Answer:
The frequencies are [tex]f_n = n (0.875 )[/tex]
Explanation:
From the question we are told that
The speed of the wave is [tex]v = 0.700 \ m/s[/tex]
The length of vibrating clothesline is [tex]L = 40.0 \ cm = 0.4 \ m[/tex]
Generally the fundamental frequency is mathematically represented as
[tex]f = \frac{v}{2 L }[/tex]
=> [tex]f = \frac{ 0.700 }{2 * 0.4 }[/tex]
=> [tex]f = 0.875 \ Hz[/tex]
Now this other frequencies of vibration experience by the clotheslines are know as harmonics and they are obtained by integer multiple of the fundamental frequency
So
The frequencies are mathematically represented as
[tex]f_n = n * f[/tex]
=> [tex]f_n = n (0.875 )[/tex]
Where n = 1, 2, 3 ....
A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. Water flows out from the bottom through a small hole. How fast is this water moving?
Answer:
The water is flowing at the rate of 28.04 m/s.
Explanation:
Given;
Height of sea water, z₁ = 10.5 m
gauge pressure, [tex]P_{gauge \ pressure}[/tex] = 2.95 atm
Atmospheric pressure, [tex]P_{atm}[/tex] = 101325 Pa
To determine the speed of the water, apply Bernoulli's equation;
[tex]P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2[/tex]
where;
P₁ = [tex]P_{gauge \ pressure} + P_{atm \ pressure}[/tex]
P₂ = [tex]P_{atm}[/tex]
v₁ = 0
z₂ = 0
Substitute in these values and the Bernoulli's equation will reduce to;
[tex]P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 = P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 = P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + \rho gz_1 = \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} + \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} + \rho gz_1)}{\rho} }[/tex]
where;
[tex]\rho[/tex] is the density of seawater = 1030 kg/m³
[tex]v_2 = \sqrt{ \frac{2(2.95*101325 \ + \ 1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s[/tex]
Therefore, the water is flowing at the rate of 28.04 m/s.
Answer:
C. effusion because there is a movement of a gas through a small opening into a larger volume
Explanation:
Edge2020
Have a great day y'all :)
Two parallel very long straight wires carrying current of 5A each are kept at a separation of 1m. If the currents are in the same direction, the force per unit length between them is __________
Answer:
The force per unit between the two parallel wires with same current flowing in the same direction is 5 x 10⁻⁶ N/m repulsive force.
Explanation:
Given;
current though the two parallel wires, I₁ and I₂ = 5A
distance between the two wires, R = 1 m
The force per unit of the wires is calculated as;
[tex]\frac{F}{L} = \frac{\mu_o I_1I_2}{2\pi R}[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
Substitute in the given values into the equation and determine the force per unit length (F/L).
[tex]\frac{F}{L} = \frac{\mu_o I_1I_2}{2\pi R} \\\\ \frac{F}{L} = \frac{4\pi *10^{-7}*5*5}{2\pi *1}\\\\ \frac{F}{L} = 5*10^{-6} \ N/m \ (repulsive)[/tex]
Therefore , the force per unit between the two parallel wires with same current flowing in the same direction is 5 x 10⁻⁶ N/m repulsive force.
If you converted 0.000013 to scientific notation, what would the prefix be to the correct number of significant digits?
Answer:
1.3
Explanation:
it will taken to as in from of standard form
Answer:
1.3 * 10^-5
Explanation:
We are learning about scientific notation.
When a number becomes a decimal followed before with zeroes, we know that the value of that number is decreasing. So instead of usually doing a positive exponent, we will do a negative exponent indicating we are going back.
So let's not only count the amount of zeroes followed before 13, but the decimal.
0.000013
The original number "1.3" went back 5 spaces, therefore making our exponent 5.
1.3 * 10^-5
Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of the system by a factor of
Answer:
The frequency changes by a factor of 0.27.
Explanation:
The frequency of an object with mass m attached to a spring is given as
[tex]f[/tex] = [tex]\frac{1}{2\pi } \sqrt{\frac{k}{m} }[/tex]
where [tex]f[/tex] is the frequency
k is the spring constant of the spring
m is the mass of the substance on the spring.
If the mass of the system is increased by 14 means the new frequency becomes
[tex]f_{n}[/tex] = [tex]\frac{1}{2\pi } \sqrt{\frac{k}{14m} }[/tex]
simplifying, we have
[tex]f_{n}[/tex] = [tex]\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }[/tex]
[tex]f_{n}[/tex] = [tex]\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }[/tex]
if we divide this final frequency by the original frequency, we'll have
==> [tex]\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }[/tex] ÷ [tex]\frac{1}{2\pi } \sqrt{\frac{k}{m} }[/tex]
==> [tex]\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }[/tex] x [tex]2\pi \sqrt{\frac{m}{k} }[/tex]
==> 1/3.742 = 0.27
What type of information is available to scientists through a Global Positioning System (GPS) device?
Answer:
GPS receivers provide location in latitude, longitude, and altitude. They also provide the accurate time. GPS includes 24 satellites that circle Earth in precise orbits.
What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.
Answer:
Density = 1.1839 kg/m³
Mass = 227.3088 kg
Specific Gravity = 0.00118746 kg/m³
Explanation:
Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³
Now, from tables, density of air at 25°C is 1.1839 kg/m³
Now formula for density is;
ρ = mass(m)/volume(v)
Plugging in the relevant values to give;
1.1839 = m/192
m = 227.3088 kg
Formula for specific gravity of air is;
S.G_air = density of air/density of water
From tables, density of water at 25°C is 997 kg/m³
S.G_air = 1.1839/997 = 0.00118746 kg/m³
1. The most likely injuries in an Anatomy class are (circle all that apply)
a. chemical spill
b. cut from scalpel
c, burn from
open
flame
d. foreign object or splash in eye
e, animal bite
Zoning laws establish _______. a. what types of buildings can be built in an area b. the uses an area of land can be put to c. who can live in an area d. the types of business that can occupy a building Please select the best answer from the choices provided A B C D
Answer:
its B
Explanation:
Answer:
It's B
Explanation: hope it helps ^w^
A uniform crate with a mass of 22 kg must be moved up along the 15° incline without tipping. The force P is horizontal. Determine the corresponding magnitude of force P.
Answer:
[tex]F_x=208.25\ N[/tex]
Explanation:
Given that,
Mass of a crate is 22 kg
It moved up along the 15 degrees incline without tipping.
We need to find the corresponding magnitude of force P. The force P is acting in horizontal direction.
It means that the horizontal component of force is given by :
[tex]F_x=F\cos\theta\\\\F_x=mg\cos\theta\\\\F_x=22\times 9.8\times \cos(15)\\\\F_x=208.25\ N[/tex]
So, the horizontal component of force is 208.25 N.
Statement I: At the same temperature lighter gas molecules have a higher average velocity than heavier gas molecules.
Statement II: At the same temperature lighter gas molecules have a higher average kinetic energy than heavier gas molecules.
a) Statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1
b) Both the statement 1 and statement 2 are correct and statement 2 is not the correct explanation of statement 1
c) Statement 1 is correct but statement 2 is not correct
d) Statement 1 is not correct but statement 2 is correct
e) Both the statement 1 and statement 2 is not correct
Answer:
Statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1
Explanation:
Both the velocity and kinetic energy of a gas molecule depends on its relative molecular mass according to Graham's law of diffusion in gases. Hence, the greater the relative molecular mass of the gas, the lesser its average velocity and kinetic energy.
Hence we can see that statement 2 vividly explains the postulation of statement 1 and makes the points more easily comprehensible.
Please help and if you have the answer if you can please explain how you got it :)!
Answer:
The mass of ball C is greater than the mass of ball A but less than the mass of ball B.
Explanation:
From Newton's second law, net force = mass × acceleration.
Using the data for ball B, the acceleration of gravity near the surface of the moon is:
∑F = ma
9.6 N = (6 kg) a
a = 1.6 m/s²
Therefore, the mass of ball C is:
∑F = ma
6.6 N = m (1.6 m/s²)
m = 4.1 kg
Consider the waveform expression. y(x,t)=ymsin(801t+3.38+0.503x) The transverse displacement ( y ) of a wave is given as a function of position ( x in meters) and time ( t in seconds) by the expression. Determine the wavelength, frequency, period, and phase constant of this waveform.
Answer:
f = 127.48 Hz , T = 7.844 1⁻³ s , Ф = 3.38 , λ = 12.49 m
Explanation:
The general equation for the motion of a wave in a string is
y = A sin (kx -wt + fi)
the expression they give is
y = ym sin (0.503x + 801 t + 3.38 )
the veloicda that accompanies time is
w = 801 rad / s
angular velocity is related to frequency
w = 2π f
f = w / 2π
f = 801 / 2π
f = 127.48 Hz
The period is the inverse of the frequency
T = 1 / f
T = 1 / 127.48
T = 7.844 10⁻³ s
the csntnate of phase fi is the independent term
Ф = 3.38
the wave vector accompanies the position k = 0.503 cm
ka = 2pi /λ
λ = 2 π / k
λ = 2 π / 0.503
λ = 12.49 m
You've recently read about a chemical laser that generates a 20.0-cm-diameter, 26.0 MW laser beam. One day, after physics class, you start to wonder if you could use the radiation pressure from this laser beam to launch small payloads into orbit. To see if this might be feasible, you do a quick calculation of the acceleration of a 20.0-cm-diameter, 110 kg, perfectly absorbing block.
Required:
a. What speed would such a block have if pushed horizontally 100 m along a frictionless track by such a laser?
b. Does this seem like a promising method for launching satellites?
Answer :
(a). The speed of the block is 0.395 m/s.
(b). No
Explanation :
Given that,
Diameter = 20.0 cm
Power = 26.0 MW
Mass = 110 kg
diameter = 20.0 cm
Distance = 100 m
We need to calculate the pressure due to laser
Using formula of pressure
[tex]P_{r}=\dfrac{I}{c}[/tex]
[tex]P_{r}=\dfrac{P}{Ac}
Put the value into the formula
[tex]P_{r}=\dfrac{26.0\times10^{6}}{\pi\times(10\times10^{-2})^2\times3\times10^{8}}[/tex]
[tex]P_{r}=2.75\ N/m^2[/tex]
We need to calculate the force
Using formula of force
[tex]F=P\times A[/tex]
[tex]F=P\times \pi r^2[/tex]
Put the value into the formula
[tex]F=2.75\times\pi (0.01)^2[/tex]
[tex]F=0.086\ N[/tex]
We need to calculate the acceleration
Using formula of force
[tex]F=ma[/tex]
Put the value into the formula
[tex]0.086=110\times a[/tex]
[tex]a=\dfrac{0.086}{110}[/tex]
[tex]a=0.000781\ m/s^2[/tex]
[tex]a=7.81\times10^{-4}\ m/s^2[/tex]
(a). We need to calculate speed of the block
Using equation of motion
[tex]v^2=u^2+2ad[/tex]
Put the value into the formula
[tex]v=\sqrt{2\times7.81\times10^{-4}\times100}[/tex]
[tex]v=0.395\ m/s[/tex]
(b). No because the velocity is very less.
Hence, (a). The speed of the block is 0.395 m/s.
(b). No
Light with an intensity of 1 kW/m2 falls normally on a surface with an area of 1 cm2 and is completely absorbed. The force of the radiation on the surface is
Answer:
The force of the radiation on the surface is 3.33 X 10⁻¹⁰ N
Explanation:
Given;
intensity of light, I = 1 kw/m²
area of the surface, A = 1 cm² = 1 x 10⁻⁴ m²
Power of the incident light, P = I x A
Power of the incident light, P = (1 kw/m²) x (1 x 10⁻⁴ m²)
Power of the incident light, P = 1 x 10⁻⁴ kW = 0.1 W
Power of the incident light is given by;
P = Fv
where;
F is the force of the radiation on the surface
v is the speed of light = 3 x 10⁸ m/s
F = P/ v
F = (0.1) / (3 x 10⁸)
F = 3.33 X 10⁻¹⁰ N
Therefore, the force of the radiation on the surface is 3.33 X 10⁻¹⁰ N
Explain Cheetah how force, velocity, and acceleration are related.
Explanation:
A cheetahs force, velocity, and acceleration are related because velocity goes by seconds/mile per hour, force goes by strength and energy, and acceleration goes by how the velocity changes the speed.
Assume that helium behaves as an ideal monatomic gas. If 2 moles of helium undergo a temperature increase of 100 K at constant pressure, how much energy has been transferred to the helium as heat
Answer:
6235.5J
Explanation:
Using ( nစ)p= ncp x change in temp
But cp= ( 1+ f/2)R
So cp= ( 1+ 3/2R
Cp= 5R/2
So = n x 5R/2x 150k
= 2 x 5/2x 8.314 x150
= 6235.5J
Answer:
2500 J
Explanation:
Q=(3/2)nRΔT
Q=(3/2)*2 mol*(8.314 J/mol*k)*100 k
Q=2494 J
Find the sum of 46 and -46
Answer:
0
Explanation:
Air in a 124 km/h wind strikes head-on the face of a building 42 m wide by 73 m high and is brought to rest. If air has a mass of 1.3 kg per cubic centimeter, determine the average force of wind on the building.
Answer:
The average force of wind on the building is 4.728 x 10¹² N
Explanation:
Given;
speed of the air wind, v = 124 km/h
dimension of the building, 42 m wide by 73 m high
density of the air, ρ = 1.3 kg/cm³ =
speed of the air in m/s = 124/3.6 = 34.44 m/s
Area of the building, A = 42 m x 73 m = 3066 m²
density of the air in (k.g/m³);
[tex]\rho = \frac{1.3 \ kg}{cm^3} *(\frac{100\ cm}{1 \ m} )^3\\\\\rho = \frac{1.3 \ kg}{cm^3} *\frac{10^6\ cm^3}{1 \ m^3} = \frac{1.3*10^6 \ kg}{m^3}[/tex]
The average force of wind on the building;
F = mass flow rate x velocity
F = (ρvA) x V
F = ρAv²
F = 1.3 x 10⁶ x 3066 x (34.44)²
F = 4.728 x 10¹² N
Therefore, the average force of wind on the building is 4.728 x 10¹² N
Trial 1: Get a textbook and put a sheet of paper on top of it. Fold the paper as needed to keep the paper from sticking over the edge of the book.Hold the textbook with the paper on top, horizontally about waist high.Drop the book and paper so that they hit the floor flat. Record your observations.Trial 2: With the book in one hand and the paper in the other, drop the book and paper simultaneously from the same height. Record your observations.
Answer:
1) the two objects reach the floor at the same time.
2)the book reaches the floor much earlier than the foil
In conclusion, the difference in motion between the two systems subjected to the same acceleration depends on the weight of the body and friction force, when the body has less weight, the friction of the air affects it more
Explanation:
This interesting experiment has the following results
1) first case. Sheet on top of book
In this case the two objects reach the floor at the same time.
This shows that the acceleration in the two objects is the same and we call it the acceleration of gravity.
The speed of the body increases as it goes down linearly.
This occurs because the book that receives air resistance is much heavier, so the resistance has almost no effect on its movement, the sheet does not have the air resistance because it goes down next to the book.
2) second case. Book and sheet next to each other.
In this case the book reaches the floor much earlier than the foil.
This is because the resisting force of the air has almost no effect on the book and its movement is little affected by this force.
In the case of the blade, it has very little weight, therefore as its speed increases, the resistance force of the air rapidly equals the weight of the blade.
W_sheet - fr = 0
so after this, since the acceleration is zero, it goes down at constant speed, this speed is called the terminal velocity.
In conclusion, the difference in motion between the two systems subjected to the same acceleration depends on the weight of the body and friction force, when the body has less weight, the friction of the air affects it more.
A 22-g bullet traveling 240 m/s penetrates a 2.0-kg block of wood and emerges going 150 m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?
Answer:
After the bullet emerges the block moves at 0.99 m/s
Explanation:
Given;
mass of bullet, m₁ = 22 g = 0.022 kg
initial speed of the bullet, u₁ = 240 m/s
final speed of the bullet, v₁ = 150 m/s
mass of block, m₂ = 2.0 kg
initial speed of the block, u₂ = 0
Let the final speed of the block = v₂
Apply principles of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.022 x 240 + 2 x 0 = 0.022 x 150 + 2v₂
5.28 = 3.3 + 2v₂
5.28 - 3.3 = 2v₂
1.98 = 2v₂
v₂ = 1.98 / 2
v₂ = 0.99 m/s
Therefore, after the bullet emerges the block moves at 0.99 m/s
The body moves at a speed of 2.61m/s after the bullet emerges.
According to the law of collision which states that the momentum of the body before the collision is equal to the momentum of the body after the collision.
The formula for calculating the collision of a body is expressed as:
p = mv
m is the mass of the body
v is the velocity of the body
Based on the law above;
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
v is the final velocity of the body after the collision
Substitute the given parameters into the formula as shown:
[tex]0.022(240) + 2(0) = (0.022+2)v\\ 5.28 = 2.022v\\v=\frac{5.28}{2.022}\\v= 2.611m/s[/tex]
This shows that the body moves at a speed of 2.61m/s after the bullet emerges.
Learn more here: https://brainly.com/question/9537044
A hydraulic press has one piston of diameter 4.0 cm and the other piston of diameter 8.0 cm. What force must be applied to the smaller piston to obtain a force of 1600 N at the larger piston
Answer:
400 N
Explanation:
Pressure is equal on both pistons.
P = P
F / A = F / A
F / (πd²/4) = F / (πd²/4)
F / d² = F / d²
1600 N / (8.0 cm)² = F / (4.0 cm)²
F = 400 N
The force that should be applied to the smaller piston is 400 N.
Given that,
A hydraulic press has one piston of diameter 4.0 cm and the other piston of diameter 8.0 cm. Pressure is equal on both pistons.Based on the above information, the calculation is as follows:
[tex]1600 N \div (8.0 cm)^2 = F \div (4.0 cm)^2[/tex]
F = 400 N
Learn more: brainly.com/question/17429689
Narrow, bright fringes are observed on a screen behind a diffraction grating. The entire experiment is then immersed in water. Do the fringes on the screen get closer together, get farther apart, remain the same, or disappear? Explain.
Answer:
n (a sin θ) = m λ₀
n> 1, therefore the fringes move away from each other
Explanation:
The diffraction experiment the constructive interference fringes is described by
a sin θ = m λ₀
in this equation it is assumed that the experiment emptied the air n = 1
When the same experiment is performed in water, the wavelength changes
λₙ = λ₀ / n
execution for constructive interference
a sin θ = m λₙ
we substitute
a sin θ = m λ / n
n (a sin θ) = m λ₀
the refractive index of water is n = 1.33, so for the same wavelength the separation of the spectrum is multiplied by n> 1, therefore the fringes move away from each other
In the previous part, you determined the maximum angle that still allows the crate to remain at rest. If the coefficient of friction is less than 0.7, what happens to this angle? A. The maximum angle increases.B. The maximum angle decreases.C. The maximum angle remains the same.D. Simulation q not sure if needed.
Answer:
B. The maximum angle decreases
Explanation:
If θ be the maximum angle of a slope that allows a crate placed on it to remain at rest , following condition exists .
tanθ = μ , θ is called angle of repose . μ is coefficient of static friction .
So the tan of angle of repose θ is proportional to coefficient of static friction.
If coefficient of static friction is less than .7 , naturally angle of repose will also become less ,ie, it at lower angle of inclination , the object will start slipping .
#1 A boy pushes forward a cart of groceries with a total mass of 40 kg. What is
the acceleration of the cart if the net force on the cart is 60 N?
Explanation:
∑F = ma
60 N = (40 kg) a
a = 1.5 m/s²
What is the mass of a rock lifted 2 meters off the ground that has 196 J of potential energy?
Answer:
10kg
Explanation:
Let PE=potential energy
PE=196J
g(gravitational force)=9.8m/s^2
h(change in height)=2m
m=?
PE=m*g*(change in h)
196=m*9.8*2
m=10kg
sound source of frequency f moves with constant velocity (less than the speed of sound) through a medium that is at rest. A stationary observer hears a sound whose frequency is appreciably different from f because
Answer:
A static observer hears a sound whose frequency is appreciably different from actual frequency because here the change in frequency of the sound due to doppler effect.
Explanation:
Given that,
Frequency = f
We know that,
The sound source of frequency f moves with constant velocity through a medium that is at rest.
A static observer hears a sound whose frequency is significantly different from actual frequency due to doppler effect.
We know that,
The doppler effect is defined as
[tex]f=f_{0}(\dfrac{v+v_{0}}{v-v_{s}})[/tex]
Where, f₀ = actual frequency
f = observe frequency
v = speed of sound
[tex]v_{o}[/tex] = speed of observer
[tex]v_{s}[/tex] = speed of source
Hence, A static observer hears a sound whose frequency is significantly different from actual frequency because here the change in frequency of the sound due to doppler effect.
What is the direction of the magnetic force on the current in each of the six cases?
Answer:
is the equation for magnetic force on a length l of wire carrying a current I in a uniform magnetic field B, as shown in Figure 2. If we divide both sides of this expression by l, we find that the magnetic force per unit length of wire in a uniform field is F l=IBsinθ.
Explanation:
Turning the barrel of a 50-mm-focal-length lens on a manual-focus camera moves the lens closer to or farther from the sensor to focus on objects at different distances. The lens has a stated range of focus from 0.70 m infinity.
How far does the lens move between these two extremes?
Answer:
Explanation:
To focus object at .7m , the image distance can be measured as follows
object distance u = .7m
focal length f = .05 m
image distance v = ?
from lens formula
[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f}[/tex]
[tex]\frac{1}{v} +\frac{1}{.7} = \frac{1}{.05}[/tex]
[tex]\frac{1}{v} =\frac{1}{.05} -\frac{1}{.7}[/tex]
v = .054 m
= 54 mm
when the object is at infinity , image is formed at focus ie at distance of
50 mm .
So lens position from sensor where image is formed , varies from 54 mm to 50 mm .
Water enters a student's house 10.0 m above the ground through a pipe with a cross section area of 1.00 x 10-4m2 at ground. Inside the house the pipe's cross section area is 0.50 x 10-4m2. The student in the house want to know the water pressure inside the pipe at the ground level. He first measured the volume of the bath tank that equals to 45.0 L. Then he fill the tank (the tank is 10 meters above the ground) inside the house with 90.0 seconds. The pipe inside the house is open with the sea level pressure The density of water is 1000 kgm3.
(a) Calculate the water speed at the ground pipe with larger cross section area and the water speed inside the house with smaller cross section area.
(b) Calculate the water pressure in the pipeline at the ground level.
Answer:
(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)
(b). P₂ = 236500 Pa
Explanation:
This is quite straight-forward so let us begin by defining the terms given.
Given that;
The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.
Let us make the velocity of water inside the house be V₁
such that the Volume of water entering the per second is = A₁V₁
Therefore, in 90sec:
45 L = 90 A₁V₁
V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴
V₁ = 10m/s (velocity of water inside the house)
From the continuity equation we have that;
A₁V₁ = A₂V₂
0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂
V₂ = 5m/s (velocity at ground level)
(b). We are told to calculate the water pressure in the pipeline at the ground level.
Using Bernoulli's equation;
P₁ + pgh₁ + 1/2PV₁² (inside) = P₂ + pgh₂ + 1/2PV₂² (ground level)
1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²
P₂ (pressure) = 1.01*10⁵Pa
Therefore we have;
101000 + 98000 + 50000 = P₂ + 12500
P₂ = 236500 Pa
cheers I hope this helped !!
J. Henry Alston was the first African American to publish his research findings on the perception of heat and cold in a major US psychology journal. Please select the best answer from the choices provided T F
Answer:
True
Explanation:
J. Henry Alston was known as a famous African American psychologist. He was known through his thorough study of the sensations of heat and cold.
He thereby became the first African American to publish his research findings on the perception of heat and cold in a major US psychology journal and was an important figure in the field.