Answer:
9.42*10^-4 m^3/s
Explanation:
d=3/100 m
=0.03 m
A=3.14*0.03^2/6
=4.71*10^-4 m^2
Volume flow rate V = A * s
= 4.71*10^-4 * 2
= 9.42*10^-4
So, the volume flw rate of fluid is 9.42*10^-4 m^3/s
A brick of mass M has been placed on a rubber cushion of mass m. Together they are sliding to the right at constant velocity on an ice-covered parking lot. (a) Draw a free-body diagram of the brick and identify each force acting on it. (b) Draw a free-body diagram of the cushion and identify each force acting on it. (c) Identify all of the action–reaction pairs of forces in the brick–cushion–planet system.
A) The free-body diagram of the forces acting on the brick is attached.
B) The free-body diagram of the forces acting on the rubber cushion is attached.
C) The action and reaction forces of the entire brick–cushion–planet system has been enumerated below.
A) The brick has a Mass M placed on top of a rubber cushion of mass m.
This means that there will be a normal force acting acting upwards on the brick and also a gravitational force acting downward. These forces are denoted as;
Normal force of rubber cushion acting on brick = [tex]n_{cb}[/tex]
Gravitational force acting on brick = Mg
Find attached the free body diagram.
B) The forces acting on the cushion will be;
Normal force of parking lot pavement on rubber cushion = [tex]n_{pc}[/tex]
Gravitational force of earth acting on cushion = mg
Force of brick acting on the rubber cushion = [tex]F_{bc}[/tex]
C) The action pairs of forces are;
i) Force; Normal force of rubber cushion acting on brick = [tex]n_{cb}[/tex]
Reaction Force; Force of brick acting on the rubber cushion = [tex]F_{bc}[/tex]
ii) Action Force; Gravitational force acting on brick = Mg
Reaction; Gravitational force of brick acting on the earth
iii) Action Force; Normal force of parking lot pavement on rubber cushion = [tex]n_{pc}[/tex]
Reaction; Force of rubber cushion on parking lot pavement
iv) Action Force; Gravitational force of earth acting on rubber cushion = mg
Reaction Force; Gravitational force of rubber cushion on the earth.
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Astronauts increased in height by an average of approximately 40 mm (about an inch and a half) during the Apollo-Soyuz missions, due to the absence of gravity compressing their spines during their time in space. Does something similar happen here on Earth
Answer:
Yes. Something similar occurs here on Earth.
Explanation:
Gravity tends to pull objects perpendicularly to the ground. In space, the absence of this force means there is no compression on the spine due to gravity trying to pull it down. This means that astronauts undergo an increase in height in space.
Here on Earth, we experience gravity pull on our spine during the day. At night when we sleep, we lie down with our spine parallel to the ground, which means that our spine is no longer under compression from gravity force. The result is that we are a few centimetres taller in the morning when we wake up, than we are before going to bed at night. The increase is not much pronounced here on Earth because there is a repeated cycle of compression and decompression of our spine due to gravity, unlike when compared to that of astronauts that spend long duration in space, all the while without gravity forces on their spine
A long, current-carrying solenoid with an air core has 1800 turns per meter of length and a radius of 0.0165 m. A coil of 210 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system
Answer:
The mutual inductance is [tex]M = 0.000406 \ H[/tex]
Explanation:
From the question we are told that
The number of turns per unit length is [tex]N = 1800[/tex]
The radius is [tex]r = 0.0165 \ m[/tex]
The number of turns of the solenoid is [tex]N_s = 210 \ turns[/tex]
Generally the mutual inductance of the system is mathematically represented as
[tex]M = \mu_o * N * N_s * A[/tex]
Where A is the cross-sectional area of the system which is mathematically represented as
[tex]A = \pi * r^2[/tex]
substituting values
[tex]A = 3.142 * (0.0165)^2[/tex]
[tex]A = 0.0008554 \ m^2[/tex]
also [tex]\mu_o[/tex] is the permeability of free space with the value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
So
[tex]M = 4\pi * 10^{-7} *1800 * 210 * 0.0008554[/tex]
[tex]M = 0.000406 \ H[/tex]
2. A solid plastic cube of side 0.2 m is submerged in a liquid of density 0.8 hgm calculate the
upthrust of the liquid on the cube.
Answer:
vpg = 0.064 N
Explanation:
Upthrust = Volume of fluid displaced
upthrust liquid on the cube g=10ms−2
vpg =0.2 x 0.2 x 0.2 x0.8 x 10= 0.064N
vpg = 0.064 N
hope it helps.
Following a collision between a large spacecraft and an asteroid, a copper disk of radius 28.0 m and thickness 1.20 m, at a temperature of 850°C, is floating in space, rotating about its axis with an angular speed of 20.0 rad/s. As the disk radiates infrared light, its temperature falls to 20.0°C. No external torque acts on the disk.
A) Find the change in kinetic energy of the disk.
B) Find the change in internal energy of the disk.
C) Find the amount of energy it radiates.
Answer:
A. 9.31 x10^10J
B. -8.47x10 ^ 12J
C. 8.38x 10^12J
Explanation:
See attached file pls
Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where low-frequency sound is originating from.
(a) Suppose a low-frequency sound source is placed to the right of a person, whose ears are approximately 20 cm apart, and the speed of sound generated is 340 m/s. How long (in s) is the interval between when the sound arrives at the right ear and the sound arrives at the left ear?
(b) Assume the same person was scuba diving and a low-frequency sound source was to the right of the scuba diver. How long (in ) is the interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s? S
(c) What is significant about the time interval of the two situations?
Answer:
(a) 0.59 ms
(b) 0.15 ms
(c) The significance is that the speed of sound in different media determines the time interval of perception by the ears, which are at constant distance apart.
Explanation:
(a) distance between ears = 20 cm = 0.2 m
speed of sound generated = 340 m/s
time = ?
speed = [tex]\frac{distance covered}{time taken}[/tex]
⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]
= [tex]\frac{0.2}{340}[/tex]
= 5.8824 × [tex]10^{-4}[/tex]
= 0.59 ms
The time interval of the arrival of the sound at the right ear to the left ear is 0.59 ms.
(b) distance between ears = 20 cm = 0.2 m
speed of sound in water = 1530 m/s
time = ?
speed = [tex]\frac{distance covered}{time taken}[/tex]
⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]
= [tex]\frac{0.2}{1530}[/tex]
= 1.4815 × [tex]10^{-4}[/tex]
= 0.15 ms
The sound heard by the right ear of the diver would arrive at the left 0.15 ms latter.
(c) The significance is that the speed of sound in different media, determines the time interval of perception by the ears, which are at constant distance apart.
A) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear is; t = 0.588 × 10⁻³ seconds
B) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s is; t = 0.131 × 10⁻⁵ seconds
C) The significance about the time interval of the two situations is that;
Transmission of sound varies with different mediums.
A) We are given;
Distance between the two ears; d = 20 cm = 0.2 m
Speed of sound; v = 340 m/s
Since the sound source is placed at the right ear, the time interval for it to get to the left ear is;
t = d/v
t = 0.2/340
t = 0.588 × 10⁻³ seconds
B) We are now told that the speed of sound in water is 1530 m/s. Thus;
t = 0.2/1530
t = 0.131 × 10⁻⁵ seconds
C) We can see that in answer A and B, the time interval is different even when the distance remained the same. This means that, the time interval of hearing a sound changes with respect to the medium of transmission.
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At the first minimum adjacent to the central maximum of a single-slit diffraction pattern the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit is:
Answer:
Explanation:
The whole wave front may be divided into two halves , the upper half and the lower half . Waves coming from top of the slit or top of upper half and top of lower half or from the mid point of slit can form minima at given point only when there is phase difference of π radian or path difference of λ or one wavelength. Every other point in upper half and corresponding point in lower half will interfere destructively at that point and will form dark spot at the given point . In this way minima will be formed at that point
Hence the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit at first minima is π radian .
Tom is climbing a 3.0-m-long ladder that leans against a vertical wall, contacting the wall 2.5 m above the ground. His weight of 680 N is a vector pointing vertically downward. (Weight is measured in newtons, abbreviated N).
A) What is the magnitude of the component of Tom's weight parallel to the ladder?
B) What is the magnitude of the component of Tom's weight perpendicular to the ladder?
Answer: A) [tex]P_{x}[/tex] = 564.4 N
B) [tex]P_{y}[/tex] = 374 N
Explanation: The ladder forms with the wall a right triangle, with one unknown side. To find it, use Pythagorean Theorem:
[tex]hypotenuse^{2} = side^{2} + side^{2}[/tex]
[tex]side = \sqrt{hypotenuse^{2} - side^{2}}[/tex]
side = [tex]\sqrt{3^{2} - 2.5^{2}}[/tex]
side = 1.65
Tom's weight is a vector pointing downwards. Since he is at an angle to the floor, the gravitational force has two components: one that is parallel to the floor ([tex]P_{x}[/tex]) and othe that is perpendicular ([tex]P_{y}[/tex]). These two vectors and weight, which is gravitational force, forms a right triangle with the same angle the ladder creates with the floor.
The image in the attachment illustrates the described above.
A) [tex]P_{x}[/tex] = P sen θ
[tex]P_{x} = P.\frac{oppositeside}{hypotenuse}[/tex]
[tex]P_{x}[/tex] = 680.[tex]\frac{1.65}{3}[/tex]
[tex]P_{x}[/tex] = 564.4 N
B) [tex]P_{y}[/tex] = P cos θ
[tex]P_{y} = P.\frac{adjacentside}{hypotenuse}[/tex]
[tex]P_{y}[/tex] = 680. [tex]\frac{1.65}{3}[/tex]
[tex]P_{y}[/tex] = 374 N
Consider a long rod of mass, m, and length, l, which is thin enough that its width can be ignored compared to its length. The rod is connected at its end to frictionless pivot.
a) Find the angular frequency of small oscillations, w, for this physical pendulum.
b) Suppose at t=0 it pointing down (0 = 0) and has an angular velocity of 120 (that is '(t = 0) = 20) Note that 20 and w both have dimensions of time-1. Find an expression for maximum angular displacement for the pendulum during its oscillation (i.e. the amplitude of the oscillation) in terms of 20 and w assuming that the angular displacement is small.
Answer:
Explanation:
The rod will act as pendulum for small oscillation .
Time period of oscillation
[tex]T=2\pi\sqrt{\frac{l}{g} }[/tex]
angular frequency ω = 2π / T
= [tex]\omega=\sqrt{\frac{g}{l} }[/tex]
b )
ω = 20( given )
velocity = ω r = ω l
Let the maximum angular displacement in terms of degree be θ .
1/2 m v ² = mgl ( 1 - cosθ ) ,
[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]
.5 ( ω l )² = gl( 1 - cos θ )
.5 ω² l = g ( 1 - cosθ )
1 - cosθ = .5 ω² l /g
cosθ = 1 - .5 ω² l /g
θ can be calculated , if value of l is given .
A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 390 kV. The secondary of this transformer is being replaced so that its output can be 515 kV for more efficient cross-country transmission on upgraded transmission lines. (a) What is the ratio of turns in the new secondary compared with turns in the old secondary
Answer:
1.32 is the turns ratio
Explanation:
Note that The transformer steps up the voltage from 12000 V to 390000V
12000 V is the primary and in the secondary it is 390000 V in old transformer
If n₁ be no of turns in primary coil and n₂ be no of turns in secondary coils
the formula is
n₂ / n₁ = voltage in secondry / voltage in primary
n₂ / n₁ = 390000 / 12000
ratio of turns in old transformer is 32.5
ratio of turns in new transformer
n₃ / n₁ = 515 / 12 ( n₃ is no of turns in the secondary of new transformer )
= 42.9
T he ratio of turns in the new secondary compared with the old secondary
n₃ / n₂ = 42.9 / 32.5
= 1.32
A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the satellite). The mass of the satellite is 200 kg.
About how much energy (in kJ) was converted to heat?
Answer:
The energy coverted to heat is 200 kilojoules.
Explanation:
GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:
[tex]m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v[/tex]
Where:
[tex]m_{M}[/tex], [tex]m_{S}[/tex] - Masses of the small meteorite and the communication satellite, measured in kilograms.
[tex]v_{M}[/tex], [tex]v_{S}[/tex] - Speeds of the small meteorite and the communication satellite, measured in meters per second.
[tex]v[/tex] - Final speed of the satellite-meteorite system, measured in meters per second.
The final speed of the satellite-meteorite system is cleared:
[tex]v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}[/tex]
If [tex]m_{M} = 1\times 10^{-3}\,kg[/tex], [tex]m_{S} = 200\,kg[/tex], [tex]v_{M} = 20000\,\frac{m}{s}[/tex] and [tex]v_{S} = 0\,\frac{m}{s}[/tex], the final speed is now calculated:
[tex]v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}[/tex]
[tex]v = 0.1\,\frac{m}{s}[/tex]
Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:
[tex]K_{S} + K_{M} - K - Q_{disp} = 0[/tex]
[tex]Q_{disp} = K_{S}+K_{M}-K[/tex]
Where:
[tex]K_{S}[/tex], [tex]K_{M}[/tex] - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.
[tex]K[/tex] - Kinetic energy of the satellite-meteorite system, measured in joules.
[tex]Q_{disp}[/tex] - Dissipated heat, measured in joules.
The previous expression is expanded by using the definition for the translational kinetic energy:
[tex]Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}][/tex]
Given that [tex]m_{M} = 1\times 10^{-3}\,kg[/tex], [tex]m_{S} = 200\,kg[/tex], [tex]v_{M} = 20000\,\frac{m}{s}[/tex], [tex]v_{S} = 0\,\frac{m}{s}[/tex] and [tex]v = 0.1\,\frac{m}{s}[/tex], the dissipated heat is:
[tex]Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right][/tex][tex]Q_{disp} = 200000\,J[/tex]
[tex]Q_{disp} = 200\,kJ[/tex]
The energy coverted to heat is 200 kilojoules.
a car brakes and stops at 10 [m]. While stopping, the friction force of the wheels on the pavement is 400 [N]. Calculate the work done.
Explanation:
Work = force × distance
W = (400 N) (10 m)
W = 4000 J
Bright and dark fringes are seen on a screen when light from a single source reaches two narrow slits a short distance apart. The number of fringes per unit length on the screen can be doubled:______.
a. if the distance between the slits is doubled.
b. if the wavelength is changed to λ = λ/2.
c. if the distance between the slits is quadruple the original distance and the wavelength is changed to λ = 2λ.
d. if any of the above occurs.
e. only if the width of the slits is changed to w = w/2
Answer:
d. if any of the above occurs
Explanation:
That is The number of fringes per unit length on the screen can be doubled if
if the distance between the slits is doubled.
if the wavelength is changed to λ = λ/2.
And if the distance between the slits is quadruple the original distance and the wavelength is changed to λ = 2λ
A box of mass 0.8 kg is placed on an inclined surface that makes an angle 30 above
the horizontal, Figure 1. A constant force of 18 N is applied on the box in a direction 10°
with the horizontal causing the box to accelerate up the incline.
The coefficient of
kinetic friction between the block and the plane is 0.25.
Show the free body diagrams
(a) Calculate the block's
acceleration as it moves up the incline. (6 marks)
(b) If the block slides down at a constant speed, find the value of force applied.
(4 marks)
Answer:
a) a = 17.1 m / s², b) F = 3.04 N
Explanation:
This is an exercise of Newton's second law, in this case the selection of the reference system is very important, we have two possibilities
* a reference system with the horizontal x axis, for this selection the normal and the friction force have x and y components
* a reference system with the x axis parallel to the plane, in this case the weight and the applied force have x and y components
We are going to select the second possibility, since it is the most used in inclined plane problems, let's analyze the angle of the applied force (F) it has an angle 10º with respect to the x axis, if we rotate this axis 30º the new angle is
θ = 10 -30 = -20º
The negative sign indicates that it is in the fourth quadrant. Let's use trigonometry to find the components of the applied force
sin (-20) = F_{y} / F
cos (-20) = Fₓ / F
F_{y} = F sin (-20)
Fₓ = F cos (-20)
F_y = 18 sin (-20) = -6.16 N
Fₓ = 18 cos (-20) = 16.9 N
The decomposition of the weight is the customary
sin 30 = Wₓ / W
cos 30 = W_y / W
Wₓ = W sin 30 = mg sin 30
W_y = W cos 30 = m g cos 30
Wₓ = 0.8 9.8 sin 30 = 3.92 N
W_y = 0.8 9.8 cos 30 = 6.79 N
Notice that in the case the angle is measured with respect to the axis y perpendicular to the plane
Now we can write Newton's second law for each axis
X axis
Fₓ - fr = m a
Y Axis
N - [tex]F_{y}[/tex] - Wy = 0
N =F_{y} + Wy
N = 6.16 + 6.79
They both go to the negative side of the axis and
N = 12.95 N
The friction force has the formula
fr = μ N
we substitute
Fₓ - μ N = m a
a = (Fₓ - μ N) / m
we calculate
a = (16.9 - 0.25 12.95) / 0.8
a = 17.1 m / s²
b) now the block slides down with constant speed, therefore the acceleration is zero
ask for the value of the applied force, we will suppose that with the same angle, that is, only its modulus was reduced
Newton's law for the x axis
Fₓ -fr = 0
Fₓ = fr
F cos 20 = μ N
F = μ N / cos 20
we calculate
F = 0.25 12.95 / cos 20
F = 3.04 N
this is the force applied at an angle of 10º to the horizontal
After a long walk in the 128C outdoors, a person wearing glasses enters a room at 258C and 55 percent relative humidity. Determine whether the glasses will become fogged.
Answer:
Yes the glasses will be fogged
Explanation:
See attacher file
Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
PLZ HURRY WILL MARK BRAINLIEST IF CORRECT
Answer:
Option A
Explanation:
Acceleration will be obviously zero when Force = 0
That is how:
Force = Mass * Acceleration
So, If force = 0
0 = Mass * Acceleration.
Dividing both sides by Mass
Acceleration = 0/Mass
Acceleration = 0 m/s²
Answer:
[tex]\boxed{\mathrm{A. \: It \: will \: be \: 0 \: meters \: per \: second \: per \: second. }}[/tex]
Explanation:
[tex]\mathrm{force=mass \times acceleration}[/tex]
The force is given 0 newtons.
[tex]\mathrm{force=0 \: N}[/tex]
Plug force as 0.
[tex]\mathrm{0=mass \times acceleration}[/tex]
Divide both sides by mass.
[tex]\mathrm{\frac{0}{mass} =acceleration}[/tex]
[tex]\mathrm{0 =acceleration}[/tex]
[tex]\mathrm{acceleration= 0\: m/s/s}[/tex]
Suppose there is a uniform electric field pointing in the positive x-direction with a magnitude of 5.0 V/m. The electric potential is measured to be 50 V at the position x = 10 m. What is the electric potential at other positions?
Position [m] = (−20)--- (0.00) ---(10)--- (11)--- (99)
Electric Potential [V]=
Answer:
Electric potential at position, x = -20 m, = -100 V
Electric potential at position, x = 0 m, = 0
Electric potential at position, x = 10 m, = 50 V
Electric potential at position, x = 11 m, = 55 V
Electric potential at position, x = 99 m, 495 V
Explanation:
Given;
magnitude of electric field, E = 5.0 V/m
at position x = 10 m, electric potential = 50 V
Electric potential at position, x = -20 m
V = Ex
V = 5 (-20)
V = -100 V
Electric potential at position, x = 0 m
V = Ex
V = 5(0)
V = 0
Electric potential at position, x = 10 m
V = Ex
V = 5(10)
V = 50 V
Electric potential at position, x = 11 m
V = Ex
V = 5(11)
V = 55 V
Electric potential at position, x = 99 m
V = Ex
V = 5(99)
V = 495 V
How would the interference pattern produced by a diffraction grating change if the laser light changed from red to blue?
Answer
fringe separation l distance between maxima decreases
Explanation:
Because the wavelength of blue light is smaller than that if red light
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string
Answer:
The time interval is [tex]t = 5.48 *10^{-3} \ s[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]l = 3.00 \ m[/tex]
The mass of the string is [tex]m = 5.00 \ g = 5.0 *10^{-3}\ kg[/tex]
The tension on the string is [tex]T = 500 \ N[/tex]
The velocity of the pulse is mathematically represented as
[tex]v = \sqrt{ \frac{T}{\mu } }[/tex]
Where [tex]\mu[/tex] is the linear density which is mathematically evaluated as
[tex]\mu = \frac{m}{l}[/tex]
substituting values
[tex]\mu = \frac{5.0 *10^{-3}}{3}[/tex]
[tex]\mu = 1.67 *10^{-3} \ kg /m[/tex]
Thus
[tex]v = \sqrt{\frac{500}{1.67 *10^{-3}} }[/tex]
[tex]v = 547.7 m/s[/tex]
The time taken is evaluated as
[tex]t = \frac{d}{v}[/tex]
substituting values
[tex]t = \frac{3}{547.7}[/tex]
[tex]t = 5.48 *10^{-3} \ s[/tex]
At what frequency f, in hertz, would you have to move the comb up and down to produce red light, of wavelength 600 nm
Answer:
f = 500 x 10^12Hz
Explanation:
E=hc/wavelength
E=hf
hc/wavelength =hf
c/wavelength =f
f = 3 x 10^8 / 600 x 10^-9 = 500 x 10^12Hz
Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in its secondary coil, and an input voltage of 120 V. Randomized Variables Δ 33%
Part (a) What is the voltage output Vs, in volts, of the transformer used for to charge the batteries? Grade Summar Deductions Potential sin tan) ( Submissions Attempts remain coso cotan) asin) acos() atan acotan)sinh( cosh)tanhcotanh0 % per attempt detailed view END Degrees Radians DEL CLEAR Submit Hint I give up! Hints:% deduction per hint. Hints remaining:I Feedback: 1% deduction per feedback. - 쇼 33%
Part (b) what input current ,. İn milliamps, is required to produce a 3.2 A output current? 33%
Part (c) What is the power input, in watts?
Answer:
a) 0.72 V
b) 19.2 mA
c) 2.304 Watts
Explanation:
A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.
number of primary turns = [tex]N_{p}[/tex] = 500 turns
input voltage = [tex]V_{p}[/tex] = 120 V
number of secondary turns = [tex]N_{s}[/tex] = 3 turns
output voltage = [tex]V_{s}[/tex] = ?
using the equation for a transformer
[tex]\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }[/tex]
substituting values, we have
[tex]\frac{V_{s} }{120 } = \frac{3 }{500} }[/tex]
[tex]500V_{p} = 120*3\\500V_{p} = 360[/tex]
[tex]V_{p}[/tex] = 360/500 = 0.72 V
b) by law of energy conservation,
[tex]I_{P}V_{p} = I_{s}V_{s}[/tex]
where
[tex]I_{p}[/tex] = input current = ?
[tex]I_{s}[/tex] = output voltage = 3.2 A
[tex]V_{s}[/tex] = output voltage = 0.72 V
[tex]V_{p}[/tex] = input voltage = 120 V
substituting values, we have
120[tex]I_{p}[/tex] = 3.2 x 0.72
120[tex]I_{p}[/tex] = 2.304
[tex]I_{p}[/tex] = 2.304/120 = 0.0192 A
= 19.2 mA
c) power input = [tex]I_{p} V_{p}[/tex]
==> 0.0192 x 120 = 2.304 Watts
Exercise 1 - Questions 1. Hold the grating several inches from your face, at an angle. Look at the grating that you will be using. Record what details you see at the grating surface. 0 Words 2. Hold the diffraction grating up to your eye and look through it. Record what you see. Be specific. 0 Words 3. Before mounting the diffraction grating, look through the opening that you made for your grating. Record what you see across the back of your spectroscope.
Answer:
1) on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.
2)If the angle is zero we see a bright light called undispersed light
For different angles we see the colors of the spectrum
3) must be able to see the well-collimated light emission source
Explanation:
1) A diffraction grating (diffraction grating) is a surface on which a series of indentations are drawn evenly spaced.
These crevices or lines are formed by copying a standard metal net when the plastic is melted and after hardening is carefully removed, or if the nets used are a copy of the master net.
The network can be of two types of transmission or reflection, in teaching work the most common is the transmission network, on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.
The number of lines per linear mm determines which range of the spectrum a common value can be observed to observe the range of viable light is 600 and 1200 lines per mm.
2) when looking through the diffraction grating what we can observe depends on the relative angle between the eye and the normal to the network.
If the angle is zero we see a bright light called undispersed light
For different angles we see the colors of the spectrum, if it is an incandescent lamp we see a continuum with all the colors in the visible range and if it is a gas lamp we see the characteristic emission lines of the gas.
3) Before mounting the grid on the spectrometer, we must be able to see the well-collimated light emission source, this means that it is clearly observed.
The spectrometers have several screws to be able to see the lamp clearly, this is of fundamental importance in optical experiments.
A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its location at time t = 3.01.
Answer:
New location at time 3.01 is given by: (7.49, 2.11)
Explanation:
Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:
[tex]V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11[/tex]
With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:
[tex]distance=v\,*\, t[/tex]
[tex]distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11[/tex]
Therefore, adding these displacements in component form to the original particle's position, we get:
New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)
Consider an electromagnetic wave where the electric field of an electromagnetic wave is oscillating along the z-axis and the magnetic field is oscillating along the x-axis.
Required:
In what directions is it possible that the wave is traveling?
Answer:
The wave is traveling in the y axis direction
Explanation:
Because the wave will always travel in a direction 90° to the magnetic and electric components
A light bulb is completely immersed in water. Light travels out in all directions from the bulb, but only some light escapes the water surface. What happens to the fraction (f) of light that escapes the water's surface as the bulb is moved deeper into the water?
Answer:
The fraction of light that escapes the water surface as the water moves deeper into the water will decrease.
Explanation:
The speed of light in water is small compared to the speed of light in air, and a larger part of the light energy is absorbed in water than in air. When the bulb is immersed in water, some of the light energy is absorbed by the mass of water. When the light bulb is further moved deeper into the water, the fraction of light that escapes decreases, because more mass of water is made available to absorb more of the light energy from the bulb.
A dust particle on a phonograph record rotates at a speed of 45 revolutions per minute if the particle is 10 cm from the axis of rotation. Find. 1) its linear speed and linear acceleration.
Explanation:
ω = 45 rev/min × (2π rad/rev) × (1 min / 60 s) = 4.71 rad/s
r = 10 cm = 0.10 m
1) The linear speed is:
v = ωr
v = (4.71 rad/s) (0.10 m)
v = 0.471 m/s
2) The linear acceleration in the tangential direction is 0.
The linear acceleration in the radial direction is:
a = v² / r
a = (0.471 m/s)² / (0.10 m)
a = 2.22 m/s²
Jane is collecting data for a ball rolling down a hill. she measure out a set of different distances and then proceeds to use a stopwatch to find the time it takes the ball to roll each distance
Answer:
The Independent variable in this experiment is the time taken by the ball to roll down each distance.
The dependent variable is the distance through which the ball rolls
The control variables are: slope of hill, weight, of the ball, size of ball, wind speed, surface characteristics of the ball.
Explanation:
The complete question is
Jane is collecting data for a ball rolling down a hill. She measures out a set of different distances and then proceeds to use a stop watch to find the time it takes the ball to roll. What are the independent, dependent, and control variables in this experiment?
Independent variable have their values not dependent on any other variable in the scope of the experiment. The time for the ball to roll down the hill is not dependent on any other variable in the experiment. Naturally, some common independent variables are time, space, density, mass, fluid flow rate.
A dependent variable has its value dependent on the independent variable in the experiment. The value of the distance the ball rolls depends on the time it takes to roll down the hill.
The relationship between the dependent and independent variables in an experiment is given as
y = f(x)
where y is the output or the dependent variable,
and x is the independent variable.
Control variables are those variable that if not held constant could greatly affect the results of an experiment. For an experiment to be more accurate, control variables should be confined to a given set of value throughout the experiment.
Two buses are moving in opposite directions with velocities of 36 km/hr and 108
km/hr. Find the distance between them after 20 minutes.
Explanation:
It is given that,
Speed of bus 1 is 36 km/h and speed of bus 2 is 108 km/h. We need to find the distance between bus 1 and 2 after 20 minutes.
Time = 20 minutes = [tex]\dfrac{20}{60}\ h=\dfrac{1}{3}\ h[/tex]
As the buses are moving in opposite direction, then the concept of relative velocity is used. So,
Distance, [tex]d=v\times t[/tex]
v is relative velocity, v = 108 + 36 = 144 km/h
So,
[tex]d=144\ km/h \times \dfrac{1}{3}\ h\\\\d=48\ km[/tex]
So, the distance between them is 48 km after 20 minutes.
Describe the orientation of magnetic field lines by drawing a bar magnet, labeling the poles, and drawing several lines indicating the direction of the forces.
Answer:
A field is a way of mapping forces surrounding any object that can act on another object at a distance without apparent physical connection. The field represents the object generating it. Gravitational fields map gravitational forces, electric fields map electrical forces, and magnetic fields map magnetic forces.
Explanation:
A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is the velocity of the exiting water? Ignore all orificelosses.
Answer:
4.75 m/s
Explanation:
The computation of the velocity of the existing water is shown below:
Data provided in the question
Tall = 2 m
Inside diameter tank = 2m
Hole opened = 10 cm
Bottom of the tank = 0.75 m
Based on the above information, first we have to determine the height which is
= 2 - 0.75 - 0.10
= 2 - 0.85
= 1.15 m
We assume the following things
1. Compressible flow
2. Stream line followed
Now applied the Bernoulli equation to section 1 and 2
So we get
[tex]\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2[/tex]
where,
P_1 = P_2 = hydrostatic
z_1 = 0
z_2 = h
Now
[tex]\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\ \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}[/tex]
= 4.7476 m/sec
= 4.75 m/s