Answer:
[tex]\triangle P=1.95*10^{-4}[/tex]
Explanation:
Mass [tex]m=0.001[/tex]
Diameter [tex]d=1.2m[/tex]
Length [tex]l=10m[/tex]
Generally the equation for Volume flow rate is mathematically given by
[tex]Q=AV[/tex]
[tex]V=\frac{Q}{\pi/4D^2}[/tex]
[tex]V=\frac{0.001}{\pi/4(1.2)^2}[/tex]
[tex]V=8.84*10^{-4}[/tex]
Generally the equation for Friction factor is mathematically given by
[tex]F=\frac{64}{Re}[/tex]
Where Re
Re=Reynolds Number
[tex]Re=\frac{pVD}{\mu}[/tex]
[tex]Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}[/tex]
[tex]Re=1040[/tex]
Therefore
[tex]F=\frac{64}{Re}[/tex]
[tex]F=\frac{64}{1040}[/tex]
[tex]F=0.06[/tex]
Generally the equation for Friction factor is mathematically given by
[tex]Head loss=\frac{fLv^2}{2dg}[/tex]
[tex]H=\frac{0.06*10*(8.9*10^-4)^2}{2*1.2*9.81}[/tex]
[tex]H=19.9*10^{-9}[/tex]
Where
[tex]H=\frac{\triangle P}{\rho g}[/tex]
[tex]\triangle P=\frac{19.9*10^{-9}}{10^3*(9.81)}[/tex]
[tex]\triangle P=H*\rho g[/tex]
[tex]\triangle P=1.95*10^{-4}[/tex]
If the open body is postively charged and another body is negatively charged ,free electrons tend to
Answer:
The free electrons tend to Negatively Charged body to Positively Charged body
Explanation:
As they have different charges, Electrons are more attracted to the Positive or it's opposite charge.
In higher mass stars, repeating cycles of fusion will create heavier elements in layers
until which element is created at the center of the core?
hydrogen
iron
uranium
helium
what is the main function of a step-up and step-down transformers
Answer:
Explanation:
The difference between step-up and step-down transformer is given below:-
Step-up transformer
i) It increases the voltage.
ii) It decreases the current strength.
iii) It has a greater number of turns in secondary as compared to the primary.
iv) The wire of primary is thicker than that of the secondary coil.
Step-down transformer
i) It decreases the voltage.
ii) It increases the current strength of the secondary.
iii) It has less number of turns in secondary as compared to the primary.
iv) The wire of the secondary coil is thicker than that of the primary coil.
In another version of the "Giant Swing", the seat is connected to two cables as shown in the figure (Figure ), one of which is horizontal. The seat swings in a horizontal circle at a rate of 39.3 rev / m * i * n
If the seat weighs 268 N and a 896-N person is sitting in it, find the tension in the horizontal cable
If the seat weighs 268 N and a 896-N person is sitting in it, find the tension in the inclined cable
Answer:Solution to 47E Step 1 Angular velocity of the swing=32rpm Weight of the seat =255N Weight of the person =825N Total weight =255+825=1080N Radius =7.5m
Explanation:
Our solar system is made up of the Sun, 8 planets, and other bodies such as
asteroids orbiting the Sun. The solar system is very large compared to anything we see on
Earth. The distance between planets is measured in astronomical units (AU). One AU is
equal to 149.6 million kilometers, the average distance between the Sun and Earth. Scale
models are useful for helping us understand the size of the solar system.
Mr. Wilson’s science class made a scale model of the solar system. They went out to the
school’s football field, and they used the chart shown below to mark out the scale distance
from the Sun to each planet
The average orbital radius for a planet is its distance from the Sun. Which statement BEST
describes the relationship between the planets and their average orbital radii?
A. The planets are evenly spaced in the solar system.
B. Closer to the Sun, the planets are regularly spaced apart.
C. The planets closest to each other are the ones farthest from the sun.
D. Farther from the Sun, the planets are spaced farther apart from each other.
the answer to the question is D
The v-t graph of a moving body is given below. The distance covered by the body in
the first 40s is _________and its acceleration during the last 40s is _____________.
a) 3200m; 2ms-1
b) 700m; -2ms-2
c) 900m; - 2ms-2
d) 900m; 2ms-2
Answer: The correct answer is a) 3200 m, [tex]-2m/s^2[/tex]
Explanation:
Speed is defined as the ratio of distance travelled to the time taken. The equation follows:
[tex]\text{Speed}=\frac{\text{Distance travalled}}{\text{Time taken}}[/tex]
From the graph:
Speed for the first 40 s, v = 80 m/s
Time taken, t = 40 s
Putting values in above equation, we get:
[tex]\text{Distance travalled}=(80 m/s\times 40s)=3200 m[/tex]
Acceleration is defined as the ratio of change of velocity to the change of time. The equation follows:
[tex]\text{Acceleration}=\frac{\Delta v}{\Delta t}=\frac{v_2-v_1}{t_2-t_1}[/tex]
From the graph, for the last 40 sec:
Initial velocity, [tex]v_1[/tex] = 80 m/s
Final velocity, [tex]v_2[/tex] = 0 m/s
Initial time, [tex]t_1[/tex] = 40 s
Final time, [tex]t_2[/tex] = 80 s
Putting values in above equation, we get:
[tex]\text{Acceleration}=\frac{(0-80)m/s}{(80-40)s}\\\\\text{Acceleration}=\frac{-80m/s}{40s}=-2m/s^2[/tex]
Hence, the correct answer is a) 3200 m, [tex]-2m/s^2[/tex]
How would an observer on train A, which is moving at nearly the speed of light, view a clock on train B, which is moving at the same speed and in the same direction?
A. The clock on train B would appear narrower and run more slowly.
B. The clock on train B would appear to be the same width and to run at the same rate. C. The clock on train B would appear narrower and run faster.
D. The clock on train B would appear wider and run more slowly.
Answer:
B.
Explanation:
If both train A and train B are moving in the same direction and the exact same speed then from an observer's viewpoint within either train everything would seem as though it is not moving. Therefore, the clock on train B would appear to be the same width and to run at the same rate. In order for this to be the case the speed of both trains would need to be exactly the same, any difference in speed will cause the clock on the opposite train to appear distorted and run either faster or slower depending on the speed of the train you are on.
Answer:
The trains are moving at the same speed, so the answer is A.
The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00\times 10^32.00×10 ^3 N with an effective perpendicular lever arm of 3.00 cm, producing an angular acceleration of the forearm of 120 rad/s^2 .
What is the moment of inertia of the boxer's forearm?
Answer:
Explanation:
From the given information:
The torque produced due to the force can be expressed as:
[tex]\tau = F \times r[/tex]
where;
[tex]\tau[/tex] = torque
F = force exerted
r = lever's arm radius
[tex]\tau[/tex] = [tex]2.00 \times 10^3 \times 0.03 m[/tex]
[tex]\tau[/tex] = 60 N.m
However, equating the torque with the moment of inertia & angular acceleration, we use the equation:
[tex]\tau[/tex] = I∝
60 Nm = I × 120 rad/s²
I = 60 Nm/120 rad/s²
I = 0.5 kg.m²
A balloon pops, making a loud noise that startles you. What kind of energy best describes this experience?
A. Thermal Energy
B. Sound Energy
C. Gravitational Energy
D. Radiant Energy
6. (6P) Formula Based Problem: A car is speeding down a highway at 30 m/s. A stray dog
runs into the road ahead of the driver. The driver hits the brake decelerating at a rate of
10m/s. The driver managed to stop the car right as it was about to hit the car. How far
away was the dog when the driver first hit the brake?
Answer:
45 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 30 m/s
Deceleration (a) = –10 m/s²
Final velocity (v) = 0 m/s
Distance (s) =?
The distance of the dog from the car before the driver hits the brake can be obtained as follow:
v² = u² + 2as
0² = 30² + (2 × –10 × s)
0 = 900 + (–20s)
0 = 900 – 20s
Collect like terms
0 – 900 = –20s
–900 = –20s
Divide both side by –20
s = –900 / –20
s = 45 m
Therefore, the dog was 50 m away when the driver first hit the brake
A Michelson interferometer operating at a 400 nm wavelength has a 3.70-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atm pressure is 1.00028.
How many bright-dark-bright fringe shifts are observed as the cell fills with air?
Answer:
[tex]m=42\ fringes[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lambda=400nm[/tex]
Length of cell arm [tex]h=3.70cm[/tex]
Refraction of air at at 1.00 atm pressure [tex]n=1.00028.[/tex]
Generally the equation for Number of shifts is mathematically given by
[tex]m=N-N_o[/tex]
Since
[tex]N_0=\frac{2t}{\lambda_0}[/tex]
Therefore
[tex]m=\frac{2t}{\lambda_0/n}-\frac{2t}{\lambda_0}[/tex]
[tex]m=\frac{2t}{\lambda_0} n-1[/tex]
[tex]m=\frac{2(3.7*10^{-2})}{400*10^{-9}}*(1.00028-1)[/tex]
[tex]m=51.8[/tex]
[tex]m=42\ fringes[/tex]
Two water-slide riders, A and B, start from rest at the same time and same height, h but on differently shaped slides.
Required:
a. Which rider is traveling faster at the bottom?
b. Which rider makes it to the bottom first? Ignore friction and assume both slides have the same path length.
Answer:
a. None
b. Both
Explanation:
a. Which rider is traveling faster at the bottom?
Since both riders fall from the same height, h, their potential energy, U at the top equals their kinetic energy, K at the bottom.
U = mgh and K = 1/2mv²
Since U is he same for both water-slide riders, then K will be the same and thus their speed at the bottom will be the same. This is shown below.
K = U
1/2mv² = mgh
v² = 2gh
v =√(2gh) where v = speed of rider at the bottom, g = acceleration due to gravity and h = height of slide.
Since the height is the same, so their speed at the bottom is the same. So, none of the riders travels faster than the other since they have the same speed at the bottom.
b. Which rider makes it to the bottom first? Ignore friction and assume both slides have the same path length.
Since the path length of the water slides are the same and friction is neglected, both water-slide rider get to the bottom at the same time since the distance moved is the same and they both start from rest.
So, both riders make it to the bottom at the same time.
The acceleration due to gravity acts vertically downwards, and the component of gravity acceleration is larger when the slope is steeper.
a. Rider Bb. Rider B
Reasons:
The acceleration of the riders are due to gravity
The component acceleration due to gravity acting on a slope is a = g·sin(θ)
As the steepness of the slope increase, the angle, θ, and sin(θ) increases, therefore, the acceleration increases.
Rider A is on a slide with gentle slope, such that if the slide is flat, rider A will be stationary.
The shape of the water slide rider B is on is steeper, and therefore, rider B is accelerating more than rider A. The higher acceleration of rider B, gives rider B a higher speed than rider, such that rider B, is riding faster than rider ATherefore;
a. The rider that is travelling faster at the bottom is rider B
b. Given that friction is ignored, and the path have the same length to the
bottom, the rider that makes to the bottom first is the rider that is moving
faster, which is rider B
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g Is a nucleus that absorbs at 4.13 δ more shielded or less shielded than a nucleus that absorbs at 11.45 δ? _________ Does the nucleus that absorbs at 4.13 δ require a stronger applied field or a weaker applied field to come into resonance than the nucleus that absorbs at 11.45 δ?
Answer: A nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded and a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field
Explanation:
While interpreting the data in NMR, the positions of signals are studied.
The nucleus/ protons having a higher value of [tex]\delta[/tex] are said to be less shielded. They are said to be upfield.
The nucleus/protons having a lower value of [tex]\delta[/tex] are said to be more shielded. They are said to be downfield.
So, a nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded by the nucleus that absorbs at [tex]4.13\delta[/tex]
Also, the less shielded nucleus/protons will require a weak applied field to come into resonance than the more shielded nucleus/protons
So, a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field to come into resonance than the nucleus that absorbs at [tex]11.45\delta[/tex]
You throw a stone straight down from the top of a tall tower. It leaves your hand moving at 8.00 m/s, Air resistance can be neglected. Take the positive -direction to be upward, and choose y 0 to be the point where the stone leaves your hand. Find the stone's position 1.50s after it leaves your hand.
Express your answer with the appropriate units.
Find the y-component of the stone's velocity 1.50 s after it leaves your hand. Express your answer with t0he appropriate units.
Answer:
The velocity after 1.5 s is 22.7 m/s downwards.
Explanation:
Initial velocity = - 8 m/s
acceleration, a = - 9.8 m/s2
time, t = 1.5 s
Use first equation of motion
v = u + at
v = - 8 - 9.8 x 1.5
v = - 8 - 14.7
v = - 22.7 m/s
Thus, the velocity after 1.5 s is 22.7 m/s downwards.
factors that favour mining in South Africa
Answer:
According to the data, factors influencing mining investment in South Africa's favour are the availability of labour and skills, the quality of the country's infrastructure, the quality of its geological database, and the State's environmental regulations.
A typical ceiling fan running at high speed has an airflow of about 2.00 ✕ 103 ft3/min, meaning that about 2.00 ✕ 103 cubic feet of air move over the fan blades each minute.
Determine the fan's airflow in m3/s.
Answer:
0.94 m³/s
Explanation:
From the question given above, the following data were obtained:
Air flow (in ft³/min) = 2×10³ ft³/min
Air flow (in m³/s) =.?
Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:
35.315 ft³/min = 1 m³/min
Therefore,
2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min
2×10³ ft³/min = 56.63 m³/min
Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:
1 m³/min = 1/60 m³/s
Therefore,
56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min
56.63 m³/min = 0.94 m³/s
Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.
Suppose 4 J of work are needed to stretch a spring 14 cm from its natural position. How much work is needed to stretch it 20 cm from its natural position
Answer:
8.16 J
Explanation:
Applying,
W = ke²/2................. Equation 1
Where W = work done in stretching the spring, k = spring constant, e = extension
make k the subject of the equation
k = 2W/e²................ Equation 2
From the question,
Given: W = 4 J, e = 14 cm = 0.14 m
Substitute these values into equation 2
k = 2(4)/0.14²
k = 8/0.0196
k = 408.16 N/m
If stretchd 20 cm from its natural length,
Therefore,
e = 20 cm = 0.2 m
W = 408.16(0.2²)/2
W = 8.16 J
The decibel level of a jackhammer is 125 dB relative to the threshold of hearing. Determine the decibel level if three jackhammers operate side by side.
Answer:
130 dB
Explanation:
The equation for decibel level is given by:
[tex]D=10log(\frac{I}{I_n} )\\\\Where\ D\ is\ the \ decibel\ level\ in\ dB, I\ is\ the\ intensity\ in \ W/m^2, \\I_n\ is\ threshold\ intensity\ to\ the\ human\ ear=1*10^{-12}W/m^2\\\\Given\ that\ D=125dB, hence:\\\\125=10log(\frac{I}{1*10^{-12}} )\\\\12.5=log(\frac{I}{1*10^{-12}} )\\\\I=3.2\ W/m^2[/tex]
The intensity for 1 jack hammer is 3.2 W/m², therefore for 3 jack hammers, the intensity = 3 * 3.2 = 9.6 W/m²
[tex]D=10\ log(\frac{I}{I_n} )\\\\D=10*log(\frac{9.6}{1*10^{-12}} )\\\\D=130\ dB[/tex]
44.7
When Xavier places his hands near a light bulb, he notices that certain areas around the light bulb are warmer than
others. Which best explains this?
The areas to the sides of the light bulb are warmest because of conduction,
O The areas to the sides of the light bulb are warmest because of convection,
The area directly above the light bulb is warmest because of conduction,
The area directly above the light bulb is warmest because of convection.
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Answer:
The area directly above the light bulb is warmest because of convection.
Explanation:
if all the sides of the bulb are equally close to the light source inside the bulb, all area of the bulb would be equally heated by conduction. however, convection heating mainly heats up the surface above the light source. in convection heating, the air above the surface of the light source get heated by the light source and expands, casuing it to be less dense and rise to the top of the bulb. colder denser air at the top of the bulb sink to the light source adn gain heat and expands, becoming less dense. this process repeats and the surface above the light source becomes the warmest due to convection heating
A rod of 3.0-m length and a square (2.0 mm X 2.0 mm) cross section is made of a material with a resistivitYof 6.0 X 10-8 Ω m. If a potential difference of 0.60 V is placed across the ends of the rod, at what rate is heat generated in the rod in watt.
Select one:
a. 24
b. 12
c. 4
d. 8
A house has well-insulated walls. It contains a volume of 105 m3 of air at 305 K.
Consider heating it at constant pressure. Calculate the energy required to increase the temperature of this diatomic ideal gas by 0.7
Answer: [tex]85.46\ kJ[/tex]
Explanation:
Given
Volume of air [tex]V=105\ m^3[/tex]
Temperature of air [tex]T=305\ K[/tex]
Increase in temperature [tex]\Delta T=0.7^{\circ}C[/tex]
Specific heat for diatomic gas is [tex]C_p=\dfrac{7R}{2}[/tex]
Energy required to increase the temperature is
[tex]\Rightarrow Q=nC_pdT\\\\\Rightarrow Q=n\times \dfrac{7R}{2}\times \Delta T\\\\\Rightarrow Q=\dfrac{7}{2}nR\Delta T\\\\\Rightarrow Q=\dfrac{7}{2}\times \dfrac{PV}{T}\times \Delta T\quad [\text{using PV=nRT}][/tex]
Insert the values
[tex]\Rightarrow Q=\dfrac{7}{2}\times \dfrac{1.01325\times 10^5\times 105}{305}\times 0.7\\ \text{Assuming air pressure to be atmospheric P=}1.01325\times 10^5\ N/m^2\\\\\Rightarrow Q=0.8546\times 10^5\\\Rightarrow Q=85.46\ kJ[/tex]
Can somebody please help
Answer:
Explanation:
part A: C
part B: B
A magnetic field of 0.276 T exists in the region enclosed by a solenoid that has 517 turns and a diameter of 10.5 cm. Within what period of time must the field be reduced to zero if the average magnitude of the induced emf within the coil during this time interval is to be 12.6 kV
Answer:
The period the field must be reduced to zero is 9.81 x 10⁻⁵ s
Explanation:
Given;
initial value of the magnetic field, B₁ = 0.276 T
number of turns of the solenoid, N = 517 turns
diameter of the solenoid, d = 10.5 cm = 0.105 m
induced emf, = 12.6 kV = 12,600 V
when the field becomes zero, then the final magnetic field value, B₂ = 0
The induced emf is given by Faraday's law;
[tex]emf = -\frac{NA\Delta B}{t} \\\\emf = -\frac{NA (B_2 -B_1)}{t} \\\\t = -\frac{NA (B_2 -B_1)}{emf}\\\\t = \frac{NA (B_1 -B_2)}{emf}\\\\where;\\\\t \ is \ the \ time \ when \ B = 0 \ \ (i.e\ B_2 = 0)\\\\A \ is \ the \ area \ of \ the \ coil\\\\A = \frac{\pi d^2}{4} = \frac{\pi (0.105)^2}{4} = 0.00866 \ m^2\\\\t= \frac{(517) \times (0.00866)\times (0.276 -0)}{12,600}\\\\t = 9.81 \times 10^{-5} \ s[/tex]
Therefore, the period the field must be reduced to zero is 9.81 x 10⁻⁵ s
what is simple definition of democracy
it's a form of government where people elect their representatives
Answer:
The word democracy itself means rule by the people.
Which of the following would experience induced magnestism mostly easily
Answer:
Copper would experience induced magnestism mostly easily.
Explanation:
Permalloy would experience induced magnetism most easily.
I don't know the options but that is correct too!!
A cheetah is running in a straight line in pursuit of prey. The cheetah's mass is 63.9 kg. Through its running motion, assume the cheetah experiences a constant forward force of 609.1 N. Also, assume the cheetah experiences a constant 107.9 N air resistance force that is opposite its motion. What is the magnitude of the cheetah's acceleration, in units of m/s2?
Answer:
a = 7.84 m / s²
Explanation:
For this exercise let's use Newton's second law
F - fr = m a
indicate that the force is F = 609.1 N and the friction force is fr = 107.9 N and is constant
a = [tex]\frac{F - fr}{m}[/tex]
let's calculate
a = [tex]\frac{ 609.1 - 107.9 }{63.9}[/tex]
a = 7.84 m / s²
An alpha particle consists of two protons and two neutrons. How will the emission of an alpha particle affect the following:
a. The atomic number of the atom: _______Atomic Number will go down_______
b. The mass number of the atom: ___Mass number will go down____________
Answer:
a) The atomic number decreases by 2 units
b) the atomic mass decreases by 4 units
Explanation:
An alpha particle is a helium nucleus, which is why it is made up of two positive protons and two negative neutrons, the mass of the particles is almost the same.
a) The atomic number decreases by 2 units
b) the atomic mass decreases by 4 units
What is the minimum value of the friction coefficient between the boxes that will keep them from slipping when the 100 N force is applied
Answer:
The friction coefficient's minimum value will be "0.173".
Explanation:
The given query seems to be incomplete. Below is the attached file of the complete question.
According to the question,
(a)
The net friction force's magnitude will be:
⇒ [tex]F_{net}=ma[/tex]
[tex]=5\times 1.7[/tex]
[tex]=8.5 \ N[/tex]
(b)
For m₃,
⇒ [tex]ma=\mu m_3 g[/tex]
Or,
⇒ [tex]\mu=\frac{a}{g}[/tex]
[tex]=\frac{1.7}{9.8}[/tex]
[tex]=0.173[/tex]
resolve the vector shown below into its components
Answer:
ans: option A
Explanation:
components along x- axis is -3x and along x- axes is -2y
now use triangle law of vector addition
Which of the following statements regarding the warming of the troposphere (i.e., the lower atmosphere) are TRUE (choose all that apply):
a. Conduction only warms the shallow layer of air within a few centimeters of the ground
b. The troposphere is heated mainly from above
c. Atmospheric gases do not readily absorb solar radiation
d. The troposphere is heated mainly by direct molecule-to-molecule contact
e. Condensation of water vapor plays a role in heating the lower atmosphere
f. Oxygen and nitrogen readily absorb and emit terrestrial radiation
g. Convection distributes heat vertically
Answer: • Conduction only warms the shallow layer of air within a few centimeters of the ground
• Atmospheric gases do not readily absorb solar radiation
• Condensation of water vapor plays a role in heating the lower atmosphere.
• Convection distributes heat vertically
Explanation:
The statements regarding the warming of the troposphere that are true include:
• Conduction only warms the shallow layer of air within a few centimeters of the ground
• Atmospheric gases do not readily absorb solar radiation
• Condensation of water vapor plays a role in heating the lower atmosphere.
• Convection distributes heat vertically.
Therefore, the correct options are A, C, E and G.
• Conduction only warms the shallow layer of air within a few centimeters of the ground
• Atmospheric gases do not readily absorb solar radiation
• Condensation of water vapor plays a role in heating the lower atmosphere.
• Convection distributes heat vertically
What is troposphere?The lowest layer of the atmosphere of the earth is the troposphere. Most of the mass (about 75-80%) of the atmosphere is in the troposphere.
Most types of clouds are found in the troposphere, and almost all weather occurs within this layer.
The statements regarding the warming of the troposphere that are true include:
• Conduction only warms the shallow layer of air within a few centimeters of the ground
• Atmospheric gases do not readily absorb solar radiation
• Condensation of water vapor plays a role in heating the lower atmosphere.
• Convection distributes heat vertically.
Thus a,c,e,g are the correct statements regarding the warming of the troposphere
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