The most prevalent in the human body by mass is oxygen.
To find the answer, we need to know more about the elemental composition in human body.
What is the elemental composition in human body?The element that is most prevalent in the human body by mass is oxygen. This makes sense if you think about it because the majority of the human body is made up of water, or H2O. The bulk of the human body is composed of 61–65% oxygen. Your body has considerably more hydrogen than oxygen atoms, but each oxygen atom is 16 times heavier than a hydrogen atom.The second prevalent element in the body is carbon. It is about 18%.Thus, we can conclude that, the most prevalent in the human body by mass is oxygen.
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Answer:B I think
Explanation:
A mass of 1.0 kg is initially held in place on a ramp of angle 45o at height of 20. meters above the ground. The mass is released and slides to the bottom of the ramp (h=0). There is friction between the mass and the ramp. At the bottom of the ramp, the object has a speed of 10. m/s. The thermal energy was generated by the friction as the mass slides down the ramp is closest to which value?
Answer:
The thermal energy generated by the friction as the mass slides down the ramp is [tex]\bf{146~J}[/tex].
Explanation:
Given:
The mass of the object is, [tex]m = 1.0~kg[/tex]
The angle of the ramp is, [tex]\theta = 45^{0}[/tex]
The initial height of the object on the ramp is, [tex]h = 20~m[/tex]
The final velocity of the object is, [tex]v = 10~m/s[/tex]
When the object is at rest on the ramp, its total energy is potential energy. When it moves down the ramp its kinetic energy is increased and potential energy is decreased and a part of its energy is lost to overcome the force of friction. Finally, when it is at the bottom of the ramp, its total energy becomes only kinetic energy.
The total energy of the object at a height [tex]20~m[/tex] on the ramp is given by
[tex]E_{1} &=& mgh\\~~~~&=& (1.0~kg)(9.8~m/s^{2})(20~m)\\~~~~&=& 196~J[/tex]
When the object is at the bottom of the ramp, its total energy is given by
[tex]E_{2} &=& \dfrac{1}{2}mv^{2}\\~~~~&=& \dfrac{1}{2}(1.0~kg)(10~m/s)^{2}\\~~~~&=& 50~J[/tex]
So, the energy that is lost as thermal energy is given by
[tex]E &=& E_{1} - E_{2}\\~~~~&=& 196~J - 50~J\\~~~~&=& 146~J[/tex]
A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark. From this information, the mass of the meterstick is _________.
A) 1/4 kg.B) 1/2 kg.C) 3/4 kg.D) 1 kg.E) none of the above
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.
[tex]9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}[/tex]
W = 3.266 N
The mass of the meters stick is :
[tex]m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg[/tex]
So, the mass of the meter stick is 0.333 kg.
The mass of the meterstick is calculated using the principle of torque balance. The meterstick must have a mass that creates an equal opposite torque to the one created by the 1-kg rock. The calculation shows that the meterstick has a mass of 1/3 kg, which is not listed in the options, so the answer is (E) none of the above.
Explanation:The mass of the meterstick can be found using the principle of moments, also known as the principle of leverage, which states that for an object to be in rotational equilibrium, the sum of the clockwise moments about the pivot (fulcrum) must equal the sum of the counterclockwise moments. In this case, the 1-kg rock is hanging from the 0-cm mark and the meterstick balances when the fulcrum is at the 12.5-cm mark. The torque created by the 1-kg mass is given by its mass multiplied by its distance from the fulcrum, or 1 kg × 12.5 cm.
For the meterstick to balance, its center of mass must be located directly above the fulcrum. Since the meterstick is uniform, its center of mass is in the middle, at the 50-cm mark. This means that the mass of the meterstick acts 50 cm - 12.5 cm = 37.5 cm away from the fulcrum. Let's denote the mass of the meterstick as 'M'. The counterbalance torque provided by the meterstick is given by M kg × 37.5 cm.
Setting the torques equal for a balanced system:
M kg × 37.5 cm = 1 kg × 12.5 cm, which simplifies to M = 1/3 kg. Since 1/3 kg is equivalent to approximately 0.33 kg, the correct answer is (E) none of the above, as none of the provided options matches this result.
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Any measurement that includes both magnitude and direction is called
A measurement that includes both magnitude and direction is called a vector, which is essential for analyzing motion and forces in physics.
Any measurement that includes both magnitude and direction is called a vector. Vectors are physical quantities that have both an amount, known as magnitude, and a specified direction in space. For example, velocity is a vector because it describes not only how fast an object is moving, but also the direction of movement.
This contrasts with a scalar, which is a quantity that has only magnitude and no direction, such as mass or time. In physics, understanding vectors is crucial for analyzing motion, forces, and other concepts that depend on both the amount and the direction of a quantity.
Suppose you want to make a nested function call (i.e. a call to a function from inside of another function) using a jal rather than a call for performance reasons. How would the push and pop pseudo-ops be proprely ordered along with the jal so that the previous return address isn't lost?
a) pop $ra
jal nested_function_label
nop
push $ra
b) push $ra
jal nested_function_label
nop
pop $ra
c) push $ra
pop $ra
jal nested_function_label
nop
d) jal nested_function_label
nop
pop $ra
push $ra
Answer:
As we need to use a nested loop in our function,hence push $ra
pop $ra
jal nested_function_label
nop is the correct option.
Recently, astronomers have observed stars and other objects that orbit the center of the Milky Way Galaxy farther out than our Sun, but move around faster than we do. How do astronomers think such an observation can be explained?
Answer:
That scenario can be explained by the idea of the contribution of dark matter on that point.
Explanation:
It can be explained through the idea of dark matter, this one was born to explain why stars (or any object) that were farther for the supermassive black hole in the center of the Milky Way galaxy didn't decrease it rotational velocity as it was expected according to equation 1.
[tex]v = \sqrt{\frac{G M}{r}}[/tex] (1)
Where v is the rotational velocity, G is the gravitational constant, M is the mass of the supermassive black hole, and r is the orbital radius.
Notice, that If the distance increases the orbital speed decreases (inversely proportional).
Final answer:
Astronomers explain the high velocities of stars orbiting the Milky Way's center outside our Sun's orbit by the gravitational influence of dark matter. This halo of invisible matter adds mass to the galaxy, affecting the orbital speeds in accordance with Kepler's third law.
Explanation:
Astronomers have observed that stars and other objects orbiting the center of the Milky Way Galaxy farther out than our Sun move around faster than we do, which is an unexpected observation according to Kepler's third law. Typically, as per Kepler's third law, we would expect objects that are farther from the center of mass to orbit more slowly. However, the presence of invisible matter, which we now understand as dark matter, affects the gravitational pull and thus the orbital velocities of these outer objects, leading them to move at unexpectedly high speeds.
The Milky Way is surrounded by a halo of this invisible matter, which does not emit or reflect light, making it undetectable through traditional means. Instead, its presence is inferred from its gravitational effects on the motions of stars and gas in the galaxy. The observations mentioned, where objects beyond the luminous part of the Milky Way are moving faster than expected, provide evidence for the existence of dark matter, as this additional mass can account for the higher orbital velocities.
Quasar spectra often show many absorption lines that all appear to be as a result of the same electron transition (such as level 1 to level 2 in hydrogen) but that fall at different wavelengths in the spectrum. Why do we think this is the case?
Answer: Because of different redshift of cloud.
Explanation:
We are seeing absorption lines from clouds of gas that lie between us and the quasar, and therefore each cloud has a different redshift.
A quasar's spectrum is hugely redshifted. And most astronomers think this large redshift tells us about the distance to the quasar.
Which statement best describes what happens when a sound source is moving? (1 point)
Grupo de opciones de respuesta
As the sound source approaches the observer, the pitch decreases and continues to do so as it passes the observer.
As the sound approaches the observer, the pitch increases; as it passes the observer, the pitch decreases.
As the sound source approaches the observer, the pitch decreases; as it passes the observer, the pitch increases.
As the sound source approaches the observer, the pitch increases and continues to do so as it passes the observer.
Answer:
As the sound approaches the observer, the pitch increases; as it passes the observer, the pitch decreases.
Explanation:
The crests come closer when it approaching, and go farther when the source passes the observer.
A man driving a car traveling at 30m/s slams on the brakes and accelerates in the negative direction at 4.75 m/s2, how far does the car travel before it stops?
Answer:
94.73 meters.
Explanation:
Using SUVAT.
s = ?
u = 30
v = 0
a = -4.75
t = ?
[tex]v^{2}=u^2+2as\\0=30^2+2(-4.75)s\\\\900=9.5s\\s = 94.73[/tex]
The car will travel approximately 94.74 meters before it stops.
To answer this question, we can use the equations of motion.
When the car stops, its final velocity will be 0 m/s.
We can use the equation v^2 = u^2 + 2as,
where v is the final velocity,
u is the initial velocity,
a is the acceleration, and
s is the distance.
Plugging in the values, we have (0)^2 = (30)^2 + 2(-4.75)s.
Solving for s, we get s = 900/9.5, which is approximately 94.74 meters.
Therefore, the car will travel approximately 94.74 meters before it stops.
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what does a electric field of two positive charges look like
Hope this helped you
:)
Have a great day!
A helium balloon containing 5m³ of gas at a pressure of 30kPa is released into the air. Assuming that the temperature is constant calculate the volume the balloon would have when the pressure inside the balloon has fallen to 5kPa. Answer in m³.
Answer:
30 m³
Explanation:
Parameters given:
Initial volume of helium, V1 = 5 m³
Initial pressure in balloon, P1 = 30 kPa
Final pressure, P2 = 5 kPa
To find the volume of the balloon at that volume, we apply Boyle's law.
It states that at constant temperature, the pressure of a gas is inversely proportional to the volume of the gas.
Mathematically:
P = k / V
Where k = constant of proportionality
This implies that:
P * V = k
This means that if the pressure or volume of the gas changes at the same temperature, the product of the pressure and volume would be the same:
Hence:
P1 * V1 = P2 * V2
Hence, to find the final volume:
30 * 5 = 5 * V2
=> V2 = (30 * 5) / 5
V2 = 30 m³
The volume of the gas when the pressure is 5 kPa is 30 m³.
What are the similarities and differences between novae and supernovae?
Answer:
Explanation:
A novae in astronomy means an explosion in the white dwarf star which had tapped enough gas from a companion star,hence it releases an incredible amount of energy which is Over a million times brighter than it normal stars.
A super novae on the other hand is a cosmic explosion that can be a billion times brighter than the normal.
From this one can see that a perculiar similarity between a novae and super novae is that both generate huge explosion and bright Ness, and a major difference is super novae release huge amount of brightness and energy more than the novae
How much time would it take to exert 700W of power while doing 700J of work?
Answer:
The answer to your question is time = 1 s
Explanation:
Data
time = ?
Power = 700 W
Work = 700 J
Power is defined as the work done per unit of time.
Formula
Power = Work / Time
-Solve for time
Time = Work/Power
-Substitution
Time = 700 / 700
-Result
Time = 1 s
Conclusion
In 1 second and with 700 J there will be produced 700 watts
"Describe how increasing the stimulus frequency affected the force developed by the isolated whole skeletal muscle in this activity. How well did the results compare with your prediction
Answer is seen below
Explanation: Stimulus frequency refers to the rate that stimulating voltage pulses are applied to an isolated whole skeletal muscle.
When a stimulus frequency is at the lowest ( let's say 50stimuli/second) the force will be at its lowest level out of all of the experiments. As the stimulus frequency was increased to 130 stimuli/second the force increased slightly but fused tetanus( tetanus refers to a sustained muscle tension due to very frequent stimuli) developed at the higher frequency. When the stimulus frequency was increased to the amounts of 146-150 stimuli/second, a maximum tetanic tension occurred, where no further increases in force occur from additional stimulus frequency.
By increasing the stimulus frequency if it resulted in increasing the muscle tension generated by each successive force and it had limit that was eventually reached. Then the results equaled to your prediction.
When different resistors are connected in parallel across an ideal battery, we can be certain that...
Answer:
Explained Below.
Explanation:
Whenever the different resistors are fixed parallel along with an ideal battery , we can be certain that it carries the same potential difference across each of them.
And potential difference is nothing but the energy that charge moves with. It can also be denoted as (p.d.) and potential difference is been calculated as volts that is been denoted as (V).
he graph shows all of the stable isotopes of elements according to the numbers of protons and neutrons in their nuclei.
A graph titled zone of nuclear stability with number of protons from 0 to 80 on the x-axis with number of neutrons on the y-axis from 0 to 140. There are two lines, one with a slope of 1 from (0, 0). The other line has points of (20, 24), (40, 52) and (60, 120) labeled stability zone.
Which statements are supported by the graph? Check all that apply.
A. For large atoms, more neutrons than protons are needed to be stable.
B. Nuclei that have 90 or greater protons are always radioactive.
C. Atoms must have equal numbers of protons and neutrons to be stable.
D. Atoms that have less than 10 protons do not need neutrons to be stable.
Answer:
A and B i think
Explanation:
(A) For large atoms, more neutrons than protons are needed to be stable.
(B) Nuclei that have 90 or greater protons are always radioactive.
Stability of nucleus of an atom
A stable nucleus of an atom must have nutron-to-proton ratio of at least one. This implies that the nucleus must have more neutrons than protons.
From the graph we can conclude the following about stability of nucleus of an atom;
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A source of emf is connected by wires to a resistor, and electrons flow in the circuit. The wire diameter is the same throughout the circuit. Compared to the drift speed of the electrons before entering the source of emf, the drift speed of the electrons after leaving the source of emf is __________.A. faster. B. slower. C. the same. D. either A or B depending on circumstances. E. any of A, B, or C depending on circumstances.
Answer:
C. the same.
Explanation:
The drift speed of electrons in a circuit is the same all through the circuit.
The drift speed of the electrons remains the same before and after they pass through the source of emf because the current is constant throughout the circuit assuming a steady state situation. The emf source does not affect the drift speed despite functioning like a pump that maintains the potential difference.
Explanation:The drift speed of the electrons after leaving the source of emf is the same as before entering it. This is so because, in a steady-state situation where the current is constant throughout a circuit, the rate at which electrons enter any given section of the circuit must equal the rate at which they exit. The drift speed is the average speed of the electrons as they move through the conductor, and it remains constant throughout the entirety of the circuit assuming the diameter of the wire is consistent.
It's important to understand that the emf source acts like a pump, helping move the electrons through the circuit and maintaining a potential difference. However, this doesn't result in an increase in the electron drift speed, as the overall velocity of the electron stays steady. Even in the presence of an electric field, the phenomenon known as drift velocity remains unaffected due to the near-random movements of the free electrons caused by their collisions with atoms and other free electrons.
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How many steps in glycolysis have atp as a substrate or product?
Answer:
4
Explanation:
1. Phosphorylation of Glucose
2. Production of Fructose-6 Phosphate
3. Production of Fructose 1, 6-Diphosphate
4. Splitting of Fructose 1, 6-Diphosphate
5. Interconversion of the Two Sugars
6. Formation of NADH and 1,3-Diphoshoglyceric acid
7. Production of ATP and 3-Phosphoglyceric Acid
8. Relocation of Phosphorus Atom
9. Removal of Water
10. Creation of Pyruvic Acid and ATP
The process of breaking down glucose into pyruvate while oxygen is present is known as glycolysis.
Thus, It is a multi-step process that the cytoplasm's many enzymes catalyze. It is recognized as the initial stage of the cellular respiration process that all living things go through.
Glucose is converted to glyceraldehyde-3-phosphate during the preliminary phase, and glyceraldehyde-3-phosphate is converted to pyruvate during the pay-off phase.
The glycolysis process produces the chemicals NADH and ATP. The ten steps of glycolysis include the partial oxidation of glucose to pyruvic acid.
Thus, The process of breaking down glucose into pyruvate while oxygen is present is known as glycolysis.
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Dan is gliding on his skateboard at 3.00 m/s. He suddenly jumps backward off the skateboard, kicking the skateboard forward at 8.00 m/s. Dan's mass is 70.0 kg and the skateboard's mass is 6.00 kg. How fast is Dan going as his feet hit the ground?
Answer:
The velocity of Dan is 2.57m/s as his feet hit the ground
Explanation:
The impact experience by Dan and the skate board is an elastic collision
Collision is elastic when the kinetic energy is not conserved and if there is rebound after collision
Given that
U= initial velocity of Dan and the skate board 3m/s
M1 =mass of Dan 70kg
M2= mass of Skate board 6kg
V1= final velocity of Dan?
V2= Final velocity of skate board 8m/s
The expression for Dan and the skate board collision can be expressed as
Momentum before impact momentum after impact
(M1+M2)U=M1V1+M2V2
Substituting our data we have
(70+6)3=(70*V1)+(6*8)
228=70V1+48
Solving for V1
228-48=70V1
V1=180/70
V1=2.57m/s
Calculate the electrical energy expended in a device across which the circuit voltage drops by 20.0 volts in moving a charge of 4 coulombs.
A)5 J
B)80 J
C)0.2 J
Answer:
The answer is 80 Joules
Explanation:
Electrical energy = Q x V
Energy = 2 x 40
= 80
I just took the test and it was right :D
The correct answer is B.
I hope this helped! :D
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Through how many revolutions does the tub turn during this 22 s interval? Assume constant angular acceleration while it is starting and stopping.
Answer:
22 revolutions
Explanation:
2 rev/s = 2*(2π rad/rev) = 12.57 rad/s
The angular acceleration when it starting
[tex]\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2[/tex]
The angular acceleration when it stopping:
[tex]\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2[/tex]
The angular distance it covers when starting from rest:
[tex]\omega^2 - 0^2 = 2\alpha_a\theta_a[/tex]
[tex]\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad[/tex]
The angular distance it covers when coming to complete stop:
[tex]0 - \omega^2 = 2\alpha_o\theta_o[/tex]
[tex]\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad[/tex]
So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions
I need some help with my science homework about energy, work, and power. It would be greatly appreciated :)
Answer:
1. B
Explanation:
Work = Force × Distance
Work = 4N × 1.5M
Consider a coin which is tossed straight up into the air. After it is released it moves upward, reaches its highest point and falls back down again. Air resistance can be neglected. What force acts on the coin while it moves upward after it is released
Answer:
GRAVITATIONAL FORCE
Explanation:
We may have noticed that a body thrown upward in air falls back down again after attaining a particular height. The object was able to fall down back due to the effect of gravity acting on it. If there are no force of gravity acting on the body, the body will not fall back but rather disappears into the thin air.
A coin tossed upward in the air which falls back down when released is therefore under the influence of gravity i.e GRAVITATIONAL FORCE while it moves upward after it is released
Technician A says a pull to the right during braking could be caused by a defective metering valve. Technician B says a pull to the left could be caused by a defective proportioning valve. Which technician is correct
Answer:
The correct option is;
Neither A nor B
Explanation:
The common cause of a vehicle pulling to the right or to the left during braking is due to a contaminated breaking surface or a faulty caliper. Other causes include tire size variation or worn out suspension components.
Other causes include;
Leaking break fluid
Piston frozen in wheel cylinder or caliper
Adjuster of rear break
Tire fault.
As soon as it is observed that the vehicle pulls on one side when braking, the vehicle should should be taking for checks by a qualified mechanic as soon as possible.
Final answer:
Both Technician A and B are incorrect, as a pull during braking is more likely due to uneven brake pad wear, sticking calipers, or collapsed brake lines rather than a defective metering or proportioning valve.
Explanation:
The question relates to whether a pull to the right during braking could be caused by a defective metering valve, and whether a pull to the left could be caused by a defective proportioning valve. Neither of these assertions is typically accurate.
A metering valve is designed to ensure that the rear brakes engage just after the front brakes, to prevent rear wheel lockup during early brake application. A proportioning valve adjusts the pressure between the front and rear brakes to prevent the rear wheels from locking up during heavy braking. Neither valve is designed to balance the braking force from side to side.
A pull to one side during braking is more likely to be caused by uneven brake pad wear, sticking calipers, or a collapsed brake line, rather than issues with the metering or proportioning valves. Therefore, both Technician A and Technician B are incorrect regarding the cause of a pull during braking.
A speed skater is travelling at 2 m/s and accelerates uniformly to 4 m/s in 5 seconds. What is her acceleration?
Answer:
The answer to your question is a = 0.4 m/s²
Explanation:
Data
speed 1 = 2 m/s
speed 2 = 4 m/s
time = 5 s
Acceleration measures the change of speed over a unit of time.
Formula
Acceleration = (speed 2 - speed 1) / time
Substitution
Acceleration = (4 - 2) / 5
Simplification
Acceleration = 2/5
Result
Acceleration = 0.4 m/s²
Answer: her acceleration is 0.4 m/s^2
If a wave has a speed of 100 m/s and a wavelength of 20 meters, what is the frequency?
Answer:
Frequency 5hz
Explanation:
Final answer:
To calculate the frequency of a wave with a speed of 100 m/s and a wavelength of 20 meters, divide the speed by the wavelength, resulting in a frequency of 5 Hz.
Explanation:
To find the frequency of a wave when you know its speed and wavelength, you can use the formula: speed = (wavelength) x (frequency). Rearranging this formula to solve for frequency gives us frequency = speed / wavelength. In this case, with a wave speed of 100 m/s and a wavelength of 20 meters, the calculation would be frequency = 100 m/s / 20 m.
Therefore, the frequency of the wave is 5 Hz. This means the wave completes 5 cycles per second.
The speed of propagation equals the frequency times the wavelength.
O
True
O
False
Answer:
true
Explanation:
because I say it's true
Answer:help me with my physics please
Explanation:
The function x = (5.2 m) cos[(5πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 5.3 s, what are the (a) displacement, (b) velocity, (c) acceleration, and (d) phase of the motion? Also, what are the (e) frequency and (f) period of the motion?
Answer:
(a) Displacement = - 3.0576 m
(b) Velocity [tex]=-66.48[/tex] m/s
(c)Acceleration = -753.39 m²/s
(d)The phase motion is 26.7 [tex]\pi[/tex].
(e)Frequency =2.5 Hz.
(f)Time period =0.4 s
Explanation:
Given function is
[tex]x= (5.2 m)cos[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
(a)
The displacement includes the parameter t, so,at time t=5.3 s
[tex]x|_{t=5.3}= (5.2 m)cos[ (5\pi \ rad/s)5.3+ \frac\pi5][/tex]
[tex]= (5.2 m)cos[ 26.5\pi+ \frac\pi5][/tex]
=(5.2)(-0.588)m
= - 3.0576 m
(b)
[tex]x= (5.2 m)cos[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
To find the velocity of simple harmonic motion, we need to find out the first order derivative of the function.
[tex]v=\frac{dx}{dt}[/tex]
[tex]=\frac{d}{dt} (5.2 m)cos[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
[tex]= (5.2 m)(-5\pi)sin[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
[tex]= -26\pi sin[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
Now we can plug our value t=5.3 into the above equation
[tex]v= -26\pi sin[ (5\pi \ rad/s)5.3\ s+ \frac\pi5][/tex]
[tex]=-66.48[/tex] m/s
(c)
To find the acceleration of simple harmonic motion, we need to find out the second order derivative of the function.
[tex]v= -26\pi sin[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
[tex]a=\frac{d^2x}{dt^2}[/tex]
[tex]=\frac{dv}{dt}[/tex]
[tex]=\frac{d}{dt}( -26\pi sin[ (5\pi \ rad/s)t+ \frac\pi5])[/tex]
[tex]= -26\pi (5\pi)cos[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
[tex]= -130\pi^2cos[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
Now we can plug our value t=5.3 into the above equation
[tex]a= -130\pi^2cos[ (5\pi \ rad/s)5.3 \ s+ \frac\pi5][/tex]
= -753.39 m²/s
(d)
The general equation of SHM is
[tex]x=x_mcos(\omega t+\phi)[/tex]
[tex]x_m[/tex] is amplitude of the displacement, [tex](\omega t+\phi)[/tex] is phase of motion, [tex]\phi[/tex] is phase constant.
So,
[tex](\omega t+\phi)=5\pi t+\frac\pi5[/tex]
Now plugging t=5.3s
[tex](\omega t+\phi)=5\pi \times 5.3+\frac\pi5[/tex]
=26.7 [tex]\pi[/tex]
The phase motion is 26.7 [tex]\pi[/tex].
The angular frequency [tex]\omega = 5\pi[/tex]
(e)
The relation between angular frequency and frequency is
[tex]\omega =2\pi f[/tex]
[tex]\therefore f=\frac{\omega}{2\pi}[/tex]
[tex]=\frac{5\pi}{2\pi}[/tex]
[tex]=\frac52[/tex]
= 2.5 Hz
Frequency =2.5 Hz.
(f)
The relation between frequency and time period is
[tex]T=\frac1 f[/tex]
[tex]=\frac1{2.5}[/tex]
=0.4 s
Time period =0.4 s
Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the man across the floor using a rope that is angled 30 degree above the floor. The tension is a constant 29.0 N and the coefficient of friction is 0.210. Use work and energy to find Paul's speed after being pulled 2.90 m.
Answer:
Paul speed after being pulled 2.9 m is 2.68m/s.
Explanation:
The work energy theorem, change in kinetic energy of the object from initial position to the final position is equal to the work done on the object ie when the force is applied on the object the object changes its position and work is done on the object.
According to the law of conservation of energy ,
ΔE = W, eqn 1
where ΔE is the change in object energy
W is the all the work done on the object.
Work done is written as W =F dcosθ
Where F is the force,
d is the distance,
θ is the angle between the force and displacement vector.
From the figure given below,
The work of friction is given by W₁ = F₁ d cos180°
The work of pulling force is given by W₂ =F₂ dcos 30°
Change in object energy ΔE = mv²/2.
Applying Newton first law along Y axis,
Fsin30° + N =mg
Normal force N =mg - Fsin30°
Frictional Force F₁ =μN =μ(mg - Fsin30°)
Substituting in eqn 1
mv²/2 = F₂ dcos 30°+ μ(mg - Fsin30°)d cos180°
=[tex]\frac{\sqrt{3} }{2}[/tex] F₂ d - μ(mg - [tex]\frac{F}{2}[/tex])d
v² = [tex]\sqrt{3}[/tex] [tex]\frac{F}{m}[/tex]d - 2μgd +
here m = 12 kg,
d = 2.9 m.
μ = 0.21
F = 29 N
Sub all values,
v² = 7.2
v = 2.68m/s
Paul speed after being pulled 2.9 m is 2.68m/s.
The photoelectric effect can be thought of as a simple function: light is an input, and electrons, which produce electricity, are the output. The model in this interactive is missing a few parts to harness this electricity effectively, what are they?
Answer:
(1) Conductive medium (2) Load.
Explanation:
Electricity is essentially Flow of electrons in conductive medium such as wire and to harness it there needs to be a load in the circuitry to use up the energy of the moving electrons through the wire.
this is how solar cells work, in them there are Silicon P and N Junctions connected in series and parallel ( quite alot of them) forming a large plate , which then can be connected to a load through wires to harness electricity produced.
Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much less than the masses of the stars.
In System A , Planet A of mass Mp orbits Star A of mass Ms in a circular orbit of radius R .
In System B , Planet B of mass 4Mp orbits Star B of mass Ms in a circular orbit of radius R .
In System C , Planet C of mass Mp orbits Star C of mass 4Ms in a circular orbit of radius R .
(a) The gravitational force exerted on Planet A by Star A has a magnitude of F0 . Determine the magnitudes of the gravitational forces exerted in System B and System C .
___ Magnitude of gravitational force exerted on Planet B by Star B
___ Magnitude of gravitational force exerted on Planet C by Star C
(b) How do the tangential speeds of planets B and C compare to that of Planet A ? In a clear, coherent paragraph-length response that may also contain equations and/or drawings, provide claims about
why the tangential speed of Planet B is either greater than, less than, or the same as that of Planet A , and
why the tangential speed of Planet C is either greater than, less than, or the same as that of Planet A .
a) 4F0
b) Speed of planet B is the same as speed of planet A
Speed of planet C is twice the speed of planet A
Explanation:
a)
The magnitude of the gravitational force between two objects is given by the formula
[tex]F=G\frac{m_1 m_2}{r^2}[/tex]
where
G is the gravitational constant
m1, m2 are the masses of the 2 objects
r is the separation between the objects
For the system planet A - Star A, we have:
[tex]m_1=M_p\\m_2 = M_s\\r=R[/tex]
So the force is
[tex]F_A=G\frac{M_p M_s}{R^2}=F_0[/tex]
For the system planet B - Star B, we have:
[tex]m_1 = 4 M_p\\m_2 = M_s\\r=R[/tex]
So the force is
[tex]F=G\frac{4M_p M_s}{R^2}=4F_0[/tex]
So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.
For the system planet C - Star C, we have:
[tex]m_1 = M_p\\m_2 = 4M_s\\r=R[/tex]
So the force is
[tex]F=G\frac{M_p (4M_s)}{R^2}=4F_0[/tex]
So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.
b)
The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:
[tex]G\frac{mM}{r^2}=m\frac{v^2}{r}[/tex]
where
m is the mass of the planet
M is the mass of the star
v is the tangential speed
We can re-arrange the equation solving for v, and we find an expression for the speed:
[tex]v=\sqrt{\frac{GM}{r}}[/tex]
For System A,
[tex]M=M_s\\r=R[/tex]
So the tangential speed is
[tex]v_A=\sqrt{\frac{GM_s}{R}}[/tex]
For system B,
[tex]M=M_s\\r=R[/tex]
So the tangential speed is
[tex]v_B=\sqrt{\frac{GM_s}{R}}=v_A[/tex]
So, the speed of planet B is the same as planet A.
For system C,
[tex]M=4M_s\\r=R[/tex]
So the tangential speed is
[tex]v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A[/tex]
So, the speed of planet C is twice the speed of planet A.