Answer:
[tex]-0.5\times 10^{-4} A/s[/tex]
Explanation:
We are given that
[tex]\frac{dV}{dt}=-0.01 V/s[/tex]
R=600 ohms
I=0.04 A
[tex]\frac{dR}{dt}=0.5ohm/s[/tex]
[tex]V=IR[/tex]
[tex]\frac{dV}{dt}=\frac{\partial V}{dI}\frac{dI}{dt}+\frac{\partial V}{dR}\frac{dR}{dt}[/tex]
[tex]\frac{dV}{dt}=R\frac{dI}{dt}+I\frac{dR}{dt}[/tex]
Substitute the values
[tex]-0.01=600\times \frac{dI}{dt}+0.04\times 0.5[/tex]
[tex]-0.01-0.04\times 0.5=600\frac{dI}{dt}[/tex]
[tex]-0.03=600\frac{dI}{dt}[/tex]
[tex]\frac{dI}{dt}=\frac{-0.03}{600}=-0.5\times 10^{-4}A/s[/tex]
Final answer:
The student's question concerns finding the rate of change of current in a circuit as described by Ohm's law. By finding partial derivatives of the voltage with respect to current and resistance and applying the provided rates, you solve for the rate of change of current.
Explanation:
The student is asking how the current I is changing over time in a circuit with a given resistance R, current I, and rates of change dR/dt and dV/dt according to Ohm's law. To solve this, we need to find the partial derivatives of voltage V with respect to current I and resistance R, and then use the given differential equation dV dt = ∂V ∂I dI dt + ∂V ∂R dR dt.
Using Ohm's law, we can express the voltage as V = IR. The partial derivative of V with respect to I is R, and with respect to R is I. Substituting these derivatives and the given values into the differential equation, we have: -0.01 = 600 dI dt + 0.04 * 0.5. Solving for dI dt, we get the rate at which the current is changing over time, which is the answer the student is looking for.
"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.
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:
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.
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.
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).
You stop for a cappuccino at a coffee shop and notice that the tiny white bubbles of steamed milk remain on the surface of the coffee. These air-filled bubbles stay where they are, rather than descending into the coffee or rising into the air, because they are:
A. more dense than the coffee but less dense than the air above thecoffee.
B. thicker than the coffee but less thick than the air above thecoffee.
C. less dense than the coffee but more dense than the air above thecoffee.
D. lighter than the cup of coffee but heavier than the column ofair above the coffee.
Answer:
C. less dense than the coffee but more dense than the air above thecoffee.
Explanation:
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
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
what does a electric field of two positive charges look like
Hope this helped you
:)
Have a great day!
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
why is it painful to lift a heavy load with a thin piece of string?
Explanation:
A thin piece of string has lower force than of a heavy load.
When the load is being lifted, the force exerted by the load is much more greater than the string, which eventually hits our hands.
For example, a heavy bag. You tie a thin string to it and try to lift it. The force exerted by the bag will hit the string harder, reaching for your plams. As the thin string has less force, its reaction force is not enough to hit back the greater force. Also, the less the surface area, the more difficult the grip gets. But, if you attach a thick rope or belt, the force exerted my the bag is automatically minimized as the reaction force of the thick rope is near about the action force. Hence, greater the surface area, better the grip.
*action force: force exerted by the bag
*reaction force: the force hit back by the rope
It is painful to lift a heavy load with a thin piece of string because the string has a small cross-sectional area and cannot withstand a large amount of tension without breaking.
What is tensile strength?Tensile strength is the maximum amount of tensile stress that a material can withstand before it breaks or undergoes permanent deformation. It is a measure of the material's ability to resist external forces that try to pull it apart or elongate it.
Tensile strength is an important mechanical property of materials, and it is commonly used in engineering and design to select materials for specific applications. The tensile strength of a material is typically determined through a tensile test, where a sample of the material is subjected to a gradually increasing tensile force until it breaks.
The tensile strength of a material depends on its composition, microstructure, and processing conditions. For example, materials that have a high degree of crystallinity and are free from defects or impurities generally have a higher tensile strength than materials with a less ordered microstructure or defects. In addition, the processing conditions, such as temperature and strain rate, can also affect the tensile strength of a material.
The tensile strength is typically reported in units of force per unit area, such as megapascals (MPa) or pounds per square inch (psi). The higher the tensile strength of a material, the greater its ability to withstand tensile stress without breaking or undergoing permanent deformation.
Here in the Question,
It is painful to lift a heavy load with a thin piece of string because the string has a small cross-sectional area and cannot withstand a large amount of tension without breaking. When a heavy load is attached to the string and lifted, the weight of the load creates a tension force in the string that is equal to the weight of the load. If the tension force in the string exceeds its maximum tensile strength, it will break.
The maximum tensile strength of a string depends on its material properties and cross-sectional area. When a thin piece of string is used to lift a heavy load, the tension force in the string can easily exceed its maximum tensile strength, causing it to break. This sudden release of tension can also cause the load to drop suddenly, which can lead to injury.
This can be explained using the concept of stress and strain. Stress is the force applied to an object per unit area, while strain is the deformation of the object due to the applied stress. When a heavy load is attached to a thin piece of string, the weight of the load creates a large stress in the string, which can cause it to undergo plastic deformation and break. This is because the string has a small cross-sectional area, which means that the stress is concentrated over a small area, leading to a high level of strain.
Therefore, a thicker piece of string or rope has a larger cross-sectional area, which means that the stress is spread out over a larger area. This allows it to withstand a greater amount of tension before breaking. Therefore, it is important to use a string or rope with a sufficient cross-sectional area when lifting heavy loads to prevent injury and damage.
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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.
5. Suppose a cup of cocoa cooled from 90° C to 60° C after 10 minutes in a room whose temperature was 20° C . Use Newton’s law of cooling [T − Ts = (T0 − Ts )⋅ e−k⋅t ] to answer the following questions: a) How much longer would it take the cocoa to cool to 35° C ? b) Instead of being left to stand in a room, the cup with initial temperature 90° C is placed in a freezer whose temperature is −15° C . How long will it take the cocoa to cool from 90° C to 35° C ?
Answer:
a) t = 1051.6 sec = 17.5 min
b) t = 795.5 sec = 13.25 min
Explanation:
First of all we use the initial data to find out constant 'K'.
T - Ts = (T₀ - Ts) e^(-kt)
Here, we have:
T = Final Temperature = 60° C
Ts = Surrounding Temperature = 20° C
T₀ = Initial Temperature = 90° C
t = time = 10 min = 600 sec
k = constant = ?
Therefore,
60° C - 20° C = (90° C - 20° C).e^(-k600)
40° C/70° C = e^(-k600)
ln (0.57142) = -600k
k = 9.327 x 10⁻⁴ sec⁻¹
a)
Now, for this case we have:
T = Final Temperature = 35° C
Ts = Surrounding Temperature = 20° C
T₀ = Initial Temperature = 60° C
t = time = ?
k = constant = 9.327 x 10⁻⁴ sec⁻¹
Therefore,
35° C - 20° C = (60° C - 20° C).e^(-9.327 x 10⁻⁴ sec⁻¹ x t)
15° C/40° C = e^(-9.327 x 10⁻⁴ sec⁻¹ x t)
ln (15/40) = - 9.327 x 10⁻⁴ sec⁻¹ x t
t = 1051.6 sec = 17.5 min
b)
Now, for this case we have:
T = Final Temperature = 35° C
Ts = Surrounding Temperature = -15° C
T₀ = Initial Temperature = 90° C
t = time = ?
k = constant = 9.327 x 10⁻⁴ sec⁻¹
Therefore,
35° C + 15° C = (90° C + 15° C).e^(-9.327 x 10⁻⁴ sec⁻¹ x t)
50° C/105° C = e^(-9.327 x 10⁻⁴ sec⁻¹ x t)
ln (50/105) = - 9.327 x 10⁻⁴ sec⁻¹ x t
t = 795.5 sec = 13.25 min
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
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
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.
How do the three opsin molecules respond to different wavelengths of light when their retinal molecular structures are identical?
Answer:
The three opsin molecules respond to different wavelengths of light because of the different structure of the protein bound to the opsin molecules.
Explanation:
Opsin is a protein that is released by the action of light and forms part of the visual pigment rhodopsin. There are two groups of protein termed opsins, type I opsins, which are employed by prokaryotes and by some algae and fungi, and type II opsins which are used by animals.
Though the retinal molecular structures of the opsin molecules are identical, they respond to different wavelengths of light because of the different structure of the protein bound to the opsin molecules.
A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the assembly expands from an initial volume of 0.90 L to a final volume of 5.92 L against an external pressure of 1.0 atm. Use the wattage of the lightbulb and the time it is on to calculate ΔE in joules (assume that the cylinder and lightbulb assembly is the system and assume two significant figures). Calculate w and q.
Answer:
[tex]w = - 508.53[/tex] joules
[tex]q = - 3091.47[/tex] joules
Explanation:
Let us convert the time in hours into seconds
[tex]0.010* 3600\\= 36[/tex]
Change in internal energy
[tex]\delta E = p * \delta t[/tex]
where E is the internal energy in Joules
p is the power in watts
and t is the time in seconds
[tex]\delta E = - 100 * 36\\[/tex]
[tex]\delta E = - 3600[/tex] Joules
Amount of work done by the system
[tex]w = - P * \delta V[/tex]
where P is the pressure and V is the volume
Substituting the given values in above equation, we get -
[tex]w = - 1 * ( 5.92 -0.90)\\[/tex]
[tex]w = -5.02[/tex] liter-atmospheres
Work done in Joules
[tex]- 5.02 * 101.3\\[/tex][tex]= 508.53[/tex]Joules
[tex]q = \delta E - w\\[/tex]
Substituting the given values we get -
[tex]q = - 3600 - (-508.53)\\q = - 3091.47[/tex]
Thus
[tex]w = - 508.53[/tex] joules
[tex]q = - 3091.47[/tex] joules
The change in energy (ΔE) is 1 joule (J), the work (w) is -41 J, and the heat (q) is 42 J.
Explanation:To calculate the change in energy (ΔE) in joules, we can use the formula ΔE = power (P) x time (t). In this case, the power of the lightbulb is 100 W and the time it is on is 0.010 hour. Therefore, ΔE = 100 W x 0.010 hour = 1 joule (J).
To calculate work (w), we can use the equation w = -PΔV, where ΔV is the change in volume of the assembly. Here, ΔV = final volume - initial volume = 5.92 L - 0.90 L = 5.02 L. Given that the external pressure is 1.0 atm, we can convert the volume to liters-atmospheres (L·atm) by multiplying by 0.0821. Therefore, w = -100 W x (5.02 L x 0.0821) = -41 J.
Finally, to calculate heat (q), we can use the first law of thermodynamics, which states that q = ΔE - w. Substituting the values, q = 1 J - (-41 J) = 42 J.
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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
Water behind a dam has a certain amount of stored energy that can be released as the water falls over the top of the dam. It may be enough energy to turn a mill wheel or an electricity-generating turbine. Choose the term that best describes the type of energy stored in the water at the top of the dam.
potential energy
kinetic energy
mechanical energy
Answer:
The answer is potential energy
Explanation:
The potential energy is the energy possessed by a body by virtue of it position
For example the water at the top of the dam is being held at a height h above the bottom of the dam
Then the potential energy
PE= weight of the water* the height
PE= m*g*h
A 2.9 kg ball strikes a wall with a velocity of 8.6 m/s to the left. The ball bounces off with a velocity of 7.3 m/s to the right. If the ball is in contact with the wall for 0.33 s, what is the constant force exerted on the ball by the wall? Answer in units of N.
Answer:
The constant force exerted by the wall is F=11.4N
Explanation:
The problem bothers on the impulse of a force
Which is given as
Ft=mv-mu
Ft=m(v-u)
Given data
mass of ball m =2.9kg
Final speed v=8.6m/s
Initial speed u=7.3m/s
Time t= 0.33s
Substituting to find F
F*0.33=2.9(8.6-7.3)
F*0.33=2.9*1.3
F=3.77/0.33
F=11.4N
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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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.
Conventional current is the rate at which positive charge flows in a circuit. In atoms, only the electrons are free to move. What can you conclude from these statements?
A) Electrons must be positively charged.
B) The definition of current must be changed.
C) Electrons do not flow in electric circuits.
D) Charges actually flow opposite the conventional current.
Answer:
In a metal circuit, the charges which are free electrons flow opposite to the flow of conventional current(which is assumed as the flow of positive charges)
Explanation:
Conventional current is defined as the direction that positive charge would flow, which is opposite to the actual flow of electrons in a circuit.
From the statements given, we can conclude that charges actually flow opposite the conventional current. This is because conventional current assumes that positive charge is moving in the direction of the electric field, whereas in actuality, in metal wires, it is the electrons— which have a negative charge—that are moving.
Electrons flow in a direction opposite to the defined conventional current. The designation of the direction of conventional current dates back to Benjamin Franklin's time, and despite later discovery that electrons are the primary charge carriers in circuits, the convention has remained the same.
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;
For large atoms, more neutrons than protons are needed to be stable.Nuclei that have 90 or greater protons are always radioactive.Learn more about stability of nucleus here: https://brainly.com/question/1729336
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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]
Un cuerpo de 10N de peso esta apoyado sobre una superficie horizontal. Se le ata una cuerda y se tira de él con una fuerza de 15 N que forma un ángulo de 60° con la horizontal. ¿Cuál es la fuerza normal sobre el cuerpo?
La fuerza normal sobre el cuerpo es de [tex]\(5 \, N\)[/tex].
Explanation:El problema implica un cuerpo de [tex]\(10 \, N\)[/tex] de peso que está siendo sometido a una fuerza de tracción de [tex]\(15 \, N\)[/tex] a un ángulo de [tex]\(60^\circ\)[/tex] con la horizontal. La fuerza normal es la componente perpendicular de la fuerza peso, ya que no hay movimiento vertical. Utilizamos la relación trigonométrica [tex]\(F_{\text{normal}} = F_{\text{peso}} \cdot \cos(\theta)\)[/tex], donde [tex]\(F_{\text{peso}}\)[/tex] es el peso del cuerpo y [tex]\(\theta\)[/tex] es el ángulo entre la fuerza peso y la horizontal.
En este caso, la fuerza peso [tex]\(F_{\text{peso}}\) es \(10 \, N\)[/tex] hacia abajo, y el ángulo [tex]\(\theta\)[/tex] es [tex]\(60^\circ\)[/tex]. Aplicando la fórmula, obtenemos [tex]\(F_{\text{normal}} = 10 \, N \cdot \cos(60^\circ)\)[/tex]. Calculando esto, encontramos [tex]\(F_{\text{normal}} = 10 \, N \cdot 0.5 = 5 \, N\)[/tex].
La fuerza normal es, por lo tanto, [tex]\(5 \, N\)[/tex], lo que significa que la superficie horizontal ejerce una fuerza hacia arriba de [tex]\(5 \, N\)[/tex] para equilibrar la componente vertical de la fuerza aplicada.
Este resultado es consistente con el principio de equilibrio en el plano horizontal. La fuerza normal contrarresta la componente vertical de la fuerza aplicada, manteniendo el cuerpo en equilibrio sin movimiento vertical. Este enfoque, basado en las leyes de la trigonometría y el equilibrio, proporciona una solución clara y precisa para el problema.
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.
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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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