The maximum rate at which energy can be added to the circuit element mathematically given as
[tex]MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}[/tex]
What is the maximum rate at which energy can be added to the circuit element?Generally, the equation for P is mathematically given as
[tex]P=\ln s \frac{\Delta T}{\Delta t}[/tex]
Therefore
[tex]Rate\ of\ Change\ of\ Temp =\frac{p}{lnS}[/tex]
[tex]\frac{p}{lnS}=\frac{7.4 \times 10^{-3}}{23 \times 10^{-6} \times 705}[/tex]
[tex]\frac{p}{lnS}=0.456^{\circ \mathrm{c}} / \mathrm{sec}[/tex]
Max temp Change
[tex]MaxT=5.6^{\circ} \mathrm{C}[/tex]
[tex]\text { time }=3 \times 60[/tex]
t=180s
In conclusion, Max Energy Rate
[tex]MER =23 \times 10^{-6} \times \frac{301 \times 5.6}{180}[/tex]
[tex]MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}[/tex]
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Please help!! will give brainliest if correct!
B: Scientists assume the universe is a vast single system in which basic laws are consistent.
I ____________ with this statement
Explain your answer, using an example from your investigation about habitable worlds.
Scientists assume the universe is a vast single system in which basic laws are consistent. I agree with this statement
What is the name of the theory that states that our universe is from a single point?The Big Bang Theory is known as one of the greatest consensus that was said to have occurred among scientists, astronomers and also that of cosmologists.
Note that the theory is one that states the Universe as we all know it was formed as a result of a massive explosion that is not only formed by the majority of matter, but also with the use of the physical laws that works in our ever-expanding cosmos.
Therefore, In the case above, i do agree with the fact that Scientists assume the universe is a vast single system in which basic laws that are consistent.
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How much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km?
1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km
What is Speed ?Speed is the distance travelled per time taken. It is a scalar quantity. And the S.I unit is meter per second. That is, m/s
In the given question, we want to find how much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 10^5 km.
What are the parameters to consider ?
The parameters are;
The distance S = 3.85 × [tex]10^{5}[/tex] kmThe Speed of Light C = 3 × [tex]10^{8}[/tex] m/sThe time taken t = ?Speed = distance S ÷ Time t
Convert kilometer to meter by multiplying it by 1000
C = S/t
3 × [tex]10^{8}[/tex] = 3.85 × [tex]10^{8}[/tex] / t
Make t the subject of formula
t = 3.85 × [tex]10^{8}[/tex] / 3 × [tex]10^{8}[/tex]
t = 1.2833
t = 1.3 s
Therefore, 1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km
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Respond to the following prompts in your initial post:
1. Is Felisha's argument sound or unsound? Use the concepts discussed in your textbook: a valid
structure, true premises, and a sound or unsound argument.
2. Using the fallacies discussed in chapter 11 of your book, identify one fallacy (by name) in
Felisha's argument and explain why this fallacy is an illogical way to argue her conclusion.
3. Describe a time when you successfully argued a point of view and explain why you believe
you were successful.
Based on the textbook, Felisha's argument is known to be unsound because not all the premises that were given was true.
2. She stated that the “The Cowboys are popular and using only their huge fan base to tell is not right and if they are are better as she claim, why are they not the best but she stated that they are among the best.
3. A time when when i successfully argued a point of view was during my high school debate on the topic that doctors are better than lawyers and i gave a lot of valid reasons why the world cannot do without doctors.
What is fallacies and its types?The term fallacies are seen as flawed or deceptive and false kind of arguments that is known to be be proven wrong with the aid of reasoning.
Note that there are two key types of fallacies which are the formal fallacy and an informal fallacy.
Therefore, Based on the textbook, Felisha's argument is seen as an unsound argument because not all the premises that were given was true and they were not standard enough.
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Consider the f(x) = cos(x) + D function shown in the figure in blue color. What is the value of parameter D?
As a reference the g(x) = cos(x) function is shown in red color, and green tick marks are drawn at integer multiples of π.
The value of parameter D is -3.
What is angular frequency of a wave?Angular frequency is the angular displacement of any element of the wave per unit time.
f(x) = Acos(x) + D
where;
A is amplitude of the waveB is phase difference of the waveWhat is amplitude of a wave?The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.
From the blue colored graph, y = -2 at x = 0
-2 = cos(0) + D
-2 = 1 + D
-2 - 1 = D
-3 = D
Thus, the value of parameter D is -3.
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A constant net force F acts on a body during a time interval t. If u and v are the initial and final velocity of the body respectively, the impulse Ft of this force is given by the equation Ft = mv-mu.
Answer:
Impulse is defined as change in momentum of an object divided by time interval.
at t= 0s
initial velocity = u , initial momentum = mu
at some time t .
final velocity = V, final momentum= mv.
now, change in momentum= ( final - initial) = ( mv-mu)
time interval = (t-0) = t
impulse force = (mv-mu)/ ( t)
Ft = (mv-mu) proved .
this law is known as Newton's second law.
*please refer to photo attached* The figure below shows a small, charged sphere, with a charge of q = +44.0 nC, that moves a distance of d = 0.189 m from point A to point B in the presence of a uniform electric field E of magnitude 300 N/C, pointing right.
What is the magnitude (in N) and direction of the electric force on the sphere?
magnitude_______N
Direction?
- toward the right
- toward the left
- the magnitude is zero
(b) What is the work (in J) done on the sphere by the electric force as it moves from A to B?
__________J
(c) What is the change of the electric potential energy (in J) as the sphere moves from A to B? (The system consists of the sphere and all its surroundings.)
PEB − PEA = ______J
(d) What is the potential difference (in V) between A and B?
VB − VA = ________V
(a) The magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.
(b) The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.
(c) The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.
(d) The potential difference between A and B is 56.7 V.
Electric force on the sphere
The electric force on the sphere is calculated as follows;
F = Eq
where;
E is electric fieldq is the chargeF = 300 x (44 x 10⁻⁹)
F = 1.32 x 10⁻⁵ N
The direction of the force is towards the right.
Work done on the sphereW = Fd
W = 1.32 x 10⁻⁵ N x 0.189 m
W = 2.5 x 10⁻⁶ J
Change of the electric potential energyThe change in the electric potential energy (in J) as the sphere moves from A to B is equal to work done in moving the charge = 2.5 x 10⁻⁶ J.
Potential difference between A and BVB − VA = Ed
VB − VA = 300 N/C x 0.189 m
VB − VA = 56.7 V
Thus, the magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.
The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.
The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.
The potential difference between A and B is 56.7 V.
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a dog runs back and forth between its two owners, who are walking toward one another. the dog starts running when the owners are 10 m apart. if the dog runs with a speed of 3 m/s and the owners each walk with a speed of 1.3 m/s how far has the dog traveled when the owners meet
Answer:
11.54 meters
Explanation:
The owners take 10m / (1.3 + 1.3 m/s) = 3.85 seconds to meet
The dog is running at constant speed of 3 m/s for these 3.85 seconds
3 m/s * 3.85 sec = 11.54 meters
The Kennedy Space Center was in charge of the Apollo 13 mission.
True
False.
Standing 38.6 m away from a rock wall, you yell. How much time in seconds will it take you to hear your echo to two significant digits? Make sure to account for the travel from you to the wall and from the wall back to you.
It will take you 0.23 seconds to hear your echo
What is Echo ?Echo can be simply defined as the reflection of sound wave.
Given that you are 38.6 m away from a rock wall, you yell. To know how much time in seconds it will take you to hear your echo to two significant digits, You must make sure to account for the travel from you to the wall and from the wall back to you.
The speed of sound V = 340 m/sThe distance D to and fro = 2 x 38.6 = 77.2 mThe time taken T = ?Speed V is the distance per time
V = D / T
Substitute all the parameters into the formula
340 = 77.2 / T
Make T the subject of formula
T = 77.2 / 340
T = 0.2270
T = 0.23 s to two significant digits
Therefore, it will take you 0.23 seconds to hear your echo.
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A centrifuge in a medical laboratory rotates at an angular speed of 3,400 rev/min. When switched off, it rotates through 52.0 revolutions before coming to rest.
Find the constant angular acceleration (in rad/s2) of the centrifuge.
______rad/s2
The constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².
Constant angular acceleration
Apply the following kinematic equation;
ωf² = ωi² - 2αθ
where;
ωf is the final angular velocity when the centrifuge stops = 0ωi is the initial angular velocity θ is angular displacementα is angular accelerationωi = 3400 rev/min x 2π rad/rev x 1 min/60s = 356.05 rad/s
θ = 52 rev x 2π rad/rev = 326.7 rad
0 = ωi² - 2αθ
α = ωi²/2θ
α = ( 356.05²) / (2 x 326.7)
α = 194.02 rad/s²
Thus, the constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².
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A 100 N force pulls a box but it doesn’t move. How much work is done?
A. 0 J
B. 100 J
C. 50 J
D. Not enough info
Answer:
A
Explanation:
Work = force * distance
= 100 n * 0 = 0 work done
**please look at attached photo for proper understanding* Inside a cathode ray tube, an electron is in the presence of a uniform electric field with a magnitude of 295 N/C.
What is the magnitude of the acceleration of the electron (in m/s2)?
_____m/s2
The electron is initially at rest. What is its speed (in m/s) after 8.50 ✕ 10−9 s?
________m/s
The magnitude of the acceleration of the electron is 5.187 x 10¹³ m/s².
The speed of the electron at the given time is 4.41 x 10⁵ m/s.
Acceleration of the electronThe magnitude of the acceleration of the electron is calculated as follows;
Force on the electron, F = ma = Eq
ma = Eq
a = Eq/m
where;
E is electric fieldq is charge of electronm is mass of electrona = (295 x 1.6 x 10⁻¹⁹) / (9.1 x 10⁻³¹)
a = 5.187 x 10¹³ m/s²
Speed of the electronThe speed of the electron at the given time is calculated as follows;
v = at
v = ( 5.187 x 10¹³)(8.5 x 10⁻⁹)
v = 4.41 x 10⁵ m/s
Thus, the magnitude of the acceleration of the electron is 5.187 x 10¹³ m/s².
The speed of the electron at the given time is 4.41 x 10⁵ m/s.
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A transverse mechanical wave is traveling along a string lying along the x-axis. The displacement of the string as a function of position and time, y(x,t), is described by the following equation:
[tex]y(x,t)=0.0480[/tex]×[tex]sin(5.40x-128t)[/tex]
where x and y are in meters and the time is in seconds.
1. What is the wavelength of the wave?
2. What is the velocity of the wave? (Define positive velocity along the positive x-axis.)
3. What is the maximum speed in the y-direction of any piece of the string? (Give a positive answer for speed.)
The wavelength of the wave is 1.16m and the velocity is 23.64m/s.
To find the answer, we have to know more about the Transverse waves.
How to find different parameters of a wave?The displacement of the string as a function of position and time, y(x,t), when the wave traveling along a string lying along the x-axis is given as,[tex]y(x,t)=0.048sin(5.40x-128t)[/tex]
Comparing this with the general form of wave equation, we get,[tex]amplitude, a=0.048m\\wave vector, k=5.40\\angular frequency,w=128Hz[/tex]
We have to find the wavelength of the wave, for this, we have the expression as,[tex]k=\frac{2\pi }{wavelength} \\\\wavelength=\frac{2\pi }{k} =\frac{2*3.14}{5.40} \\\\wavelength=1.16m[/tex]
We have to find the velocity of the wave,[tex]v=frequency*wavelength\\2\pi f=w, thus,\\f=\frac{w}{2\pi } =\frac{128}{2*3.14}=20.38s^{-1}\\\\v=1.16*20.38=23.64m/s[/tex]
Thus, we can conclude that, the wavelength of the wave is 1.16m and the velocity is 23.64m/s
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A converging lens has a focal length of 20cm. find graphically the image location for an object at each of the following distances from the lens: (a) 50cm (b) 20cm (c) 15cm (d) -40cm: determine the magnification in each case. Check your results by calculating the image position and lateral magnification from Eqs. (10) and (11), respectively.
Answer:
4
Explanation:
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*example of previous incorrect/correct answer on similar question) Inside a vacuum tube, an electron is in the presence of a uniform electric field with a magnitude of 330 N/C.
(a) What is the magnitude of the acceleration of the electron (in m/s²)?
__________m/s²
(b) The electron is initially at rest. What is its speed (in m/s) after 8.00 ✕ 10−⁹ s?
______m/s
The final speed of the electron is 4.64 * 10^5 m/s.
What is the speed of the electron?Given that the mass of the electron is obtained as 9.11 * 10^-31 Kg, we have that the charge of the electron is 1.6 * 10^-19 C.
E= F/q
F = Eq
F = 330 N/C * 1.6 * 10^-19 C
F = 5.28 * 10^-17 N
F = ma
a = F/m = 5.28 * 10^-17 N/ 9.11 * 10^-31 Kg
a = 5.8 * 10^13 m/s^2
Using
v = u + at
u = 0 m/s because the electron was initially at rest
v = at
v = 5.8 * 10^13 * 8.00 ✕ 10−⁹ s
v = 4.64 * 10^5 m/s
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In the diagram, q1 = +6.39*10^-9 C and
q2 = +3.22*10^-9 C. What is the electric
field at point P? Include a + or - sign to
indicate the direction.
P
91
0.424 m-
0.636 m
-
92
(Remember, E points away from + charges,
and toward charges.)
(Unit = N/C)
E =+823.12N/C is the electric field at point P
Each point in space has an electric field associated with it when a charge of any kind is present. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field. in the diagram, q1 = +6.39*10^-9 C and
q2 = +3.22*10^-9 C hence E = 823.12 N/C
A region of space surrounding andd P Include a + or - sign toindicate the direction.P910.424 m-0.636 m- electrically charged particle or object known as an electric field is one in which an electric charge would experience force. A vector quantity called an electric field can be represented by arrows pointing in the direction of or away from charges. The force per unit charge exerted on a positive test charge that is at rest at a given position is the force per unit charge that is used to define the electric field analytically.
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What is the mass of the sun plus all the planets in solar system?
The distance between a carbon atom ( m = 12 u ) and an oxygen atom ( m = 16 u ) in the CO molecule is 1.13 × 10-10 m .
How far from the carbon atom is the center of mass of the molecule?
Express your answer to two significant figures and include the appropriate units.
6.45 ×[tex]10^{-} ^{11}[/tex] m is far from the carbon atom and is the center of mass of the molecule.
The sum of the products of the masses of the two particles and their respective position vectors equals the product of the total mass of the system and the position vector of the center of mass.
The center of mass of CO molecule will be on the line joining C and O atoms. Let the CO molecule be along the X- axis, the C atm being at the origin (x=0). The center of mass relative to the C atom is given by
[tex]x_{cm} = \frac{m_{c}x_{c} + m_{o} x_{o} }{m_{c}+ m_{o} }[/tex]
Where, [tex]m_{c}[/tex] and [tex]m_{o}[/tex] are the respective masses of C and O atoms, [tex]x_{c}[/tex] and [tex]x_{o}[/tex] their distances relative to C atom.
Here, [tex]m_{c}[/tex] = 12 u , [tex]m_{o}[/tex] = 16 u and [tex]x_{o} =1.13[/tex]×[tex]10^-^{10}[/tex] m
[tex]x_{cm} = \frac{m_{c}x_{c} + m_{o} x_{o} }{m_{c}+ m_{o} }[/tex]
[tex]x_{cm} =[/tex] (12 × 0 + 16 × 1.13 × [tex]10^{-} ^{10}[/tex] ) / (12 + 16 )
[tex]x_{cm}[/tex] = 0.64571 × [tex]10^{-} ^{10}[/tex]
[tex]x_{cm}[/tex] = 6.45 ×[tex]10^{-} ^{11}[/tex] m
Therefore, 6.45 ×[tex]10^{-} ^{11}[/tex] m is far from the carbon atom is the center of mass of the molecule.
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A 92-kg fullback is running at 3.6 m/s to the east and is stopped in 0.85 s by a head-on tackle by a tackler running due west.
a)Calculate the original momentum of the fullback.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the momentum is to the east, and negative value if the direction of the momentum is to the west.
b)Calculate the impulse exerted on the fullback.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the impulse is to the east, and negative value if the direction of the impulse is to the west.
c)Calculate the impulse exerted on the tackler.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the impulse is to the east, and negative value if the direction of the impulse is to the west.
d)Calculate the average force exerted on the tackler.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the force is to the east, and negative value if the direction of the force is to the west.
a) The original momentum of the fullback is 331.2 kgm/s.
b) The impulse exerted on the fullback is - 331.2 kgm/s
c) The impulse exerted on the tackler is 331.2 kgm/s
d) The average force exerted on the tackler is 389.64 N
Given:Mass of the fullback, m = 92 kg
Initial velocity of the fullback , u = 3.6 m/s
Time of the motion of the fullback , t = 0.85s
The original momentum of the fullback ;
[tex]P_{i} = mv\\P_{i} = (92) (3.6)\\P_{i} = 331.2 kg m/s[/tex]
The impulse exerted on the full-back;
J = ΔP = [tex]m v_f - m v_i[/tex]
J = [tex]m ( v_f - v_i)[/tex]
J = 92 ( 0- 3.6)
J = - 331.2 kgm/s
The impulse exerted on the tackler;
[tex]J _1 = - J _2\\J_2 = - J_1\\J_2 = -( - 331.2)\\J_2 = 331.2 kgm/s[/tex]
The average force exerted on the tackler;
F= Δmv/t
[tex]F =\frac{331.2}{0.85} \\F = 389.64 N[/tex]
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Have my equation written out but struggling to solve. Can someone help me solve!
The solution of this system is (i₁, i₂, i₃) = (- 10.852 A, 8.479 A, - 2.374 A). The negative signs indicate that real direction of the current is opposite than supposed.
How to find the missing current in a circuit
In this question we must make use of Kirchhoff's laws to find the values of the missing currents in the circuit presented in the picture. There are two rules according to Kirchhoff's laws:
The sum of currents found at nodes of circuits is always equal to zero.The net sum of voltages in a closed loop of a circuit is always equal to zero.Based on the information given by the picture, we have the following system of linear equations that describes the entire circuit:
i₃ = i₁ + i₂
- i₁ · R₁ + ε₁ - i₁ · r₁ - i₁ · R₅ + ε₂ - i₂ · (r₂ + R₂) = 0
ε₂ - i₂ · (r₂ + R₂) - i₃ · r₄ - ε₄ - i₃ · r₃ + ε₃ - i₃ · R₃ = 0
- i₁ - i₂ + i₃ = 0 (1)
(R₁ + r₁ + R₅) · i₁ + (r₂ + R₂) · i₂ = ε₁ + ε₂ (2)
(r₂ + R₂) · i₂ + (r₄ + r₃ + R₃) · i₃ = ε₂ + ε₃ - ε₄ (3)
If we know that R₁ = 5 Ω, r₁ = 0.10 Ω, R₅ = 20 Ω, r₂ = 0.50 Ω, R₂ = 40 Ω, r₄ = 0.20 Ω, r₃ = 0.05 Ω, R₃ = 78 Ω, ε₁ = 22 V, ε₂ = 49 V, ε₃ = 10.5 V and ε₄ = 33 V, then the currents flowing in the circuit are:
- i₁ - i₂ + i₃ = 0
25.1 · i₁ + 40.5 · i₂ = 71
25.1 · i₂ + 78.25 · i₃ = 26.5
The solution of this system is (i₁, i₂, i₃) = (- 10.852 A, 8.479 A, - 2.374 A). The negative signs indicate that real direction of the current is opposite than supposed.
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A person is losing thermal energy through the skin at a rate of 120 W when his skin temperature is 30°C. He puts on a sweater, and his skin temperature rises to 33°C. The effective thermal conductivity between his core and the environment changes from 0.22 W/m·K to 0.18 W/m·K. At what rate (J/s) is he now losing thermal energy?
The rate at which the person is now losing thermal energy is 99 J/s.
Thickness of the person's skin
The thickness of the person's skin is calculated as follows;
Q = k(ΔT)h
where;
k is thermal conductivityh is thickness of the person's skinΔT difference in temperature across the skin = 30 °C = 303 Kh = Q/k(ΔT)
h = (120) / (0.22 x 303)
h = 1.8 m
Rate at which the person is now losing thermal energyQ = k(ΔT)h
where;
ΔT is new temperature difference across the skin = 33 °C = 306 KQ = (0.18)(306)(1.8)
Q = 99 J/s
Thus, the rate at which the person is now losing thermal energy is 99 J/s.
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The CM of an empty 1300-kg car is 2.45 m behind the front of the car.
How far from the front of the car will the CM be when two people sit in the front seat 2.70 m from the front of the car, and three people sit in the back seat 3.65 m from the front? Assume that each person has a mass of 60.0 kg .
Express your answer to three significant figures and include the appropriate units.
The center of mass of the car with all people seated will be 3.20 m far from the front of the car.
Taking the front of the car as reference:
m₁x₁ + m₂x₂ + m₃x₃ = mx
where,
m₁ = mass of empty car = 1300 kg
x₁ = distance of center of mass of empty car from front = 2.45 m
m₂ = mass of 2 people sitting in front = 2 x 60 kg = 120 kg
x₂ = distance of center of mass of 2 people sitting in front from the front = 2.7 m
m₃ = mass of 3 people sitting in back seat = 3 x 60 kg = 180 kg
x₃ = distance of center of mass of 3 people sitting in the back seat from the front = 3.65 m
m = total mass of car with all people = 1300 kg + 120 kg + 180 kg = 1600kg
x = distance of center of mass of the car from the front when all are seated
Therefore,
(1300 kg)(2.45 m) + (120 kg)(2.7 m) + (180kg)(3.65 m) = (1300 kg)x
x = 4166 kg.m/1300 kg
x = 3.20 m
Therefore, the center of mass of the car with all people seated will be 3.20 m far from the front of the car.
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The graph shows the US Department of Labor noise regulation for working without ear protection. A machinist is in an environment where the ambient sound level is of 85 dB, i.e., corresponding to the 8 Hours/day noise level. The machinist likes to listen to music, and plays a Boom Box at an average level of 86.0 dB.
1. Calculate the INCREASE in the sound level from the ambient work environment level (in dB).
2. A student at a rock concert has a seat close to the speaker system, where the average sound intensity corresponds to a sound level of 116 dB. By what factor does that sound intensity exceed the 1.5- Hours/day intensity limits from the graph?
(1) The INCREASE in the sound level from the ambient work environment level (in dB) is 1 dB.
(2) The factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.
Increase in the sound levelThe INCREASE in the sound level from the ambient work environment level (in dB) is calculated as follows;
Increase in sound level = final sound level - original sound level
Increase in sound level = 86 dB - 85 dB = 1 dB
Factor of sound level increasefrom the graph at 1.5 hours/day, sound level = 97 dB
Increase in sound intensity = final sound level - original sound level
Increase in sound intensity = 116 dB - 97 dB = 19 dB
Factor increase = 19/97 = 0.196 = 19.6%
Thus, the factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.
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An object has a gravitational
potential energy of 750 Joules and is
sitting on a shelf 3.0 meters above
the ground. What is the object's
mass?
[?] kilograms
Answer:
potential energy = (mass × height × acceleration due to gravity.)
750 = mass × 3 × 10
hence, mass = 750/30 = 25 kg.
A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface. Neglect any effects due to air resistance.
a) What is the kinetic energy of the projectile when it reaches the highest point in its trajectory?
b) How much work was done in firing the projectile?
(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.
(b) The work done in firing the projectile is 2,500 J.
Kinetic energy of the projectile at maximum height
The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;
K.E = ¹/₂mv₀ₓ²
where;
m is mass of the projectilev₀ₓ is the initial horizontal component of the velocity at maximum heightNote: At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.
K.E = (0.5)(2)(30²)
K.E = 900 J
Work done in firing the projectileBased on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.
W = K.E(i) = ¹/₂mv²
where;
v is the resultant velocityv = √(30² + 40²)
v = 50 m/s
W = (0.5)(2)(50²)
W = 2,500 J
Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.
The work done in firing the projectile is 2,500 J.
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1. If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision? Is it possible for one to be at rest after the collision? Explain.
Answer:
(a)If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision?
No. Because if you have initial momentum P⃗ ≠0 , if both of the objects were at rest after the collision the total momentum of the system would be P⃗ =0 , which violates conservation of momentum
(b)Is it possible for only one to be at rest after the collision?
Yes, that is perfectly possible. It characteristically, happens when both objects are of the same mass. When two objects of the same mass collide and Kinetic energy is conserved (Perfectly Elastic collision) then the two objects interchange velocities.
In the Roman soldier model for refraction, what happens to the distance between the rows of each soldier once they enter the stream?
A. Not enough info
B. They are closer together
C. They stay the same distance
D. They are further apart
They pick up speed and move away from the usual.
Using meter sticks, a group of pupils makes a straight line (shoulder to shoulder) and connects to their closest neighbor. Two media are separated in the room by a strip of masking tape. Students can be seen moving normally in one of the media (on one side of the tape). The kids in the other medium (or on the opposite side of the tape) move very slowly and in baby stages. The class moves in the union in a straight line in the direction of the diagonal piece of masking tape. As they get close to the masking tape, the pupils stay in line. Each pupil immediately changes the speed of her or his walk as she or he reaches the tape.
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An old grindstone, used for sharpening tools, is a solid cylindrical wheel that can rotate about its central axle with negligible friction. The radius of the wheel is 0.330 m. A constant tangential force of 200 N applied to its edge causes the wheel to have an angular acceleration of 0.844 rad/s2.
What is the moment of inertia of the wheel (in kg · m2)?
________kg · m2
What is the mass (in kg) of the wheel?
_______kg
The wheel starts from rest and the tangential force remains constant over a time period of 4.00 s. What is the angular speed (in rad/s) of the wheel at the end of this time period?
_______rad/s
(a) The moment of inertia of the wheel is 78.2 kgm².
(b) The mass (in kg) of the wheel is 1,436.2 kg.
(c) The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.
Moment of inertia of the wheel
Apply principle of conservation of angular momentum;
Fr = Iα
where;
F is applied forcer is radius of the cylinderα is angular accelerationI is moment of inertiaI = Fr/α
I = (200 x 0.33) / (0.844)
I = 78.2 kgm²
Mass of the wheelI = ¹/₂MR²
where;
M is mass of the solid cylinderR is radius of the solid cylinderI is moment of inertia of the solid cylinder2I = MR²
M = 2I/R²
M = (2 x 78.2) / (0.33²)
M = 1,436.2 kg
Angular speed of the wheel after 4 secondsω = αt
ω = 0.844 x 4
ω = 3.376 rad/s
Thus, the moment of inertia of the wheel is 78.2 kgm².
The mass (in kg) of the wheel is 1,436.2 kg.
The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.
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a)Calculate the angular velocity of the Earth in its orbit around the Sun.
Express your answer using three significant figures.
b) Calculate the angular velocity of the Earth about its axis.
Express your answer using three significant figures.
The earth's orbital angular velocity around the Sun is 0.986° per day, and its axial angular velocity is 15.041° per hour.
Calculation:
The speeds, not the velocities, are what we require in this case, therefore that is how I will approach it.
Regular speed is calculated by dividing the distance traveled by the travel time.
So, speed = [tex]\frac{distance -travelled}{time}[/tex]
The angular velocity is essentially the same.
Hence, it's divided by the amount of time it took to turn the angle.
(a) Calculation of Earth's orbit around the Sun :
The Earth orbits the Sun in a circle that is 2 radians in diameter (360 degrees). It also takes a year, as we are aware (approx 365 days)
360°/365.25636 days = 0.986°/day ≈ 1° per day
(b) Next, we have to calculate Earth's angular velocity on its axis:
determining the Earth's angular velocity as it fulfills a complete revolution on its axis (a solar day)-
This one requires more precision because a day does not always have 24 hours. Depending on how we define a day,If the day is defined as the time between the Sun's highest and lowest points in the sky, a year's worth of data equals an average of 24 hours.If we define a day as the duration of time it requires for a planet to get to the same location in the sky the following night, then it is 23 hours 56 minutes 4.09 seconds.According to this, the angular speed of rotation = 360°/ 23h 56min 4.09s = 15.041° / hour.
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In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.430 m (1.41 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 15.00 millibar. The atmospheric pressure outside is 940 millibar.
1. Calculate the force required to pull the two hemispheres apart. [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
2. Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1450N (i.e., about 326 lbs), what is the minimum number of horses required?
The force required to pull the two hemispheres is 46622.72N
Calculation and Parameters( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
The contact area between the hemispheres is (pi x 0.400^2) = 0.5024m^2.
Pressure difference = (940 - 12)
= 928 millibars.
(928 x 100)
= 92,800N/m^2.
Therefore, the required force to pull the two hemispheres is
(92800 x 0.5024)
= 46622.72N.
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