Answer:
If the stone is thrown from the ground, the correct answer is A. If it is thrown from a height h, the correct answer is D.
Explanation:
Hi there!
I can´t see the drawing but let´s assume that initially, the stone is on the ground level. If that is the case, initially, the potential energy will be zero and when it returns to Earth it will also be zero. The potential energy depends on the height of the stone. If the final and initial height of the stone is zero, then the change in potential energy will also be zero:
ΔPE = final PE - initial PE
ΔPE = m · g · hf - m · g · hi (where hf and hi are the final and initial height respectively)
ΔPE = m · g (hf - hi)
ΔPE = m · g (0)
ΔPE = 0
Initially, the kinetic energy (KE) of the stone is the following:
KE = 1/2 · m · v0²
As the stone goes up, the kinetic energy is transformed into potential energy; but as the stone starts to fall, the acquired potential energy is transformed again into kinetic energy, so that the final and initial kinetic energy of the stone is the same.
Then:
ΔKE = final KE - initial KE = 0 (because final KE = initial KE).
Then, the correct answer is A.
Always ΔKE = -ΔPE due to the conservation of energy. Potential energy can´t be acquired by the stone if there is no loss of kinetic energy and vice-versa.
Let´s assume now that the stone is thrown from a height hi to the ground.
The final potential energy will be zero (becuase h = 0) but the initial PE will be:
PE = m · g · h1
Then:
ΔPE = final PE - initial PE = 0 - m · g · h1
Then ΔPE will be negative.
The initial kinetic energy will be:
KE = 1/2 · m · v0²
But the final kinetic energy will be equal to the initial kinetic energy plus the loss of potential energy (remember: if potential energy decreases, another type of energy has to increase, in this case, kinetic energy and vice-versa):
ΔKE = final KE - initial KE
ΔKE = 1/2 · m · v0² + m · g · h1 - 1/2 · m · v0²
ΔKE = m · g · h1
Then ΔKE will be positive and the correct answer would be D.
You’re responsible for server room security. You’re concerned about physical theft of computers. Of the following, which would best be able to detect theft or attempted theft?
A) Motion-sensor activated cameras
B) Smart card access to the server rooms
C) Strong deadbolt locks for server rooms
D) Logging everyone who enters the server room
Answer:
A) Motion-sensor activated cameras
Explanation:
Motion sensor activated cameras are the ones which actuate a trigger when any sort of motion is detected by them. They can play alarming sound, send e-mails and zoom the frame of motion as a response to the detected motion.
Motion sensor cameras can be of two types:
1. Fixed view camera
2. Rotating view camera
Fixed view cameras are in fixed position and they always point in the same direction for the view while the rotating cameras are the ones which can change their direction of view.
For a small and high security space we can use the fixed type camera so that we do not miss any suspicious movements. For wide areas and distant vision we use rotating cameras.
Transmission of Borrelia burgdorferi is most likely to occur after a tick feeds a minimum of _______ hours
Answer:
36 hours
Explanation:
Transmission of Borrelia burgdorferi is most likely to occur after a tick feeds a minimum of 36 hours.
The infected ticks are most likely to transmit infection after approximately 2 or more days after feeding. If it remains unchecked, the B. Burgdorferi may trigger systemic infection by passing through the bloodstream and being rooted in different body tissues.
If you experience __________, do not use brakes, concentrate on steering, slow down gradually, brake softly when the car is under control, and pull completely off the pavement.
Answer:
Tire blowout
Explanation:
Tire Blowout
A tire experiences a blowout when it rapidly and explosively loses the inflation pressure. It can happen when some sharp object cuts or tears the surface and the air is suddenly released, making things worse.
The driver frequently loses control of the traveling direction and can be involved in serious accidents. It is recommended not to use brakes during or after the blowout because it could bring little help or make things worse.
Focusing on steering and let the vehicle softly brake.
When light energy hits the retina, the retinal changes from a _____ to a _____ configuration.
Answer:
Cis, Trans.
Explanation:
Rhodopsin also known as visual purple, pigment which contains sensory protein that helps to convert light into an electrical signal. Rhodopsin present in wide range of organisms from bacteria to vertebrates.
Rhodopsin is composed of opsin, and 11-cis-retinaldehyde which is derived from vitamin A. When the eye contact with light the 11-cis component converted to all trans-retinal, which results in the changes in configuration fundamental in the rhodopsin molecule.
Some substances have the same chemical composition, but their atoms are arranged differently. A classic example is carbon. A pencil lead is made of graphite, which contains only carbon atoms. A diamond also contains only carbon atoms. However, the carbon atoms in a diamond are packed more closely together. Would you expect the density of graphite and a diamond to be the same? Explain your answer.
Answer: No, because the atoms are arranged differently. Looking at 100 atoms in each sample, the volume would be smaller in a diamond, because the atoms are packed more closely together. The density of a diamond would be higher.
Explanation:
Final answer:
Graphite and diamond have different densities due to the varied arrangements of carbon atoms in their structures, with diamonds being denser and harder due to a three-dimensional bond network.
Explanation:
The difference in density between graphite and diamond can be attributed to the variation in how carbon atoms are arranged in these two allotropes of carbon. Graphite has a layered structure with weak bonds between each layer, allowing the sheets to easily slip past each other, leading to its softer structure and lower density.
In contrast, diamonds have carbon atoms bonded together in all three dimensions, creating a dense, strong lattice, which is why diamonds are very hard and have a higher density compared to graphite.
On the planet Zorb, the acceleration due to gravity is 10 meters per second squared. If you were to launch a projectile at an angle of 30 degrees with an initial velocity of 10 meters per second, in seconds, how long would it take for the projectile to fall to the ground?
Answer:
The time taken by the projectile to fall to the ground is 1 second.
Explanation:
It is given that,
Acceleration due to gravity on the planet Zorb, [tex]a=10\ m/s^2[/tex]
Angle of projectile is 30 degrees
Initial velocity of the projectile, u = 10 m/s
Let t is the time taken for the projectile to fall to the ground. Using second equation of motion to find it
[tex]h=u\ sin\theta t+\dfrac{1}{2}at^2[/tex]
When it fall to the ground, h = 0 and a = -g
[tex]10\ sin(30)t-\dfrac{1}{2}\times 10t^2=0[/tex]
On solving the above quadratic equation, we get the value of t as,
t = 1 second
So, the time taken for the projectile to fall to the ground is 1 second. Hence, this is the required solution.
A series combination of two resistors of resistance 12 Ω and 8 Ω is connected across a source of emf 24 V. What is the potential difference across 8 Ω resistors?
Answer:
The Potential difference across the 8 Ω resistor = 9.6 V
Explanation:
In a series combination of resistor, The total resistance
Rt = R₁ + R₂............... Equation 1
Where Rt = total resistance, R₁ = Resistance of the first resistor, R₂ = resistance of the second resistor.
Given: R₁ = 12Ω, R₂ = 8Ω.
Substituting these vales into equation 1
Rt = 12 + 8
Rt = 20 Ω.
Since both resistors are connected in series, The same current flows through both resistors.
from ohms law,
V = IR..................... Equation 3
I = V/R....................... Equation 2
where I = current flowing through both resistors, V= Emf, R = total resistance of the circuit
Also given : V = 24 V, R = 20 Ω,
Substituting these values into equation 3,
I = 24/20
I = 1.2 A.
Note: Since both resistors are connected in series, The same current flows through both resistors.
The Potential difference across the 8 Ω resistor
V = 1.2×8
V = 9.6 V
The Potential difference across the 8 Ω resistor = 9.6 V
Worker A works at a constant rate and, working alone, can complete a job in 6 hours. Worker B works at a constant rate and, working alone, can complete the same job in 5 hours. Worker C works at a constant rate and, working alone, can complete the same job in 3 hours.
Answer:
[tex]R= \frac{1job}{1.429 hours}[/tex]
So then we will have that the 3 working together will complete 1 job in approximately 1.429 hours for this case.
Explanation:
If we want to express the situation in math terms and find the number of hours that takes to complete 1 job with the 3 at the same time, we can do this.
For this case we have the following rates:
[tex]R_A =\frac{1job}{6hours}[/tex]
[tex]R_B =\frac{1 job}{5 hours}[/tex]
[tex]R_C=\frac{1job}{3 hours}[/tex]
And we know that working together the rate would be the addition of the rates like this:
[tex]R=R_A +R_B +R_C = \frac{1job}{6hours}+\frac{1job}{5hours}+\frac{1job}{3hours} =\frac{7 jobs}{10hours}[/tex]
And if we divide the numerator and denominator by 7 we got:
[tex]R= \frac{1job}{1.429 hours}[/tex]
So then we will have that the 3 working together will complete 1 job in approximately 1.429 hours for this case.
If you hold a bar magnet in each hand and bring your hands together, will the force be attractive or repulsive if the magnets are held:
a) with the two north poles together?
b) with a north pole and south pole together?
a) The force is repulsive
b) The force is attractive
Explanation:
Every magnet has a magnetic field around it. It is possible to distinguish two different poles in the magnet, according to the direction of the magnetic field: in particular, the lines of the field go out from the North pole and go into the South Pole. When a magnet is broken, two new magnets are formed, each of them having its own north and south pole.
The force between two magnets can be either attractive or repulsive, depending on which poles are facing each other. We have the following situation:
The magnetic force between two like poles (north-north and south-south) is repulsiveThe magnetic force between two opposite poles (north-south) is attractiveTherefore, we have the following situations in this problem:
a)
Here we are holding the two north poles together: since they are like poles, they repel each other, so the force in this case is repulsive
b)
Here we are holding a north pole and a south pole together: since they are opposite poles, they attract each other, so the force in this case is attractive
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A lead bullet at 27°C just melts when stopped by an obstacle. Assuming that 25 % of heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is 27^{\circ}C. [For lead, melting point = 327^{\circ}C, specific heat = 0.03 cal/g-^{\circ}C, latent heat of fusion = 6 cal/g and J = 4.2 J/cal.]
Answer:
409.87803 m/s
Explanation:
v = Velocity of bullet
L = Latent heat of fusion = 6 cal/g
c = Specific heat of lead = 0.03 cal/g°C
[tex]\Delta T[/tex] = Change in temperature = (327-27)
m = Mass of bullet
[tex]1\ J=4.2\ J/cal[/tex]
The heat will be given by the kinetic energy of the bullet
[tex]Q=\dfrac{1}{2}mv^2[/tex]
According to the question
[tex]Q=0.75\dfrac{1}{2}mv^2[/tex]
This heat will balance the heat going into the obstacle
[tex]Q=mc\Delta T+mL\\\Rightarrow 0.75\dfrac{1}{2}mv^2=m(c\Delta T+L)\\\Rightarrow v^2=\dfrac{2}{0.75}\times (0.03\times (327-27)+6)\\\Rightarrow v^2=40\ kcal\\\Rightarrow v^2=40\times 4.2\times 10^3\\\Rightarrow v^2=168000\ m^2s^2\\\Rightarrow v=\sqrt{168000}\\\Rightarrow v=409.87803\ m/s[/tex]
The speed of the bullet is 409.87803 m/s
a person can throw a 300 gram ball at 25 m/s the maximum height the person can throw the ball on Earth?
Answer:
Elevation =31.85[m]
Explanation:
We can solve this problem by using the principle of energy conservation. This consists of transforming kinetic energy into potential energy or vice versa. For this specific case is the transformation of kinetic energy to potential energy.
We need to first identify all the input data, and establish a condition or a point where the potential energy is zero.
The point where the ball is thrown shall be taken as a reference point of potential energy.
[tex]E_{p} = E_{k} \\where:\\E_{p}= potential energy [J]\\ E_{k}= kinetic energy [J][/tex]
m = mass of the ball = 300 [gr] = 0.3 [kg]
v = initial velocity = 25 [m/s]
[tex]E_{k}=\frac{1}{2} * m* v^{2} \\E_{k}= \frac{1}{2} * 0.3* (25)^{2} \\E_{k}= 93.75 [J][/tex]
[tex]93.75=m*g*h\\where:\\g = gravity = 9.81 [m/s^2]\\h = elevation [m]\\replacing\\h=\frac{E_{k}}{m*g} \\h=\frac{93.75}{.3*9.81} \\h=31.85[m][/tex]
An 899-kg (1987 lb) dragster, starting from rest, attains a speed of 26.3 m/s (58.9 mph) in 0.59 s. (a) Find the average acceleration of the dragster during this time interval. 44.58 Correct: Your answer is correct. m/s2 (b) What is the magnitude of the average net force on the dragster during this time
Explanation:
Mass, m = 899 kg
Initial velocity, u = 0 m/s
Final velocity, v = 26.3 m/s
Time, t = 0.59 s
a) We have equation of motion v = u + at
Substituting
v = u + at
26.3 = 0 + a x 0.59
a = 44.58 m/s²
Acceleration of dragster = 44.58 m/s²
b) Force = Mass x Acceleration
F = ma
F = 899 x 44.58 = 40074.07 N
F = 40.07 kN
Average net force on the dragster during this time is 40.07 kN
A child's toy consists of a spherical object of mass 50 g attached to a spring. One end of the spring is fixed to the side of the baby's crib so that when the baby pulls on the toy and lets go, the object oscillates horizontally with a simple harmonic motion. The amplitude of the oscillation is 6 cm and the maximum velocity achieved by the toy is 3.2 m/s . What is the kinetic energy K of the toy when the spring is compressed 5.1 cm from its equilibrium position?
Problem-Solving Strategy: Simple Harmonic Motion II: Energy
The energy equation, E=12mvx2+12kx2=12kA2, is a useful alternative relationship between velocity and position, especially when energy quantities are also required. If the problem involves a relationship among position, velocity, and acceleration without reference to time, it is usually easier to use the equation for simple harmonic motion, ax=d2xdt2=−kmx (from Newton’s second law) or the energy equation above (from energy conservation) than to use the general expressions for x, vx, and ax as functions of time. Because the energy equation involves x2 and vx2, it cannot tell you the sign of x or of vx; you have to infer the sign from the situation. For instance, if the body is moving from the equilibrium position toward the point of greatest positive displacement, then x is positive and vx is positive.
IDENTIFY the relevant concepts
Energy quantities are required in this problem, therefore it is appropriate to use the energy equation for simple harmonic motion.
SET UP the problem using the following steps
Part A
The following is a list of quantities that describe specific properties of the toy. Identify which of these quantities are known in this problem.
Select all that apply.
Select all that apply.
maximum velocity vmax
amplitude A
force constant k
mass m
total energy E
potential energy U at x
kinetic energy K at x
position x from equilibrium
Part B
What is the kinetic energy of the object on the spring when the spring is compressed 5.1 cm from its equilibrium position?
Part C
What is the potential energy U of the toy when the spring is compressed 5.1 cm from its equilibrium position?
Answer:
Part A
Mass = 50g
Vmax = 3.2m/s
Amplitude= 6cm
Position x from the equilibrium= 5.1cm
Part B
Kinetic energy = 0.185J
Part C
Potential energy = 0.185J
Explanation:
Kinetic energy = 1/2mv×2
Vmax = wa
w = angular velocity= 53.33rad/s
Kinetic energy = 1/2mv^2×r^2 = 0.185J
Part c
Total energy = 1/2m×Vmax^2= 0.256J
1/2KA^2= 0.256J
K= 142.22N/m (force constant)
Potential energy = 1/2kx^2
=1/2×142.22×0.051^2
= 0.185J
To find the kinetic energy of the toy, we need to use the energy equation for simple harmonic motion and the relationship between velocity and position. We can then substitute the known values to calculate the kinetic energy.
Explanation:In this problem, we are given the amplitude (A) of the oscillation and the maximum velocity (vmax) achieved by the toy. We need to find the kinetic energy (K) of the toy when the spring is compressed 5.1 cm from its equilibrium position.
To solve for the kinetic energy, we can use the energy equation for simple harmonic motion: K = 1/2mvx2, where m is the mass of the object and vx is the velocity of the object at position x. The mass of the object is given as 50 g, which is equal to 0.05 kg.
Since we know the maximum velocity (vmax = 3.2 m/s), we can use the relationship between velocity and position in simple harmonic motion to find the velocity (vx) at a displacement of 5.1 cm from the equilibrium position. The velocity and position in simple harmonic motion are related by vx = ±ω√(A2 - x2), where ω is the angular frequency of the motion.
Substituting the known values into the equations, we can calculate the kinetic energy of the toy.
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if five joules were required to move a crate in 3.7 seconds, what power was applied?
Answer:
The answer to your question is 1.35 Watts
Explanation:
Data
Work = W = 5 J
time = t = 3.7 s
Power = P = ?
Formula
Power is a rate in which work is done or energy is transferred over time
P = [tex]\frac{W}{t}[/tex]
Substitution
[tex]P = \frac{5}{3.7}[/tex]
Result
P = 1.35 W
How many of following statements about the photoelectric effect are true? (a) The greater the frequency of the incident light is, the greater is the stopping potential. (b) The greater the intensity of the incident light is, the greater is the cutoff frequency. (c) The greater the work function of the target material is, the greater is the stopping potential.
Answer:
(a) The greater the frequency of the incident light is, the greater is the stopping potential.
Explanation:
The stopping potential is crucial to terminate the braking of the photoemitted electrons, stopping the current completely.
Since the kinetic energy of electrons depends on the light's incidence frequency (rather than the intensity), therefore, the stopping potential is proportional to the light's incidence frequency.
The cutoff frequency, in turn, is a limiting frequency below which no photoelectric effect occurs. The cutoff frequency depends on the material from which it is made the emitting surface (and its work function).
A 57 kg skier starts from rest at a height of H = 27 m above the end of the ski-jump ramp. As the skier leaves the ramp, his velocity makes an angle of 28° with the horizontal. Neglect the effects of air resistance and assume the ramp is frictionless.
(a) What is the maximum height h of his jump above the end of the ramp?
(b) If he increased his weight by putting on a backpack, would h then be greater, less or, the same?
Answer:
(a)[tex]h=5.95m[/tex]
(b) h is the same
Explanation:
According to the law of conservation of energy:
[tex]E_i=E_f\\U_i+K_i=U_f+K_f[/tex]
The skier starts from rest, so [tex]K_i=0[/tex] and we choose the zero point of potential energy in the end of the ramp, so [tex]U_f=0[/tex]. We calculate the final speed, that is, the speed when the skier leaves the ramp:
[tex]mgH=\frac{mv^2}{2}\\v=\sqrt{2gH}\\v=\sqrt{2(9.8\frac{m}{s^2})(27m)}\\v=23\frac{m}{s}[/tex]
Finally, we calculate the maximum height h above the end of the ramp:
[tex]v_f^2=v_i^2-2gh\\[/tex]
The initial vertical speed is given by:
[tex]v_i=vsin\theta[/tex]
and the final speed is zero, solving for h:
[tex]h=\frac{v_i^2}{2g}\\h=\frac{((23\frac{m}{s})sin(28^\circ))^2}{2(9.8\frac{m}{s^2})}\\h=5.95m[/tex]
(b) We can observe that the height reached does not depend on the mass of the skier
A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle. The takeoff jump was inclined at 53.0 degrees, the river was 40.0 m wide, and the far bank was 15.0m lower than the top of the ramp. The river itself was 100 m below the ramp. You can ignore air resistance.
a. What should his speed have been at the top of the ramp to have just made it to the edge of the far bank?
b. If his speed was only half the value found in (a), where did he land?
Answer:
a) 17.8 m/s
b) 28.3 m
Explanation:
Given:
angle A = 53.0°
sinA = 0.8
cosA = 0.6
width of the river,d = 40.0 m,
the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,
The river itself was 100 m below the ramp H = 100 m,
(a) find speed v
vertical displacement
[tex]-h= vsinA\times t-gt^2/2[/tex]
putting values h=15 m, v=0.8
[tex]-15 = 0.8vt - 4.9t^2[/tex] ............. (1)
horizontal displacement d = vcosA×t = 0.6×v ×t
so v×t = d/0.6 = 40/0.6
plug it into (1) and get
[tex]-15 = 0.8\times40/0.6 - 4.9t^2[/tex]
solving for t we get
t = 3.734 s
also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s
(b) If his speed was only half the value found in (a), where did he land?
v = 17.8/2 = 8.9 m/s
vertical displacement = [tex]-H =v sinA t - gt^2/2[/tex]
⇒ [tex]4.9t^2 - 8.9\times0.8t - 100 = 0[/tex]
t = 5.30 s
then
d =v×cosA×t = 8.9×0.6×5.30= 28.3 m
For every _______ over 50 mph, your chances of being seriously injured, disfigured, or killed are doubled.
Answer:
10 mph
Explanation:
Exceeding speed limit is one of the major causes of road accident.Crash severity increases with the speed of the vehicle at impact.The probability of death, disfigurement, or debilitating injury grows with higher speed at impact.The above consequences double for every 10 mph over
50 mph that a vehicle travels.
Answer:
10 mph
Explanation:
Exceeding speed limit is one of the major causes of road accident.
Crash severity increases with the speed of the vehicle at impact.
The probability of death, disfigurement, or debilitating injury grows with higher speed at impact.
The above consequences double for every 10 mph over
50 mph that a vehicle travels
Explanation:
how many seconds of space should you keep between you if you're driving a 100 foot truck 30 mph?
Answer:
The space in seconds that will be kept = 2.27 seconds
Explanation:
S = d/t..................... Equation 1
making t the subject of formula in the equation above,
t = d/S.................... Equation 2
Where S = speed, d = distance, t = time.
Conversion: (i)if 1 mph = 0.44704 m/s,
then, 30 mph = 30×0.44704
= 13.41 m/s
(ii) If 1 foot = 0.3048 m
then, 100 foot = 30.48 m.
Given: S = 30 mph = 13.41 m/s, d = 100 foot = 30.48 m
Substituting these values into equation 2
t = 30.48/13.41
t = 2.27 seconds.
Therefore the space in seconds that will be kept = 2.27 seconds
If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)
Final answer:
The question is related to determining the critical value of x, or xcritical, that ensures a bar with a heavy block remains stable by balancing torques around the pivot point.
Explanation:
The question asks about stability conditions in a physical system involving a block on a bar and requires the application of concepts such as the center of mass, force equilibrium, and tension. When the mass of the block is too large and positioned too close to the left end of the horizontal bar, the system may become unstable, potentially causing the bar to tilt. To determine the smallest value of x, denoted as xcritical, we must set up equations considering the torques about the pivot point, ensuring that the sum of torques must equal zero for stability. This involves calculating the torques due to the weight of the bar, the weight of the block, and any other forces acting on the system. The position of the center of mass of the bar is crucial as it influences the bar's propensity to rotate and the distribution of weight. Hence, xcritical is that specific distance at which the torques balance each other out, and the system remains in a stable, horizontal state.
In what state must matter exist for fusion reactions to take place
Answer:
Plasma
Explanation:
For a fusion reaction to take place, there must be conditions in which the particles have extreme thermal kinetic energies, in this way the collisions that cause the nuclear fusion are generated. Therefore, it is necessary to reach very high temperatures, in which the state of matter will necessarily be plasma.
Difference between global warming and climate change
Answer:
Global warming refers only to the Earth’s rising surface temperature, while climate change includes warming and the “side effects”
Find the rate of change for x3. You need to work out the change in f(x)=x3 when x is increased by a small number h to x+h. So you will work out f(x+h)-f(x). Then do some algebra to simplify this. Then divide this by h to get the average rate of change of f(x) between x and x+h.
Answer:
Explanation:
Given
[tex]F(x)=x^3[/tex]
Rate of change of F(x) is given by
[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{F(x+h)-F(x)}{x+h-x}[/tex]
[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{(x+h)^3-x^3}{h}[/tex]
[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{x^3+h^3+3x^2h+3xh^2-x^3}{h}[/tex]
[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{h^3+3x^2h+3xh^2}{h}[/tex]
[tex]F'(x)=\lim_{h\rightarrow 0}\\h^2+3x^2+3xh[/tex]
Putting limits
[tex]F'(x)=3x^2[/tex]
The average rate of change will be "3x²".
Average rate of changeAccording to the question,
The function, f(x) = x³
then,
f(x + h) = (x + h)³
Now,
→ f(x + h) - f(x) = (x + h)³ - x³
= x³ + h³ + 3x²h + 3xh² - x³
= h³ + 3x²h + 3xh²
and,
→ [tex]\frac{f(x+h) -f(x)}{h}[/tex] = [tex]\frac{h^3+3x^2h+3xh^2}{h}[/tex]
= [tex]\frac{h[h^2+3x^2+3xh]}{h}[/tex]
= h² + 3x² + 3xh
By applying the limit, we get
→ [tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex] = [tex]\lim_{h \to 0}[/tex] h² + 3x² + 3xh
By substituting the values,
= 0² + 3x² + 3x(0)
= 3x²
Thus the above approach is correct.
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Which of the following forms an interference pattern when directed toward two suitably-spaced slits?
a. electrons
b. Sound
c. light
d. all of these
enone of these
Answer:
Option d
Explanation:
Interference is the phenomenon in which two or more waves having the same frequency combines or cancel out each other to form a resulting wave with the amplitude equals the sum of the amplitudes of the waves that are combined.
Thus it is clear that the particle with the wave nature can produce interference pattern. Hence, electron, light, sound, all of these produces interference pattern when directed towards the slits that are suitably spaced.
Transform active margins are associated with which type of boundary? Convergent Transform Divergent
Answer:
Transform active margins are associated with which type of boundary?
Transform boundary
Explanation:
The transform boundary is a boundary where one plates(crust) slides past another plate horizontally. This kind of plate movement have been detected to exist between the interaction of the North pacific plates(continental plate) and the pacific plates(oceanic plates) .
At the transform margin the crust are usually broken. But overall crust are neither created nor destroyed . The transform margin region are active as it is marked by shallow-focus earthquakes .
Along the fractured zone where this transform movement occurs is known to create extensive transform faults .Notable transform fault that exist in this kind of boundary(transform) is the San Andrea fault and Alpine Fault.
The motion of this plates can occur on a single fault or on a group of faults.
what are the charge and the charge density on the surface of a conducting sphere of radius 0.15 m whose potantial is 200v?
Answer:
The charge and the charge density on the surface of a conducting sphere is [tex]3.34\times 10^{-9}\ C[/tex] and [tex]1.18\times 10^{-8}\ C/m^2[/tex] respectively.
Explanation:
It is given that,
Radius of the conducting sphere, r = 0.15 m
Potential, V = 200 V
Potential on the surface of sphere is given by :
[tex]V=\dfrac{kq}{r}[/tex]
q is the charge on the sphere
[tex]q=\dfrac{Vr}{k}[/tex]
[tex]q=\dfrac{200\times 0.15}{9\times 10^9}[/tex]
[tex]q=3.34\times 10^{-9}\ C[/tex]
Charge per unit area is called charge density on the surface. it is given by :
[tex]\sigma=\dfrac{q}{A}[/tex]
[tex]\sigma=\dfrac{q}{4\pi r^2}[/tex]
[tex]\sigma=\dfrac{3.34\times 10^{-9}}{4\pi (0.15)^2}[/tex]
[tex]\sigma=1.18\times 10^{-8}\ C/m^2[/tex]
So, the charge and the charge density on the surface of a conducting sphere is [tex]3.34\times 10^{-9}\ C[/tex] and [tex]1.18\times 10^{-8}\ C/m^2[/tex] respectively. Hence, this is the required solution.
Final answer:
The charge on the conducting sphere with a radius of 0.15 m and a potential of 200V is approximately 3.34 × 10⁻⁶ C, and the surface charge density is approximately 1.18 × 10⁻⁵ C/m².
Explanation:
To find the charge (q) and the surface charge density (σ) on the surface of a conducting sphere with a given potential (V), we can use the relationship between potential, charge, and radius of a sphere, along with the equation that relates charge to surface charge density.
The potential (V) of a conducting sphere is given by V = (k × q)/R, where k is Coulomb's constant (k = 8.99 × 10⁹ N.m²/C²), q is the charge on the sphere, and R is the radius of the sphere.
In this case, V = 200V and R = 0.15m. Using the formula, we can find the charge q on the sphere:
V = (k × q)/R
200V = ((8.99 × 109 N.m²/C²) × q)/0.15m
q = (200V × 0.15m)/(8.99 × 10⁹ N.m²/C²)
q ≈ 3.34 × 10⁻⁶ C
Once we have q, we can use the equation q = σ(4πR2) to find the surface charge density σ:
σ = q / (4πR²)
σ ≈ 3.34 × 10⁻⁶ C / (4π × (0.15m)²)
σ ≈ 1.18 × 10⁻⁵ C/m²
Kenneth ran a marathon (26.2 miles) in 5.5 hours. What was Kenneth’s average speed? (Round your answer to the nearest tenth.)
Answer:
Explanation:
Speed: Speed can be defined as the ratio of distance to time. The S. I unit of speed is m/s. And it is expressed mathematically as,
speed = distance/time
S = d/t............................. Equation 1
Conversion: (i) If 1 miles = 1609.344 m,
then, 26.2 miles = 1609.344× 26.2
= 42164.813 m.
(i) if 1 hours = 3600 seconds,
then, 5.5 hours = 5.5×3600
=19800 seconds.
Given: d = 26.7 miles= 42164.813 m, t = 5.5 hours = 19800 seconds
Substituting these values into equation 1
S = 42164.813/19800
S = 2.1 m/s.
Therefore Kenneth's average speed = 2.1 m/s
Answer:4.8 mph on edge
Explanation: have a great day :)
A Ping-Pong ball moving East at a speed of 4 m/s collides with a stationary bowling ball. The Ping-Pong ball bounces back to the West, and the bowling ball moves very slowly to the East. Which object experiences the greater magnitude impulse during the collision?A car hits another and the two bumpers lock together during the collision. Is this an elastic or inelastic collision?In a game of pool, the white cue ball hits the #5 ball and stops, while the #5 ball moves away with the same velocity as the cue ball had originally. The type of collision is...In an inelastic collision, the final total momentum is...When a cannon fires a cannonball, the cannon will recoil backward because the.....
Answer:
Explanation:
During any collision whether elastic or inelastic, force of action and reaction are generated at the point of touch for a brief period during which they remain in touch with each other. These action and reaction forces are equal and opposite. Impulse is defined by force multiplied by time duration . Hence we can say that during any collision , the impulse generated are equal and opposite .
In inelastic collision , colliding objects coalesce with each other . Hence in the given case , collision is inelastic.
In elastic collision , when an object collides with a similar object at rest , there is exchange of velocity . In the given case also , similar objects are colliding and exchange of velocity is taking place.So it is an example of elastic collision.
In an inelastic collision , momentum is conserved. So final total momentum is equal to initial total momentum.
When a cannon fires a cannonball, the cannon will recoil backward because the total momentum is conserved.
In an experiment to determine ΔH for the reaction between HBr and NaOH , 43.6 mL of 1.08 M HBr at 19.95 °C is placed in a coffee cup calorimeter which has a heat capacity of 7.99 J/°C. 43.6 mL of 1.08 M NaOH at 19.95 °C is added to the acid solution in the calorimeter and quickly mixed. A final temperature of 26.88 °C was recorded. What is ΔH for this reaction, in kJ? Assume that the specific heats of all solutions are the same as for pure water (4.18 J g-1 °C-1), and that all solution densities are 1.00 g/mL. Don't forget to put the proper sign on your ΔH value.
Answer:
ΔH = -57.78kj/mol
Explanation:
Assumptions
These are solutions, not pure water. The specific heat of water is 4.184 J/goC. Assume that these solutions are close enough to being like water that their specific heats are also 4.1984 J/goC.
the heat evolved will be
q = mcΔt
first of all convert ml to grams
(43.6 mL + 43.6 mL ) = 87.2mL of solution.
density of water=1g/ml
87.2 mL X 1 g/ml = 87.2 grams of solution.
(mass = Volume X Density)
Find the temperature change.
Δt =tfinal - tinitial = 26.88oC - 19.95oC = 6.93oC
q = mcΔt
= 87.2 grams X 4.184 J/goC X 6.9oC
= 2.52 X 10^3 J
This is the heat gained by the water, but in fact it is the heat lost by the reacting HBr and NaOH, therefore q = -2.52 x 10^3 J.
i.e. it is an exothermic reaction, heat was lost to the water and it got warmer
to find the how much of HBr that was used in mol
43.6 mL of HBr X 1.00 mol HBr/ 1000 mL HBr = 0.0436 mol HBr
same quantity of base NaoH was used
. molar enthalpy = J/mol = -2.52 x 10^3 J / 0.0436 mol
-57.78kj/mol
Therefore, for the neutralization of HBrl and NaOH, the enthalpy change, often called the enthalpy of reaction is ΔH = -57.78kj/mol
Someone plans to float a small, totally absorbing sphere 0.500 m above an isotropic point source of light,so that the upward radiation force from the light matches the downward gravitational force on the sphere. The sphere’s density is 19.0 g/cm3, and its radius is 2.00 mm. (a) What power would be required of the light source?
Answer:
468449163762.0812 W
Explanation:
m = Mass = [tex]\rhoV[/tex]
V = Volume =[tex]\dfrac{4}{3}\pi r^3[/tex]
r = Distance of sphere from isotropic point source of light = 0.5 m
R = Radius of sphere = 2 mm
[tex]\rho[/tex] = Density = 19 g/cm³
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
A = Area = [tex]\pi R^2[/tex]
I = Intensity = [tex]\dfrac{P}{4\pi r^2}[/tex]
g = Acceleration due to gravity = 9.81 m/s²
Force due to radiation is given by
[tex]F=\dfrac{IA}{c}\\\Rightarrow F=\dfrac{\dfrac{P}{4\pi r^2}{\pi R^2}}{c}\\\Rightarrow F=\dfrac{PR^2}{4r^2c}[/tex]
According to the question
[tex]F=mg\\\Rightarrow \dfrac{PR^2}{4r^2c}=\rho \dfrac{4}{3}\pi R^3g\\\Rightarrow P=\dfrac{16r^2\rho c\pi Rg}{3}\\\Rightarrow P=\dfrac{16\times 0.002\times 19000\times \pi\times 0.5^2\times 9.81\times 3\times 10^8}{3}\\\Rightarrow P=468449163762.0812\ W[/tex]
The power required of the light source is 468449163762.0812 W
Final answer:
To calculate the power required of the light source, you need to find the upward radiation force exerted on the sphere, which should match the downward gravitational force acting on the sphere. The power is given by P = I * A.
Explanation:
To calculate the power required of the light source, we first need to find the upward radiation force exerted on the sphere, which should match the downward gravitational force acting on the sphere. The radiation force is given by the formula:
Fr = (P / c)A
where Fr is the radiation force, P is the power of the light source, c is the speed of light, and A is the cross-sectional area of the sphere.
Since the sphere is totally absorbing, all the light incident on it is absorbed, so the intensity of the light is given by:
I = P / A
where I is the intensity of the light. Rearranging the equation, we get:
P = I * A
Substituting the value of A for the area of the sphere, we can find the power required.