The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.
How to determine the friction factorUsing the formula
μ = viscosity = 0. 06 Pas
d = diameter = 120mm = 0. 12m
V = velocity = 1m/s and 3m/s
ρ = density = 0.9
a. Velocity = 1m/s
friction factor = 0. 52 × [tex]\frac{0. 06}{0. 12* 1* 0. 9}[/tex]
friction factor = 0. 52 × [tex]\frac{0. 06}{0. 108}[/tex]
friction factor = 0. 52 × 0. 55
friction factor [tex]= 0. 289[/tex]
b. When V = 3mls
Friction factor = 0. 52 × [tex]\frac{0. 06}{0. 12 * 3* 0. 9}[/tex]
Friction factor = 0. 52 × [tex]\frac{0. 06}{0. 324}[/tex]
Friction factor = 0. 52 × 0. 185
Friction factor [tex]= 0.096[/tex]
Loss When V = 1m/s
Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter
Head loss = 0. 289 × [tex]\frac{1}{2*6. 6743 * 10^-11}[/tex] × [tex]\frac{1^2}{0. 120}[/tex] × [tex]\frac{1}{100}[/tex]
Head loss = 1. 80 × 10^8
Head loss When V = 3m/s
Head loss = [tex]0. 096[/tex] × [tex]\frac{1}{1. 334 *10^-10}[/tex] × [tex]\frac{3^2}{0. 120}[/tex] × [tex]\frac{1}{100}[/tex]
Head loss = 5. 3× 10^8
Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.
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'The wave' is a particular type of pulse that can propagate through a large crowd gathered at a sports arena to watch a soccer, hockey, or CFL game. The elemets of the medium are the spectators, with zero position corresponding to their being seated and maximum position corresponding to their standing and raising their arms. When a large fraction of the spectators participate in the wave motion, a somewhat stable pulse shape can develop. The wave speed depends on people's reaction time, which is typically on the order of 0,1s. Estimate the order of magnitude, in minutes, of the time required for such a pulse to make one circuit around BC Place Stadium in Vancouver.
State all the assumptions that you've made.
Information about BC Place: dimensions are approximately 100 m X 85 m.
The total time required is 1 minute and 0.3 seconds.
Assumption: The distance between the people is 1 m and the stadium is a circle with a radius of 100m.
Here, the time taken by the person is 0.1 seconds.
Total distance covered by the wave = Circumference of the circle
So, total distance = 2 πr = 2 × 3.14 × 100 = 618 m
As the distance between each of the people is 1 m.
So, the number of personal interactions is 618.
Time taken by each person is 0.1 seconds.
So, total time, t = 0.1 × 618 = 61.8 seconds.
So, the order of the magnitude of the time required is 1 minute and 0.3 seconds for such a pulse to make one circuit around BC Place Stadium in Vancouver.
Hence, the total time required is 1 minute and 0.3 seconds.
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Which statement about the accuracy and precision of measuring tools is
true?
OA. A tool can be both accurate and precise.
о B. Tools that are precise must also be accurate.
о C. Tools cannot be both accurate and precise.
O D. A tool that is accurate lacks precision.
The true statement about the accuracy and precision of measuring tools is Tools cannot be both accurate and precise. The correct option is C.
What is accuracy?When in an experiment, a value is measured 5 times, then if the values measured are same for most of the time or like three times out of five, it said to be accurate. The phenomenon is accuracy.
Precision is about comparing the values to each other then find them near to each other.
Accuracy compares the experimental value to the theoretical value.
So, when all the values are close to each other but not nearest to the theoretical value, then it is said to be precise but not accurate.
Thus, the correct option is C.
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the temprerture of an object is 40°c.what is this tempreture in farhenhite scale?
This is an exercise of thermal scales.
Data:T = 40 °C
T = °F ¿?
To convert degrees Celsius to degrees Fahrenheit, we have the formula:
[tex]\bf{^{\circ}F=\dfrac{9}{5} \ ^{\circ}C+32 \ \ \to \ \ \ Formula }[/tex]
We substitute our data into the formula.
[tex]\bf{^{\circ}F=\dfrac{9}{5} \times 40+32}[/tex]
First we divide 9/5 multiplied by 40.
[tex]\bf{=75+32}[/tex]
Add 75 plus 32
[tex]\bf{=104 \ ^{\circ}F}[/tex]
Therefore, the temperature change of 40 °C on the °F scale is equal to 104.
[tex]\huge \red{\boxed{\green{\boxed{\boldsymbol{\purple{Piscis04}}}}}}[/tex]
DON'T ADD LINKS!
Imagine that a single cell below goes through four cell divisions.
How many cells are produced? To find out, draw the parent cell
and cells present after each division. Then write the number of
cells in the space below each drawing. (Hint: the number of cells
doubles with each division.)
Two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.
How many cells are produced?When the body cell passes through cell division process, it divides into two daughter cells. In the second cell division, two cells divides into four cells. In the third cell division, the four cells divides into eight cells. In the fourth cell division, the eight cells divides and sixteen cells are produced.
So we can conclude that two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.
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A cannonball is shot horizontally off a high castle wall at 47.4 m/s. What is the magnitude of the cannonball's velocity after 1.23 s? (Ignore direction)
The magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.
What is velocity?The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is
u is the initial velocity of canonball= 47.4 m/sec
g is the acceleration of free fall = 9.81 m/sec²
v is the velocity after 1.23 s
According to Newton's first equation of motion,
[tex]\rm v=u +gt \\\\ v= 47.4 \ m/sec + 9.81 \ m/sec^2 \times 1.23 sec \\\\\ v=59.46 \ m/sec[/tex]
Hence, the magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.
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A ball of mass 500g is dropped from a height 1.5m . It rebounds the floor and reached the height 1.2m. Calculate the impulse given to the bàll by the floor
The impulse given to the ball by the floor is 0.2865 kg.m/s.
What is impulse?The change in momentum is equal to the product of impact force applied while colliding and time for that impact.
Impulse F. t = m (Vf -Vi)
where, Vf is the final velocity and Vi is the initial velocity.
A ball of mass 500g is dropped from a height 1.5m . It rebounds the floor and reached the height 1.2m.
The initial velocity u = √2x 9.81 x 1.5 = 5.425 m/s
The final velocity v = √2x 9.81 x 1.2 = 4.852 m/s
Substitute the values into the expression, we get
Impulse = m(v- u)
Impulse=0.5 x (4.852- 5.425 )
Impulse = - 0.2865 kg.m/s
Thus, the magnitude of impulse given to the ball by the floor is 0.2865 kg.m/s.
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Which of the following can be contracted from contact with bloodborne pathogens?
HIV can be contracted from contact with bloodborne pathogens.
Other bloodborne diseases are HBV, malaria, syphilis and brucellosis
What are bloodborne pathogens?Bloodborne pathogens can be defined as those microorganisms or pathogenic organisms that cause disease and are present in human blood.
Blood borne pathogens can also be contacted through the following means
Se- xual contactNeedle contactIn conclusion; HIV can be contracted from contact with bloodborne pathogens.
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Question 3 of 15
Which statement describes an advantage of using nuclear fission to produce
electricity?
A. It uses a fuel made of atoms with large nuclei that will be
destroyed.
B. It is much more efficient than any other process used to generate
power.
C. It happens in large buildings that are expensive to build.
D. It produces radioactive waste products that must be stored safely.
Answer:
it is much more efficient than any other process to generate power
an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall, what is the height of the object?
The height of the object will be -5.19 cm
A concave mirror's reflecting surface curves inward and away from the light source. Light is reflected inward to a single focus point via concave mirrors. Concave mirrors, in contrast to convex mirrors, produce a variety of images depending on the object's to the mirror.
Given an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall
So let,
v = Image distance from the mirror = -33.5 cm
u = object distance from the mirror (concave) = 24 cm
hi = Image height = 7.25 cm
h = height of the object = ?
Using below formula to find height of the object
-v/u = hi/h
Putting all value in the formula we get
-(-33.5)/(-24) = 7.25/h
h = -5.19 cm
Therefore the height of the object will be -5.19 cm
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The wavelength of the wave pictured below is... Select one: a. 1 cm b. 1.5 cm c. 2 cm d. 3 cm
Answer:
D. 3 cm
Wavelength is the distance between two crests or two troughs.
In the given picture, the distance between two adjacent crests is 3 cm.
Hence, the wavelength is 3 cm.
Please help me!
If ball C is 3 times the volume of ball D and ball D has 1/3 the mass of ball C, which has the greater density?
A. Ball C
B. Ball D
C. The Densities are equal
C. The Densities are equal.
What is density?
Density is mass per unit volume or mass of a unit volume of a material substance.
If m1, V1 and D1 = mass, volume and density respectively of ball C
m2, V2 and D2 = mass, volume and density respectively of ball D
According to the Question ,
[tex]V_{1} = 3V_{2} , m_{2} = \frac{1}{3} (m_{1} ) \\ \\= m_{1} = 3m_{2}[/tex]
Therefore,
[tex]\frac{D_{1} }{D_{2} } = (\frac{m_{1} }{V_{1} } )* (\frac{m_{2} }{V_{2} } )\\ \\= (\frac{3m_{2} }{3V_{2} })*(\frac{V_{2} }{m_{2} }) \\\\= 1[/tex]
Hence, D1 = D2
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NGC 300 is the name of a spiral galaxy that looks similar to our Milky Way galaxy. It contains billions of stars. How is NGC 300 different from our solar system?
NGC 300 is different from our solar system because of its smaller size.
How is NGC 300 different from our solar system?NGC 300 galaxy is different from our solar system because of its size. NGC 300 is smaller than our solar system which means that it has less number of stars as compared to our solar system.
So we can conclude that NGC 300 is different from our solar system because of its smaller size.
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A 3,204 kg tree positioned on the edge of a cliff 247 m above the ground breaks away and falls into the valley below which is considered zero potential energy. If the tree’s mechanical energy is conserved, what is the speed of the tree just before it hits the ground in meters/sec?
Let's see
PE is turned to KE as per law of conservation of energy[tex]\\ \rm\Rrightarrow mgh=\dfrac{1}{2}mv^2[/tex]
[tex]\\ \rm\Rrightarrow 2gh=v^2[/tex]
[tex]\\ \rm\Rrightarrow 2(10)(247)=v^2[/tex]
[tex]\\ \rm\Rrightarrow v²=4940[/tex]
[tex]\\ \rm\Rrightarrow v=70.3ms^{-1}[/tex]
Water's specific heat capacity is 4.184 J/(g · °C). How does this compare to the heat capacities of most common substances?
Answer:
Explanation:
Comment
It's high. It takes quite a bit of heat to get water to change its heat content. Most substances don't reach much more the 0.5 J/(gram*oC)
What is the gravitational potential energy of a 15.0kg object that is 5.00m above the ground relative to a point 8.00m above the ground?
-441J
735J
-735J
441J
Explanation:
an object's gravitational potential energy Eg is m×g×h where:
m=mass
g=9.8m/s²
h=height relative to the closest object below it (because it cannot potentially fall through it
so Eg = 15×9.8×5=735J
imagine your lifting a toy car above your head and then holding it stationary above your head. While it is stationary what forces do you think act on the toy car
Gravity and normal forces act on the toy car.
What is force?A push or pull that an object experiences as a result of interacting with another item are known as a force. Every time two items interact, a force is exerted on each of the objects. The force is no longer felt by the two objects when the interaction ends. Only when there is interaction do forces actually exist.Both contact forces and forces coming from the action at a distance can be used to categorize all forces (interactions) between objects.Frictional forces, tensional forces, normal forces, air resistance forces, and applied forces are a few examples of contact forces.Gravitational forces are examples of action-at-a-distance forces.The standard metric unit known as the Newton is used to quantify force. "N" stands for Newton in abbreviation.To learn more about force refer to:
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Which refers to a material that causes a wave to bounce off of it?
A. absorber
B. reflector
C. conductor
D. insulator
Answer:
the answer to the question is a reflector
1. If the amplitude of a vibrating mass attached to the free end of an ideal spring is
25 cm, then the mass in one period travels 'a distance:
a) 1m
b) 0
c) 25m
The mass in one period travels 'a distance less than 25 cm which will be 0.
What is amplitude of a pendulum?
The amplitude of a pendulum is the maximum displacement of the pendulum.
If the maximum displacement of the pendulum is 25 cm, the mass in one period travels a distance less that 25 cm.
Thus, then the mass in one period travels 'a distance less than 25 cm which will be 0.
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It takes a minimum distance of 76.50 m to stop a car moving at 15.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 32.0 m/s.
The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.
Acceleration of the car
The acceleration of the car before stopping at the given distance is calculated as follows;
v² = u² + 2as
when the car stops, v = 0
0 = u² + 2as
0 = 15² + 2(76.5)a
0 = 225 + 153a
-a = 225/153
a = - 1.47 m/s²
Distance traveled when the speed is 32 m/sIf the same force is applied, then acceleration is constant.
v² = u² + 2as
0 = 32² + 2(-1.47)s
2.94s = 1024
s = 348.3 m
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A trolley is moving at constant speed in a friction compensated track. Some plasticine is dropped on the trolley and sticks on it. State with a reason what is observed about the motion of the trolley.
Answer:
Since the momentum of the body remains constant ( conserved) the trolley slows down (its velocity reduces) since its mass increases.
Aude slipped on ice going 7.6 m/s .What was her velocity before she hit the wal?
Answer:
Define universal gravitational constant. Why is newton's law of gravitation called universal law?
loade Brack Figure below show a circuit diagram of a device for controlling the temperature in a room. Explain how the Iron Heating element S loade Brack Figure below show a circuit diagram of a device for controlling the temperature in a room . Explain how the Iron Heating element S
Answer:
suna bro i am nepali and nepali are pro
Using what you learned above, try to predict the missing entries in the table below. Use the simulation to check your answer and type in the correct answer.
there is no pic or whatever to show
I was having trouble converting this p-t graph into a velocity-time graph and acceleration-time graph. Can someone convert it and send a picture of it?
We can obtain the velocity from the graph and plot is against time to obtain a velocity time graph.
What is a velocity time graph?The velocity time graph shows the change of velocity with time. Recall that velocity is the change of position with time.
As such, when we have a position time graph as shown, we can obtain the velocity from the graph and plot is against time to obtain a velocity time graph.
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A contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake as shown in Figure 4.29(a). The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.
a) Calculate the minimum force F he must exert to get the block moving.
N
(b) What is its acceleration once it starts to move, if that force is maintained?
m/s2
Answer:
54.0 x0.1=5.4 x0.03=0.162
kinetic force
If a contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03, then the minimum force F he must exert to get the block moving would be 5.2974 Nwrons.
What is friction?
Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.
As given in the problem, If a contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03,
The force required to get the block moving = μMg
= 0.01×54×9.81
= 5.2974 Newtons
Thus, the minimum force required to move the block would be 5.2974 Newtons.
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In a recent study of how mice negotiate turns, the mice ran around a circular 90∘ turn on a track with a radius of 0.10 m. The maximum speed measured for a mouse (mass = 18.5 g) running around this turn was 1.39 m/s. Part A What is the minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed?
The minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed without sleeping is 1.97.
Minimum coefficient of frictionThe minimum coefficient of friction is calculated as follows;
F = μmg = mac
μmg = mac
where;
ac is centripetal accelerationμg = ac
μg = v²/r
μ = v²/rg
where;
μ is the minimum coefficient of frictionv is velocityr is radius of the pathg is acceleration due to gravityμ = (1.39²)/(0.1 x 9.8)
μ = 1.97
Thus, the minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed without sleeping is 1.97.
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How many minutes would be required for a 300.0 W (i.e., power=energy per second=300 J/s)
immersion heater to heat 0.25 kg of water from 20.0°C to 100.0°C? (Note: The specific heat of water
is 4184 J/kg°C)
Hence, the time required is approximately equal to 4 minutes and 39 seconds.
Given, Power, P = 300 W= 300J/s
Power= Energy per second
Mass, m= 0.25 kg
Initial Temperature= 20.0°C
Final Temperature= 100.0°C
Temperature difference, T =( 100-20)°C= 80.0°C
Specific heat of water, S= 4184 J/kg°C
Energy can be represented as
E=mST
where E is the energy, m is the mass, S is the specific heat of water and T is the temperature difference.
As Energy can be written as the product of time and power, E=Pt
Pt=mST
So,
t=(msT)/P
t=(0.25×4184×80)/300
Time, t = 278.934 seconds= 4 minutes and 39 seconds.
Hence, 4 minutes and 39 seconds will be required for a 300.0 W immersion heater to heat 0.25 kg of water from 20.0°C to 100.0°C.
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Guillaume Amontons first took a stab at measuring absolute zero in 1702. What would be the most reasonable way it would have been done then?
A) Measure the pressure of a gas in a container at various temperatures, say from 100 °C down to maybe −20 °C, then extrapolate to the temperature at which the pressure would reach zero.
B) Measure the speed of molecules as the temperature gets colder until the temperature is measured at which motion stops.
C) Measure the pressure of a gas and cool it until the pressure is zero.
He would measure the speed of molecules as the temperature gets colder until the temperature is measured at which motion stops(option A).
What is the absolute zero?The absolute zero refers to the temperature at which the gas seems to stop moving. Recall that temperature is the measure of the average kinetic energy of the molecules of a gas. We now have to consider the options and see the one that connects temperature and speed of molecules.
Hence, if Guillaume Amontons first took a stab at measuring absolute zero in 1702, he would measure the speed of molecules as the temperature gets colder until the temperature is measured at which motion stops.
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Two small plastic spheres are given positive electric charges. When they are 15 cm apart the repulsive force between them has magnitude 0.22 N. What is the charge on each sphere?
The charge on each sphere is 742 n
What is Charge?In physics, Charge, also known as electric charge is a electrical charge, or electrostatic charge and symbolized q, is a characteristic of a unit of matter that expresses the extent to which it has more or fewer electrons than protons.
According to the question,
Distance = 15cm or 0.15m
Force = 0.22N
[tex]K=9*10^{9} kgm^2/c^2[/tex] ( this is a constant value of K)
By using the formula,
F = [tex]\frac{kQq}{d^{2} }[/tex]
0.220 N = 8.99
N·m²/C² * Q² / (0.15m)²
Q = 7.42e-7
C = 742 n
C is the charge of each sphere .
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What is the mass of an object if it has 915J of energy when it is 6.50m above the ground?
A.851kg
B0.0696kg
C.583kg
D.14.4kg
Answer:
D
Explanation:
GPE = mgh
GPE / gh = m
m = GPE/gh
m = 915 / 6.5 x 9.8
m = 915 / 63.7
m = 14.4kg
