Answer:
The vision becomes blurred due to the refractive defects of the eye. There are mainly three common refractive defects of vision. These are (i) myopia or near-sightedness, (ii) Hypermetropia or far – sightedness, and (iii) Presbyopia. These defects can be corrected by the use of suitable spherical lenses.
A horizontal board of negligible thickness and area 1.0 m2 hangs from a spring scale that reads 50 N when a 4.0 m/s wind moves above the board. The air below the board is stationary. What does the scale read when the wind stops? The density of air is 1.25 kg/m3 .
what is liquid pressure? and tell me its si unit please
The SI unit of pressure is the pascal: 1Pa=1N/m2 1 Pa = 1 N/m 2 . Pressure due to the weight of a liquid of constant density is given by p=ρgh p = ρ g h , where p is the pressure, h is the depth of the liquid, ρ is the density of the liquid, and g is the acceleration due to gravity.
At a department store, you adjust the mirrors in the dressing room so that they are parallel and 6.2 ft apart. You stand 1.8 ft from one mirror and face it. You see an infinite number of reflections of your front and back.(a) How far from you is the first "front" image? ft (b) How far from you is the first "back" image? ft
Answer:
a) 3.6 ft
b) 12.4 ft
Explanation:
Distance between mirrors = 6.2 ft
difference from from the mirror you face = 1.8 ft
a) you stand 1.8 ft in front of the mirror you face.
According to plane mirror rules, the image formed is the same distance inside the mirror surface as the distance of the object (you) from the mirror surface. From this,
your distance from your first "front" image = 1.8 ft + 1.8 ft = 3.6 ft
b) The mirror behind you is 6.2 - 1.8 = 4.4 ft behind you.
the back mirror will be reflected 3.6 + 4.4 = 8 ft into the front mirror,
the first image of your back will be 4.4 ft into the back mirror,
therefore your distance from your first "back" image = 8 + 4.4 = 12.4 ft
within which type of system is the total mass conserved but not the total energy
In a closed system the mass is conserved, but the energy is not conserved.
To find the answer, we have to study about different systems in thermodynamics.
What is thermodynamic system?A system, which can be expressed in terms of thermodynamic coordinates is called Thermodynamic system.Open system: System can exchange both energy and matter, thus, both energy and matter is not conserved here.Closed system can exchange energy with its surroundings (as heat or work), but not matter.Isolated system: A system that is open to the environment can interchange energy and matter, but a system that is insulated from it cannot.Thus, we can conclude that, in closed system the mass is conserved, but the energy is not conserved.
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Light of wavelength 520 nm is used to illuminate normally two glass plates 21.1 cm in length that touch at one end and are separated at the other by a wire of radius 0.028 mm. How many bright fringes appear along the total length of the plates.
Answer:
The number is [tex]Z = 216 \ fringes[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 520 \ nm = 520 *10^{-9} \ m[/tex]
The length of the glass plates is [tex]y = 21.1cm = 0.211 \ m[/tex]
The distance between the plates (radius of wire ) = [tex]d = 0.028 mm = 2.8 *10^{-5} \ m[/tex]
Generally the condition for constructive interference in a film is mathematically represented as
[tex]2 * t = [m + \frac{1}{2} ]\lambda[/tex]
Where t is the thickness of the separation between the glass i.e
t = 0 at the edge where the glasses are touching each other and
t = 2d at the edge where the glasses are separated by the wire
m is the order of the fringe it starts from 0, 1 , 2 ...
So
[tex]2 * 2 * d = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]
=> [tex]2 * 2 * (2.8 *10^{-5}) = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]
=>
[tex]m = 215[/tex]
given that we start counting m from zero
it means that the number of bright fringes that would appear is
[tex]Z = m + 1[/tex]
=> [tex]Z = 215 +1[/tex]
=> [tex]Z = 216 \ fringes[/tex]
Monochromatic light of wavelength, λ is traveling in air. The light then strikes a thin film having an index of refraction n1 that is coating a material having an index of refraction n2. If n1 is larger than n2, what minimum film thickness will result in minimum reflection of this light?
Answer:
tmin= lambda/2
Explanation:
See attached file pls
Which of these cannot be a resistor in a series or parallel circuit?
A)switch
B) battery
C) light bulb
D) all of these are resistors
Answer:
it is going to D. all of these are resistors
A long solenoid consists of 1700 turns and has a length of 0.75 m.The current in the wire is 0.48 A. What is the magnitude of the magnetic field inside the solenoid
Answer:
1.37 ×10^-3 T
Explanation:
From;
B= μnI
μ = 4π x 10-7 N/A2
n= number of turns /length of wire = 1700/0.75 = 2266.67
I= 0.48 A
Hence;
B= 4π x 10^-7 × 2266.67 ×0.48
B= 1.37 ×10^-3 T
Which unbalanced force accounts for the direction of the net force of the rocket?
a. Air resistance
b. Friction
c. Gravity
d. Thrust of rocket engine
It depends on what stage of the mission you're talking about.
==> While it's sitting on the pad before launch, the forces on the rocket are balanced, so there's no net force on it.
==> When the engines ignite, their thrust (d) is greater than the force of gravity. So the net force on the rocket is upward, and the spacecraft accelerates upward.
==> After the engines shut down, the net force acting on the rocket is due to Gravity (c).
. . . If the rocket has enough vertical speed, it escapes the Earth completely, and just keeps going.
. . . If it has enough horizontal speed, it enters Earth orbit.
. . . If it doesn't have enough vertical or horizontal speed, it falls back to Earth.
A rocket will preserve to speed up so long as there's a resultant pressure upwards resulting from the thrust of the rocket engine.
What unbalanced force bills for the course of the internet pressure of the rocket?A rocket launches whilst the pressure of thrust pushing it upwards is greater than the burden force because of gravity downwards. This unbalanced pressure reasons a rocket to accelerate upwards. A rocket will maintain to hurry up so long as there's a resultant force upwards resulting from the thrust of the rocket engine.
What's the net pressure of unbalanced?
If the forces on an item are balanced, the net pressure is zero. If the forces are unbalanced forces, the results do not cancel each difference. Any time the forces acting on an object are unbalanced, the net pressure is not 0, and the movement of the item modifications.
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550 J of heat is added to the gas in an isothermal process. As the gas expands, pushing against the piston, how much work does it do
Answer:
The work done by the system is 550 J
Explanation:
Given;
heat added to the system, Q = 550 J
Apply the first law of thermodynamics;
ΔU = Q - W
Where;
ΔU is change in internal energy
Q is the heat added to the system
W is the work done by the system
During an isothermal process, the temperature of the system is constant for the entire process. During this process, the change in the internal energy is zero.
0 = Q - W
W = Q
W = 550 J
Therefore, the work done by the system is 550 J
Describe the change in motion and kinetic energy of the particles as thermal energy is removed from a liquid. Which change of state might happen?
Please I will give brainliest!!!!
Answer:
Change in state from liquid to solid.
Explanation:
Decrease in thermal energy will decrease energy and particles will slow down. Change in state from liquid to solid.
Calculate the kinetic energy of a mass 2kg moving with a velocity of 0.1m/s
ANSWER-:
1/2 mv²
K.E = 1/2 mv²
K.E = 0.01 J.
Hence, the kinetic energy of a body is 0.01 Joule.
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A fan is turned off, and its angular speed decreases from 10.0 rad/s to 6.3 rad/s in 5.0 s. What is the magnitude of the angular acceleration of the fan?
A) 0.37 rad/s2
B) 11.6 rad/s2
C) 0.74 rad/s2
D) 0.86 rad/s2
E) 1.16 rad/s2
Answer:
chk photo
Explanation:
In France, the wall sockets provide an AC voltage with Vrms = 230 V. You want to use an appliance designed to operate in the United States (Vrms = 120 V) and decide to build a transformer to convert the power line voltage in France to the value required by your appliance.
(a) Should you use a "step-down" transformer (to make Vrms smaller) or a "step-up" transformer (which makes Vrms larger)?
a "step-up" transformer
a "step-down" transformer
(b) If the input coil of your transformer has 2760 turns, how many turns should the output coil have?
_____ turns
Answer:
a)step-down" transformer
b) 1440 turns
Explanation:
There are two types of transformers; step up transformers and step down transformers. A step down transformer converts a higher voltage to a lower voltage.
In a stepdown transformer, there are more turns in the primary coil than in the secondary coil, the turns ratio Ns/Np is less than 1 for a stepdown transformer.
If
Number of turns in primary coil Np= 2760
Number of turns in secondary coil Ns= unknown
Voltage in primary coil Vp= 230 V
Voltage in secondary coil Vs= 120 V
Ns/Np= Vs/VP
NsVp= NpVs
Ns= NpVs/VP = 2760 × 120/230
Ns= 1440 turns
A circular conducting loop of radius 31.0 cm is located in a region of homogeneous magnetic field of magnitude 0.700 T pointing perpendicular to the plane of the loop. the loop is connected in series with a resistor of 265 ohms. The magnetic field is now increased at a constant rate by a factor of 2.30 in 29.0 s.
Calculate the magnitude of induced emf in the loop while the magnetic field is increasing.
With the magnetic field held constant a ts its new value of 1.61 T, calculate the magnitude of its induced voltage in the loop while it is pulled horizontally out of the magnetic field region during a time interval of 3.90s.
Answer:
(a) The magnitude of induced emf in the loop while the magnetic field is increasing is 9.5 mV
(b) The magnitude of the induced voltage at a constant magnetic field is 124.7 mV
Explanation:
Given;
radius of the circular loop, r = 31.0 cm = 0.31 m
initial magnetic field, B₁ = 0.7 T
final magnetic field, B₂ = 2.3B₁ = 2.3 X 0.7 T = 1.61 T
duration of change in the field, t = 29
(a) The magnitude of induced emf in the loop while the magnetic field is increasing.
[tex]E = A*\frac{\delta B}{\delta t} \\\\[/tex]
[tex]E = A*\frac{B_2 -B_1}{\delta t}[/tex]
Where;
A is the area of the circular loop
A = πr²
A = π(0.31)² = 0.302 m²
[tex]E = A*\frac{B_2 -B_1}{\delta t} \\\\E = 0.302*\frac{1.61-0.7}{29} \\\\E = 0.0095 \ V\\\\E = 9.5 \ mV[/tex]
(b) the magnitude of the induced voltage at a constant magnetic field
E = A x B/t
E = (0.302 x 1.61) / 3.9
E = 0.1247 V
E = 124.7 mV
Therefore, the magnitude of the induced voltage at a constant magnetic field is 124.7 mV
which is example of radiation
Answer:
Ultraviolet light from the sun.
Explanation:
This is an example of radiation.
Answer:
X-Ray
Explanation:
x-Ray is an example of radiation.
Which of the following explains why a “control” is important in a case-control study of a disease? The researchers need to control the bias that those who contracted the disease may create when they talk to others. The researchers need to compare those who contracted the disease to those who did not. The researchers need to compare those who contracted the disease to those who contracted previous diseases. The researchers need to control the disease so that it is not spread further.
The researchers need to compare those who contracted the disease to those who did not.
Suppose we want to calculate the moment of inertia of a 56.5 kg skater, relative to a vertical axis through their center of mass.
Required:
a. First calculate the moment of inertia (in kg-m^2) when the skater has their arms pulled inward by assuming they are cylinder of radius 0.11 m.
b. Now calculate the moment of inertia of the skater (in kg-m^2) with their arms extended by assuming that each arm is 5% of the mass of their body. Assume the body is a cylinder of the same size, and the arms are 0.875 m long rods extending straight out from their body being rotated at the ends.
Answer:
a. 0.342 kg-m² b. 2.0728 kg-m²
Explanation:
a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m
I = 1/2MR²
= 1/2 × 56.5 kg × (0.11 m)²
= 0.342 kgm²
So the moment of inertia of the skater is
b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)
I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m
I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²
= 0.1802 kg-m² + 0.6852 kg-m²
= 0.8654 kg-m²
The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².
So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²
a) The moment of inertia as the skater pulled his/her arm inward by assuming he/she is a cylinder is 0.3418kg-m².
b) If the body of the skater is assumed to be a cylinder of the same size, and the arms are rods extending straight out from his/her body being rotated at the ends, the moment of inertia is 1.7495kg-m².
Given the data in the question;
Mass of skater; [tex]M = 56.5kg[/tex]
a)
When the skater has his arms pulled inward by assuming they are cylinder of radius; [tex]R = 0.11 m[/tex]
Moment of inertia; [tex]I = \ ?[/tex]
From Parallel axis theorem; Moment of Inertia for a cylindrical body is expressed:
[tex]I = \frac{1}{2}MR^2[/tex]
Where M is the mass and R is the radius
We substitute our given values into the equation
[tex]I = \frac{1}{2}\ *\ 56.5kg\ *\ (0.11m)^2\\\\I = \frac{1}{2}\ *\ 56.5kg\ *\ 0.0121m^2\\\\I = 0.3418kg.m^2[/tex]
Therefore, the moment of inertia as the skater pulled his/her arm inward by assuming he/she is a cylinder is 0.3418kg-m²
b)
With the skater's arms extended by assuming that each arm is 5% of the mass of their body
Mass of each arm; [tex]M_a = \frac{5}{100} * M = \frac{5}{100} * 56.5kg = 2.825kg[/tex]
Remaining mass; [tex]M_b = M - 2M_a = 56.5kg - 2(2.825kg) = 50.85kg[/tex]
Assume the body is a cylinder of the same size and the arms are 0.875 m long rods extending straight out from their body being rotated at the ends.
Length of arm; [tex]L = 0.875 m[/tex]
From Parallel axis theorem; Moment of Inertia about vertical axis is expressed as:
[tex]I = \frac{1}{2}M_bR^2 + \frac{2}{3}M_aL^2[/tex]
We substitute in our values
[tex]I = \frac{1}{2}*50.85kg*(0.11m)^2 + \frac{2}{3}*2.825kg*(0.875m)^2\\\\I = [\frac{1}{2}*50.85kg * 0.0121m^2] + [\frac{2}{3}*2.825kg*0.765625m^2]\\\\I = 0.3076kg.m^2 + 1.4419kg.m^2\\\\I = 1.7495kg.m^2[/tex]
Therefore, if the body of the skater is assumed to be a cylinder of the same size, and the arms are rods extending straight out from his/her body being rotated at the ends, the moment of inertia is 1.7495kg-m².
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Find the average magnitude of the induced emf if the change in shape occurs in 0.125 ss and the local 0.504-TT magnetic field is perpendicular to the plane of the loop.
Complete Question
An emf is induced in a conducting loop of wire 1.12m long as its shape is.
changed from square to circular. Find the average magnitude of the induced emf if the change in shape occurs in 0.125 ss and the local 0.504-TT magnetic field is perpendicular to the plane of the loop.
Answer:
The induced emf is [tex]\epsilon = 0.0863 \ V[/tex]
Explanation:
From the question we are told that
The time taken is [tex]\Delta t = 0.125 \ s[/tex]
The magnitude of the magnetic field is B = 0.504 T
The length of the loop wire is [tex]l = 1.12 \ m[/tex]
Generally the circumference of the wire when in circular form is
[tex]C = 2 \pi r[/tex]
=> [tex]l = 2 \pi r[/tex]
=> [tex]r =[/tex][tex]\frac{l}{2 \pi}[/tex]
=> [tex]r =[/tex][tex]\frac{1.12}{2 * 3.142}[/tex]
=> [tex]r =[/tex][tex]0.1782 \ m[/tex]
Now the area of the wire as a circle is
[tex]A = \pi r^2[/tex]
=> [tex]A = 3.142 * (0.1782)^2[/tex]
=> [tex]A = 0.0998 \ m^2[/tex]
The length of one side of the square is
[tex]b = \frac{l}{4}[/tex]
[tex]b = \frac{1.12}{4}[/tex]
[tex]b = 0.28 \ m[/tex]
Now the area of the wire as a square is
[tex]A_s = b^2[/tex]
=> [tex]A_s =(0.28 )^2[/tex]
[tex]A_s = 0.0784 \ m^2[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = \frac{B * [A - A_s ]}{\Delta t }[/tex]
=> [tex]\epsilon = \frac{0.504 * [0.0998 - 0.0784 ]}{0.125 }[/tex]
=> [tex]\epsilon = 0.0863 \ V[/tex]
There was a major collision of an asteroid with the Moon in medieval times. It was described by monks at Canterbury Cathedral in England as a red glow on and around the Moon. How long after the asteroid hit the Moon, which is 3.84 x 10^5 km away, would the light first arrive on Earth?
Answer:
c = 3.00E108 m/s = 3.00E5 km/s
t = S / v = 3.84E5 / 3.00E5 = 1.28 sec
1.
(a)
P
center
Figure 1
A ball is released at point P with a tangential velocity of 5 ms to move in a circular track in a
vertical plane as shown in the Figure 1. Can the ball reach the highest point of the circular track
of radius 1.0 m? Give reasons. (4 marks]
Answer:
No.
Explanation:
Given the following :
Velocity (V) of ball = 5m/s
Radius = 1m
Can the ball reach the highest point of the circular track
of radius 1.0 m?
The highest point in the track could be considered as the diameter of the circle :
Radius = diameter / 2;
Diameter = (2 * Radius) = (2*1) = 2
Maximum height which the ball can reach :
Using the relation :
Kinetic Energy = Potential Energy
0.5mv^2 = mgh
0.5v^2 = gh
0.5(5^2) = 9.8h
0.5 * 25 = 9.8h
12.5 = 9.8h
h = 12.5 / 9.8
h = 1.2755
h = 1.26m
Therefore maximum height which can be reached is 1.26m.
Since h < Diameter
If a transformer has 50 turns in the primary winding and 10 turns on the secondary winding, what is the reflected resistance in the primary if the secondary load resistance is 250 W?
Answer:
The reflected resistance in the primary winding is 6250 Ω
Explanation:
Given;
number of turns in the primary winding, [tex]N_P[/tex] = 50 turns
number of turns in the secondary winding, [tex]N_S[/tex] = 10 turns
the secondary load resistance, [tex]R_S[/tex] = 250 Ω
Determine the turns ratio;
[tex]K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5[/tex]
Now, determine the reflected resistance in the primary winding;
[tex]\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms[/tex]
Therefore, the reflected resistance in the primary winding is 6250 Ω
A 17.0 g bullet traveling horizontally at 785 m/s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m/s.
What is the maximum temperature increase that the water could have as a result of this event? (in degrees)
Answer:
The maximum temperature increase is [tex]\Delta T = 0.0497 \ ^oC[/tex]
Explanation:
From the question we are told that
The mass of the bullet is [tex]m = 17.0 \ g =0.017 \ kg[/tex]
The speed is [tex]v_1 = 785 \ m/s[/tex]
The mass of the water is [tex]m_w = 13.5 \ kg[/tex]
The velocity it emerged with is [tex]v_2 = 534 \ m/s[/tex]
Generally due to the fact that energy can nether be created nor destroyed but transferred from one form to another then
the change in kinetic energy of the bullet = the heat gained by the water
So
The change in kinetic energy of the water is
[tex]\Delta KE = \frac{1}{2} m (v_1^2 - v_2 ^2 )[/tex]
substituting values
[tex]\Delta KE =0.5 * 0.017 * (( 785)^2 - (534) ^2 )[/tex]
[tex]\Delta KE = 2814.1 \ J[/tex]
Now the heat gained by the water is
[tex]Q = m_w* c_w * \Delta T[/tex]
Here [tex]c_w[/tex] is the specific heat of water which has a value [tex]c_w = 4190 J/kg \cdot K[/tex]
So since [tex]\Delta KE = Q[/tex]
we have that
[tex]2814.1 = 13.5 * 4190 * \Delta T[/tex]
[tex]\Delta T = 0.0497 \ ^oC[/tex]
The number of daylight hours, D, in the city of Worcester, Massachusetts, where x is the number of days after January 1 (), may be calculated by the function: What is the period of this function? N/A What is the amplitude of this function? 12 What is the horizontal shift? What is the phase shift? What is the vertical shift? How many hours of sunlight will there be on February 21st of any year?
Answer:
a. 365; b. 3; c. 78; d. 1.343 rad; e. 12; f. 10.66
Explanation:
Assume that the function is
[tex]D(x) = 3 \sin \left (\dfrac{2\pi}{365}(x - 78) \right ) + 12[/tex]
The general formula for a sinusoidal function is
y = A sin(B(x - C))+ D
|A| = amplitude
B = frequency
2π/B = period, P
C = horizontal shift (phase shift)
D = vertical shift
By comparing the two formulas, we find
|A| = 3
B = 2π/365
C = 78
D = 12
a. Period
P = 2π/B = 2π/(2π/365) = 2π × 365/2π = 365
The period is 365.
b. Amplitude
|A| = 3
The amplitude is 3.
c. Horizontal shift
C= 78
The horizontal shift is 78.
d. Phase shift (φ)
Ths phase shift is the horizontal shift expressed in radians.
φ = C × 2π/365 = 78 × 2π/365 ≈ 1.343
The phase shift is 1.343 rad.
e. Vertical shift
D = 12
The vertical shift is 12.
f. Hours of sunlight on Feb 21
Feb 21 is the 52nd day of the year, so x = 51 (the number of days after Jan 1),
[tex]\begin{array}{rcl}D(x) &=& 3 \sin \left (\dfrac{2\pi}{365}(x - 78) \right ) + 12\\\\&=& 3 \sin (0.01721(51 - 78) ) + 12\\&=& 3\sin(-0.4648) + 12\\&=& 3(-0.4482) + 12\\\&=& -1.345 + 12\\& = & \textbf{10.66 h}\\\end{array}[/tex]
There will be 10.66 h of sunlight on Feb 21 of any given year.
The figure below shows the graph of the function from 0 ≤ x ≤ 365.
A bug on the surface of a pond is observed to move up and down a total vertical distance of 6.5 cm , from the lowest to the highest point, as a wave passes. If the ripples decreaseto 4.7 cm, by what factor does thebug's maximum KE change?
Answer:
factor that bug maximum KE change is 0.52284
Explanation:
given data
vertical distance = 6.5 cm
ripples decrease to = 4.7 cm
solution
We apply here formula for the KE of particle that executes the simple harmonic motion that is express as
KE = (0.5) × m × A² × ω² .................1
and kinetic energy is directly proportional to square of the amplitude.
so
[tex]\frac{KE2}{KE1} = \frac{A2^2}{A1^2}[/tex] .............2
[tex]\frac{KE2}{KE1} = \frac{4.7^2}{6.5^2}[/tex]
[tex]\frac{KE2}{KE1}[/tex] = 0.52284
so factor that bug maximum KE change is 0.52284
The factor does the bug's maximum KE change should be considered as the 0.52284.
Calculation of the factor:Since
vertical distance = 6.5 cm
ripples decrease to = 4.7 cm
So here we apply the given formula
KE = (0.5) × m × A² × ω² .................1
here,
kinetic energy is directly proportional to square of the amplitude.
So,
= 4.7^2/ 6.5^2
= 0.52284
hence, The factor does the bug's maximum KE change should be considered as the 0.52284.
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A cylindrical container with a cross-sectional area of 66.2 cm2 holds a fluid of density 856 kg/m3 . At the bottom of the container the pressure is 119 kPa . Assume Pat = 101 kPa
A) What is the depth of the fuild?
B) Find the pressure at the bottom of the container after an additional 2.35×10−3 m3 of this fluid is added to the container. Assume that no fluid spills out of the container.
Answer:
A. h = 2.15 m
B. Pb' = 122 KPa
Explanation:
The computation is shown below:
a) Let us assume the depth be h
As we know that
[tex]Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h[/tex]
After solving this,
h = 2.15 m
Therefore the depth of the fluid is 2.15 m
b)
Given that
height of the extra fluid is
[tex]h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}[/tex]
h' = 0.355 m
Now let us assume the pressure at the bottom is Pb'
so, the equation would be
[tex]Pb' - Pat = d \times g \times (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000[/tex]
Pb' = 122 KPa
(A) The depth of the fluid is 2.14 m.
(B) The new pressure at the bottom of container is 121972 Pa.
Given data:
The cross-sectional area of the container is, [tex]A =66.2 \;\rm cm^{2}=66.2 \times 10^{-4} \;\rm m^{2}[/tex].
The density of fluid is, [tex]\rho = 856 \;\rm kg/m^{3}[/tex].
The container pressure at bottom is, [tex]P=119 \;\rm kPa=119 \times 10^{3} \;\rm Pa[/tex].
The atmospheric pressure is, [tex]P_{at}=101 \;\rm kPa=101 \times 10^{3}\;\rm Pa[/tex].
(A)
The given problem is based on the net pressure on the container, which is equal to the difference between the pressure at the bottom and the atmospheric pressure. Then the expression is,
[tex]P_{net} = P-P_{at}\\\\\rho \times g \times h= P-P_{at}[/tex]
Here, h is the depth of fluid.
Solving as,
[tex]856\times 9.8 \times h= (119-101) \times 10^{3}\\\\h=\dfrac{ (119-101) \times 10^{3}}{856\times 9.8}\\\\h= 2.14 \;\rm m[/tex]
Thus, the depth of the fluid is 2.14 m.
(B)
For an additional volume of [tex]2.35 \times 10^{-3} \;\rm m^{3}[/tex] to the liquid, the new depth is,
[tex]V=A \times h'\\\\h'=\dfrac{2.35 \times 10^{-3}}{66.2 \times 10^{-4}}\\\\h'=0.36 \;\rm m[/tex]
Now, calculate the new pressure at the bottom of the container as,
[tex]P'-P_{at}= \rho \times g \times (h+h')\\\\\P'-(101 \times 10^{3})= 856 \times 9.8 \times (2.14+0.36)\\\\P'=121972 \;\rm Pa[/tex]
Thus, we can conclude that the new pressure at the bottom of container is 121972 Pa.
Learn more about the atmospheric pressure here:
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What is the speed of light (in m/s) in air? (Enter your answer to at least four significant figures. Assume the speed of light in a vacuum is 2.997 ✕ 108 m/s.) m/s What is the speed of light (in m/s) in polystyrene? m/s
Answer:
The speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.
Explanation:
To find the speed of light in air and in polystyrene we need to use the following equation:
[tex] c_{m} = \frac{c}{n} [/tex]
Where:
[tex]c_{m}[/tex]: is the speed of light in the medium
n: is the refractive index of the medium
In air:
[tex]c_{a} = \frac{c}{n_{a}} = \frac{2.997 \cdot 10^{8} m/s}{1.0003} = 2.996 \cdot 10^{8} m/s[/tex]
In polystyrene:
[tex]c_{p} = \frac{c}{n_{p}} = \frac{2.997 \cdot 10^{8} m/s}{1.6} = 1.873 \cdot 10^{8} m/s[/tex]
Therefore, the speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.
I hope it helps you!
The ancient Greek Eratosthenes found that the Sun casts different lengths of shadow at different points on Earth. There were no shadows at midday in Aswan as the Sun was directly overhead. 800 kilometers north, in Alexandria, shadow lengths were found to show the Sun at 7.2 degrees from overhead at midday. Use these measurements to calculate the radius of Earth.
Answer:
The radius of the earth is [tex]r = 6365.4 \ km[/tex]
Explanation:
From the question we are told that
The distance at Alexandria is [tex]d_a = 800 \ km = 800 *10^{3} \ m[/tex]
The angle of the sun is [tex]\theta = 7.2 ^o[/tex]
So we want to first obtain the circumference of the earth
So let assume that the earth is circular ([tex]360 ^o[/tex])
Now from question we know that the sun made an angle of [tex]7.2 ^o[/tex] so with this we will obtain how many [tex](7.2 ^o)[/tex] are in [tex]360^o[/tex]
i.e [tex]N = \frac{360}{7.2}[/tex]
=> [tex]N = 50[/tex]
With this value we can evaluate the circumference as
[tex]c = 50 * 800[/tex]
[tex]c = 40000 \ km[/tex]
Generally circumference is mathematically represented as
[tex]c = 2\pi r[/tex]
[tex]40000 = 2 * 3.142 * r[/tex]
=> [tex]r = 6365.4 \ km[/tex]
The resistor used in the procedures has a manufacturer's stated tolerance (percent error) of 5%. Did you results from Data Table agree with the manufacturer's statement? Explain.
Resistor Measured Resistance
100 99.1
Answer:
e% = 0.99% this value is within the 5% tolerance given by the manufacturer
Explanation:
Modern manufacturing methods establish a tolerance in order to guarantee homogeneous characteristics in their products, in the case of resistors the tolerance or error is given by
e% = | R_nominal - R_measured | / R_nominal 100
where R_nominal is the one written in the resistance in your barcode, R_measured is the real value read with a multimeter and e% is the tolerance also written in the resistors
let's apply this formula to our case
R_nominal = 10 kΩ = 10000 Ω
R_measured = 100 99 Ω
e% = | 10000 - 10099.1 | / 10000 100
e% = 0.99%
this value is within the 5% tolerance given by the manufacturer
What is the angle between a wire carrying an 8.40 A current and the 1.20 T field it is in, if 50.0 cm of the wire experiences a magnetic force of 2.55 N? ° (b) What is the force (in N) on the wire if it is rotated to make an angle of 90° with the field? N
Answer:
A. 30.38°
B 5.04N
Explanation:
Using
F= ILBsin theta
2 .55N= 8.4Ax 0.5mx 1.2T x sintheta
Theta = 30.38°
B. If theta is 90°
Then
F= 8.4Ax 0.5mx 1.2x sin 90°
F= 5.04N