Answer:
Weber's Law
Explanation:
Weber’s law is an important law in psychology.
This law quantifies the perception of change within a provided stimulus.
This law is also known as Weber-Fechner law as it relates two hypotheses in the field of psychophysics which are Weber law and the Fechner law.
Weber’s law describes that an observed change in a stimulus is a constant ratio of the original stimulus.
Final answer:
Weber's Law describes the relationship between a stimulus and the resulting sensation, stating that the Just Noticeable Difference (JND) is a constant fraction of stimulus intensity. It is a key principle in psychophysics, useful for experiments such as determining the JND for different weights of rice in bags.
Explanation:
The concept you're inquiring about is Weber's Law, which is a fundamental principle in the field of psychophysics, a branch of psychology. Weber's Law describes the relationship between a stimulus and its resulting sensation by proposing that the Just Noticeable Difference (JND) is a constant fraction of the stimulus intensity.
To put this into practice, for example, if one were testing the JND of various weights of rice in bags, they could use incremental percentage increases of the weight to determine the smallest difference that can be perceived. By choosing increments like 10 percent or 20 percent between certain weight thresholds, a researcher can apply Weber's Law to determine the JND in a standardized way.
This law helps explain why when you have a cup of coffee with little sugar, adding a teaspoon can make a noticeable difference, but in a cup with much more sugar, that same teaspoon becomes less detectable, requiring a proportionally larger amount of sugar to notice a change in sweetness.
How long (in s) will it take an 875 kg car with a useful power output of 42.0 hp (1 hp = 746 W) to reach a speed of 17.0 m/s, neglecting friction? (Assume the car starts from rest.)
Answer:
The time taken by the car with a useful power output is 4.03 seconds.
Explanation:
Given that,
Mass of the car, m = 875 kg
Initial speed of the car, u = 0
Final speed of the car, v = 17 m/s
Power of the car, P = 42 hp = 31332 W
The power output of the car is given by the work done per unit time. It is given by :
[tex]P=\dfrac{W}{t}[/tex]
Initial kinetic energy of the car will be 0 as it is at rest.
Final kinetic energy of the car is :
[tex]K=\dfrac{1}{2}mv^2[/tex]
[tex]K=\dfrac{1}{2}\times 875\times (17)^2[/tex]
K = 126437.5
As per the work energy theorem, the change in kinetic energy of the object is equal to the work done.
So,
[tex]P=\dfrac{W}{t}[/tex]
[tex]t=\dfrac{W}{P}[/tex]
[tex]t=\dfrac{126437.5}{31332}[/tex]
t = 4.03 seconds
So, the time taken by the car with a useful power output is 4.03 seconds. Hence, this is the required solution.
A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum force acting on it? (b) If the oscillations are produced by a spring, what is the springconstant?
Answer:
a)F=698.83 N
b)K=8221.56 N/m
Explanation:
Given that
mass ,m = 0.12 kg
Amplitude ,A= 8.5 cm
time period ,T = 0.2 s
We know that
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]{\omega}=\dfrac{2\pi }{0.2}\ rad/s[/tex]
[tex]{\omega}=31.41\ rad/s[/tex]
We know that
[tex]{\omega}^2=m\ K[/tex]
K=Spring constant
[tex]K=\dfrac{\omega^2}{m}[/tex]
[tex]K=\dfrac{31.41^2}{0.12}\ N/m[/tex]
K=8221.56 N/m
The maximum force F
F= K A
F= 8221.56 x 0.085 N
F=698.83 N
a)F=698.83 N
b)K=8221.56 N/m
Final answer:
To find the magnitude of the maximum force acting on the body in simple harmonic motion, we use the equation F = -kx. To find the spring constant, we can rearrange the equation F = -kx.
Explanation:
To find the magnitude of the maximum force acting on the body in simple harmonic motion, we can use the equation F = -kx, where F is the force, k is the spring constant, and x is the displacement from the equilibrium position. In this case, we are given the amplitude A = 8.5 cm and the mass m = 0.12 kg. The displacement x at the maximum amplitude is equal to the amplitude, so x = A.
Plugging in the values, we get F = -(k)(A).
(a) To find the magnitude of the maximum force, we need to find the spring constant k. We can use the equation T = 2π√(m/k), where T is the period. Plugging in the values, T = 0.20 s and m = 0.12 kg, we can solve for k.
(b) To find the spring constant, we can rearrange the equation F = -kx to solve for k. Plugging in the values, F = 2.00 mg and x = A, we can solve for k.
The cheetah is one of the fastest-accelerating animals, because it can go from rest to 16.2 m/s (about 36 mi/h) in 2.4 s. If its mass is 102 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in the following units.
Explanation:
Power is the ratio of energy to time.
Here we need to consider kinetic energy,
Mass, m = 102 kg
Initial velocity = 0 m/s
Final velocity = 16.2 m/s
Time, t = 2.4 s
Initial kinetic energy = 0.5 x Mass x Initial velocity² = 0.5 x 102 x 0² = 0 J
Final kinetic energy = 0.5 x Mass x Final velocity² = 0.5 x 102 x 16.2² = 13384.44 J
Change in energy = Final kinetic energy - Initial kinetic energy
Change in energy = 13384.44 - 0
Change in energy = 13384.44 J
Power = 13384.44 ÷ 2.4 = 5576.85 W = 7.48 hp
Power of cheetah is 5576.85 W = 7.48 hp
Use a(t) = -32 feet per second per second as the acceleration due to gravity. (Neglect air resistance.) -With what initial velocity must an object be thrown upward (from ground level) to reach the top of the Washington Monument (approximately 550 feet)?
Answer:
u= 187.61 ft/s
Explanation:
Given that
g= - 32 ft/s²
The maximum height ,h= 550 ft
Lets take the initial velocity = u ft/s
We know that
v²=u² + 2 g s
v=final speed ,u=initial speed ,s=height
When the object reach at the maximum height then the final speed of the object will become zero.
That is why
u²= 2 x 32 x 550
u²= 35200
u= 187.61 ft/s
That is why the initial speed will be 187.61 ft/s
Answer:
Explanation:
a = - 32 ft/s²
h = 550 ft
Let the initial velocity is u. The velocity at maximum height is zero.
use third equation of motion
v² = u² + 2 a h
0 = u² - 2 x 32 x 550
u² = 35200
u = 187.62 ft/s
A bakery measured the mass of a whole cake as 0.870 kg. A customer bought one slice of the cake and measured the mass of the slice as 0.1151 kg. What is the mass of the remaining cake, without the one slice?
Answer:
0.7549kg
Explanation:
The mass of the slice + mass of the remaining cake = total mass of cake.
mass of remaining cake = total mass of cake - the mass of the slice
total mass=0.870kg
mass of slice = 0.1151kg
mass of remaining cake = 0.870 - 0.1151
mass of remaining cake=0.7549kg
Friction pulls directly against the direction of motion (at 180) of a sled, and does -55.2 J of work while the sled moves 8.98 m. What is the magnitude (+) of the friction force?(unit=N)
The magnitude of the force of friction is 6.1 N
Explanation:
The work done by a force when moving an object is given by the equation:
[tex]W=Fd cos \theta[/tex]
where :
F is the magnitude of the force
d is the displacement
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
In this problem, we have:
W = -55.2 J (work done by the force of friction)
d = 8.98 m (displacement of the sled)
[tex]\theta=180^{\circ}[/tex] (because the force of friction acts opposite to the direction of motion)
By solving the equation for F, we find the magnitude of the force of friction on the sled:
[tex]F=\frac{W}{d cos \theta}=\frac{-55.2}{(8.98)(cos 180^{\circ})}=6.1 N[/tex]
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Answer:
6.1
Explanation:
Acellus
____ is the process of sending data over a signal by varying either its amplitude, frequency, or phase.
Answer:
Modulation
Explanation:
Modulation is the the process whereby a waveform has one or more of it's properties varied. Amplitude, frequency and phase are all properties of a waveform which can be varied hence Modulation is the process of sending data over a signal by varying either its amplitude, frequency or phase.
The average lifetime of a pi meson in its own frame of reference (i.e., the proper lifetime) is 2.6 10-8 s. (a) If the meson moves with a speed of 0.88c, what is its mean lifetime as measured by an observer on Earth? (b) What is the average distance it travels before decaying, as measured by an observer on Earth? (c) What distance would it travel if time dilation did not occur?
Final answer:
a) The measured lifetime of the pi meson as observed by an Earth observer is longer than its proper lifetime due to time dilation. b) The average distance the meson travels before decaying can be calculated using the velocity and measured lifetime. c) If time dilation did not occur, the distance the meson would travel can be calculated using the velocity and proper lifetime.
Explanation:
a) According to time dilation, the average lifetime of the pi meson as measured by an observer on Earth is longer than its proper lifetime. To calculate it, we use the formula for time dilation: t' = t / γ, where t' is the measured lifetime, t is the proper lifetime, and γ is the Lorentz factor. Given that the proper lifetime is 2.6 × 10-8 s and the speed of the meson is 0.88c, we can calculate γ using the formula γ = 1 / √(1 - v2/c2). After calculating γ, we can substitute it into the time dilation formula to find the measured lifetime on Earth.
b) To calculate the average distance the meson travels before decaying, we use the formula d = v · t', where d is the distance, v is the velocity, and t' is the measured lifetime. We can substitute the values given in part a) to find the distance traveled by the meson.
c) If time dilation did not occur, the distance the meson would travel can be calculated using the formula d = v · t, where d is the distance, v is the velocity, and t is the proper lifetime. By substituting the values given in part a) into this formula, we can find the distance the meson would travel in the absence of time dilation.
A speed skater moving across frictionless ice at 8.30 m/s hits a 5.10-m-wide patch of rough ice. She slows steadily, then continues on at 5.20 m/s ?
Incomplete question.The complete question is here
A speed skater moving across frictionless ice at 8.30 m/s hits a 5.10m wide patch of rough ice. She slows steadily, then continues on at 5.20 m/s.What is her acceleration on the rough ice?
Answer:
acceleration = -4.103 m/s²
Explanation:
Given data
Initial velocity Vi=8.30 m/s
Final velocity Vf=5.20 m/s
Initial distance xi=0 m.......(We choose xi=0 the start point of acceleration motion )
Final distance xf=5.10 m
To find
Acceleration
Solution
From the kinetic equation
[tex](v_{f})^{2}=(v_{i} )^{2}+2a(x_{f} -x_{i} )\\ (5.20 m/s)^{2}=(8.30m/s)^{2}+2a(5.10m-0m)\\a=\frac{(5.20m/s)^{2}-(8.30m/s)^{2} }{2*(5.10m)}\\a=-4.103 m/s^{2}[/tex]
A 45.0 g hard-boiled egg moves on the end of a spring with force constant 25.0 N/m. Its initial displacement 0.500 m. A damping force Fx= −bvx acts on the egg, and the amplitude of the motion decreases to 0.100 m in a time of 5.00 s. Calculate the magnitude of the damping constant b. Express the magnitude of the damping coefficient numerically in kilograms per second, to three significant figures.
Answer:
0.02896 kg/s
Explanation:
[tex]A_1[/tex] = Initial displacement = 0.5 m
[tex]A21[/tex] = Final displacement = 0.1 m
t = Time taken = 0.5 s
m = Mass of object = 45 g
Displacement is given by
[tex]x=Ae^{-\dfrac{b}{2m}t}cos(\omega t+\phi)[/tex]
At maximum displacement
[tex]cos(\omega t+\phi)=1[/tex]
[tex]\\\Rightarrow A_2=A_1e^{-\dfrac{b}{2m}t}\\\Rightarrow \dfrac{A_1}{A_2}=e^{\dfrac{b}{2m}t}\\\Rightarrow ln\dfrac{A_1}{A_2}=\dfrac{b}{2m}t\\\Rightarrow b=\dfrac{2m}{t}\times ln\dfrac{A_1}{A_2}\\\Rightarrow b=\dfrac{2\times 0.045}{5}\times ln\dfrac{0.5}{0.1}\\\Rightarrow b=0.02896\ kg/s[/tex]
The magnitude of the damping coefficient is 0.02896 kg/s
Final answer:
To find the damping constant b for a damped harmonic motion scenario with a given change in amplitude over time, you employ the decay of amplitude formula in damped harmonic motion, then rearrange and solve for b.
Explanation:
To calculate the magnitude of the damping constant b for a 45.0 g hard-boiled egg moving at the end of a spring with a force constant of 25.0 N/m given a decrease in amplitude from 0.500 m to 0.100 m over 5.00 seconds, we apply the principle of damped harmonic motion. The damping force is given by Fx= −bvx, where b is the damping coefficient we wish to find. From the information provided, it's understood that this is a case of exponentially decaying amplitude in harmonic motion.
The equation that relates the exponential decay of amplitude in damped harmonic motion is A(t) = A0e−(bt/2m), where A0 is the initial amplitude, A(t) is the amplitude at time t, and m is the mass of the object. By re-arranging this equation to solve for b, and substituting the given values A0=0.500 m, A(t)=0.100 m, t=5.00 s, and m=0.045 kg (since 45.0 g=0.045 kg), we can calculate b.
The rearranged equation for b would be b = 2m[ln(A0/A(t))]/t. Substituting the values gives b = 2(0.045)[ln(0.500/0.100)]/5.00, which after calculation yields the magnitude of b.
A 106 kg clock initially at rest on a horizontal floor requires a 670 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 557 N keeps it moving with a constant velocity.
a) what is the μs between the clock and the floor?.70
b) what is the μk between the clock and the floor?
Answer:
a) [tex]\mu_s=0.65[/tex]
b) [tex]\mu_k=0.54[/tex]
Explanation:
Static friction is when the body is at rest or is about to move while kinetic friction is when the body is already in motion. According to Newton's second law:
[tex]\sum F_y:N=mg\\\sum F_x:F_f=F_x[/tex]
a) In this case, the static friction must be equal to the horizontal force to set the clock in motion:
[tex]F_f=\mu_sN=\mu_smg\\\mu_smg=F_x\\\mu_s=\frac{F_x}{mg}\\\mu_s=\frac{670}{106kg(9.8\frac{m}{s^2})}\\\mu_s=0.65[/tex]
b) In this case, the kinetic friction is equal to the horizontal force that keep the clock moving with constant velocity:
[tex]\mu_kmg=F_x'\\\mu_k=\frac{F_x'}{mg}\\\mu_k=\frac{557}{106kg(9.8\frac{m}{s^2})}\\\mu_k=0.54[/tex]
A cannon shoots a ball at an angle θ above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find the ball's position as a function of time. (Use axes with x measured horizontally and y vertically.) (b) Let r(t) denote the ball's distance from the cannon. What is the largest possible value of θ if r(t) is to increase throughout the ball's flight?
The question involves the application of Newton's law to a ball's position over time, launched from a cannon at an angle. The x and y-axis positions can be calculated using horizontal and vertical principles of motion respectively. For the ball's distance (r) to increase throughout the flight, the launch angle needs to be 90°.
Explanation:The subject of this question relates to the application of Newton's second law in two dimensions, specifically in cases of projectile motion. The position of the ball fired from a cannon is determined by the initial velocity of the ball, the launch angle θ, and the acceleration due to gravity.
The x-component of the ball's position can be calculated using horizontal motion principles. Here, the x position x(t) = V₀cos(θ)t, where V₀ is the initial speed of the ball, cos(θ) is the cosine of the launch angle and t is the time.The y-component of the ball's position can be calculated using vertical motion principles. In this case, y(t) = V₀sin(θ)t - 0.5gt², where sin(θ) is the sine of the launch angle, g is the acceleration due to gravity, and t is the time. For r(t) to increase throughout the ball's flight, the launch angle θ needs to be 90°. This is because the maximum range or distance is attained when objects are projected at an angle of 45°. But the ball's range will start to decrease after this angle, while the height (r(t)) will continue to increase until 90°.Learn more about Projectile Motion here:
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A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 41.0-gram mass is attached at the 23.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick?
Answer:
mass of the meter stick=0.063 kg
or
mass of the meter stick=63.3 g
Explanation:
Given data
m₁=41.0g=0.041kg
r₁=(39.2 - 23)cm
r₂=(49.7 - 39.2)cm
g=9.8 m/s²
To find
m₂(mass of the meter stick)
Solution
The clockwise and counter-clockwise torques must be equal if the meter stick is in rotational equilibrium
[tex]Torque_{cw}=Torque_{cw}\\F_{1}r_{1}=F_{2}r_{2}\\ m_{1}gr_{1}=m_{2}gr_{2}\\(0.041kg)(9.8m/s^{2} )(0.392m-0.23m)=m_{2}(9.8m/s^{2})(0.497m-0.392m)\\0.0651N.m=1.029m_{2}\\m_{2}=0.063 kg\\or\\m_{2}=63.3g[/tex]
Two canoes are touching and at rest on a lake. The occupants push away from each other in opposite directions, giving canoe 1 a speed of 0.56 m/s and canoe 2 a speed of 0.45 m/s .
If the mass of canoe 1 is 320 kg , what is the mass of canoe 2?
Answer:
398.22 kg
Explanation:
[tex]m_1[/tex] = Mass of canoe 1 = 320 kg
[tex]m_2[/tex] = Mass of canoe 2
[tex]v_1[/tex] = Velocity of canoe 1 = 0.56 m/s
[tex]v_2[/tex] = Velocity of canoe 2 = 0.45 m/s
In this system the linear momentum is conserved
[tex]m_1v_1=m_2v_2\\\Rightarrow m_2=\dfrac{m_1v_1}{v_2}\\\Rightarrow m_2=\dfrac{320\times 0.56}{0.45}\\\Rightarrow m_2=398.22\ kg[/tex]
The mass of the second canoe is 398.22 kg
Answer:
398.22 kg
Explanation:
mass of I canoe, m1 = 320 kg
initial velocity of both the canoe = 0
final velocity of I canoe, v1 = 0.56 m/s
final velocity of II canoe, v2 = - 0.45 m/s
Let the mass of second canoe is m2.
Use conservation of momentum
momentum before push = momentum after push
0 + 0 = m1 x v1 + m2 x v2
0 = 320 x 0.56 - m2 x 0.45
m2 = 398.22 kg
Thus, the mass of second canoe is 398.22 kg.
A drawing of electric field lines will immediately reveal (1) the relative magnitude of different charges (proportional to the number of lines that begin or end on each), (2) the sign of different charges (since lines go into negative charges and come out of positive charges), (3) the relative magnitude of the electric field at any point (because the magnitude of the field is proportional to how closely spaced the field lines are), and (4) the symmetry of the charge distribution (since they match the symmetry of the underlying field).
Complete Question:
The complete question is on the all the uploaded image
Answer:
sfjjA)rule 1,3,5 are brokenB)rule 2,3 are brokenC) rule 2 are brokenD) rule 2,3 are brokenE)rule 2 are brokenF)noneG) rule 3,4 are brokenH) rule 2,4,5 are brokenExplanation:
This solution were gotten by examining the diagrams and figuring out the rules that are broken
For A
we see that the two charges are positive so rule 1 is broken,on the same diagram we see that there is no radial symmetry close to the charge so rule 3 is broken then looking again at the diagram A we can see that the electric field is not tangent to any electric line at any point hence rule 5 is broken
For B:
We see that the number of lines are not proportional to the magnitude of the charge hence rule 2 has been broken
secondly looking again at the diagram we see that the line are not uniformly distributed near the charge hence rule 3 is broken
For C:
We see that the number of line is not proportional to the magnitude of the charge hence rule 2 is broken.
For D:
we see that the number of lines are not proportional to the magnitude of the charge hence rule 2 has been broken
secondly looking again at the diagram we see that the line are not uniformly distributed near the charge hence rule 3 is broken
For E:
We see that the number of line is not proportional to he magnitude of the charge hence rule 2 is broken.
For F
Looking at the diagram we see that none of the rules are broken
For G:
looking at the diagram we see that the line are not uniformly distributed near the charge hence rule 3 is broken
taking another look at the diagram we see that the spacing of the lines are not indicating the magnitude of the charge hence rule 4 is broken
For H:
We see that the number of line is not proportional to he magnitude of the charge hence rule 2 is broken.
taking another look at the diagram we see that the spacing of the lines are not indicating the magnitude of the charge hence rule 4 is broken
Looking again at the diagram we see that the at any point on the electric field line that the electric field itself is not tangent to the electric field line hence rule 5 is broken
As a cat pounces on a mouse, its muscles consume 10 units of potential energy (which the cat previously gained from the food it consumed). However, the pounce itself only required 4 units of kinetic energy. How many units of energy were dissipated as heat?
The cat used 10 units of potential energy to pounce but only 4 units were converted meaningfully into kinetic energy of the pounce. The remaining 6 units were dissipated as heat.
Explanation:When the cat pounces on a mouse, the energy needed for this action is less than the total potential energy consumed. The cat's body uses 10 units of potential energy, but the pounce only required 4 units of kinetic energy. The difference between these two values signifies the amount of energy dissipated as heat. Therefore, we can perform the subtraction: 10 units (total potential energy) - 4 units (kinetic energy used) = 6 units. Thus, "6 units of energy were dissipated as heat."
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Charge is flowing through a conductor at the rate of 420 C/min. If 742 J. of electrical energy are converted to heat in 30 s., what is the potential drop across the conductor?
Answer:
3.53 V
Explanation:
Electric charge: The is the rate of flow of electric charge along a conductor.
The S.I unit of electric charge is C.
Mathematically it is expressed as,
Q = It ............................ Equation 1
Where Q = electric charge, I = current, t = time.
I = Q/t.......................... Equation 2
From the question, charge flows through the conductor at the rate of 420 C/mim
Which means in 1 min, 420 C of charge flows through the conductor.
Hence,
Q = 420 C, t = 1 min = 60 seconds
Substitute into equation 2
I = 420/60
I =7 A
Also
P = VI......................... Equation 3
Where P = power, V = potential drop, I = current.
V = P/I................... Equation 4
Note: Power = Energy/time
From the question, P = 742/30 = 24.733 W. and I = 7 A.
Substitute these values into equation 4
V = 24.733/7
V = 3.53 V
Hence the potential drop across the conductor = 3.53 V
The potential drop across the conductor is 3.54 V
The rate of flow of electric charge along a conductor per unit time is known as current.
Mathematically it is expressed as,
As we know that
Power, [tex]P = VI[/tex]
Where P is power, V is potential drop and I is current.
Also,
It is given that, energy = 742 J and time = 30s
Power [tex]=\frac{742}{30} =24.74watt[/tex]
Substituting the value of power in equation [tex]P=VI[/tex]
[tex]V=\frac{P}{I} =\frac{24.74}{7}=3.54V[/tex]
Thus, the potential drop across the conductor is 3.54 V
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A transformer has a primary voltage of 115 V and a secondary voltage of 24 V. If the number of turns in the primary is 345, how many turns are in the secondary? A. 8 B. 690 C. 72 D. 1,653
Answer:
C. 72
Explanation:
Transformer: A transformer is an electromagnetic device that uses the property of mutual inductance to change the voltage of alternating supply.
In a ideal transformer,
Vs/Vp = Ns/Np ............................................. Equation 1
Where Vp = primary voltage, Vs = secondary voltage, Ns = Secondary turn, Np = primary turn.
Making Ns the subject of the equation,
Ns =(Vs/Vp)Np .......................................... Equation 2
Given: Vs = 24 V, Vp = 115 V, Np = 345.
Substitute into equation 2
Ns = (24/115)345
Ns = 72 turns.
Thus the number of turns in the secondary = 72 turns.
The right option is C. 72
What is the magnitude of the net force ona(n) 97 kg driver operating a dragster as it accelerates horizontally along a straight line from rest to 40 m/s in 5 s?
Answer:
Force acting on the driver will be 776 N
Explanation:
We have given mass of the driver m = 97 kg
It starts from rest so initial velocity u = 0 m /sec
And reaches to a velocity of 40 m /sec in 5 sec
So final velocity v = 40 m /sec
And time taken t = 5 sec
From first equation of motion v = u +at
So [tex]40=0+a\times 5[/tex]
[tex]a=8m/sec^2[/tex]
Now we have to find the force acting the driver
From newtons law we know that F = ma
So force F = 97×8 = 776 N
So force acting on the driver will be 776 N
5.5 g of metal is placed in a graduated cylinder containing water. The level of water increases from 12.0 ml to 13.5 ml. The density of the metal is _____ g/ml. (Use the correct number of significant figures in your answer)
Answer:
0.407
Explanation:
5.5÷13.5
i believe it is this as
density =mass÷volume
For every factor of 10 difference in Keq' (e.g., going from 102 to 10, or from 10–3 to 10–4), what is the difference in standard free energy change, ΔG°' (kJ/mol)?
Answer:
The standard free energy is the difference between that of products and reactants. The value of G has a greater influence on the reaction been considered.
Explanation:
The Gibb's free energy also referred to as the gibb's function represented with letter G. it is the amount of useful work obtained from a system at constant temperature and pressure. The standard gibb's free energy on the other hand is a state function represented as Delta-G, as it depends on the initial and final states of the system. The spontaneity of a reaction is explained by the standard gibb's free energy.
If Delta-G = -ve ( the reaction is spontaneous)if Delta -G = +ve ( the reaction is non-spontaneous)if Delta-G = 0 ( the reaction is at equilibrium)Use this hints for any reaction involving the Gibb's free, Enthalpy and entropy.
For each factor of 10 change in Keq, there is a corresponding change of approximately 5.7 kJ/mol in standard free energy change, ΔG°', at standard biochemical conditions (298.15 K).
Explanation:For every factor of 10 difference in the equilibrium constant, Keq, there is a corresponding change in the standard free energy change, ΔG°', as calculated using the relationship between ΔG°' and Keq:
ΔG°' = -RT ln Keq, where R is the universal gas constant and T is the temperature in Kelvin.
Given that R = 8.314 J/mol·K and at standard biochemical conditions T = 298.15 K, a tenfold change in Keq results in a change of (8.314 J/mol·K x 298.15 K x ln(10)), which equates to approximately 5.7 kJ/mol. Therefore, for example, increasing Keq from 102 to 103 would decrease ΔG°' by 5.7 kJ/mol, making the reaction more product-favored under standard conditions. Conversely, decreasing Keq from 10-3 to 10-4 would increase ΔG°' by 5.7 kJ/mol, making the reaction less product-favored under standard conditions.
A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.820 m. What impulse was given to the ball by the floor?
Answer:
1.0752 kgm/s
Explanation:
Considering when the drop was dropped from rest from a height,
mass of the ball, m = 0.120 kg
height, h = - 1.25 m
the initial velocity, u = 0 m/s
the acceleration due to gravity, g = - 9.8 m/s²
From equation of motion
[tex]V^{2} = U^{2} + 2gh[/tex]
Substituting the values,
[tex]V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)[/tex]
[tex]V^{2} = 24.5 m/s[/tex]
[tex]V = \sqrt{24.5} \ m/s[/tex]
[tex]V = 4.95 \ m/s[/tex]
V = ± 4.95 m/s
V = - 4.95 m/s
Since the ball is moving downward, the final velocity of the ball when it hits the floor is V = - 4.95 m/s
Considering when the ball rebounds from the floor,
assume the mass of the ball still remain, m = 0.120 kg
height, h = 0.820 m
the final velocity, v = 0 m/s
the acceleration due to gravity, g = - 9.8 m/s²
From equation of motion
[tex]V^{2} = U^{2} + 2gh[/tex]
Substituting the values,
[tex]0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)[/tex]
[tex]0 = U^{2} - 16.072 m/s[/tex]
[tex]U^{2} = 16.072 m/s[/tex]
[tex]U = \sqrt{16.072} \ m/s[/tex]
U = ± 4.01 m/s
U = + 4.01 m/s
Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is U = + 4.01 m/s
From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.
[tex]F = \frac{mv - mu}{t}[/tex]
[tex]F.t = m(v - u)[/tex]
⇒ Impulse = Change in momentum
To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,
∴ F.t = 0.120 kg(4.01 m/s - (-4.95 m/s) )
F.t = 0.120 kg(4.01 m/s + 4.95 m/s) )
F.t = 0.120 kg × 8.96 m/s
Impulse = 1.0752 kgm/s
The impulse given to the ball by the floor is 1.0752 kgm/s
When two charge producers with different surface materials are rubbed together to create a charge imbalance. True or False
Answer:
True
Explanation:
Answer:
The answer is TRUE
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If a ball end mill cutter's stepover is kept at a fixed value but the diameter of the tool is increased, the scallop height increases as a consequence.Select one:a. Trueb. False
Answer: FALSE
Explanation:Ball End Mills are hemispherical tip used to cut and and shape/ milling rounded objects, such as the metal bearing grooves, slotting and pocketing.
Ball End Mill also called Ball nose end mills, the scallop height will not increase as a result of the increase in diameter. The diameter is the distance between one end of the round tool to another when taken from the middle.
A positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away from the rod we can conclude: A) the object is positively charged B) the object is negatively charged C) the object is an insulator D) the object is a conductor E) none of the above Two small charged objects repel each other with a force F when separated by a distance d. If the charge on each object is reduced to one-fourth of its original value and the distance between them is reduced to d/2 the force becomes: A) F/16 B) F/8 C) F/4 D) F/2 E) F
Answer
A) Positively charged two insulating rod are brought closed to an object they repel each other. It means the Object is positively charged. Because similar charge repel each other.
The correct answer is Option A.
B) we know force between to charges is calculated using Formula
[tex]F = \dfrac{kQ_1Q_2}{r^2}[/tex].......(1)
form the given condition in the question
[tex]F' = \dfrac{k\dfrac{Q_1}{4}\dfrac{Q_2}{4}}{(\dfrac{r}{2})^2}[/tex]
[tex]F' = \dfrac{k\dfrac{Q_1}{4}\dfrac{Q_2}{4}}{(\dfrac{r^2}{4})}[/tex]
[tex]F' = \dfrac{4}{16}\dfrac{kQ_1Q_2}{r^2}[/tex]
from equation (1)
[tex]F' = \dfrac{F}{4}[/tex]
hence, the correct answer is Option C.
The object is positively charged if repelled by a positively charged rod. The force between two charged objects reduces to F/16 when both charges are one-fourth and the distance is halved.
Explanation:Charged Objects and Electrostatic Forces
If the object suspended by a string is repelled away from a positively charged insulating rod, we can conclude that the object is positively charged (A). This is because like charges repel each other, according to electrostatic principles.
Regarding the force between two charged objects, Coulomb's law states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. If the charge on each object is reduced to one-fourth of its original value and the distance between them is reduced to d/2, the new force is calculated as:
New charge = Original charge / 4
New distance = d / 2
New force = (1/4 × 1/4) / (1/2 × 1/2)^2 = 1/16 / 1/4 = F/16 (A)
Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical brass spacers. Knowing that the average normal stress must not exceed 216 MPa in the bolts and 143 MPa in the spacers, determine the outer diameter of the spacers that yields the most economical and safe design. The outer diameter of the spacers that yields the most economical and safe design is 34.854 mm
To determine the outer diameter of the spacers that yields the most economical and safe design, consider the average normal stress in the bolts and spacers. The outer diameter of the spacers that satisfies the conditions is approximately 34.854 mm.
Explanation:To determine the outer diameter of the spacers that yields the most economical and safe design, we need to consider the average normal stress in the bolts and spacers. The average normal stress in the bolts must not exceed 216 MPa, and in the spacers, it must not exceed 143 MPa. Given that the bolts have a diameter of 16 mm and the spacers fit snugly inside, we can use the equation for calculating stress in a cylindrical object: stress = force/area.
Let's assume the outer diameter of the spacers is 'd'. The area of the spacers can be calculated as follows: area = pi/4 * (d^2 - (d-16)^2), where 'd-16' is the inner diameter of the spacers. To achieve the most economical and safe design, we want to maximize the area of the spacers while keeping the normal stress within the limits.
By substituting the given stress limits and solving for 'd', we can find the outer diameter that satisfies the conditions. After calculation, the outer diameter of the spacers that yields the most economical and safe design is found to be approximately 34.854 mm.
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The economic and safe design requires an outer spacer diameter of 34.854 mm.
To determine the outer diameter of the spacers that yields the most economical and safe design, we need to ensure that the average normal stress does not exceed the maximum allowable stress in both the bolts and the spacers.
The diameter of the bolt is 16 mm, so the cross-sectional area (Abolt) is:
Abolt = π/4 * d² = π/4 * (0.016 m)² Abolt = 2.01 × 10⁻⁴ m²Given the maximum normal stress is 216 MPa:
216 MPa = 216 × 10⁶ N/m²The maximum force (Fbolt) applied on the bolt:
Fbolt = stress × area = 216 × 10⁶ N/m² * 2.01 × 10⁻⁴ m² Fbolt = 43.416 kNThe outer diameter (D) of the spacer is what we need to determine.
The cross-sectional area of the spacer (Aspacer) should ensure that the normal stress does not exceed 143 MPa:
143 MPa = 143 × 10⁶ N/m²Using the same maximum force from the bolt (as it transfers to the spacer):
Aspacer = F / stress = 43.416 kN / 143 × 10⁶ N/m² Aspacer = 3.036 × 10⁻⁴ m²The cross-sectional area
Aspacer = π/4 * (D2 - inner diameter2) Aspacer = 3.036 × 10⁻⁴ m2Since the inner diameter (d) is 16 mm (bolt diameter),
Aspacer = π/4 * (D2 - (0.016 m)²)After solving for D using the above relationship:
D = 0.034854 m (or 34.854 mm)Therefore, the most economical and safe outer diameter for the spacers is 34.854 mm.
If a proton and an electron are released when they are 6.50×10⁻¹⁰ m apart (typical atomic distances), find the initial acceleration of each of them.
a) Acceleration of electron
b) Acceleration of proton
Answer:
(a) Acceleration of electron= 5.993×10²⁰ m/s²
(b) Acceleration of proton= 3.264×10¹⁷ m/s²
Explanation:
Given Data
distance r= 6.50×10⁻¹⁰ m
Mass of electron Me=9.109×10⁻³¹ kg
Mass of proton Mp=1.673×10⁻²⁷ kg
Charge of electron qe= -e = -1.602×10⁻¹⁹C
Charge of electron qp= e = 1.602×10⁻¹⁹C
To find
(a) Acceleration of electron
(b) Acceleration of proton
Solution
Since the charges are opposite the Coulomb Force is attractive
So
[tex]F=\frac{1}{4(\pi)Eo }\frac{|qp*qe|}{r^{2} }\\ F=(8.988*10^{9}Nm^{2}/C^{2})*\frac{(1.602*10^{-19})^{2} }{(6.50*10^{-10} )^{2} } \\F=5.46*10^{-10}N[/tex]
From Newtons Second Law of motion
F=ma
a=F/m
For (a) Acceleration of electron
[tex]a=F/Me\\a=(5.46*10^{-10} )/9.109*10^{-31}\\ a=5.993*10^{20}m/s^{2}[/tex]
For(b) Acceleration of proton
[tex]a=F/Mp\\a=(5.46*10^{-10} )/1.673*10^{-27} \\a=3.264*10^{17}m/s^{2}[/tex]
Acceleration of the body is rate of change of the increasing velocity with respect to the time. is The acceleration of the electron is [tex]5.992\times10^{20}[/tex] meter per second squared and acceleration of the proton is [tex]3.265\times10^{17}[/tex] meter per second squared.
Given information-A proton and an electron are released when they are 6.50×10⁻¹⁰ m apart.
The the distance of proton and electron is 6.50×10⁻¹⁰ m.
Now it is known that,
Mass of the electron,[tex]m_e=9.109\times10^{-31}[/tex]
Mass of the proton,[tex]m_p=1.673\times10^{-27}[/tex]
Charge of the electron,[tex]Q_e=-1.602\times10^{-19}[/tex]
Charge of the proton,[tex]Q_e=1.602\times10^{-19}[/tex]
AccelerationAcceleration of the body is rate of change of the increasing velocity with respect to the time. Acceleration of a body is the ratio of the force to the mass of the body.
The coulomb force between the two charges can be given as,
[tex]F=k\dfrac{q_1q_2}{r^2} [/tex]
Here k is the coulombs constant, r is the distance and q is the charge of proton and electron.Put the values,
[tex]F=9\times10^9\times\dfrac{(1.602\times10^{-19})^2}{(6.5\times10^{-10})^2} [/tex]
[tex]F=5.46\times10^{-10}[/tex]
Now it is known that the acceleration of a body is the ratio of the force to the mass of the body.
Acceleration of electron[tex]a_e=\dfrac{F}{m_e} [/tex]
[tex]a_e=\dfrac{5.46\times10^{-10}}{9.109\times10^{-31}} [/tex]
[tex]a_e=5.992\times10^{20}[/tex]
Acceleration of Proton[tex]a_p=\dfrac{F}{m_p} [/tex]
[tex]a_p=\dfrac{5.46\times10^{-10}}{1.673\times10^{-27}} [/tex]
[tex]a_e=3.625\times10^{17}[/tex]
Thus the acceleration of the electron is [tex]5.992\times10^{20}[/tex] meter per second squared and acceleration of the proton is [tex]3.265\times10^{17}[/tex] meter per second squared.
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A plane leaves the airport in Galisteo and flies 160 km at 66.0 ∘ east of north; then it changes direction to fly 260 km at 49.0 ∘ south of east, after which it makes an immediate emergency landing in a pasture?
The question doesn't have any particular requirement, but we'll compute the displacement of the plane from its initial and final landing point in the pasture
Answer:
[tex]\displaystyle |\vec{r}|=321.464\ km[/tex]
[tex]\displaystyle \theta =-19.395^o[/tex]
Explanation:
Displacement
The vector displacement [tex]\vec r[/tex] is a measure of the change of position of a moving object. The displacement doesn't depend on the path followed, only on the final and initial positions. Its scalar counterpart, the distance, does measure the total space traveled and considers all the changes in the direction taken by the object. To find the displacement, we must add all the particular displacements by using vectors.
The plane first flies 160 km at 66° east of north. To find the vector expression of this displacement, we must know the angle with respect to the East direction or North of East. Knowing the angle East of North is 66°, the required angle is 90°-66°=34°
The first vector is expressed as
[tex]\displaystyle \vec{r_1}=\left \langle 160^o\ cos34^o, 160\ sin34^o \right \rangle[/tex]
[tex]\displaystyle \vec{r_1}=\left \langle 132.646, 89.471 \right \rangle[/tex]
The second displacement is 260 km at 49° South of East. This angle is below the horizontal respect to the reference, thus we use -49°.
The second vector is expressed as:
[tex]\displaystyle \vec{r_2}=\left \langle 260\ cos(-49^o), 260\ sin(-49^o)\right \rangle[/tex]
[tex]\displaystyle \vec{r_2}=\left \langle 170.575,-196.224\right \rangle[/tex]
The total displacement is computed as the vectorial sum of both vectors
[tex]\displaystyle \vec{r}=\vec{r_1}+\vec{r_2}=\left \langle 132.646+170.575\right \rangle+ \left \langle89.471-196.224\right \rangle[/tex]
[tex]\displaystyle \vec{r}=\left \langle 303.221,-106.753\right \rangle\ km[/tex]
The magnitude of the total displacement is
[tex]\displaystyle |\vec{r}|=\sqrt{303.221^2+(-106,753)^2}[/tex]
[tex]\displaystyle |\vec{r}|=321.464\ km[/tex]
And the direction is
[tex]\displaystyle tan\ \theta =\frac{-106.753}{303.221}=-0.352[/tex]
[tex]\displaystyle \theta =-19.395^o[/tex]
A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.
(a) Find the magnitude of the electrostatic force that one charge exerts on the other.
(b) Is the force attractive or repulsive?
Explanation:
Given that,
Charge 1, [tex]q_1=6.75\ nC=6.75 \times 10^{-9}\ C[/tex]
Charge 2, [tex]q_2=4.46\ nC=4.46\times 10^{-9}\ C[/tex]
The distance between charges, r = 1.99 m
To find,
The electrostatic force and its nature
Solution,
(a) The electric force between two charges is given by :
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
[tex]F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}[/tex]
[tex]F=6.84\times 10^{-8}\ N[/tex]
(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.
Therefore, this is the required solution.
(a) The magnitude of the electrostatic force that one charge exerts on the other is 6.83×10⁻⁸ N.
(b) The force is repulsive.
(a) To calculate the magnitude of the electrostatic force that one charge exerts on the other, we use the formula below.
Formula:
F = kqQ/r².................. Equation 1Where:
F = Force of between the chargesq = First chargeQ = second charger = distance between the chargesk = coulomb's constant.From the question,
Given:
q = 6.75 nC = 6.75×10⁻⁹ CQ = 4.46 nC = 4.46×10⁻⁹ Cr = 1.99 mk = 8.98×10⁶ Nm²/C²Substitute these values into equation 1
F = (6.75×10⁻⁹)(4.46×10⁻⁹)(8.98×10⁹)/1.99²F = 6.83×10⁻⁸ N(b) The force is repulsive because both charges a the same (positive).
Hence, (a) The magnitude of the electrostatic force that one charge exerts on the other is 6.83×10⁻⁸ N (b) The force is repulsive.
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A frog is at the bottom of a 17-foot well. Each time the frog leaps, it moves up 3 feet. If the frog has not reached the top of the well, then the frog slides back 1 foot before it is ready to make another leap. How many leaps will the frog need to escape the well?
Answer:
The frog takes 8 jumps to reach top of well
Explanation:
Given data
Frog at bottom=17 foot
Each time frog leaps 3 feet
Frog has not reached the top of the well, then the frog slides back 1 foot
To Find
Total number of leaps the frog needed to escape from well
Solution
in 1 jump distance jumped=3+(-1)
=2 feet
=2×1 feet
The "-1" is because the frog goes back
Now After 2 jumps the distance jumped as:
Distance Jumped=2+2
Distance Jumped=2*2
=4 feet
Similarly after 7 jumps
Distance Jumped=2+2+......+2
Distance Jumped=2*7
=14 feet
Now after 8th jump the frog climbs but doesnot slide back as it is reached to the top of well.
So
Distance Jumped=(Distance Jumped after 7 jumps)+3
=14+3
=17 feet
The frog takes 8 jumps to reach top of well
The frog needs 8 leaps each of 2 feet and a final leap of 3 feet to escape a 17-foot well. Therefore, it takes the frog a total of 9 leaps to escape the well.
Explanation:In this mathematical problem, we need to determine how many leaps it takes for a frog to escape a 17-foot well, given that each leap propels the frog 3 feet up, but it slides back 1 foot before the next leap. Given thefa, each leap results net gain of 2 feet (3 feet up minus 1 foot slide back). However, for the last leap, the frog won't slide back, so the final leap has a net gain of 3 feet.
Therefore, the frog need to leap 15 feet using jumps with a net gain of 2 feet and then make a final leap of 3 feet out of the well. Each of the first leaps covers 2 feet, so the number of such leaps needed is 15/2 = 7.5. Since it's not possible to make half a jump, we round up to 8 jumps. Then, the frog makes its final jump of 3 feet. So in total, the frog needs 9 leaps to escape the well.
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