Answer:
Magnification = 1
Explanation:
given data
radius of curvature r = - 0.983 m
image distance u = - 0.155
solution
we get here first focal length that is
Focal length, f = R/2 ...................1
f = -0.4915 m
we use here formula that is
[tex]\frac{1}{v} + \frac{1}{u} + \frac{1}{f}[/tex] .................2
put here value and we get
[tex]\frac{1}{v} = \frac{1}{0.155} - \frac{1}{4915}[/tex]
v = 0.155 mso
Magnification will be here as
m = [tex]- \frac{v}{u}[/tex]
m = [tex]\frac{0.155}{0.155}[/tex]
m = 1Answer:
The magnification is 1.5.
Explanation:
radius of curvature, R = - 0.983 m
distance of object, u = - 0.155 m
Let the distance of image is v.
focal length, f = R/2 = - 0.492 m
Use the mirror equation
[tex]\frac{1}{f}=\frac{1}{v}+\frac {1}{u}\\\\\frac{-1}{0.492}=\frac{1}{v}-\frac{1}{0.155}\\\\\frac{1}{v}=\frac{1}{0.155}-\frac{1}{0.492}\\\\\frac{1}{v}=\frac{0.492-0.155}{0.155\times 0.492}\\\\\frac{1}{v}=\frac{0.337}{0.07626}\\ \\v = 0.226 m[/tex]
The magnification is given by
m = - v/u
m = 0.226/0.155
m = 1.5
A smooth circular hoop with a radius of 0.400 m is placed flat on the floor. A 0.325-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 8.50 m/s. After one revolution, its speed has dropped to 5.50 m/s because of friction with the floor.
(a) Find the energy transformed from mechanical to internal in the particle "hoop" floor system as a result of friction in one revolution.
(b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion.
Answer:
a) W = - 6.825 J, b) θ = 1.72 revolution
Explanation:
a) In this exercise the work of the friction force is negative and is equal to the variation of the kinetic energy of the particle
W = ΔK
W = K_f - K₀
W = ½ m v_f² - ½ m v₀²
W = ½ 0.325 (5.5² - 8.5²)
W = - 6.825 J
b) find us the coefficient of friction
Let's use Newton's second law
fr = μ N
y-axis (vertical) N-W = 0
fr = μ W
work is defined by
W = F d
the distance traveled in a revolution is
d₀ = 2π r
W = μ mg d₀ = -6.825
μ = [tex]\frac{ -6.825}{d_o \ mg}[/tex]
The total work as the object stops the final velocity is zero v_f = 0
W = 0 - ½ m v₀²
W = - ½ 0.325 8.5²
W = - 11.74 J
μ mg d = -11.74
we subtitle the friction coefficient value
( [tex]\frac{-6.8525 }{d_o mg}[/tex]) m g d = -11.74
6.825 [tex]\frac{d}{d_o}[/tex] = 11.74
d = 11.74/6.825 d₀
d = 1.7201 2π 0.400
d = 4.32 m
this is the total distance traveled, the distance and the angle are related
θ = d / r
θ = 4.32 / 0.40
θ = 10.808 rad
we reduce to revolutions
θ = 10.808 rad (1rev / 2π rad)
θ = 1.72 revolution
1 hallar el trabajo mecanico de un cuerpo que tiene una fuerza de 250 newton y recorre 750 metros
2 hallar la potencia necesaria para levantar un transformador de masa 2500kg,una altura de 4 metros en un tiempo de 30 segundos
porfa es para hoy
Answer: TRACK
Explanation:
27. The part of the Earth where life exists .
Mesosphere
Stratosphere
Troposphere
Biosphere
Answer:
Biosphere is the part of the earth where life exists.
In the diagram, the amplitude of the wave is shown by:
A
B
C
D
Answer:
A.
Explanation:
Amplitude measures how much a wave rises or falls. This is illustrated by A.
In the diagram, the amplitude of the wave is shown by A.
What is Amplitude?The amplitude of a periodic variable is a measure of its change in a single period. The amplitude of a non-periodic signal is its magnitude compared with a reference value.There are various definitions of amplitude, which are all functions of the magnitude of the differences between the variable's extreme values.
The amplitude of a variable is simply a measure of change relative to its central position. In contrast, magnitude is a measure of the distance or quantity of a variable irrespective of its direction.
Amplitude is a property that is unique to waves and oscillations.
Therefore, in the diagram, the amplitude of a wave is shown by A.
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The mass flow rate through a centrifugal compressor is 1 kg/s. If air enters at 1 bar and 288k and leaves at 200 kN/m² and 370k, determine the power of the compressor. Take Cp = 1.103 kJ (kg.K), R = 287 kJ (kg.k)
We have that the power of the compressor is
[tex]H_p=24.242hp[/tex]
From the question we are told that:
Flow rate [tex]W=1kg/s[/tex]
inlet Pressure [tex]P_1=1 bar[/tex]
inlet Temperature [tex]T_1= 288k[/tex]
Outlet Temperature [tex]T_2= 370k[/tex]
Outlet Pressure [tex]P_2=200 kN/m^2=2bars[/tex]
[tex]Cp = 1.103 kJ[/tex]
[tex]R = 287 kJ (kg.k)[/tex]
Generally, the equation for Adiabatic head is mathematically given by
[tex]H=\frac{ZRT_1}{Cp-1/K}[\frac{P_2}{P_1}^{(Cp-1)/Cp}-1][/tex]
Where
[tex]Z=Compressibility\ factor[/tex]
[tex]Z=0.99[/tex]
Therefore
[tex]H=\frac{(0.99)(287)(288)}{(1.103)-1/(1.103)}[\frac{(200)}{(1)}^{((1.103)-1)/(1.103)}-1][/tex]
[tex]H=560925.5958 J/kg[/tex]
[tex]H=5.6*10^5J/kg[/tex]
Generally, the equation for centrifugal compressor power is mathematically given by
[tex]H_p=\frac{WH}{E*33000}[/tex]
Where
E is efficiency (adiabatic)
[tex]E=70\%=0.7[/tex]
Therefore
[tex]H_p=\frac{(1)(5.6*10^5)}{0.7*33000}[/tex]
[tex]H_p=24.242hp[/tex]
In conclusion
The power of the compressor is
[tex]H_p=24.242hp[/tex]
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A racing car going a 20 m/s stops in a distance of 20 m.What is its acceleration?
step by step
formular v^2 = u^2+2as
stop v = 0
0 = 400+2a(20)
-400=40a
a = -10 m/s^2
ans affect acceleration is 10 m/s^2
calculate horizontal distance travelled by a ball travelling with a speed of 20root2 mper sec without hitting ceiling of height 20 m per sec
Calculate the horizontal distance travelled by a ball throw with a velcoity 20 sqrt 2 ms^(-1) without hitting the ceiling of an anditorium of heith 20 m. Use g= 10 ms^(-2). =(20√2)2sin2×45∘10=9800×1)10=80m .
HOPE SO IT HELPS YOU
(b) State the unit of volume in S.I. system and define it. 4.
Answer:
the cubic meter (m3)
Explanation:
Volume is the measure of the 3-dimensional space occupied by matter, or enclosed by a surface, measured in cubic units. The SI unit of volume is the cubic meter (m3),
Answer:
please answer all component's
Explanation:
please answer all component's
A chair of weight 85.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 40.0 N directed at an angle of 35.0deg below the horizontal and the chair slides along the floor.
Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.
Answer:
N = 107.94 N
Explanation:
For this exercise we must use Newton's second law.
Let's set a reference system with the x-axis parallel to the ground and the y-axis vertical
X axis
Fₓ = ma
ej and
N -F_y - W = 0
let's use trigonometry to decompose the applied force
cos -35 = Fₓ / F
sin -35 = F_y / F
Fₓ = F cos -35
F_y = F sin -35
Fₓ = 40.0 cos -35 = 32.766 N
F_y = 40.0 sin -35 = -22.94 N
we substitute
N = Fy + W
N = 22.94 + 85
N = 107.94 N
Need help with this one! Thank you xoxo!
Answer:
it's electron attraction
The answer is Electron Attraction
Liquid plastic is frozen in a physical change that increases its volume. What can be known about the plastic after the change?
(A) Its mass will increase.
(B) Its density will increase.
(C) Its mass will remain the same.
(D) Its density will remain the same.
Answer:
c
Explanation:
Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same, therefore the correct answer is option C.
What is the matter?Anything which has mass and occupies space is known as matter ,mainly there are four states of matter solid liquid gases, and plasma.
These different states of matter have different characteristics according to which they vary their volume and shape.
It is known about plastic that its mass will remain the same when liquid plastic is frozen, by increasing its volume.
Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same, therefore the correct answer is C.
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A rectangular loop of wire with sides 0.129 and 0.402 m lies in a plane perpendicular to a constant magnetic field (see part a of the drawing). The magnetic field has a magnitude of 0.888 T and is directed parallel to the normal of the loop's surface. In a time of 0.172 s, one-half of the loop is then folded back onto the other half, as indicated in part b of the drawing. Determine the magnitude of the average emf induced in the loop.
Answer:
[tex]0.2677\ \text{V/m}[/tex]
Explanation:
A = Area of loop = [tex]0.129\times0.402[/tex]
B = Magnetic field = [tex]0.888\ \text{T}[/tex]
t = Time taken = [tex]0.172\ \text{s}[/tex]
Electric field is given by
[tex]E=B\dfrac{dA}{dt}\\\Rightarrow E=0.888\times\dfrac{0.129\times 0.402}{0.172}\\\Rightarrow E=0.2677\ \text{V/m}[/tex]
The emf induced is [tex]0.2677\ \text{V/m}[/tex].
Who should you invite to the meeting?
—-
Answer
Irma
Explanation:
You should invite Irma to the meeting because she is in Charlotte and she also visited after 10/24/20
Katie swings a ball around her head at constant speed in a horizontal circle with circumference 2.1 m. What is the work done on the ball by the 34.4 N tension force in the string during one half-revolution of the ball
Answer:
the work done on the ball is 0
Explanation:
Given the data in the question;
Katie swings a ball around her head at constant speed in a horizontal circle with circumference 2.1 m.
circle circumference = 2πr = 2.1 m
radius r will be; r = 2.1 m / 2π = 0.33 m
Tension force = 34.4 N
one half revolution means, displacement of the ball is;
d = 2r = 2 × 0.33 = 0.66 m
Now, Work done = force × displacement × cosθ
we know that, the angle between the tension force on string and displacement of object is always 90.
so we substitute
Work done = 34.4 N × 0.66 m × cos(90)
Work done = 34.4 N × 0.66 m × 0
Work done = 0 J
Therefore, the work done on the ball is 0
Calculate the potential energy stored in a metal ball of a mass of 80 kg kept at a height of 15m from the earth surface.What will be the potential energy when the metal ball is kept on the earth surface.
Answer:
39200 joules
the potential energy will be zero
Explanation:
we know that potential energy is found by multiplying mass, acceleration due to gravity and height from the Earth's surface
so it will be
potential energy= mgh
80x9.8x15
= 39200 joules
the potential energy of the mental ball will be zero when kept on the Earth's surface because the height from the Earth's surface will be zero and zero multiplied to any number is zero only
I have a doubt with the second one, this is what I think it is. Consult your teacher if you think my answer for the second one is wrong
Answer:
392000 joules
Explanation:
hope it helpsss
A 10 n force is applied horizontally on a box to move it 10 m across a frictionless surface. How much work was done to move the box?
Answer:
[tex]\boxed {\boxed {\sf 100 \ J}}[/tex]
Explanation:
We are asked to calculate the work done to move a box.
Work is the product of force and distance or displacement.
[tex]W= F*d[/tex]
A 10 Newton force is applied horizontally on the box. Since the surface is frictionless, there is no force of friction, and the net force is 10 Newtons. The force moves the box 10 meters.
F= 10 N d= 10 mSubstitute the values into the formula.
[tex]W= 10 \ N * 10 \ m[/tex]
Multiply.
[tex]W= 100 \ N*m[/tex]
Let's convert the units. 1 Newton meter is equal to 1 Joule, therefore our answer of 100 Newton meters is equal to 100 Joules.
[tex]W= 100 \ J[/tex]
100 Joules of work was done to move the box.
A 80 kg bungee jumper is on a bridge that is 100 meters above a river. Attached to the jumper is a bungee cord that is 50 meters long. After jumping off of the bridge, the jumper reaches a position that is 10 meters above the river when the bungee cord is at its maximum stretch.
Required:
a. How much energy is stored in the bungee cord at that maximum stretch?
b. What is the spring force constant of the bungee cord?
Answer:
a) 70,560 J
b) 88.2 N/m
Explanation:
The spring potential will equal the change in gravity potential
PS = PE = mgh = 80(9.8)(100 - 10) = 70,560 J
PS = ½kx²
k = 2PS/x² = 2(70560)/(100 - 50 - 10)² = 88.2 N/m
The resistance of the light bulb changed as the voltage (and current) changed. Why does this resistance change occur?
00
Use base units to check whether the following equations are balance
(a) pressure = depth x density gravitational field strength,
(b) energy = mass x (speed of light).
dod to molt solid
Answer:
In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus a column of fluid, or an object submerged in the fluid, experiences greater pressure at the bottom of the column than at the top. This difference in pressure results in a net force that tends to accelerate an object upwards.
The pressure at a depth in a fluid of constant density is equal to the pressure of the atmosphere plus the pressure due to the weight of the fluid, or p = p 0 + ρ h g , p = p 0 + ρ h g , 14.4
Granite: 2.70 × 10 32.70 × 10 3
Lead: 1.13 × 10 41.13 × 10 4
Iron: 7.86 × 10 37.86 × 10 3
Oak: 7.10 × 10 27.10 × 10 2
A conductor is placed in a steady external electric field. Which of the following is FALSE?
a) All excess charge is distributed on the surface of the conductor
b) The electric field inside the conductor is the same as the external electric field
c) The electric field is zero inside the conductor
d) the electric field just outside the surface of the conductor is perpendicular to the surface
Answer:
a
Explanation:
because the electric field doesn't effect the conductor and its goes into storage for later
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.
You attach a 2.30 kg weight to a horizontal spring that is fixed at one end. You pull the weight until the spring is stretched by 0.500 m and release it from rest. Assume the weight slides on a horizontal surface with negligible friction. The weight reaches a speed of zero again 0.400 s after release (for the first time after release). What is the maximum speed of the weight (in m/s)
Answer: [tex]3.92\ m/s[/tex]
Explanation:
Given
Mass of the attached object is [tex]m=2.3\ kg[/tex]
Spring is stretched by [tex]A=0.5\ m[/tex]
Speed reaches zero after [tex]t=0.4\ s[/tex]
Speed is zero at the extremities of the S.H.M motion that is
[tex]\Rightarrow \dfrac{T}{2}=0.4\\\\\Rightarrow T=0.8\ s[/tex]
Time period of motion is [tex]0.8\ s[/tex] which can also be given by
[tex]\Rightarrow \omega T=2\pi\\\\\Rightarrow \omega=\dfrac{2\pi }{T}\\\\\Rightarrow \omega =\dfrac{2\pi }{0.8}\\\\\Rightarrow \omega=\dfrac{5\pi }{2}[/tex]
Maximum speed for S.H.M. is [tex]v_{max}=A\omega[/tex]
[tex]\Rightarrow v_{max}=0.5\times 2.5\pi\\\Rightarrow v_{max}=3.92\ m/s[/tex]
Although human beings have been able to fly hundreds of thousands of miles into outer space, getting inside the earth has proven much more difficult. The deepest mines ever drilled are only about 10 miles deep. To illustrate the difficulties associated with such drilling, consider the following: The density of steel is about 7900 kilograms per cubic meter, and its breaking stress, defined as the maximum stress the material can bear without deteriorating, is about 2.0×1092.0×109 pascals. What is the maximum length of a steel cable that can be lowered into a mine? Assume that the magnitude of the acceleration due to gravity remains constant at 9.8 meters per second per second.
Use two significant figures in your answer, expressed in kilometers.
Answer:
26 km
Explanation:
Let's say our "cable" has a cross section of 1 m²
Then each meter of cable would weight 7900(9.8) = 77420 N
A Pascal is a Newton per square meter
2 x 10⁹ / 77420 = 25840 m or about 26 km or about 16 miles
The magnetic flux that passes through one turn of a 8-turn coil of wire changes to 5.0 Wb from 8.0 Wb in a time of 0.098 s. The average induced current in the coil is 140 A. What is the resistance of the wire
Answer:
Resistance is 1.75 ohms
Explanation:
Magnetic flux:
[tex]{ \phi{ = NBA}}[/tex]
N is number of turns, N = 8
B is magnetic flux
A is area of projection.
From faradays law:
[tex]E = - \frac{ \triangle \phi}{t} [/tex]
where E is the Electro motive force.
But E = IR
where I is current and R is resistance:
[tex]IR = \frac{( \phi_{1} - \phi _{2}) }{t} \\ \\ 140 \times R = \frac{8 \times (8 - 5)}{0.098} \\ \\ R = \frac{24}{0.098 \times 140} \\ \\ resistance = 1.75 \: ohms[/tex]
Tay quay OB quay đều quanh trục cố định đi qua O với vận tốc góc không đổi ω. Con lăn A chuyển động trong rãnh thẳng đứng. Tại vị trí trên hình vẽ thì thanh OB thẳng đứng, OA có phương nằm ngang. Hãy xác định vận tốc góc thanh AB, vận tốc của con lăn A; gia tốc góc của thanh AB, gia tốc của con lăn A. Cho ω = 1,5 rad/s, r = 1 m.
Consider two closely spaced and oppositely charged parallel metal plates. The plates are square with sides of length L and carry charges Q and -Q on their facing surfaces. What is the magnitude of the electric field in the region between the plates
Answer:
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
Explanation:
In this exercise you are asked to calculate the electric field between two plates, the electric field is a vector
E_ {total} = E₁ + E₂
E_ {total} = 2 E
where E₁ and E₂ are the fields of each plate, we have used that for the positively charged plate the field is outgoing and for the negatively charged plate the field is incoming, therefore in the space between the plates for a test charge the two fields point in the same direction
to calculate the field created by a plate let's use Gauss's law
Ф = ∫ E . dA = q_{int} /ε₀
As a Gaussian surface we use a cylinder with the base parallel to the plate, therefore the direction of the electric field and the normal to the surface are parallel, therefore the scalar product is reduced to the algebraic product.
E 2A = q_{int} / ε₀
where the 2 is due to the surface has two faces
indicate that the surface has a uniform charge for which we can define a surface density
σ = q_{int} / A
q_{int} = σ A
we substitute
E 2A = σ A /ε₀
E = σ / 2ε₀
therefore the total field is
E_ {total} = σ /ε₀
let's substitute the density for the charge of the whole plate
σ= Q / L²
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
An electron is moving through a magnetic field whose magnitude is 83 x 10-5 T. The electron experiences only a magnetic force and has an acceleration of magnitude 34 x 10+13 m/s2. At a certain instant, it has a speed of 72 x 10+5 m/s. Determine the angle (less than 90°) between the electron's velocity and the magnetic field.
Answer:
the angle between the electron's velocity and the magnetic field is 19⁰
Explanation:
Given;
magnitude of the magnetic field, B = 83 x 10⁻⁵ T
acceleration of the electron, a = 34 x 10¹³ m/s²
speed of the electron, v = 72 x 10⁵ m/s
mass of electron, m = 9.11 x 10⁻³¹ kg
The magnetic force experienced by the electron is calculated as;
F = ma = qvB sinθ
where;
q is charge of electron = 1.602 x 10⁻¹⁹ C
θ is the angle between the electron's velocity and the magnetic field.
[tex]sin(\theta ) = \frac{ma}{qvB} \\\\sin(\theta ) = \frac{(9.11\times 10^{-31})(34\times 10^{13})}{(1.602\times 10^{-19})\times (72\times 10^5) \times (83 \times 10^{-5})} \\\\sin(\theta ) = 0.3235\\\\\theta =sin^{-1}(0.3235)\\\\\theta =18.9^0[/tex]
[tex]\theta \approx 19^ 0[/tex]
Therefore, the angle between the electron's velocity and the magnetic field is 19⁰
If a body travels 6km in 30 minutes in a fixed direction, calculate it's velocity.
Plz show me the process too.
We know
[tex]\boxed{\large{\sf Velocity=\dfrac{Distance}{Time}}}[/tex]
[tex]\\ \Large\sf\longmapsto Velocity=\dfrac{6}{\dfrac{1}{2}}[/tex]
[tex]\\ \Large\sf\longmapsto Velocity=6\times 2[/tex]
[tex]\\ \Large\sf\longmapsto Velocity=12km/h[/tex]
Is this the right answer??
We should keep km and min in smallest SI unit
A single force acts on a particle situated on the positive x axis. The torque about the origin is in the negative z direction. The force might be:_______.
A. in the positive y direction
B. in the negative y direction
C. in the positive x direction
D. in the negative x direction
A single force acts on a particle situated on the positive x axis. The torque about the origin is in the negative z direction. The force might be in the negative y direction. Thus, option B is correct.
To determine the force that could generate a torque in the negative z direction, we need to consider the right-hand rule for cross products. The torque vector, denoted by τ, is given by the cross product of the position vector, r, and the force vector, F:
[tex]τ = r × F[/tex]
In this case, the position vector, r, points along the positive x-axis. The negative z-direction torque indicates that the force vector must be perpendicular to both the position vector and the negative z-axis.
Using the right-hand rule, we can determine that the force vector must be in the negative y-direction, which is option B.
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During the data transmission there are chances that the data bits in the frame might get corrupted. This will require the sender to re-transmit the frame and hence it will increase the re-transmission overhead. By considering the scenarios given below, you have to choose whether the packets should be encapsulated in a single frame or multiple frames in order to minimize the re-transmission overhead.
Justify your answer with one valid reason for both the scenarios given below.
Scenario A: Suppose you are using a network which is very prone to errors.
Scenario B: Suppose you are using a network with high reliability and accuracy.
1. Based on Scenario A, multiple frames will minimize re-transmission overhead and should be preferred in the encapsulation of packets.
2. Based on Scenario B, the encapsulation of packets should be in a single frame because of the high level of network reliability and accuracy.
Justification:
There will not be further need to re-transmit the packets in a highly reliable and accurate network environment, unlike in an environment that is very prone to errors. The reliable and accurate network environment makes a single frame economically better.
Encapsulation involves the process of wrapping code and data together within a class so that data is protected and access to code is restricted.
With encapsulation, each layer:
provides a service to the layer above itcommunicates with a corresponding receiving nodeThus, in a reliable and accurate network environment, single frames should be used to enhance transmission and minimize re-transmission overhead. This is unlike in an environment that is very prone to errors, where multiple frames should rather be used to minimize re-transmission overhead.
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