Water flows through a garden hose and fills a tub of 200 Liters in 5.6 minutes. The speed of the water in the hose 0.841 meters per second. A beach ball is filled with air and has a radius of 49 cm and around 513.3 kg of mass is needed to pull the beach ball underwater in a swimming pool.
(a) To calculate the speed of water in the hose, we need to determine the flow rate. First, let's convert the volume of water from liters to cubic meters. Since 1 liter is equal to 0.001 cubic meters, we have:
Volume = 200 liters * 0.001 cubic meters/liter = 0.2 cubic meters
Next, let's convert the time from minutes to seconds:
Time = 5.6 minutes * 60 seconds/minute = 336 seconds
The flow rate (Q) can be calculated by dividing the volume by the time:
Q = \(\frac{Volume}{Time} }{}\) = \(\frac{ 0.2 }{336}\) = 0.0005952 cubic meters per second
The cross-sectional area of a circular hose can be calculated using the formula: Area =\(π * radius^2\)
Given a radius of 1.5 cm, which is 0.015 meters, we have:
Area = \(π * (0.015 meters)^2\) ≈ 0.00070686 square meters
Now we can calculate the speed (v) using the formula:
v = Q / Area = \(\frac{0.0005952}{0.00070686}\) square meters ≈ 0.841 meters per second
Therefore, the speed of the water in the hose is approximately 0.841 meters per second.
(b) The volume of a sphere can be calculated using the formula:
Volume = \((\frac{4}{3} ) * π * radius^3\)
Given a radius of 49 cm, which is 0.49 meters, we have:
Volume = \((\frac{4}{3} ) * π * 0.49^3\) ≈ 0.512 cubic meters
The density of water is approximately 1000 kg/m^3. Therefore, the weight of the water displaced by the ball is:
Weight of water displaced = Volume * Density * gravitational acceleration
= 0.512 cubic meters * \(1000 kg/m^3 * 9.8 m/s^2\)
≈ 5025.6 Newtons
To balance the buoyant force, an equal and opposite gravitational force is required. The gravitational force is given by:
Gravitational force = Mass * gravitational acceleration
To find the mass needed to balance the buoyant force, we divide the weight of water displaced by the gravitational acceleration:
Mass = Weight of water displaced / gravitational acceleration
=\(\frac{5025.6 Newtons}{9.8 m/s^2}\)
≈ 513.3 kg
Therefore, approximately 513.3 kg of mass would be needed to pull the beach ball underwater in a swimming pool.
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in what year did the mars probe mariner 4 run into a cloud of space dust?
The Mariner 4 spacecraft encountered a cloud of space dust during its mission to Mars in 1965.
Mariner 4 was the fourth spacecraft in NASA's Mariner program and the first to successfully conduct a flyby of Mars, sending back the first close-up images of the planet's surface.
On July 29, 1965, as Mariner 4 was approaching Mars, the spacecraft encountered a cloud of interplanetary dust particles. This caused some concern among mission controllers, as the dust particles could have potentially damaged the spacecraft's delicate instruments.
Fortunately, the spacecraft survived the encounter with the dust cloud and continued on to conduct its historic flyby of Mars on July 14, 1965. During the flyby, Mariner 4 captured 21 images of Mars, providing the first close-up views of the planet's surface and revolutionizing our understanding of our neighboring planet.
The encounter with the space dust cloud was just one of the many challenges faced by the Mariner 4 mission, but its successful navigation through it demonstrated the resilience and technological prowess of NASA's early space probes.
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A CAT-III certified DMM should be used whenever measuring high-voltage circuits or components. The CAT-III rating relates to ___.
Question options:
A) high voltage
B)high energy
C)high electrical resistance
D)both A and B
The CAT- III rating relates to high voltage and high energy. The correct options are Option A and Option B
What is CAT- III ?Circuit breakers, wiring, switches, and other electrical installations found in a structure are referred to as CAT III.
The transient voltage range present on the majority of distribution circuits can be handled by test equipment having a CAT III rating. These devices are mostly utilized on branch circuits or fixed primary feeds.
The primary distinction between Category II and Category III operations is that Category II offers enough visual cues to allow a manual landing at DH, whereas Category III does not and necessitates an autonomous landing system.
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A book is sitting on a table. Which of the following is true about the table? O A. It is pushing up on the book. O B. It exerts no force on the book at all. O C. It is pulling down on the book. D. It can affect the mass of the book.
Answer:
yes it does exert a force, it pushes it up
Explanation:
this is called normal force
if it didn't exert a force the book would keep going down
according to newton every force has an equal amd opposite reaction
so the book exerts a force on the table and vice versa
hope this helped
True or False. The two metals in a battery are the electrolytes.
True
False
When more energy is released when new bonds are formed. Which bonds are stronger. Please add why
1. The bonds that are formed
2. The bonds that are broken
Answer:
the bonds that are formed
Explanation:
Energy is absorbed in order to break bonds. Energy is released when new bonds are formed. Bond-making is a process of exothermic. Whether the reaction is endothermic or exothermic refers to the difference between the energy needed to break bonds and the energy released when new bonds are formed.
A metal sphere has a charge of +5.0 C. What is the net charge after 1.00 1014 electrons have been placed on it?
The net charge on the metal sphere after adding 1.00 × 10¹⁴ electrons is 3.596 × 10⁻⁵ C. This is calculated by adding the initial charge of +5.0 C to the charge added by the electrons (-1.602 × 10⁻¹⁹ C per electron).
The net charge on the sphere is the sum of the initial charge and the charge added by the electrons. The charge added by the electrons is equal to the number of electrons multiplied by the charge on each electron (which is -1.602 × 10⁻¹⁹ C).
So, the net charge Qf is:
Qf = Qi + Ne * (-e)
where Qi is the initial charge (+5.0 C), Ne is the number of electrons added (1.00 × 10¹⁴), and -e is the charge on each electron (-1.602 × 10⁻¹⁹ C).
Substituting the values, we get:
Qf = 5.0 C + 1.00 × 10¹⁴ * (-1.602 × 10⁻¹⁹ C) = 3.596 × 10⁻⁵ C
Therefore, the net charge on the sphere after 1.00 × 10¹⁴ electrons have been placed on it is 3.596 × 10⁻⁵ C.
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pls awnser short i need awnsers
Answer:
4. 0.12mm
5:
a. screw gauge
b. caliper
Answer:
0.11 mm
vernier calipers
im not sure about the sure about B
Explanation:
A helicopter flies with a ground speed of 250 km/h If the wind speed is 17 km/h southeast, what is the air speed?
The air speed of the helicopter has the ground speed of 250 km/h is 233 km/h
The ground speed of the helicopter = 250 km/h
The wind speed of the helicopter = 17 km/h
The air speed of the helicopter can be found using the formula,
G = W + A
where G is the ground speed of the helicopter
W is the wind speed of the helicopter
A is the air speed of the helicopter
Let us rearrange the equation in order to get the airspeed,
A = G - W
Let us substitute the known values in the above equation, we get
A = 250 - 17
= 233 km/h
Therefore, the air speed of the air speed is 233 km/h
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What is Carnot engine ?
A Carnot heat engine is a theoretical engine that operates on the Carnot cycle. The basic model for this engine was developed by Nicolas Léonard Sadi Carnot in 1824. ... In the process of going through this cycle, the system may perform work on its surroundings, thereby acting as a heat engine.
A voltaic cell with ni/ni2 and co/co2 half-cells has the following initial concentrations: [ni2 ] = 0. 80 m; [co2 ] = 0. 20 m. What is the initial e cell
The initial Ecell for the given voltaic cell is approximately 0.0263 V.
The rearranged calculation for the initial cell potential (Ecell) is as follows:
Given:
[Ni2+] = 0.80 M
[Co2+] = 0.20 M
E°cell = -0.26 V (Ni2+ reduction) - (-0.28 V) (Co2+ reduction) = 0.02 V
n = 2 (number of electrons involved)
F = Faraday constant = 96485 C mol^-1
R = gas constant = 8.314 J mol^-1 K^-1
T = temperature in kelvins = 298.15 K
Q = [Ni2+]³[Co2+]/(1)² = 0.80³ * 0.20 = 0.256
Ecell = E°cell - (RT/nF)lnQ
Ecell = 0.02 V - (8.314 * 298.15 / (2 * 96485)) ln(0.256)
Ecell ≈ 0.02 V - 0.0063
Ecell ≈ 0.0263 V
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Pluto has a mass of 1.27x1022 kg and a radius of 1.14x109 m.
(a) An astronaut throws a rock at 12 m/s at an angle of 25° above horizontal from a height of 3.25 m above
the surface. How far horizontally does the rock travel before hitting the surface of Pluto?
What questions do you still have about supermassive black holes after watching this Ted Talk? Do you feel that you have a deeper understanding of what they are and why they are important, like was asked of you in the third question? Explain and discuss.
After watching the Ted Talk, there were still a few questions that I had about supermassive black holes. Firstly, I wanted to know more about the event horizon and what it exactly entails. Although the speaker briefly touched upon this subject, I would have appreciated a more in-depth explanation. Additionally, I would have liked to know more about the role of supermassive black holes in the universe.
While the speaker did mention that these black holes are responsible for the creation of galaxies, I wanted to know more about how this process works and why it is so important.Despite these questions, I do feel that I have a deeper understanding of supermassive black holes and their importance.From the Ted Talk, I learned that supermassive black holes are some of the largest objects in the universe and are essential for the formation of galaxies. I also learned that these black holes are incredibly powerful and have the ability to affect the trajectory of stars and planets.Overall, I think that the Ted Talk did a great job of explaining supermassive black holes in a way that was easy to understand. While there were still a few questions that I had after watching the video, I feel that I now have a better grasp of what supermassive black holes are and why they are so important.For such more question on supermassive
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The low point of a transverse wave is called a
crest
trough
rarefaction
compression
We can determine how the density changes with radius in the Sun using a. radar observations. b. neutrino detections. c. high-energy (gamma ray) observations. d. helioseismology. e. infrared observations.
We can determine how the density changes with the radius of the sun using helioseismology.
Helioseismology is the study of the interior of the Sun through its surface oscillations, similar to how seismologists study the Earth's interior through earthquakes. By analyzing these oscillations, scientists can determine how the density changes with the radius of the Sun.
The other options are a. radar observations, b. neutrino detections, c. high-energy (gamma ray) observations, and e. infrared observations, can provide valuable information about the Sun, but they are not the most effective methods for determining changes in density with radius.
So, option d is correct.
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Help me with this homework
Answer:
4. The reading decreases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction opposite to your weight.
Explanation:
When you push on anything, Newtons Third Law comes into effect: "when two objects interact, they apply forces to each other of equal magnitude and opposite direction." So pretty much, the table you are putting some of your weight on is taking some of the downward force and in return the scale isn't going to push up with as much force.
I'm not the best at explaining, but im in physics right now and thats somewhat how my teacher explained it.
what did jeff make to convert solar energy into electricity?
Jeff has created a device that can convert solar energy into electricity. This innovative technology harnesses the power of sunlight to generate electrical energy, providing a sustainable and renewable source of power.
Determine the Jeff's device?Jeff's device utilizes photovoltaic cells to convert solar energy into electricity. Photovoltaic cells, commonly known as solar cells, are made of semiconducting materials, such as silicon, that can absorb photons from sunlight.
When sunlight hits the solar cell, it excites electrons within the material, creating an electric current. The photovoltaic cells are arranged in a panel or module, which can be connected in series or parallel to achieve the desired voltage and current levels.
To optimize the conversion of solar energy, Jeff's device may include additional components, such as a charge controller, which regulates the charging of batteries or the direct usage of electricity.
It may also feature an inverter to convert the direct current (DC) produced by the solar cells into alternating current (AC), suitable for powering appliances and feeding into the electrical grid.
Overall, Jeff's device harnesses the photovoltaic effect to transform solar energy into a usable form of electricity, providing a sustainable and renewable energy source.
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An ideal fluid, of density 0.90 ´ 103 kg/m3, flows at 6.0 m/s through a level pipe with radius of 0.50 cm. The pressure in the fluid is 1.3 ´ 105 N/m2. This pipe connects to a second level pipe, with radius of 1.5 cm. Find the speed of flow in the second pipe.
An ideal fluid, of density 0.90 ´ 103 kg/m3, flows at 6.0 m/s through a level pipe with radius of 0.50 cm. The pressure in the fluid is 1.3 ´ 105 N/m2. This pipe connects to a second level pipe, with radius of 1.5 cm. The speed of flow in the second pipe is 0.666 m/s.
The speed of flow in the second pipe, we can use the principle of conservation of mass, which states that the mass flow rate through any cross-section of a pipe must remain constant.
The mass flow rate can be expressed as:
mass flow rate = density x cross-sectional area x speed of flow
Since the fluid is ideal, its density remains constant throughout the pipes. Therefore, we can set the mass flow rate in the first pipe equal to the mass flow rate in the second pipe:
density x A1 x v1 = density x A2 x v2
where A1 and v1 are the cross-sectional area and speed of flow in the first pipe, and A2 and v2 are the cross-sectional area and speed of flow in the second pipe.
We are given the density, speed of flow, and radius of the first pipe. Using the formula for the cross-sectional area of a pipe (A = πr^2), we can calculate the cross-sectional area of the first pipe:
A1 = π(0.50 cm)^2 = 0.785 cm^2
We want to find v2, the speed of flow in the second pipe. We are given the radius of the second pipe, so we can calculate its cross-sectional area:
A2 = π(1.5 cm)^2 = 7.069 cm^2
Substituting the known values into the equation for conservation of mass, we get:
0.90 x 10^3 kg/m^3 x 0.785 cm^2 x 6.0 m/s = 0.90 x 10^3 kg/m^3 x 7.069 cm^2 x v2
Simplifying and converting units, we get:
v2 = (0.785 cm^2 x 6.0 m/s) / 7.069 cm^2 = 0.666 m/s
Therefore, the speed of flow in the second pipe is 0.666 m/s.
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Please help! Will mark Brainliest
Explanation:
I think it is the third image as marked in the diagram I kept above because as we know first class levers have the fulcrum between the force and the load
What quantities are related by Ohm's law? Check all that apply. voltage conductivity current resistance insulation
Answer: Current, resistance and voltage are the quantities which are related by Ohm's law.
Explanation:
A law which states that electric current is directly proportional to voltage and inversely proportional to resistance is called Ohm's law.
Mathematically, it is represented as follows.
\(I = \frac{V}{R}\)
where,
I = current
V = voltage
R = resistance
This means that the quantities related by Ohm's law include current, voltage and resistance.
Thus, we can conclude that current, resistance and voltage are the quantities which are related by Ohm's law.
Answer:
The answer is 1, 3, 5.
✓ voltageX conductivity✓ current✓ resistanceX insulationExplanation:
did it right
Amir bought a lamp with a design attached to it that casts a shadow when the lamp is lit, as shown below. He wants to carryout an experiment using this lamp to find out the factors that effect the size of the shadow.
The effect of which factors can he actually find out using only this lamp?
The effect of the factors which he can he actually find out using only this lamp to determine the size of shadow is the light intensity and type of object.
What is Experiment?
This forms part of the scientific methods and is referred to as a procedure which is used to support or refute an hypothesis.
The size of the shadow is dependent on factors such as the light intensity and the type of object. Since the lamp produces the light in which the intensity can be gotten and the type of material it is in contact with then it therefore the correct choices.
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The parking brake on a 1200kg automobile has broken, and the vehicle has reached a momentum of 7800kgm/s. What is it's
velocity? m/s
Answer:
the velocity is like 17 or something ion even know
Explanation:
A ball rolls off the top of the roof of a building that is 13 meters tall. Calculate the amount of time it takes for it to hit the ground.
Answer:
The ball takes \($t=1 \cdot 917sec$\) to hit the ground.
Explanation:
• The second equation of motion represents the total distances travelled by an object in a time interval of \($\Delta t$\) with an initial speed of \($u$\) and acceleration \($a$\).
• To find the time, ball takes to hit the ground, use the formula: \($$s = ut + \frac{1}{2}a{t^2}$$\)
Where, \($t$\) is time, \($u$\) is initial velocity, \($a$\) is acceleration and\($s$\) is displacement.
• In this case, \($a = g = 9 \cdot 8m/se{c^2}$\).
• Placing the value of the given initial velocity, \($u=0cm/s$\) and displacement,\($s = 13m$\) in the above formula.
\(& \therefore s = ut + \frac{1}{2}g{t^2} \\& \Rightarrow 13 = 0 \cdot t + \frac{1}{2} \times 9 \cdot 8 \times {t^2} \\\)
\(& \Rightarrow 13 = 4 \cdot 9{t^2} \\& \Rightarrow {t^2} = \frac{{13}}{{4.9}} \\& \Rightarrow t = 1 \cdot 917sec \\\end{align}\]\)
• Hence, ball takes \($t=1 \cdot 917sec$\) to hit the ground.
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Help! How do you find the Eccentricity??
Explanation:
The formula to determine the eccentricity of an ellipse is the distance between foci divided by the length of the major axis
Could someone please help me
Answer:
For the first one I Would just draw a triangle type shape going out towards the screen like light is being shined upon it. And the next question I would say the answer is yes.Explanation:
Hope this helps :)
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And have an amazing day <3
A skyjumper freely falls for 7 s. What is her final speed then?
Answer:
if we ignore air resistance .. which might slow her down a lot towards the end of the 7 seconds.. but we will just ignore that for this question
Explanation:
then V=Vo+a*t
Vo=0
V=a*t
V=9.8(7)
V=68.6 m/s just a little over 150 MPH.. but.. I think terminal velocity is close to 120... soo hmm maybe about that fast... but that with air resistance.. so this problem depends a lot on if you're going to ignore that
1
5a. What forces were acting on the objects dropped in the air? (1pt)
Answer:
GRAVITYExplanation:
When an object is dropped in the air , it experiences free fall.
Free fall is a special type of motion in which the only force acting upon an object is gravity. They do not encounter a significant force of air resistance
The acceleration of the car with the data in the table above would be , I( m)/(s^(2)). If the applied force were cut in half, what do you predict the acceleration would be? ( m)/(s^(2))
The predicted acceleration of the car if the applied force were cut in half would be 2.94 m/s^2.
Given data, Mass of car, m = 850 kg
Force applied, F = 5000 N
First, calculate the acceleration with the given data using the following formula
,F = ma Where,F = 5000 Nm = 850 kg
Using the above formula, a can be found as below: a = F / ma = 5000 / 850a = 5.88 m/s^2
The acceleration of the car with the given data would be 5.88 m/s^2.
Now, if the applied force were cut in half, the acceleration can be calculated using the same formula as follows ,F = ma
Where,F = 2500 N m = 850 kg
Using the above formula, a can be found as below:
a = F / ma = 2500 / 850a = 2.94 m/s^2
Therefore, the predicted acceleration of the car if the applied force were cut in half would be 2.94 m/s^2.
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if the plot of d vs t2 (distance travelled versus the square of the travel time) is well fit by a straight line, what should the slope of the plot be equal to?
The slope of the plot of d vs \(t^2\)should be equal to one-half of the acceleration of the object.
If a graph plotting the distance travelled (d) of an object against time squared (t2) is well-fitted by a straight line, then its slope can be used to calculate the acceleration of said object.
We can observe this relationship by rearranging the kinematic equation for distance travelled (d) with a constant acceleration (a):\(a = 2 * d / t^2.\)
In order to calculate the constant acceleration of an object, we need to draw a straight line with the slope equal to 2a on a graph of d vs \(t^2\). By rearranging the equation we get
:\(a = 2d / t^2.\)
Now, by dividing both sides of this equation by 2, we can obtain the acceleration, which is simply given by
:\(a = d / t^2.\)
\(a/2 = d / t^2\)
This tells us that the slope of the plot of d vs \(t^2\)should be equal to one-half of the acceleration of the object.
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So I have to write a motion story for physics. For the assignments you need to have:
• constant velocity
• speeding up while moving forward
• speeding up while moving backward
• object at rest
• slowing down while moving forward
That’s my story in the picture that I wrote, I just need a confirmation if the m/s and seconds are correct and the story itself is correct.
(I’ll try my best to give a brainliest if I find out how to)
Thank you!
Answer:
,
Explanation:
Answer:
Is
• speeding up while moving forward
the speed of a rollar coaster could be measured in
Answer:
pe=mgh
Explanation:
hope this helped