Answer:
P₂ = 4098.96 Pa
Explanation:
For this exercise let's use Bernoulli's equation
Let's use the subscript 1 for the point of the liquid surface and the subscript 2 for the ends (point A)
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
the velocity at the end of the tank
v₂ = w r₂
the velocity at the surface of the liquid is
v₁ - w r₁
where r₂ = 1.5 m and r₁ = 1 m
the tank pressure is P₁ = P₀ = 0.1 10³ Pa
P₂ = P₁ + ½ ρ [w² (r₁² - r₂²)] + ρ g (y₁ -y₂)
We must remember that the pressure measurements the distances are measured from the lowest part to the surface that has zero height
let's reduce the magnitudes to the SI system
w = 3 rev / min (2π rad / 1rev) (1 min / 60 s) = 0.314159 rad / s
let's calculate
P₂ = 0.1 10³ + ½ 800 0.314159² |(1² -1.5²)| + 800 9.8 |(1-1.5)|
P₂ = 0.1 103 +78.96 + 3920
P₂ = 4098.96 Pa
Explain two ways of magnetising an object
Answer:
There are two methods generally used to magnetize permanent magnets: static magnetization and pulse magnetization.
What is the momentum if the mass is 1.5 kg and the velocity is 4 m/s?
Answer:
6
Explanation:
momentum = mass * velocity
momentum = 1.5 * 4
momentum = 6
As you change the value of the battery voltage, how does this change the current through the circuit and the resistance of the resistor
Answer:
Current increases
Resistance remains same
Explanation:
As you change the value of the battery voltage
We know that the current is dependent on the supply of power , so the current increases but the resistance remains same.
A 50kg mass is sitting on a frictionless surface. An unknown constant force pushes the mass for 2 seconds until the mass reaches a velocity of 3m/s. What was the force acting on the mass?
Answer:
75N
Explanation:
a = v/t = 3/2
F = ma = 50(3/2) = 75
To understand the relationships between the parameters that characterize a wave. It is of fundamental importance in many areas of physics to be able to deal with waves. This problem will lead you to understand the relationship of variables related to wave propagation: frequency, wavelength, velocity of propagation, and related variables. Note that these are kinematic variables that relate to the wave's propagation and do not depend on its amplitude.
1. Traveling waves propagate with a fixed speed usually denoted as v (but sometimes c). The waves are called __________ if their waveform repeats every time interval T.
a. transverse
b. longitudinal
c. periodic
d. sinusoidal
2. If the velocity of the wave remains constant, then as the frequency of the wave is increased, the wavelength __________.
a. decreases
b. increases
c. stays constant
Answer:
1) c.
2) a.)
Explanation:
1)
At any wave, if its waveform repeats itself every time interval T, it is said that the wave is periodic, with a period T, which is the time needed to complete an entire cycle. The other options refer at the way in the waves propagates (longitudinal or transversal) and to the type of waveform (sinusoidal), so the right answer is c).2)
At any wave that propagates at a constant speed, there exists a fixed relationship between the velocity v, the frequency f and the wavelength λ, as follows:[tex]v = \lambda * f (1)[/tex]
So in order to v keep constant, if the frequency is increased, the wavelength will decrease in the same proportion, so a) is the right answer.help plz! what vibrates in following types of wave motion 1)light wave 2)sound waves 3)x-rays 4)water waves
Answer:
I believe it's 2) sound waves
Explanation:
With sound waves, the energy travels along in the same direction as the particles vibrate. This type of wave is known as a longitudinal wave, so named because the energy travels along the direction of vibration of the particles.
A storage tank has the shape of an inverted circular cone with height 12 m and base radius of 4 m. It is filled with water to a height of 10 m. Find the work required to empty the tank by pumping all of the water to the top of the tank. (The density of water is 1000 kg/m3. Assume g
Answer:
Work required to empty the tank by pumping all of the water to the top of the tank = 1674700 Kgm/s^2
Explanation:
Volume of Circular cone = V = (1/3)πr2h
where r is the radius in meters
and h is the height in meters
Substituting the given values in above equation, we get -
V = [tex]\frac{1}{3} * 3.14 * 4^2 * 10 = 167.47[/tex] cubic meters.
The force required will be equal to the mass of water in the cone
[tex]= 167.47 * 1000[/tex]
= 167470 Kg
Weight = Mass * g
= 167470 * 10
= 1674700 Kgm/s^2
Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.152 N when their center-to-center separation is 60.9 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0395 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other
Answer:
a) lets take the positive charge to be
q1 = 3.13 * 10^-6 C
b) lets take the negative charge to be
q2 = 3.13 * 10^-6 C
Explanation:
Given data:
electrostatic force between Two identical conducting spheres = 0.152 N
Center-to-center separation = 60.9 cm = 60.9 * 10^-2 m
When wire is removed
electrostatic force between the spheres ( repelling force ) = 0.0395 N
Determine the negative and positive charges on the identical spheres
q1, q2 = 6.26 * 10^-12 C ( positive net charge )
lets take the positive charge to be
q1 = 3.13 * 10^-6 C
lets take the negative charge to be
q2 = 3.13 * 10^-6 C
since q1,q2 = 6.26 * 10^-12 C
attached below is detailed solution
g A large tank, at 400 kPa and 450 K, supplies air to a converging-diverging nozzle of throat area 4 cm2 and exit area 5 cm2. For what range of back pressures will the flow (a) be entirely subsonic; (b) have a shock wave inside the nozzle; (c) have oblique shocks outside the exit (i.e., over-expanded exit flow); and (d ) have supersonic expansion waves outside the exit (under-expanded exit flow)
Answer:
A) ≥ 325Kpa
B) ( 265 < Pe < 325 ) Kpa
C) (94 < Pe < 265 )Kpa
D) Pe < 94 Kpa
Explanation:
Given data :
A large Tank : Pressures are at 400kPa and 450 K
Throat area = 4cm^2 , exit area = 5cm^2
a) Determine the range of back pressures that the flow will be entirely subsonic
The range of flow of back pressures that will make the flow entirely subsonic
will be ≥ 325Kpa
attached below is the detailed solution
B) Have a shock wave
The range of back pressures for there to be shock wave inside the nozzle
= ( 265 < Pe < 325 ) Kpa
attached below is a detailed solution
C) Have oblique shocks outside the exit
= (94 < Pe < 265 )Kpa
D) Have supersonic expansion waves outside the exit
= Pe < 94 Kpa
What principle does a heat engine take advantage of in order to use heat to perform work?
A. Magnets can convert motion into electrical energy
B. The speed of atoms increases when heat is added
C. Fluids expand when heat is added to them
D. Energy can be created from nothing
How do spectroscopes help with studying of distance stars?
It measures the distance of faraway stars.
It collects sound waves from distant stars.
It breaks apart lights that comes from stars faraway.
It gives a clear image of the shape of stars.
Answer:
You take the light from a star, planet or galaxy and pass it through a spectroscope, which is a bit like a prism letting you split the light into its component colours. "It lets you see the chemicals being absorbed or emitted by the light source. From this you can work out all sorts of things," says Watson
A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 21.3 m above the river, whereas the opposite side is a mere 2.3 m above the river. The river itself is a raging torrent 54.0 m wide. 1. How fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side? 2. What is the speed of the car just before it lands safely on the other side?
Answer:
Explanation:
1 ) Let the initial horizontal velocity of car be v .
For vertical displacement
vertical displacement h = 21.3 - 2.3 = 19 m
Time taken to fall by 19 m be t
19 = 1/2 x 9.8 t² ( initial downward velocity is zero )
t = 1.97 s
This is also the time taken to cover horizontal displacement of 54 m which is width of river .
horizontal speed v = 54 / 1.97 m /s
v = 27.41 m /s
2 )
At the time of landing on other side , car will have both vertical and horizontal speed .
vertical speed
v = u + gt
= 0 + 9.8 x 1.97 = 19.31 m /s
horizontal speed will remain same as the initial speed = 27.41 m /s
Resultant speed = √ ( 27.41² + 19.31² )
= √ ( 751.3 + 372.87)
= 33.52 m /s
In separate experiments, four different particles each start from far away with the same speed and impinge directly on a gold nucleus. The masses and charges of the particles are particle 1: mass m0, charge q0 particle 2: mass 2m0, charge 2q0 particle 3: mass 2m0, charge q0/2 particle 4: mass m0/2, charge 2q0 Rank the particles according to the distance of closest approach to the gold nucleus, from smallest to largest.
A. 1, 2, 3, 4
B. 4, 3, 2,1
C. 3, 1 and 2 tie, then 4
D. 4, 1 and 2 tie, then 1
Answer:
Potential energy at the beginning: Ui =Qq/R*k ≈ 0 (particles are far apart)
Kinetic energy at the beginning: KEi =[tex]mv^{2}[/tex]/2
Potential energy at the end: Uf =(Qq/r)*k
Kinetic energy at the beginning: KEf = 0
From energy conservation:
r = 2*Qk/[tex]v^{2}[/tex]*q/m
Ranking of the closest approach to the gold nucleous comes from q/m:
- particle 3 has the smallest q/m = 1/4 ⋅q0/m0 and, hence, comes the closest
- particle 4 has the largest q/m = 4⋅q0/m0 and, hence, comes the last in ranking
- particles 1 and 2 have equal q/m=q0/m0 and, hence, come tie in between particles 3 and 4
Explanation:
Potential energy at the beginning: Ui =Qq/R*k ≈ 0 (particles are far apart)
Kinetic energy at the beginning: KEi =[tex]mv^{2}[/tex]/2
Potential energy at the end: Uf =(Qq/r)*k
Kinetic energy at the beginning: KEf = 0
From energy conservation:
r = 2*Qk/[tex]v^{2}[/tex]*q/m
Ranking of the closest approach to the gold nucleous comes from q/m:
- particle 3 has the smallest q/m = 1/4 ⋅q0/m0 and, hence, comes the closest
- particle 4 has the largest q/m = 4⋅q0/m0 and, hence, comes the last in ranking
- particles 1 and 2 have equal q/m=q0/m0 and, hence, come tie in between particles 3 and 4
A painter sits on a scaffold that is connected to a rope passing over a pulley. The other end of the rope rests in the hands of the painter who wants to lift the scaffold. She plans to pull downward on the loose end of the rope, thinking that the scaffold will then rise vertically with her along for the ride. The scaffold has a mass of 52 kg, and her mass is 63 kg. The painter pulls downward on the rope with a force of 600.0 N, while she and the scaffold are hanging from the other end above the ground.
Required:
a. What is the net acceleration on the system consisting of the painter and the scaffold?
b. What is the magnitude of the normal force exerted on the painter by the scaffold?
Solution :
a). From Newtons second law,
F = ma
The total tension force is 2T.
∴ 2T - (m + M)g = (m+ M)a
Then
[tex]$a=\frac{2T-(m+M)g}{m+M}$[/tex]
[tex]$a=\frac{2\times 600-(52+63)9.8}{52+63}$[/tex]
[tex]$=0.63 \ m/s^2$[/tex]
b). From the person,
F = ma
T - Mg + N = Ma
or N = Ma + Mg - T
= (63 x 9.8) + (52 x 9.8) - 600
= 617.4 + 509.6 - 600
= 527 N
Jason leaves his dorm and drives 3 kilometers north to a fruit shop.
Then he turns and drives 4 kilometers south to a restaurant. Then he
drives 5 kilometers east to an art classroom. The total distance that
Jason traveled is kilometers.
The total distance that Jason traveled is 12 kilometers.(3+4+5)
hii guys!! ~ i've comeback again!! could you please help me with an assignment?? i have a couple of i know questions!! thank you for helping me, hehe.
tenise can run the 100 meter dash in 9.77 seconds. what is her speed??
Answer:
Speed = 10.24 m/s.
Explanation:
Given the following data;
Distance = 100m
Time = 9.77
To find her speed;
Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.
Mathematically, speed is given by the equation;
[tex]Speed = \frac{distance}{time}[/tex]
Substituting into the equation, we have;
[tex]Speed = \frac{100}{9.77}[/tex]
Speed = 10.24 meter per seconds.
Which statements accurately describe sound waves? Check all that apply.
Sound waves are transverse waves.
Sound waves require a medium to transfer energy.
Sound waves start with a vibration.
Energy from a vibration creates a mechanical wave.
Sound can be heard as soon as an object begins vibrating.
Answer:
the answers are
b) sound waves require a medium to transfer energy
c) sound waves start with a vibration
d) energy from a vibration creates a mechanical wave
Explanation:
i just did this and got 100%, edge 2021
Following statements are true for sound wave,
Sound waves require a medium to transfer energy.Sound waves start with a vibration.Energy from a vibration creates a mechanical wave.In sound waves,
Energy is transferred through vibration of air particles or particles of a solid through which the sound travels.As sound passes through air (or any fluid medium), the particles of air do not vibrate in a transverse mannerA transverse wave is a wave whose oscillations are perpendicular to the direction of the wave's advance.Learn more:
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Find the least frequency of incident light that will knock electrons out of the surface of a metal with a work function of 3eV.
Answer:
The frequency must be: [tex]725\,\,10^{12} \,Hz[/tex]
Explanation:
If the work function of the metal ([tex]\phi[/tex]) is 3 eV, then we can use the formula for the kinetic energy of an ejected electron:
[tex]KE= h*f-\phi[/tex]
considering for the minimum KE = 0, and using the Plank constant h in eV s as: 4.14 * 10 ^(-15) eV s, to solve for the frequency:
[tex]h*f=\phi\\4.14*10^{-15} * f = 3\\f=3*10^{15} /4.14\,\,\frac{1}{s} \\f=725\,10^{12} \,Hz[/tex]
The sides of a right triangle that has any given vector for the hypotenuse are called _____
A. Scalar
B. Component
C. Resultant
D. Vector
Answer:
They are the resultant vector.
I have a problem in this questions?
Answer:
8.46E+1
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = 39 C
Charge 2 (q₂) = –53 C
Force (F) of attraction = 26×10⁸ N
Electrical constant K) = 9×10⁹ Nm²/C²
Distance apart (r) =?
The distance between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
26×10⁸ = 9×10⁹ × 39 × 53 / r²
26×10⁸ = 1.8603×10¹³ / r²
Cross multiply
26×10⁸ × r² = 1.8603×10¹³
Divide both side by 26×10⁸
r² = 1.8603×10¹³ / 26×10⁸
r² = 7155
Take the square root of both side
r = √7155
r = 84.6 m
r = 8.46E+1 m
An accelerometer has a damping ratio of 0.5 and a natural frequency of 18,000 Hz. It is used to sense the relative displacement of a beam to which it is attached. (a)If an impact to the beam imparts a vibration at 4500 Hz, calculate the dynamic error and phase shift in the accelerometer output. (b)Calculateits resonance frequency.(c)What isthe maximumpossiblemagnitude ratio that the system can achieve
Answer:
A) i) Dynamic error ≈ 3.1%
ii) phase shift ≈ -12°
B) 79971.89 rad/s
Explanation:
Given data :
Damping ratio = 0.5
natural frequency = 18,000 Hz
a) Calculate the dynamic error and phase shift in accelerometer output at an impart vibration of 4500 Hz
i) Dynamic error
This can be calculated using magnitude ratio formula attached below is the solution
dynamic error ≈ 3.1%
ii) phase shift
This phase shift can be calculated using frequency dependent phase shift formula
phase shift ≈ -12°
B) Determine resonance frequency
Wr = 2[tex]\pi[/tex] ( 18000 [tex]\sqrt{0.5}[/tex] ) = 79971.89 rad/s
C) The maximum magnitude ratio that the system can achieve
How much energy is required to move 2 electrons through a potential difference of 1.0 x 10^ 2 volts?
The energy required to move 2 electrons through a potential difference of 1.0 x 10² volts is -3.2 x 10⁻¹⁷ joules.
What is the potential difference?The potential difference, also known as voltage, is the difference in electrical potential energy per unit of charge between two points in an electrical circuit or an electric field.
In simpler terms, the potential difference is the amount of energy required to move a unit of electric charge from one point to another in an electric field or an electrical circuit. It is measured in volts (V) and can be calculated using the equation:
V = W/Q
Where V = is the potential difference,
W = is the work done in moving the charge Q from one point to another,
Q =is the amount of charge that is moved.
Potential difference is an essential concept in electrical engineering and physics, as it governs the flow of electric current in a circuit and determines the behavior of electrical devices such as resistors, capacitors, and batteries.
Here in this question,
The energy required to move an electron through a potential difference is given by the formula:
E = qV
Where
E = is the energy required,
q = is the charge of the electron,
V = is the potential difference.
The charge of one electron is -1.6 x 10⁻¹⁹coulombs.
Therefore, for two electrons, the total charge is:
q = 2 x (-1.6 x 10⁻¹⁹coulombs) = -3.2 x 10⁻¹⁹ coulombs
The potential difference is given as 1.0 x 10² volts.
Now, the energy required to move 2 electrons through a potential difference of 1.0 x 10² volts is:
E = (-3.2 x 10⁻¹⁹ coulombs) x (1.0 x 10² volts) = -3.2 x 10⁻¹⁷ joules.
Note: that the negative sign indicates that the electrons lose potential energy as they move through the potential difference.
Therefore, The required energy is -3.2 x 10⁻¹⁷ joules.
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A ball accelerates at 2 m/s what is the velocity after 5sec?
Answer:
10m
Explanation:
assuming the ball was at 0 m/s before, after 5 seconds the ball would be traveling at 10m
a jogger runs around a quarter of a full track in 10 seconds what is the period of teh motion around teh track
40 seconds
Explanation:
a quarter is 1/4 so to go the full thing requires you do it 4 time, and sense it takes 10 seconds to do a 1/4 you'd have to multiply 10 by 4.
During the nineteenth century, biologists knew that South America and Africa contained similar fossils. These fossils let them to believe that a temporary land bridge between South America and Africa allowed ancient species to move between both continents. In the twentieth century, evidence showed that South America and Africa were once part of the same larger continent and that there were no land bridges between them. How should biologists respond to the new geological finding?
A. Biologists should reject the geological finding, because it is new.
B. Biologists should ask physicists what to do, because physics is a more exact science.
C. Biologists should reject the belief in ancient species, because there were no land bridges.
D. Biologists should update the history of life, because it should include the new geological finding.
Explain
Cameron is standing on the edge of a 60 m high cliff. He horizontally throws a football
with an initial velocity of 10 m/s. How long does it take for the football to hit the
ground?
Answer:34.6 m/s
Explanation: It is asking how long meaning the answer is in time
Suppose the particle is placed at the 40V equipotential surface along the line connecting the two central positive and negative charges.
(Hint: The electric field can be obtained using the 40V and 30V potential difference lines.)
1. What is the force on the charged particle if q=80μC? (mN) (m-milli 10-3)
2. What is the force on the charged particle if it is now located at the 0V potential difference line? (mN)
Answer:
Hello some part of your question is missing below is the missing part
2. What is the force on the charged particle if it is now located at the 0V potential difference line? (mN) (hint: The electric field can be obtained as above using the 0V and -10V equipotential lines.)
answer :
1) 0.8 mN
2) 0.8 mN
Explanation:
Given data:
1) Calculate the force on the charged particle
q = 80 μC , Va = 30v , Vb = 40v, ∝ = 1 m
E = ( Δv ) / ∝
= ( Vb - Va ) / ∝
F = qE
= 80 μC * ( 40 - 30 ) / 1 m
= 800 μC
F = 0.8 mN
2) Calculate the force on the charged particle when it is located at 0V
Va = -10V , Vb = 0V, q = 80 μC, ∝ = 1 m
F = qE
where E = ( 0 - ( -10 ) / 1
F = 80 μC * ( 0 - ( -10 ) / 1
= 800 μC = 0.8 mN
a. What is an electric field line?
b. What are equipotential sufaces are they necessarily physical surffaces?
c. Are they necessarily physical surfaces?
Answer:
Please see below as the answers are self-explanatory
Explanation:
a)
A electric field line is an imaginary line, which has the property that the electric field vector is tangent to it at any point. It starts from positive charges (since the electric field by convention it has the direction of the trajectory that would take a positive test charge, so it always goes away from positive charges) and ends in negative charges.b)
Since the potential difference between two points represents the work per unit charge needed for a charge to move between these points, a equipotential surface is the one over which it is not needed to do work to move a charge from any point on the surface to any other point, which means that all points are at the same potential.c)
Equipotential surfaces are not necessarily physical surfaces, they can be defined in vaccum for instance.As an example, any spherical surface concentric with a point charge, is an equipotential surface, and it can be a real surface or a fictitious one.A substance of mass 200g measures 50cm*4cm*8cm. Calculate it density giving your answer in SI unit
Answer:
0.2kg/0.0016m^3
Explanation:
200g is 0.2 kg
50*4*8 is equivalent to 1600 cubic centimeters which is also equivalent to 0.0016 cubic meters.
The SI unit for density is kg per cubic meter
So now we have 0.2kg/0.0016m^3
I hoped this helped, and if it did please give me brainliest
A 0.25 kg beach ball rolling at a speed of 7 m/s collides with a heavy exercise ball at rest. The beach ball bounces straight back with a speed of 4 m/s. That is the change in momentum of the beach ball? What is the impulse exerted on the beach ball? What is the impulse exerted on the exercise ball?
The impulse exerted on the beach ball is 2.75 kgm/s.
The impulse exerted on the exercise ball is - 2.75 kgm/s.
What is impulse?
This is the force applied to an object that acts over a period of time.
The impulse exerted on the beach ball is the change in the momentum of the ball and it is calculated as follows;
J = ΔP
J = m(v - u)
J = 0.25(7 - (-4))
J = 0.25(7 + 4)
J = 2.75 kgm/s
The impulse exerted on the exercise ball is equal in magnitude but opposite in direction to the beach ball.
Thus, the impulse exerted on the exercise ball is - 2.75 kgm/s.
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