Answer:
The capacitance is [tex]C = 3.2 9 *10^{-16} \ F[/tex]
Explanation:
From the question we are told that
The induction of the LC circuit is [tex]L = 15 mH = 15 *10^{-3} \ H[/tex]
The frequency is [tex]w = 450 \ MHz = 450 *10^{6} \ Hz[/tex]
The natural frequency is mathematically represented as
[tex]w = \frac{1}{\sqrt{LC} }[/tex]
Where C is the capacitance So
=> [tex]C = \frac{1}{L * w^2}[/tex]
substituting values
[tex]C = \frac{1}{15 *10^{-3} * [450 *10^{6}]^2}[/tex]
[tex]C = 3.2 9 *10^{-16} \ F[/tex]
The value for which the capacitance should be set to detect a 450 MHz signal is [tex]8.34 \times 10^{-24} \;F[/tex]
Given the following data:
Inductance = 15 mH = [tex]15 \times 10^{-3}\;H[/tex]Frequency = 450 MHz = [tex]450 \times 10^6 \;Hz[/tex]To determine the value for which the capacitance should be set to detect a 450 MHz signal:
Mathematically, natural frequency is given by the formula:
[tex]f_o = \frac{1}{2\pi \sqrt{LC} }[/tex]
Where:
L is the inductance.C is the capacitance.Making C the subject of formula, we have:
[tex]C = \frac{1}{(2\pi f_o)^2L} \\\\C = \frac{1}{(2\;\times \;3.142 \times \;450 \;\times\; 10^9)^2 \; \times \;15 \times 10^{-3}}\\\\C = \frac{1}{8 \times 10^{24} \;\times \;15 \times 10^{-3} } \\\\C = \frac{1}{1.2 \times 10^{23}} \\\\C= 8.34 \times 10^{-24} \;F[/tex]
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A 150 g ball rolls at 10 cm/s rightward over a frictionless surface towards a spring 75 cm away, with spring constant 200 N/m. How far does is the spring compressed when the ball is brought to rest
Answer:
x = 0.27 cm
Explanation:
given data
mass = 150 g
velocity = 10 cm/s = 0.1 m/s
spring constnat = 200 N/m
solution
as we know that here ball is moving with constant speed
so
0.5 × m × v² = 0.5 × k × x² .......................1
here x is compression in spring
so put here value and we get
0.5 × 150 × (0.1)² = 0.5 × 200 × x²
solve it we get
x = 0.27 cm
Which statement correctly describes how a bar magnet should be placed on a globe to correctly align with Earth's magnetic field?
Answer:
The answer is B. When the magnet is placed on a globe to correctly align with Earth’s magnetic field, it is considered to be suspended freely. The Earth has geographical poles as well with North and South poles. Since unlike poles attract, the South Pole of the magnet will be attracted to the geographical North.
Explanation:
B)Place the magnet vertically on the equator, with the south end facing the North pole.
What is a bar magnet?A bar magnet is a square piece of an item, made from iron, metal, or every other ferromagnetic substance or ferromagnetic composite, that indicates everlasting magnetic homes. It has two poles, a north and a south pole such that when suspended freely, the magnet aligns itself so that the northern pole factors towards the magnetic north pole of the earth.
What are the uses of a bar magnet?Bar magnets are used as stirrers in laboratories for magnetic experiments.They also find applications in medical procedures.Electronic devices such as telephones, radios, and television sets use magnets.Many industries use bar magnets for the collection of loose metals and also for retaining the magnetism of other magnets.Learn more about bar magnet here: https://brainly.com/question/18742643
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In a high school swim competition, a student takes 1.6 s to complete 2.0 somersaults. Determine the average angular speed of the diver, in rad/s, during this time interval.
Answer:
9.82 rads/sec
Explanation:
We are given;
Time taken; t = 1.6 secs
Number of somersaults = 2
Now, we know that,
1 revolution = 2π radians
And number of somersaults is the same thing as number of revolutions
So,
Total radians = 2π × 2 = 5π
Angular velocity = total number of revolutions/time period = 5π/1.6 = 9.82 rads/sec
In Young's double slit experiment if the maximum intensity of light is Imax, then the intensity at path difference λ/2 will be
A .Imax
B .2Imax
C .4Imax
D. None
Answer:
Explanation:
When the path difference is half the wave length or λ /2 , destructive interference takes place which results in reduced or zero intensity in case equal intensity waves interfere as in Young's double slit experiment
Hence dark fringe is formed at that place where intensity is zero .
Hence the right option is D
A negatively charged object is located in a region of space where the electric field is uniform and points due north. The object may move a set distance d to the north, east, or south. Write the three possible movements by the change in electric potential energy (Ue) of the object.
Answer:
the three possible movements by the change in electric potential energy (Ue) of the object are NORTH EAST SOUTH
Explanation:
This is because When the object moves south, the force is in the direction of the displacement, and positive work is done with decreasing electric potential energy.
The opposite is true if the particle moves north—that is, negative work is done with increasing electric potential energy.
No work is done and the electric potential energy is constant if the motion is perpendicular to the electric field.
The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.939 W/m2. The wave is incident upon a rectangular area, 1.5 m by 2.0 m, at right angles. How much total electromagnetic energy falls on the area during 1.0 minute
Answer:
The total energy is [tex]T = 169.02 \ J[/tex]
Explanation:
From the question we are told that
The Poynting vector (energy flux ) is [tex]k = 0.939 \ W/m^2[/tex]
The length of the rectangle is [tex]l = 1.5 \ m[/tex]
The width of the rectangle is [tex]w = 2.0 \ m[/tex]
The time taken is [tex]t = 1 \ minute = 60 \ s[/tex]
The total electromagnetic energy falls on the area is mathematically represented as
[tex]T = k * A * t[/tex]
Where A is the area of the rectangle which is mathematically represented as
[tex]A= l * w[/tex]
substituting values
[tex]A= 2 * 1.5[/tex]
[tex]A= 3 \ m^2[/tex]
substituting values
[tex]T = 0.939 * 3 * 60[/tex]
[tex]T = 169.02 \ J[/tex]
A Young'sdouble-slit interference experiment is performed with monochromatic light. The separation between the slits is 0.44 mm. The interference pattern on the screen 4.2 m away shows the first maximum 5.5 mm from the center of the pattern. What is the wavelength of the light in nm
Answer:
Explanation:
The double slit interference phonemene is described for the case of constructive interference
d sin θ= m λ (1)
let's use trigonometry to find the sinus
tan θ = y / L
in general in interference phenomena the angles are small
tan θ = sin θ / cos θ = sin θ
The double slit interference phonemene is described for the case of constructive interference
d sin θ = m lam (1)
let's use trigonometry to find the sinus
tan θ = y / L
in general in interference phenomena the angles are small
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
we substitute in equation 1
d y / L = m λ
λ = dy / L m
let's reduce the magnitudes to the SI system
d = 0.44 mm = 0.44 10⁻³ m
y = 5.5 mm = 5.5 10⁻³ m
L = 4.2m
m = 1
let's calculate
λ = 0.44 10⁻³ 5.5 10⁻³ / (4.2 1)
λ = 5.76190 10-7 m
let's reduce to num
lam = 5.56190 10-7 m (109 nm / 1m)
lam = 556,190 nmtea
we substitute
without tea = y / L
we substitute in equation 1
d y / L = m lam
lam = dy / L m
let's reduce the magnitudes to the SI system
d = 0.44 me = 0.44 10-3 m
y = 5.5 mm = 5.5 10-3
L = 4.2m
m = 1
let's calculate
lam = 0.44 10⁻³ 5.5 10⁻³ / (4.2 1)
lam = 5.76190 10⁻⁷ m
let's reduce to num
lam = 5.56190 10⁻⁷ m (109 nm / 1m)
lam = 556,190 nm
A current carrying loop of wire lies flat on a table top. When viewed from above, the current moves around the loop in a counterclockwise sense.
(a) For points OUTSIDE the loop, the magnetic field caused by this current:________.
a. points straight up.
b. circles the loop in a clockwise direction.
c. circles the loop in a counterclockwise direction.
d. points straight down.
e. is zero.
(b) For points INSIDE the loop, the magnetic field caused by this current:________.
a. circles the loop in a counterclockwise direction.
b. points straight up.
c. points straight down.
d. circles the loop in a clockwise direction.
e. is zero
Answer:
D &B
Explanation:
Using Fleming right hand rule that States that if the fore-finger, middle finger and the thumb of left hand are stretched mutually perpendicular to each other, such that fore-finger points in the direction of magnetic field, the middle finger points in the direction of the motion of positive charge, then the thumb points to the direction of the force
A cylindrical shell of radius 7.00 cm and length 2.59 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 20.1 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C.
A) Use approximate relationships to find the net charge on the shell.
B) Use approximate relationships to find the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell.
Three point charges (some positive and some negative) are fixed to the corners of the same square in various ways, as the drawings show. Each charge, no matter what its algebraic sign, has the same magnitude. In which arrangement (if any) does the net electric field at the center of the square have the greatest magnitude?
Answer:
The magnitude of the net field located at the center of the square is the same in every of arrangement of the charges.
Assume that a lightning bolt can be represented by a long straight line of current. If 15.0 C of charge passes by in a time of 1.5·10-3s, what is the magnitude of the magnetic field at a distance of 24.0 m from the bolt?
Answer:
The magnitude of the magnetic field is 8.333 x 10⁻⁷ T
Explanation:
Given;
charge on the lightening bolt, C = 15.0 C
time the charge passes by, t = 1.5 x 10⁻³ s
Current, I is calculated as;
I = q / t
I = 15 / 1.5 x 10⁻³
I = 10,000 A
Magnetic field at a distance from the bolt is calculated as;
[tex]B = \frac{\mu_o I}{2\pi r}[/tex]
where;
μ₀ is permeability of free space = 4π x 10⁻⁷
I is the current in the bolt
r is the distance of the magnetic field from the bolt
[tex]B = \frac{\mu_o I}{2\pi r} \\\\B = \frac{4\pi *10^{-7} 10000}{2\pi *24} \\\\B = 8.333 *10^{-5} \ T[/tex]
Therefore, the magnitude of the magnetic field is 8.333 x 10⁻⁷ T
Earth moves in an elliptical orbit with the sun at one of the foci. The length of half of the major axis is kilometers, and the eccentricity is . Find the minimum distance (perihelion) and the maximum distance (aphelion) of Earth from the sun.
Complete question is;
Earth moves in an elliptical orbit with the sun at one of the foci. The length of half of the major axis is 149,598,000 kilometers, and the eccentricity is 0.0167. Find the minimum distance (perihelion) and the maximum distance (aphelion) of Earth from the sun
Answer:
Minimum distance = 147,099,713.4 km
Maximum distance = 152,096,286.6 km
Explanation:
The formula for the eccentricity of an ellipse is given by;
e = c/a
where;
c is distance from the center of the ellipse to the focus of the ellipse.
a is distance from the center of ellipse to a vertex.
From the question, we are given;
a = 149,598,000
e = 0.0167
Thus;
0.0167 = c/149,598,000
c = 0.0167 × 149,598,000
c = 2,498,286.6
Now, formula for the minimum distance (perihelion) is;
Minimum distance = a - c
Minimum distance = 149598000 - 2498286.6
Thus;
Minimum distance = 147,099,713.4 km
Similarly, formula for the maximum distance (aphelion) is;
Max distance = a + c
Max distance = 149598000 + 2498286.6
Maximum distance = 152,096,286.6 km
In a system with only a single force acting upon a body, what is the relationship between the change in kinetic energy and the work done by the force?
Answer: W.D = 1/2mv^2
Explanation:
If an external force or a single force is acting on a body. Just like the first law of thermodynamics, the force acting on the body will cause work done on the system.
Work done = force × distance
And the work done on the body will cause the molecules of the body to experience motion and thereby producing kinetic energy.
The work done will be converted to kinetic energy.
W.D = 1/2mv^2
Inductance is usually denoted by L and is measured in SI units of henries (also written henrys, and abbreviated H), named after Joseph Henry, a contemporary of Michael Faraday. The EMF E produced in a coil with inductance L is, according to Faraday's law, given by
E=−LΔIΔt.
Here ΔI/Δt characterizes the rate at which the current I through the inductor is changing with time t.
Based on the equation given in the introduction, what are the units of inductance L in terms of the units of E, t, and I (respectively volts V, seconds s, and amperes A)?
What EMF is produced if a waffle iron that draws 2.5 amperes and has an inductance of 560 millihenries is suddenly unplugged, so the current drops to essentially zero in 0.015 seconds?
Answer:
Explanation:
E= −L ΔI / Δt.
L = E Δt / ΔI
Hence the unit of inductance may be V s A⁻¹
or volt s per ampere .
In the given case
change in current ΔI = - 2.5 A
change in time = .015 s
L = .56 H
E = − L ΔI / Δt.
= .56 x 2.5 / .015
= 93.33 V .
A system loses 510 J of potential energy. In the process, it does 430 J of work on the environment and the thermal energy increases by 100 J.
Required:
Find the change in kinetic energy.
Answer:
20J
Explanation:
The computation of the change in kinetic energy is shown below:
As we know that
Work was done on system = change in potential energy + change in kinetic energy + change in thermal energy
-430J = -510J + change in kinetic energy + 100J
-430J = -410J + change in kinetic energy
So, the change in kinetic energy is 20J
We simply applied the above formula to find out the change in kinetic energy
A parallel-plate capacitor in air has a plate separation of 1.31 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 255 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator.
(a) Determine the charge on the plates before and after immersion.
before pC
after pC
(b) Determine the capacitance and potential difference after immersion.
Cf = F
ΔVf = V
(c) Determine the change in energy of the capacitor.
[ ] nJ
Answer:
a) before immersion
C = εA/d = (8.85e-12)(25e-4)/(1.31e-2) = 1.68e-12 F
q = CV = (1.68e-12)(255) = 4.28e-10 C
b) after immersion
q = 4.28e-10 C
Because the capacitor was disconnected before it was immersed, the charge remains the same.
c)*at 20° C
C = κεA/d = (80.4*)(8.85e-12)(25e-4)/(1.31e-2) = 5.62e-10 F
V = q/C = 4.28e-10 C/5.62e-10 C = 0.76 V
e)
U(i) = (1/2)CV^2 = (1/2)(1.68e-12)(255)^2 = 5.46e-8 J
U(f) = (1/2)(5.62e-10)(0.76)^2 = 1.62e-10 J
ΔU = 1.62e-10 J - 5.46e-8 J = -3.84e-8 J
Determine the smallest distance x to a position where 450-nm light reflected from the top surface of the glass interferes constructively with light reflected from the silver coating on the bottom. The light changes phase when reflected at the silver coating.
A wedge of glass of refractive index 1.64 has a silver coating on the bottom, as shown in the image attached below.
Determine the smallest distance x to a position where 450-nm light reflected from the top surface of the glass interferes constructively with light reflected from the silver coating on the bottom. The light changes phase when reflected at the silver coating.
Answer:
the smallest distance x = 2.74 × 10⁻³ m or 2.74 mm
Explanation:
From the given information:
The net phase change is zero because both the light ray reflecting from the air-glass surface and silver plate undergo a phase change of [tex]\dfrac{\lambda}{2}[/tex] , as such the condition for the constructive interference is:
nΔy = mλ
where;
n = refractive index
Δy = path length (inside the glass)
So, from the diagram;
[tex]\dfrac{y}{x}=\dfrac{10^{-5} \ m}{0.2 \ m}[/tex]
[tex]\dfrac{y}{x} = 5 \times 10^{-5}[/tex]
[tex]y = 5 \times 10^{-5} x[/tex]
Now;
Δy can now be = 2 ( 5 × 10⁻⁵ [tex]x[/tex])
Δy =1 × 10⁻⁴[tex]x[/tex]
From nΔy = mλ
n( 1 × 10⁻⁴[tex]x[/tex] ) = mλ
[tex]x = \dfrac{m \lambda}{n \times 1 \times 10^{-4} }[/tex]
when the thickness is minimum then m = 1
Thus;
[tex]x = \dfrac{1 \times 450 \times 10^{-9} \ m}{1.64 \times 1 \times 10^{-4} }[/tex]
x = 0.00274 m
x = 2.74 × 10⁻³ m or 2.74 mm
Answer: B. The surface of the coating is rough, so light that shines on it gets scattered in many directions.
Explanation: On Edge!!!!!!!!!!!!!!!!!!!!
A drum rotates around its central axis at an angular velocity of 19.4 rad/s. If the drum then slows at a constant rate of 8.57 rad/s2, (a) how much time does it take and (b) through what angle does it rotate in coming to rest
Answer:
Explanation:
Using equations of motion:
(a)
v=u+at
∴0=19.4−8.57t
∴t=19.4/8.57
=2.3s
B. Using s= ut + 1/2 at²
19.4(2.3)-1/28.57(2.3)²
= 21.92rad
An astronaut out on a spacewalk to construct a new section of the International Space Station walks with a constant velocity of 2.30 m/s on a flat sheet of metal placed on a flat, frictionless, horizontal honeycomb surface linking the two parts of the station. The mass of the astronaut is 71.0 kg, and the mass of the sheet of metal is 230 kg. (Assume that the given velocity is relative to the flat sheet.)
Required:
a. What is the velocity of the metal sheet relative to the honeycomb surface?
b. What is the speed of the astronaut relative to the honeycomb surface?
Answer:
Explanation:
Let the velocity of astronaut be u and the velocity of flat sheet of metal plate be v . They will move in opposite direction , so their relative velocity
= u + v = 2.3 m /s ( given )
We shall apply conservation of momentum law for the movement of astronaut and metal plate
mu = M v where m is mass of astronaut , M is mass of metal plate
71 u = 230 x v
71 ( 2.3 - v ) = 230 v
163.3 = 301 v
v = .54 m / s
u = 1.76 m / s
honeycomb will be at rest because honeycomb surface is frictionless . Plate will slip over it . Over plate astronaut is walking .
a ) velocity of metal sheet relative to honeycomb will be - 1.76 m /s
b ) velocity of astronaut relative to honeycomb will be + .54 m /s
Here + ve direction is assumed to be the direction of astronaut .
A charge of 0.80 nC is placed at the center of a cube that measures 4.0 m along each edge. What is the electric flux through one face of the cube
Answer:
The magnetic flux is [tex]\phi = 15 \ Nm^2 /C[/tex]
Explanation:
From the question we are told that
The value of the charge is [tex]q = 0.80 \ nC = 0.80 *10^{-9} \ C[/tex]
The length of each side of the cube is [tex]d = 4.0 \ m[/tex]
Generally the magnetic flux through a closed surface is mathematically represented as
[tex]\phi = \frac{q}{\epsilon_o} * D[/tex]
Where D is the area enclosing the charge
Now a cube is made up of six faces but in this question we are considering only one face which is mathematically represented as
[tex]D = \frac{1}{6}[/tex]
So the electric flux through one face of the cube is mathematically represented as
[tex]\phi = \frac{q}{6 * \epsilon _o }[/tex]
where [tex]\epsilon _o[/tex] is the permitivity of free space with value
[tex]\epsilon_o = 8.85 *10^{-12} F/m[/tex]
substituting value
[tex]\phi = \frac{0.80 *10^{-9}}{6 * 8.85 *10^{-12} }[/tex]
[tex]\phi = 15 \ Nm^2 /C[/tex]
A record player rotates a record at 45 revolutions per minute. When the record player is switched off, it makes 4.0 complete turns at a constant angular acceleration before coming to rest. What was the magnitude of the angular acceleration (in rads/s2) of the record as it slowed down
Answer:
The angular acceleration is [tex]\alpha = 0.4418 \ rad /s^2[/tex]
Explanation:
From the question we are told that
The angular speed is [tex]w_f = 45 \ rev / minutes = \frac{45 * 2 * \pi }{60 }= 4.713 \ rad/s[/tex]
The angular displacement is [tex]\theta =4 \ rev = 4 * 2 * \pi = 25.14 \ rad[/tex]
From the first equation of motion we can define the movement of the record as
[tex]w_f ^2 = w_o ^2 + 2 * \alpha * \theta[/tex]
Given that the record started from rest [tex]w_o = 0[/tex]
So
[tex]4.713^2 = 2 * \alpha * 25.14[/tex]
[tex]\alpha = 0.4418 \ rad /s^2[/tex]
The driver of a stationary car hears a siren of an approaching police car at a frequency of 280Hz. If the actual frequency of the siren is 240Hz, find the speed of the police car (speed of sound is 343m/s).
Answer:
The speed of the police car is 294 m/s
Explanation:
Given;
frequency of the siren in air, f = 280 Hz
speed of sound in air, v = 343 m/s
Determine the wavelength of the sound in air to the stationary car:
v = fλ
where;
λ is wavelength of the sound
λ = v/f
λ = 343 / 280
λ = 1.225 m
Now, determine the speed at which the police car is approaching the stationary car;
The actual frequency of the police car, F = 240 Hz
V = Fλ
Where;
V is speed of the police car
λ is the distance between the police car and the stationary car, (wavelength)
V = 240 x 1.225
V = 294 m/s
Therefore, the speed of the police car is 294 m/s
Zack is driving past his house. He wants to toss his physics book out the window and have it land in his driveway. If he lets go of the book exactly as he passes the end of the driveway. Should he direct his throw outward and toward the front of the car (throw 1), straight outward (throw 2), or outward and toward the back of the car (throw 3)? Explain.
Answer:
Zack should direct his throw outward and toward the back of the car.
Explanation:
As the car is moving forward, the book will be thrown with a forward component. Therefore, throwing this book backwards at a constant speed would cancel the motion of the car, allowing the book to have a greater chance of ending on the driveway. I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum.
The solution is throw 3.
I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.
Which statement best applies Newton’s laws of motion?The statement that best applies Newton’s laws of motion to explain the skydiver’s motion is that an upward force balances the downward force of gravity on the skydiver. Newton's 3rd law often applies to skydiving.
When gravity is not acting upon the skydivers they would continue moving in the direction the vehicle they jumped from was moving. If no air resistance takes place, then the skydivers would still accelerating at 9.8 m/s until they hit the ground.
The skydiver after leaving the aircraft will accelerates downwards due to the force of gravity usually as there is no air resistance acting in the upwards direction, and there is a resultant force acting downwards, the skydiver will accelerates towards the ground.
Therefore, I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.
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A tennis player swings her 1000 g racket with a speed of 12 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 15 m/s. The ball rebounds at 40 m/s.
A) How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.
_________m/s
B) If the tennis ball and racket are in contact for 7.00, what is the average force that the racket exerts on the ball?
_________N
The velocity and force are required.
The speed of the racket is 8.7 m/s
The required force is 471.43 N.
[tex]m_1[/tex] = Mass of racket = 1000 g
[tex]m_2[/tex] = Mass of ball = 60 g
[tex]u_1[/tex] = Initial velocity of racket = 12 m/s
[tex]u_2[/tex] = Initial velocity of ball = -15 m/s
[tex]v_1[/tex] = Final velocity of racket
[tex]v_2[/tex] = Final velocity of ball = 40 m/s
[tex]\Delta t[/tex] = Time = 7 ms
The equation of the momentum will be
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_1=\dfrac{m_1u_1+m_2u_2-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{1\times 12+0.06\times (-15)-0.06\times 40}{1}\\\Rightarrow v_1=8.7\ \text{m/s}[/tex]
Force is given by
[tex]F=m_2\dfrac{v_2-u_2}{\Delta t}\\\Rightarrow F=0.06\times \dfrac{40-(-15)}{7\times 10^{-3}}\\\Rightarrow F=471.43\ \text{N}[/tex]
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A centrifugal pump is operating at a flow rate of 1 m3/s and a head of 20 m. If the specific weight of water is 9800 N/m3 and the pump efficiency is 85%, the power required by the pump is most nearly:
Answer:
The power required by the pump is nearly 230.588 kW
Explanation:
Flow rate of the pump Q = 1 m^3/s
the head flow H = 20 m
specific weight of water γ = 9800 N/m^3
efficiency of the pump η = 85%
First note that specific gravity of water is the product of the density of water and acceleration due to gravity.
γ = ρg
where ρ is density. For water its value is 1000 kg/m^3
g is the acceleration due to gravity = 9.81 m/s^2
The power to lift this water at this rate will be gotten from the equation
P = ρgQH
but ρg = γ
therefore,
P = γQH
imputing values, we'll have
P = 9800 x 1 x 20 = 196000 W
But the centrifugal pump that will be used will only be able to lift this amount of water after the efficiency factor has been considered. The power of pump needed must be greater than this power.
we can say that
196000 W is 85% of the power of the pump power needed, therefore
196000 = 85% of [tex]P_{p}[/tex]
where [tex]P_{p}[/tex] is the power of the pump needed
85% = 0.85
196000 = 0.85[tex]P_{p}[/tex]
[tex]P_{p}[/tex] = 196000/0.85 = 230588.24 W
Pump power = 230.588 kW
To work on your car at night, you use an extension cord to connect your work light to a power outlet near the door. How would the illumination provided by the light be affected by the length of the extension cord
Answer:
The longer the cord, the lower the illumination
Explanation:
The illumination provided by the light bulb will be reduced as the length of the extension cord increases. This is because the resistance provided by the wire increases with its length.
Long wires have more electrical resistance than shorter ones.
Let us consider this formula:
Resistance =[tex]\frac{\rho L}{A}[/tex]
From this formula, we can see that as the length increases, the resistance to current flow offered by the wire increases also provided the resistivity and cross-sectional area of the wire remain constant. As a result of this, the illumination will drop.
You push a shopping cart full of groceries. The shopping cart has a mass of 32
kg.
a. What is the weight of your shopping cart?
b. How much force must you apply to give the cart an acceleration of 1.3 m/s2?
c. If you push with a force of 200 N, what is the acceleration of the cart?
d. You are driving home from the store. If your car has an acceleration of 4.1 m/s2 and its engine is applying a forward force of 7000 N, what is the mass of your car?
Answer:
a) 320N (if you take gravity as 10ms^-2) or 313.6N (if you take gravity as 9.8ms^-2
b) 41.6N
c) 6.25ms^-2
d) 1707.317 kg
Explanation:
a) W=mg
W= 32 x 10 or 9.8
W = 320N or 313.6N
b) F=ma
F= 32 x 1.3
F= 41.6N
c) F=ma
200 = 32 x a
a= 6.25ms^-2
d) F=ma
7000= m x 4.1
m= 1707.317 kg
I am not completely sure about the d) part because I dont whether you will be taking Friction and Normal Reaction too. As per my knowledge, I think no, as no angles nor the gradient of the floor/road is mentioned here.
a) The weight of your shopping cart is 313.6N .
b) The required force is 41.6N
c) The acceleration of the car is 6.25ms^-2
d) The mass of the car is 1707.317 kg.
What is acceleration?Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).
a) The weight of the shopping cart: W=mg
W= 32 x 9.8 N
W = 313.6N
b) The required force is to be applied: F=ma
F= 32 x 1.3 N.
F= 41.6N.
c) Let the acceleration of the cart is a.
Then, force: F=ma
200 = 32 x a
a= 6.25ms^-2
Hence, the acceleration of the cart is 6.25ms^-2.
d) Let the mass of the car is m.
Force applied on the car: F=ma
7000= m x 4.1
m= 1707.317 kg
The mass of the car is 1707.317 kg.
Learn more about acceleration here:
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dandre expands 120w of power in moving a couch 15 meters in 5 seconds how much force does he exert ?
Answer:
The answer is 40 N for APX
Explanation:
A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of ρ0 is placed in a container of water. Initially the sphere floats and the water level is marked on the side of the container. What happens to the water level, when the original sphere is replaced with a new sphere which has different physical parameters? Notation: r means the water level rises in the container, f means falls, s means stays the same.
A)
The new sphere has a density of ρ = ρ0 and a mass of m < m0.
B)
The new sphere has a density of ρ = ρ0 and a radius of r > r0.
C)
The new sphere has a density of ρ < ρ0 and a mass of m = m0.
The options are r, f, and s. Rises, Falls, Stays the same.
Answer:
(a) f
(b) r
(c) s
Explanation:
There are two forces on the sphere: weight and buoyancy.
Sum of forces in the y direction:
∑F = ma
B − mg = 0
B = mg
Buoyancy is equal to the weight of the displaced fluid, or ρVg, where ρ is the density of the fluid and V is the displaced volume.
ρVg = mg
ρV = m
V = m/ρ
(a) The mass decreases, so the displaced volume decreases.
(b) The sphere's density is constant and its radius increases, which means its mass increases, so the displaced volume increases.
(c) The mass stays the same, so the displaced volume is the same.
A 22.0 cm diameter loop of wire is initially oriented perpendicular to a 1.5-T magnetic field. The loop is rotated so that its plane is parallel to the field direction in 0.20 s. What is the average induced emf in the loop?
Answer:
The average induced emf in the loop is 0.3 V
Explanation:
Given:
d = 22 cm
Magnetic field is B = 1.5 T
Change in time Δt = 0.20 sec.
Radius of loop = r = d/2 = 11 x [tex]10^{-2}[/tex] m
According to the faraday's law, Induced emf is given by e = - (ΔФ / Δt)
where Ф = magnetic flux
Ф = BAcos0 (in this occasion, Ф=0)
where area A = pi * r²
note that we have to neglect negative sign due to its lenz law...so
B * pi * r
e = ----------------
Δt
1.5 ( 3.1416) 11 x [tex]10^{-2}[/tex] )²
e = ------------------------------------
0.20
e = 0.3 V
Therefore, the average induced emf in the loop is 0.3 V