Answer:
a) 0 V
b) 10 turns
c) 4000 turns
d) 12.5 A
e) 400 W
f) 0.5 A
g) 95.4%
Explanation:
A
0
B
To solve this, we would be using the simple relationship between voltage and number of turns
V1/V2 = N1/N2
500/25 = 200/N2
20 = 200/N2
N2 = 200/20
N2 = 10 turns
C
Here also, we would be using the relationship between current and the number of turns
I1/I2 = N2/N1
500/25 = N2/20
20 = N2/20
N2 = 20 * 20
N2 = 4000 turns
D
Like in the previous question, current and the number of turn relationship is used
N1/N2 = I2/I1
400/80 = I2/2.5
5 = I2/2.5
I2 = 5 * 2.5
I2 = 12.5 A
E
The power remains unchanged at 400 W
F
Power = Voltage * Current
P = VI
I = P/V
I = 60/120
I = 0.5 A
G
95.4%
The transformer is a device used to step up or step down voltage.
Part A;
Given that;
Es/Ep = Ns/Np
Es = voltage in the secondary coil
Ep = voltage in primary coil
Ns = Number of turns in secondary coil
Np = Number of coils in primary coil
Es = Ns/Np × Ep
Es = 200/100 × 1.5 V
Es = 3 V
Part B
Ns = Es/Ep × Np
Ns = 25/500 × 200
Ns = 10 turns
Part C
Ns/Np = Ip/Is
Ns = Ip/Is × Np
Ns = 500/25 × 200
Ns = 4000 turns
Part D
Ns/Np = Ip/Is
NsIs = NpIp
Is = NpIp/Ns
Is = 400 × 2.5/80
Is =12.5 A
Part E
The power in the primary coil is the same as the power in the secondary coil. The power in the secondary coil is 400 watts.
Part F
Power supplied = 60 watts
Voltage of primary coil = 120 V
Since;
P = IV
I = P/V = 60/120 = 0.5 A
Part G
Since;
E = 100Pout/Pin
Pin = 120 V × 2 A = 240 W
Pout = 19.4 V × 11.8 A = 228.92 W
E = 100(228.92/240)
E = 95.4%
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A heat pump has a coefficient of performance of 3.85 and operates with a power consumption of 7020 W. How much energy does it deliver into a home during 1 h of continuous operation?
Answer:
97.3 MJ
Explanation:
The formula for the coefficient of Perfomance is given as
COE = Q/W, where
COE is the coefficient of Perfomance
Q is the heat provided
W serves as the work input.
Dividing both sides of the equation by a factor of time t, we get the coefficient of Perfomance in terms of heating power and input power, so we say
COE = P / P(i),
making heating power, P the subject of formula, we have
P = COE * P(i)
P = 3.85 * 7020 * 1 * 3600
P = 97297200 J
P = 97.3 MJ
Help me please,
A ball is thrown straight up in the air. What is the velocity and acceleration at the top of the path?
A) v 0m/s, = 0m/s/s
B) v = 0m/s, a 10m/s/s
C) v = 10m/s, a 10m/s/s
D) v = 10m/s, a = 0m/s/s
E) None of the above
Option B
Explanation:
no distance was given only the acceleration due to the fact that it went up (10m/s/s)
s0 it is
0 m/s and 10m/s/s (option B)
What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump?
COP =
Complete Question
A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.
What is the refrigerator's coefficient of performance? COP
(b) What If? What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump? COP
Answer:
a
[tex]COP = 8.49[/tex]
b
[tex]COP_1 = 9.49[/tex]
Explanation:
From the question we are told that
The lower operation temperature of refrigerator is [tex]T_1 = -8.00^oC = 265 \ K[/tex]
The upper operation temperature of the refrigerator is [tex]T_2 = 23.2 ^oC = 296.2 \ K[/tex]
Generally the refrigerators coefficient of performance is mathematically represented as
[tex]COP = \frac{T_1}{T_2 - T_1 }[/tex]
=> [tex]COP = \frac{265}{296.2 - 265 }[/tex]
=> [tex]COP = 8.49[/tex]
Generally if a refrigerator (operating between the same temperatures) was instead used as a heat pump , the coefficient of performance is mathematically represented as
[tex]COP_1 = \frac{T_2}{ T_2 - T_1}[/tex]
=> [tex]COP_1 = \frac{296.2}{ 296.2 - 265 }[/tex]
=> [tex]COP_1 = 9.49[/tex]
The coefficient of performance if the refrigerator is used as a heat pump is 9.5.
The given parameters;
initial temperature, T₁ = -8 ⁰C = -8 + 273 = 265 Kfinal temperature, T₂ = 23.2°C = 23.2°C + 273 = 296.2 KThe coefficient of performance if the refrigerator is used as a heat pump is calculated as follows;
[tex]COP = \frac{T_2}{T_2 - T_1}\\\\COP = \frac{296.2}{296.2 - 265} \\\\COP =9.5[/tex]
Thus, the coefficient of performance if the refrigerator is used as a heat pump is 9.5.
"Your question is not complete, it seems to be missing the following information";
A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.
What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump?
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A plumber applies a torque of 408 to a bolt using a wrench. If he moves his hand twice as far
away on the wrench's handle from the bolt (doubling the radius), but applies the same force
as before, how much torque will he now apply?
Answer:
816
Explanation:
We must remember that torque is defined as the product of a force by a distance.
This distance is measured from the pivot point of the Bolt to the point where the force is applied. in this way, we have the following equation to be able to determine the torque.
[tex]T=F*d[/tex]
Now if we double the turning distance we have the torque should also be double.
[tex]T =F*2*d[/tex]
Lab - Wave Properties in a Spring
11-05
The wave characteristics you will observe in this lab are common to all waves (water, light, sound,
etc.). Use your prior knowledge and the book to fill in the following blanks, then go in the hall and
perform the lab.
A wave is a disturbance that moves through (propagates) through empty space or through a
_____________. There are two types of waves. A _____________________ wave requires
matter to travel. List some examples of this type:
A _____________________ wave does not require a medium. Examples include:
In order to start and transmit a mechanical wave, a source of _____________ and an
_______________ medium are required. A single disturbance is referred to as a
_______________, and a series of disturbances is a wave __________.
The questions in bold are those you should observe directly. Others will be answered using the book.
A. TYPES OF MECHANICAL WAVES: In the hall, stretch the slinky on the floor until it is
stretched (but still loose). Practice sending single pulses down the slinky by popping your wrist
from the center to the side and back to the center. Then send a continuous wave train along as
your partner holds the other end still. A piece of ribbon should be tied to one coil. Watch the
motion of this ribbon (representing a particle) as the wave travels through the spring.
In this type of wave, the particles move (perpendicular, parallel)
to the direction the wave travels. This type of wave is called a __________________ wave.
Its pulses are called ________________ and ________________.
Now send a pulse by quickly pushing the spring forward and pulling
it back, as shown. This type of wave is called _______________. Watch the motion of the ribbon.
In this type, the particles move _____________ to the direction the wave travels. Its pulses
are called _____________ and _____________. Label each.
Note that all waves transfer _____________ without transferring _______________. In
mechanical waves, particles of the medium vibrate back and forth in simple harmonic motion while
the disturbance (or _____________) moves from one place to another.
B. WAVE SPEED
Send a large pulse, followed by a small one. Does one pulse catch up to the other? ______
(Hint: The person who sends these waves should watch how the waves look when they return. Make
sure that both pulses are large enough initially to make it back to the sender!) The size of the
pulse is called the __________________ of the wave. Did the size affect the speed? ______
Generate a single transverse pulse in the slinky, keeping the stretch constant. Using a stopwatch,
time the journey of the pulse from one end to the other and back again. Take the average of
several trials. _________
Without changing your positions on the floor (therefore keeping the _____________ the pulse
travels the same), pull the slinky tighter using only about 3/4 of the coils. This makes a completely
different medium through which the pulse will travel. Time the journey as before. ___________
Does the kind of medium affect the speed of the pulse? ___________
Lab – Wave Properties in a Spring ____________________
PHYSICSFundamentals
© 2004, GPB
11-06
C. WAVELENGTH AND FREQUENCY
Shake the slinky back and forth steadily to send a
transverse wave train while your partner holds the other end still. On the diagram, label wavelength
(- Greek letter lambda). The frequency of the wave depends on how fast you shake the slinky.
Shake it regularly but slowly, then regularly but rapidly.
Higher frequency waves are generated by shaking the spring (slowly, rapidly). High frequency
waves have (short, long) wavelengths, and low frequency waves have __________.
The speed of a wave in any medium is equal to the _______________ of the wave X
________________. This wave equation ___________________ shows that f and are
______________ proportional. Write the units for each of the variables in this equation.
The exercise involves filling in the gaps with the possible wave
properties that can be obtained from a spring.
How is the Wave Properties in a Spring Lab exercise correctly completed?The correctly completed exercise is presented as follows;
A wave is a disturbance that moves through a medium. There are two
types of waves. A mechanical wave requires matter to travel. List some
examples of this type: sound wave, water wave, spring waves.
A electromagnetic wave does not require a medium. Examples include: Light waves
In order to start and transmit a mechanical wave, a source of
disturbance and a physical medium are required. A single disturbance is
referred to as a pulse, and a series of disturbance is a wave train.
This type of wave is called transverse wave. Its pulses are called crest
and troughs.
Now send a pulse by quickly pushing the spring forward and pulling it
back, as shown. This type of wave is called longitudinal wave. Watch the
motion of the ribbon. In this type, the particles move parallel to the
direction the wave travels. Its pulses are called compression and
rarefactions. Note that all waves transfer energy without transferring
matter. In mechanical waves, particle of the medium vibrate back and
forth in simple harmonic motion while the disturbance (or energy)
moves from one place to another.
B. Wave speed
Does the pulse catch up to the other? yes. The size of the pulse is called
the amplitude of the wave.
Did the size of the pulse affect the speed? No.
The average time wave it takes the wave to travel
Without changing your positions therefore keeping the distance the
pulse travels the same), pull the slinky tighter using only about 3/4 of
coils. This makes a completely different medium through which the
pulse will travel. Time the journey as before time record. Does the kind
of medium affect the speed of the pulse? Yes
C. Wavelength and Frequency
High frequency waves have short wavelengths and low frequency waves
have long wavelengths.
The speed of a wave in any medium is equal to the frequency of the wave × the wavelength. This wave equation [tex]\underline{f = \dfrac{v}{\lambda } }[/tex] shows that f and λ are
inversely proportional. The units of the variables are;
Units of the frequency, f is hertz unit HzUnits of the velocity, v, is m/sUnits of the wavelength, λ, is meters (m)Learn more about waves here:
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A 1.2-kg object moving with a speed of 8.0 m/s collides perpendicularly with a wall and emerges with a speed of 6.0 m/s in the opposite direction. If the object is in contact with the wall for 2.0 ms, what is the magnitude of the average force on the object by the wall?
a. 9.8 kN.
b. 8.4 kN.
c. 7.7 kN.
d. 9.1 kN.
e. 1.2 kN.
Given that,
Mass of the object, m = 1.2 kg
Initial speed of the object, u = 8 m/s
Final speed of the object, v = -6 m/s (in opposite direction)
Time, t = 2 ms
To find,
The average force on the object by the wall.
Solution,
Let F be the force. Using Newton's second law of motion,
F = ma, a is acceleration
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{1.2\times ((-6)-8)}{2\times 10^{-3}}\\\\=8400\ N[/tex]
or
F = 8.4 N
So, the magnitude of average force in the object by the wall is 8.4 N.
A carousel at the local carnival rotates once every 45 seconds.
(a) What is the linear speed of an outer horse on the carousel, which is 2.75 m from the axis of rotation?
(b) What is the linear speed of an inner horse that is 1.75 m from the axis of rotation?
Answer:
We know that the carousel does a complete rotation in 45 seconds.
Then the frequency of this carousel will be f = 1/45 seconds.
And the angular frequency will be 2*pi times the frequency, then we have:
angular frequency = w = 2*3.14*(1/45s) = 0.1396 s^-1
Now, the linear speed of an object that rotates with a radius R, and an angular frequency W is:
S = R*W
then:
a) in this case the radius is 2.75m, then the linear speed is:
S = 2.75m*0.1396 s^-1 = 0.3839 m/s
b) in this case the radius is 1.75m, then the linear speed here is:
S = 1.75m*0.1396 s^-1 = 0.2443 m/s
(a) The linear speed of an outer horse on the carousel is 0.384 m/s.
(b) The linear speed of an inner horse on the carousel is 0.244 m/s.
Given data:
The time interval for the rotation of carousel is, t = 45 s.
The distance of the outer horse from the axis of rotation is, r = 2.75 m.
The distance of an inner horse from the axis of rotation is, r' = 1.75 m.
(a)
The linear speed in this problem can be obtained from the concept of rotational mechanic, in which the ratio of the circumference and the time gives required linear speed. So,
v = 2 π r/t
Solving as,
v = 2 π (2.75) / 45
v = 0.384 m/s
Thus, we can conclude that the linear speed of an outer horse on the carousel is 0.384 m/s.
(b)
Now similarly the linear speed of an inner horse is calculated as,
v' = 2 π r' / t
Solving as,
v' = 2 π (1.75) / 45
v' = 0.244 m/s
Thus, we can conclude that the linear speed of an outer horse on the carousel is 0.244 m/s.
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A block of mass, m, sits on the ground. A student pulls up on
the block with a tension, T, but the block remains in contact
with the ground. What is the normal force on the block?
Answer a
Explanation: a
A bicycle has a momentum of 36 kg* m/s and a very!I city of 4 m/s.What is the mass of the bicycle?
p = 36 kgm/s
v = 4m/s
we know that,
p = mv
so,
[tex]m = \frac{p}{v} [/tex]
[tex]m = \frac{36}{4} [/tex]
[tex]m = 9kg[/tex]
What happens to the molecules of water when it moves from a liquid to a gas?
A. Water molecules condense and move slower.
B. Water molecules spread out and move slower.
C. Water molecules spread out and move faster.
D. Water molecules condense and move faster.
its A or D but im not sure which one ik it moves fast
Two spheres, 1.00 kg each, whose centers are 2.00 m apart, would have what gravitational force between them? A. 3.14 X 10-17 N
B. 1.67 X 10-11 N
C. 8.17 X 10-6N
D. 5.78 X 10-6 N
Answer: B
Explanation: the teacher just told us the answer
The gravitational force between the two spheres is [tex]1.67 \times 10^{-11} \ N[/tex].
The given parameters;
mass of each sphere, m = 1.00 kgdistance between their center mass, r = 2 mThe gravitational force between the two spheres is determined by applying Newton's law of universal gravitation as shown below;
[tex]F = \frac{Gm_1 m_2 }{r^2} \\\\[/tex]
where;
G is universal gravitation constant = 6.67 x 10⁻¹¹ N/m[tex]F = \frac{(6.67\times 10^{-11})\times (1\times 1)}{2^2} \\\\F = 1.67 \times 10^{-11} \ N[/tex]
Thus, the gravitational force between the two spheres is [tex]1.67 \times 10^{-11} \ N[/tex].
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Calculate the mechanical advantage of a ramp if the box you are trying to move has a mass of 10 kilograms, the
board is 15 feet long and the height of the ramp is 5 feet .
3
300
150
45
Answer:
[tex]MA = 3[/tex]
Explanation:
Given
[tex]Box = 10kg[/tex]
[tex]Ramp\ Height = 5ft[/tex]
[tex]Ramp\ Length = 15ft[/tex]
Required
Determine the mechanical advantage
This is calculated as follows:
[tex]MA = \frac{Ramp\ Length}{Ramp\ Height}[/tex]
[tex]MA = \frac{15ft}{5ft}[/tex]
[tex]MA = 3[/tex]
Hence, the mechanical advantage is 3
A 2150 kg car, moving east at 10.0 m/s, collides and joins with a 3250 kg car. The cars move east together at 5.22 m/s. What is the 3250 kg car’s initial velocity calculated to the nearest tenth? Record your answer in the boxes below. Be sure to use the correct place value.
Answer:
2.1 m/s
Explanation:
According to law of conservation of momentum;
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the common velocity
Given
m1 = 2150kg
m2 = 3250kg
u1 = 10.0m/s
u2 = ?
v = 5.22m/s
Substitute and get u2
2150(10) + 3250u2 = (2150+3250)5.22
21,500 + 3250u2 = 5400(5.22)
3250u2 = 28,188 - 21500
3250u2 = 6688
u2 = 6688/3250
u2 = 2.1 m/s
Hence the 3250 kg car’s initial velocity has an initial velocity of 2.1 m/s
explain an experiment of the phenomenon of rainfall
Unclear/incomplete question. However, I inferred you need an explanation of the phenomenon of rainfall.
Explanation:
Basically, the phenomenon of rainfall follows a natural cycle called the water cycle. What we call 'rainfall' occurs when water condensed (in liquid form) in the atmosphere is made to fall down on the ground as tiny droplets as a result of the forces of gravity.
The water cycle makes rainfall possible:
First, water on the earth's surface is evaporated (or is absorbed into) the atmosphere.Next, it then condensed into liquid form; which later falls to the surface to the ground again. And the process continues.A radioactive nuclide of atomic number Z emits an alpha particle and the daughter nucleus then emits a beta particle. What is the atomic number of resulting nuclide?
A) Z-1
B) Z+1
C) Z-2
D) Z-3
Answer:
A) Z-1
Explanation:
when a radioactive element of atomic number Z emits an alpha particle, the mass of the new nucleus decreases by 2, i.e the new atomic number of the element = ( Z- 2).
Also, when the daughter nucleus emits a beta particle, the new nucleus increases by 1, that is the new atomic number of the element = (Z + 1).
Thus, the atomic number of resulting nuclide = Z ( - 2) + ( + 1).
= Z - 2 + 1
= Z - 1
Therefore, the atomic number of resulting nuclide is Z - 1
A motorcycle skids for a distance of 2.0 m with the icy road pushing on its tires with force of 120 N as its
brakes are applied
What is the change in kinetic energy for the motorcycle?
Round the answer to two significant digits.
Answer:
-240
Explanation:
A motorcycle skids for a distance of 2.0 m on an icy road, then the change in kinetic energy for the motorcycle will be equal to -240 J.
What is kinetic energy?The force which a moving object has is referred to as kinetic energy in physics. It is defined as the number of effort required to propel a person of a specific mass from still to a specific velocity.
Aside from slight fluctuations in speed, your body holds onto the kinetic energy it obtains during acceleration.
When the body slows down from its present level to a condition of rest, the same quantity of energy is used.
Formally, kinetic energy is any quantity that has a gradient concerning time in the Lagrangian of a system.
As per the given information in the question,
Distance, d = 2.0 m
Friction, f = 120 N
The angle between displacement and friction force, θ = 180°
Now, the change in kinetic energy for the motorcycle = Work done by the friction.
K.E = f × d(cos θ)
= 120 (2.0 m)(cos 180°)
Δ K.E = -240 J
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PLEASE HELP :(
I WILL GIVE EXTRA POINTS
1. When is the kinetic energy of an electron transformed into potential energy?
when it interacts with other electrons, decreasing its speed
when it interacts with neutrons without changing its speed
when it interacts with neutrons , increasing its speed
when it interacts with other electrons without changing its speed
2. Atoms bond to form molecules. Which structures or regions of the atoms interact in bonds ?
electric fields of particles with positive charge
electric fields of particles with no charge
electric fields of particles with negative charge
electric fields of particles with opposite charges
3. If two electrons that are apart get pushed toward each other, how does the repulsion between them change?
Initial repulsion is low and decreases as they approach .
Initial repulsion is high and decreases as they approach .
Initial repulsion is high and increases as they approach .
Initial repulsion is low and increases as they approach .
4. A positive charge of 5.0x10 ^ -5 C °is 0.040 m from a second positive charge of 2.0x10 ^ -6 C Calculate the force between the charges.
5.6x10^2 N
5.6x10^2 N
1.4X10^-2 N
2.3X10^1 N
(a) When the kinetic energy of an electron is transformed into potential energy is when it interacts with other electrons, decreasing its speed.
(b) The region of atoms that interact in bonds is electric fields of particles with negative charge.
(c) Initial repulsion is low and increases as they approach.
(d) The force between the charges is 562.5 N.
Kinetic theory of matter
This theory states that, the collision of particles (electrons) of matter is perfectly elastic. This implies that as the particles (electrons) collides with one another, kinetic energy is transferred from one electron to another.
ΔK.E = ΔP.E
Change in kinetic energy is equal to change in potential energy of the electrons.
Thus, when the kinetic energy of an electron is transformed into potential energy is when it interacts with other electrons, decreasing its speed. Decrease in speed implies decrease in kinetic energy and increase in potential energy.
Chemical bonds of moleculesChemical bond is formed from the transfer or sharing of electrons between atoms. (electrons between atoms implies negative charge to negative charge)
Thus, the region of atoms that interact in bonds is electric fields of particles with negative charge.
Coulomb's law
This law states that the force of attraction or repulsion between two charges is directly proportional to the product of the charges and inversely proportional to the distance between the charges.
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
When the distance between the electrons are large, the repulsive force is low and the distance is small, the repulsive force is high.
Force between the chargesThe force between the charges is determined by applying Coulomb's law,
[tex]F = \frac{kq_1q_2}{r^2} \\\\F = \frac{9\times 10^9\times (5 \times 10^{-5}) \times (2\times 10^{-6})}{0.04^2} \\\\F = 562.5 \ N[/tex]
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A major league pitcher can throw a baseball an excess of
How would the mass and weight of an object on the Moon compare to the mass and weight of the same object on Earth? * Mass and weight would both be less on the Moon. Mass would be the same but its weight would be less on the Moon. Mass would be less on the Moon and its weight would be the same. Mass and weight would both be the same on the Moon.
Answer:
B. Mass would be the same but its weight would be less on the Moon.
Explanation:
The mass of a body can be expressed as the quantity of matter it contains. While the weight of a body is the extent of the gravitational force impressed on the body by a massive body.
Thus, the mass of a body is constant either on the Earth or on the Moon. But the weight would be less on the Moon because the gravitational force on the Moon is far less than that on the Earth. Therefore the weight would be less on the Moon.
The appropriate option is B.
The mass will remain same on both moon and Earth, but weight will be lesser on Moon than Earth. Hence, option (B) is correct.
The prime focus to solve this problem is the mass and weight of an object. The mass of a body can be expressed as the quantity of matter it contains. While the weight of a body is the extent of the gravitational force impressed on the body by a massive body.
So, the mass of a body is constant either on the Earth or on the Moon. But the weight of an object will depend on the mass and the gravitational acceleration.
W = mg
Here, W is weight, m is mass and g is gravitational acceleration.
Weight would be less on the Moon because the gravitational force on the Moon is far less (due to lower value of g) than that on the Earth. Therefore the weight would be less on the Moon.
Thus, we can conclude that the mass will remain same on both moon and Earth, but weight will be lesser on Moon than Earth. Hence, option (B) is correct.
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A rigid tank contains an ideal gas at 300 kPa and 600 K. Now half of the gas is withdrawn from the tank and the gas is found at 100 kPa at the end of the process. Determine (a) the final temperature of the gas and (b) the final pressure if no mass was withdrawn from the tank and the same final temperature was reached at the end of the process.
The image below shows four boxes that each contain a different sample of gas. The atoms of each gas are represented by dots, 1 2 3 4 Which box contains the gas with the greatest density?
A. 1
B. 2
C. 3
D. 4
A bowling ball is 21.6 cm in diameter. What is the angular speed of these ball whenit is moving at 3.0 m/s?
Answer:
Angular speed = 27.78 rad/s (Approx)
Explanation:
Given:
Diameter = 21.6 cm
Speed = 3 m/s
Find:
Angular speed
Computation:
Radius = 21.6 / 2 = 10.8 cm = 0.108 m
Angular speed = v / r
Angular speed = 3 / 0.108
Angular speed = 27.78 rad/s (Approx)
If the social distancing length between two students is doubled from two metered to four meters, does the gravitational force between the two students increase or decrease?
Explain your reasoning!
Answer:
the gravitational force decreases
A stretched string is observed to have four equal segments in a standing wave driven at a frequency of 480 Hz. What driving frequency will set up a standing wave with five equal segments?
a) 360 Hz.b) 240 Hz.c) 600 Hz.d) 120 Hz.
Answer:
C) 600 Hz
Explanation:
The fundamental frequency can be related to the driving frequency by the expression below;
f(n) = n * f(1)
Where f(1)= fundamental frequency
f(n) = driving frequency
There are four equal segments in the standing wave , then our n= 4 and our f(n)=4, then we can get the fundamental frequency here
f(4) = 4× f(1)
480 = 4× f(1)
f(1) = 480/4
f(1)=120Hz
Hence, fundamental frequency is 120Hz
To calculate the driving frequency that will set up a standing wave with five equal segments?
n=5
f(n) = n× 120Hz
f(5) = 5×120Hz
= 600Hz.
Hence, the driving frequency that will set up a standing wave with five equal segments is 600Hz
Two motorcycles are traveling due east with different velocities. However, 3.63 seconds later, they have the same velocity. During this 3.63-second interval, motorcycle A has an average acceleration of 4.55 m/s2 due east, while motorcycle B has an average acceleration of 18.9 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 3.63-second interval, and (b) which motorcycle was moving faster
Answer:
52.095 m/s
Motorcycle a was moving faster
Explanation:
We start by using one of the equations of motion
V = u + at
If the first motorcycle starts with an initial speed of u(a) and accelerates at a value of a(a) = 4.55 m/s², then the final speed after a time of 3.63 seconds is V(a). We then represent it as
V(a) = u(a) + a(a).t
If the second motorcycle starts with an initial speed of u(b) and accelerates at a value of a(b) = 18.9 m/s², then the final speed after a time of 3.63 seconds is V(b). We then represent it as
V(b) = u(b) + a(b).t
Assuming that the final speeds v(a) = v(b), and then subtract the equation of the second motorcycle from that of the first, we have
0 = u(a) - u(b) + a(a).t - a(b).t
-u(a) + u(b) = a(a).t - a(b).t, on rearranging, we have
u(b) - u(a) = [a(a) - a(b)]t
Since we have the values for acceleration and the time, we substitute so that
u(b) - u(a) = (4.55 - 18.9)3.63
u(b) - u(a) = -14.35 * 3.63
u(b) - u(a) = -52.095, or we rearrange to get
u(a) - u(b) = 52.095 m/s
PLEASE HELP ME ASAP!! GUYSSS!! I AM IN CLASS AND DYING! LITERALLY
Billy and Ashley live in the same time zone. Billy lives in Brazil (blue smiley face on the image below). Ashley lives in Eastern Canada (yellow smiley face on the image below).
One day, Billy and Ashley are both outside at 1:32 pm. They are talking on the phone to each other. As they talk, Billy notices the sky in Brazil getting progressively darker. Eventually, it feels like it is nighttime, because it is so dark. Billy thinks the world is coming to an end. He asks Ashley if she is experiencing the same thing in Canada. Ashley has no idea what he is talking about. “It’s perfectly bright and sunny where I am,” she says.
Ashley and Billy conclude that the world is not coming to an end. They reach out to some 7th graders to figure out what is happening. The 7th graders tell Billy that he is experiencing a solar eclipse. To help Billy and Ashley understand, create a model to show why Billy is experiencing an eclipse in Brazil, but Ashley is not experiencing the eclipse in Canada.
Your model must include:
The sun, the earth, the moon, and solar energy (clearly labeled).
How accurate your scale is.
How solar energy interacts with both the moon and with Earth.
The tilt of the moon’s orbit relative to the Earth’s orbit
Why Billy is experiencing the solar eclipse and why Ashley is not.
Answer:
sounds like a you problem
Explanation:
yeah
Suppose a uniform solid sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The linear velocity of the sphere at the bottom of the incline depends on?
Answer:
None of the mass or the radius of the sphere
Explanation:
When a uniform solid sphere of any given mass, say M and any given radius, say R, rolls without slipping downwards an inclined plane that starts from rest. The linear velocity of the sphere at about the bottom of the inclined happens not to depend on either of its mass or that of the radius of its sphere.
Brandon hits a golf ball with an initial velocity of 30 m/s at an angle of 30 above the horizontal. How long is it in the air?
Given :
Brandon hits a golf ball with an initial velocity of 30 m/s at an angle of 30 above the horizontal.
To Find :
How long is it in the air.
Solution :
We know, the formula of time of flight is :
[tex]T = \dfrac{2usin\ \theta}{g}\\\\T = \dfrac{2\times 30\times sin\ 30^o}{9.8}\\\\T = 3.06\ seconds[/tex]
Therefore, the ball is in air for 3.06 seconds.
A particular engine has a power output of 2 kW and an efficiency of 27%. If the engine expels 9085 J of thermal energy in each cycle, find the heat absorbed in each cycle. Answer in units of J.
Answer:
12445 J
Explanation:
Given that
Power output, P = 5 kW
efficiency of the engine, e = 27% = 0.27
Thermal energy expelled, Q(c) = 9085 J
Heat absorbed, Q(h) = ?
Using the formula
e = W/Q(h)
e = [Q(h) - Q(c)] / Q(h)
e = 1 - Q(c)/Q(h)
Now, substituting the values into the formula, we have
0.27 = 1 - 9085/Q(h)
9085/Q(h) = 1 - 0.27
9085/Q(h) = 0.73
Q(h) = 9085 / 0.73
Q(h) = 12445 J
Thus, the heat absorbed is 12445 J
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If a rock falls for 3 seconds off of a bridge, how far will the rock fall?
-30 m
-45m
-60m
-75m