearby city, as well as a four- lane highway. The base contains housing, schools, a hospital, shops, and recreational areas for a population of 10,000. The city's options for using, selling, or leasing the land are numerous, and many groups come forward with proposals. Members of several local environmental groups propose that nearly all of the buildings be removed and the area turned into a regional hiking and camping area, with very limited vehicle access. This will preserve the natural areas while providing recreation for the city dwellers. One issue that is raised by opponents is ________.
Answer:
The lack of jobs and lack of transportation to jobs
Explanation:
As it is seen, the establishment of the camping and hiking areas will dispatch buildings which were hospitals, shops and schools.
These buildings and facilities previously employed a number of workers and their removal hence, will reduce the number of available jobs.
Also, the new constructions will prove difficult in switching from one job to another.
A glider of mass 0.170 kg moves on a horizontal frictionless air track. It is permanently attached to one end of a massless horizontal spring, which has a force constant of 13.0 N/m both for extension and for compression. The other end of the spring is fixed. The glider is moved to compress the spring by 0.180 m and then released from rest.(a) Calculate the speed of the glider at the point where it has moved 0.180 m from its starting point, so that the spring is momentarily exerting no force.m/s(b) Calculate the speed of the glider at the point where it has moved 0.250 m from its starting point.m/s
Answer:
Answer:
a. 1.594 m/s = v
b. 1.274 m/s = v
Explanation:
A) First calculate the potential energy stored in the spring when it is compressed by 0.180 m...
U = 1/2 kx²
Where U is potential energy (in joules), k is the spring constant (in newtons per meter) and x is the compression (in meters)
U = 1/2(13.0 N/m)(0.180 m)² = 0.2106 J
So when the spring passes through the rest position, all of its potential energy will have been converted into kinetic energy. K = 1/2 mv².
0.2106 J = 1/2(0.170 kg kg)v²
0.2106 J = (0.0850 kg)v²
2.808m²/s² = v²
1.594 m/s = v
(B) When the spring is 0.250 m from its starting point, it is 0.250 m - 0.180 m = 0.070 m past the equilibrium point. The spring has begun to remove kinetic energy from the glider and convert it back into potential. The potential energy stored in the spring is:
U = 1/2 kx² = 1/2(13.0 N/m)(0.070 m)² = 0.031J
Which means the glider now has only 0.2106 J - 0.031J = 0.1796 J of kinetic energy remaining.
K = 1/2 mv²
0.1796 J = 1/2(0.170 kg)v²
0.138 J = (0.0850 kg)v²
1.623 m²/s² = v²
1.274 m/s = v
To calculate the speed of the glider at different points, we can use the principle of conservation of energy. At 0.180 m from the starting point, the speed is 2.65 m/s. At 0.250 m from the starting point, the speed is 3.89 m/s.
Explanation:
To solve this problem, we can use the principle of conservation of energy. When the glider is released, all of the potential energy stored in the compressed spring is converted into kinetic energy. At the point where the glider has moved 0.180 m from its starting point, the potential energy of the spring is zero, so the total mechanical energy is equal to the kinetic energy. Using the formula for kinetic energy, we can calculate the speed of the glider:
KE = 1/2 mv^2
m = 0.170 kg (mass of the glider)
v = ? (speed of the glider)
At the point where the glider has moved 0.180 m, the potential energy of the spring is zero, so the total mechanical energy is equal to the kinetic energy:
0.5 kx^2 = 0.5 mv^2
k = 13.0 N/m (force constant of the spring)
x = 0.180 m (distance moved by the glider)
By substitute the given values into the equation, we can solve for v:
0.5 * 13.0 N/m * (0.180 m)^2 = 0.5 * 0.170 kg * v^2
Solving for v, we find that the speed of the glider at the point where it has moved 0.180 m from its starting point is 2.65 m/s.
To calculate the speed of the glider at the point where it has moved 0.250 m from its starting point, we can use the same principle of conservation of energy. The initial potential energy of the spring is converted into kinetic energy:
0.5 * 13.0 N/m * (0.250 m)^2 = 0.5 * 0.170 kg * v^2
Solving for v, we find that the speed of the glider at the point where it has moved 0.250 m from its starting point is 3.89 m/s.
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When push-starting your car, what is the 3rd Law partner to the friction force you exert on the ground?
A) The horizontal force you exert on the car.
B) The normal force you exert on the ground.
C) The normal force the ground exerts on you.
D) The friction force the ground exerts on you.
Answer:
C
Explanation:
Newton third law states that to every action force there is an equal and opposite reaction force. In the case of the car, the person pushes against the ground in other to get the car to move and consequently the ground exert a normal force opposite to the push on the person.
How does the saturation of a solution affect crystal formation? In your own words please
The term saturated solution is used in chemistry to define a solution in which no more solute can be dissolved in the solvent. It is understood that saturation of the solution has been achieved when any additional substance that is added results in a solid precipitate or is let off as a gas.
Which of the following circular rods, (given radius r and length l) each made of the same material and whose ends are maintained at the same temperature will conduct most heat?
a) r = 2r₀; l = 2l₀
b) r = 2r₀; l = l₀
c) r = r₀; l = 2l₀
d) r = r₀; l = l₀
Answer:
Explanation:
Heat flow in a circular rod is given by
[tex]Q=\frac{kAdT}{dx}[/tex]
where Q= heat flow
k=thermal conductivity
A=area of cross-section
dT=Change in temperature
dx=change in length
Also A can be written as
[tex]A=\pi r^2[/tex]
thus Q is Proportional to
[tex]Q\propto \frac{r^2}{l}[/tex]
For option (a)
[tex]Q\propto \frac{(2r_0)^2}{2l_0}[/tex]
[tex]Q\propto \frac{2r_0^2}{l_0^2}[/tex]
(b)[tex]Q\propto \frac{(2r_0)^2}{l_0}[/tex]
[tex]Q\propto \frac{4r_0^2}{l_0}[/tex]
(c)[tex]Q\propto \frac{r_0^2}{2l_0}[/tex]
(d)[tex]Q\propto \frac{r_0^2}{l_0}[/tex]
So Rod b will conduct the most Heat
If an object falling freely were somehow equipped with an odometer to measure the distance it travels, then the amount of distance it travels each succeeding second would be:_________
a) constant.
b) less and less each second.
c) greater than the second before
Answer:c
Explanation:
Given
object is falling Freely with an odometer
Suppose it falls with zero initial velocity
so distance fallen in time t is given by
[tex]h=ut+\frac{1}{2}gt^2[/tex]
here u=0 and t=time taken
[tex]h=\frac{1}{2}gt^2[/tex]
for [tex]t=1 s[/tex]
[tex]h_1=\frac{1}{2}g[/tex]
for [tex]t=2 s[/tex]
[tex]h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g[/tex]
distance traveled in 2 nd sec[tex]=2g-\frac{1}{2}g=\frac{3}{2}g[/tex]
for [tex]t=3 s[/tex]
[tex]h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g[/tex]
distance traveled in 3 rd sec[tex]=\frac{9}{2}g-2g=\frac{5}{2}g[/tex]
so we can see that distance traveled in each successive second is increasing
The amount of distance travels by free falling object in each succeeding second is greater than the second before.
What is the speed of free falling body?Speed of the free falling object is the speed, by which the body is falling downward towards the ground level.
At the Earth's surface, the speed of the free falling object will accelerate 9.841 meters per squared second.The odometer is the device, which is used the speed of the body in m/s or km/hrs.
Now we have to find out, whether the distance traveled by the object each succeeding second is constant, deceasing or increasing.
The second equation of the motion for distance can be given as,
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
Here, [tex]u[/tex] is the initial velocity of the body, [tex]g[/tex] is the acceleration due to gravity and [tex]t[/tex] is the time taken by it.
As for the free falling body the initial velocity is zero. Thus the distance traveled by it for the first second is,
[tex]s=0+\dfrac{1}{2}9.81\times1\\s=4.905\rm m[/tex]
The distance traveled by it for the 2nd second is,
[tex]s=\dfrac{1}{2}9.81\times2^2\\s=19.61\rm m[/tex]
The distance traveled by it for the 3rd second is,
[tex]s=\dfrac{1}{2}9.81\times3^2\\s=44.145\rm m[/tex]
Now the difference between the third and 2nd second of distance traveled by object is greater then the difference between the second and first second.
Thus, the amount of distance travels by free falling object in each succeeding second is greater than the second before.
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Two technicians are discussing master cylinders. Technician A says that it is normal to see fluid movement in the reservoir when the brake pedal is depressed. Technician B says a defective master cylinder can cause the brake pedal to slowly sink to the floor when depressed. Which technician is correct
Answer:
Technicians A and B are both correct
Explanation:
A quick take-up valve in a quick take-up master cylinder assembly which reduces fluid flow noise and restriction caused by fluid flowing in one mode of operation through an incipiently or substantially closed valve spring coil stack. The check valve arrangement has a coil compression limiting device and also has the outlet orifice so positioned as not to require fluid flow through the check valve spring coils while the check valve is open.
If the brake warning light is NOT on and there are no visible brake fluid leaks, your master cylinder may be worn or leaking internally allowing the brake pedal to slowly sink when pressure is applied to it. This type of condition will be most noticeable when you are holding constant pressure against the brake pedal at a stop light. If the pedal sinks or requires pumping to keep your car from creeping ahead, the master cylinder needs to be replaced.
A seated cable row is an example of which level of training in the NASM OPT model?
A. Stabilization
B. Strength
C. Power
D. Reactive
Answer:
B. Strength
Explanation:
The OPT Model, or Optimum Performance Training Model, is a "fitness training system created by the NASM. The OPT Model is contructed with scientific evidence and principles that progresses an individual through five training phases: stabilization endurance, strength endurance, hypertrophy, maximal strength and power".
On the stabilization level we have the phase 1 called stabilization endurance.
For the level strength part we have 3 phases . Phase 2: Strength endurance , Phase 3: Hypertrophy, Phase 4: Maximal strength. And we can consider the case "A seated cable row" on this the level strength since we need to have some abilities to do this but not enough to stay on the power level since this one is the advancd level.
For the power level we have the last phase called power in order to mantain and conduct high training level programs.
How much of a following distance should you allow between you and the vehicle in front of you when it is raining heavily?
Final answer:
When driving in heavy rain, the recommended following distance should be increased to at least 6 seconds due to longer stopping distances on wet pavement. Stopping distances depend on road conditions and driver reaction time, highlighting the need for increased safety margins in poor weather.
Explanation:
The amount of following distance you should allow between you and the vehicle in front of you during heavy rain should be considerably more than what you would maintain in dry conditions. The typical recommendation is to maintain at least a 3-second distance in good weather, but this should be increased to at least 6 seconds in heavy rain to accommodate for the increased stopping distances on wet pavement and reduced visibility.
Stopping distances can vary greatly depending on road conditions and driver reaction time, and heavy rain can significantly increase these distances. As braking distance increases with speed and poor weather conditions, it is important to adjust your following distance accordingly to ensure safety.
Referring to the given figures, we can deduce that for a car traveling at 30.0 m/s, the stopping distance will be much longer on wet pavement than on dry. If the driver's reaction time is assumed to be 0.500 s, the total distance traveled before the car comes to a stop will include both the reaction distance and the braking distance. When considering crossing a street, you must take into account that a safe distance is one where you are completely sure that the car can come to a full stop without reaching your crossing point, which can be roughly compared to the stopping distances shown in various figures.
Keaton is asked to solve the following physics problem:
What determines the speed at which a ball will drop from a five-story building?
He is given six balls and a tall box. Demonstrating his ability to think within the formal operational stage, Keaton will likely __________.
a. begin by climbing onto the box and dropping all of the balls
b. lower the height at which the balls are dropped
c. think about the problem first, systematically consider all factors, and form a hypothesis
d. take a vivid guess that the larger the ball, the faster the speed
Answer:
The answer is C "think about the problem first, systematically consider all factors, and form a hypothesis"
Explanation:
In physics there is some basic fomula that sir Isacc Newton proposed under the topic of motion. The three formulas are below;
1) v=u+at
2)v^2=u^2+2as
3)s=ut+(1/2)(at^2)
the variables are explained below;
u= initial velocity of the body
a=acceleration/Speed of the body
t= time taken by the body while travelling
s= displacement of the body.
Therefore to solve keatons problem, the factors(variables) in the formulas above need to be systematically considered. Since the ball was dropped from the top of the building, the initial velocity is 0 because the body was at rest. Also the acceleration will be acceleration due to gravity (9.8m/s^2)
Which of the following optical media (BD-R, Blu-Ray, DVD-RW, and DVD DS [Double-Sided] ): a) is capable of multiple rewrites and b) has the largest capacity?
Answer:
A. DVD-RW
B. BD-R
Explanation:
The RW stands for rewritable.
BD-R uses Blu-ray technology allowing capacities of up to 100GB
The DVD-RW is capable of multiple rewrites, while the Blu-Ray disc holds the largest capacity amongst the options provided, being able to store 25-50 GB of data.
Explanation:From the options provided (BD-R, Blu-Ray, DVD-RW, and DVD DS [Double-Sided]), the medium capable of multiple rewrites is DVD-RW. 'RW' in DVD-RW stands for 'ReWritable', meaning the media can be written, erased, and rewritten multiple times.
The optical media with the largest capacity is Blu-Ray. A single-layer Blu-Ray disc has a capacity of 25 GB, and a dual-layer Blu-Ray disc can hold 50 GB, more than five times the capacity of a DVD DS (Double Sided) which typically holds about 9 GB.
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