Answer:
D. a cation that has a smaller radius than the atom.
Explanation:
When electrons are removed from the outermost shell of a calcium atom, the atom becomes a cation that has a smaller radius than the atom.
When electrons are removed from the outermost shell of a calcium atom, the atom becomes: D. a cation that has a smaller radius than the atom.
An ion can be defined as an atom or molecules (group of atoms) that has either lost or gained one or more of its valence electrons, thereby, making it to have a net positive or negative electrical charge respectively.
Basically, there are two (2) types of ion and these are:
Anion.Cation.A cation is a positive electrical charge which is formed when the atom of a chemical element losses an electron.
Hence, a cation is formed when an electron is removed from the atom of a chemical element.
Generally, the atomic size of a chemical element would increase when its number of electrons increase, thereby, increasing its atomic radius.
Consequently, the radius of a cation in comparison with the radius of its neutral atom becomes smaller because it losses more electron from the outermost shell of its atom.
In conclusion, when electrons are removed from the outermost shell of a calcium atom, the atom becomes a cation that has a smaller radius than its atom.
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The power (P) required to run a motor is equal to the voltage (E) applied to that motor times the current (I) supplied to the motor. If the motor data says the motor uses 180 watts of power and the voltage applied to the motor is 120 volts, how much current will the motor require?
Final answer:
The motor will require a current of 1.5 amperes to operate at 180 watts with an applied voltage of 120 volts, based on the formula for electric power P = IV.
Explanation:
To calculate the current required by the motor, we will use the formula for electric power, which is P = IV, where P is the power in watts, I is the current in amperes, and V is the voltage in volts.
Given:
The power consumed by the motor (P) = 180 watts
The voltage applied to the motor (V) = 120 volts
We need to solve for I (current):
I = P / V
I = 180 W / 120 V
I = 1.5 A
Therefore, the motor will require 1.5 amperes of current when operating at 180 watts of power with a voltage of 120 volts.
Which of the following statements must be true if the net force on an object is zero? Choose all that apply.
a)The object must have constant velocity.
b)The object must be at rest.
c)The object must be at the origin.
d)The object must have zero acceleration.
Answer:
a) The object must have constant velocity.
d) The object must have zero acceleration.
Explanation:
We can solve the problem by using Newton's second law, which states that the net force acting on an object is equal to the product between mass and acceleration:
[tex]F = ma[/tex]
where
F is the net force
m is the mass of the object
a is the acceleration
In this problem, the net force on the object is zero:
F = 0
This means that the acceleration of the object is also zero, according to the previous equation:
a = 0
So statement (d) is correct. Moreover, acceleration is defined as the rate of change of velocity:
[tex]a=\frac{\Delta v}{\Delta t}[/tex]
Which means that [tex]\Delta v=0[/tex], so the velocity is constant. Therefore, statement (a) is also correct. The other two statements are false because:
b)The object must be at rest. --> false, the object can be moving at constant velocity, different from zero
c)The object must be at the origin. --> false, since the object can be in motion
If the net force on an object is zero, the object might have constant velocity or zero acceleration. It's not necessary that the object be at rest or at the origin.
Explanation:The subject of your question is related to the principles of Physics, specifically Newton's First Law of Motion. If the net force on an object is zero, it means that the object is in a state of equilibrium. From the given options:
a) The object must have constant velocity. This can be true, as an object under no net force will continue moving at a constant velocity or stay at rest, according to Newton's First Law. b) The object must be at rest. This is not necessarily true. The object can be at rest or still be moving at a constant velocity. c) The object must be at the origin. This statement is false as the position of the object does not depend on the net force. d) The object must have zero acceleration. This is absolutely true. According to Newton's Second Law, if net force is zero, acceleration is also zero because acceleration is directly proportional to the net force.Learn more about Net Force
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An airplane in a wide “outside” loop can create an apparent zero weight inside the aircraft cabin. What must be the radius of curvature of the flight path for an aircraft moving at 350 km/h to create a condition of weightlessness inside the aircraft? Assume the acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s
Answer:
R = 964.5 m
Explanation:
When plane is moving in vertical loop then at the condition of free fall then force of gravity on the passengers will be balanced by the the pseudo force on them.
In ground frame we can say that normal force on the passengers will become zero
So we have
[tex]mg = \frac{mv^2}{r}[/tex]
[tex]m(9.8) = \frac{mv^2}{R}[/tex]
here we know that
v = 350 km/h = 97.22 m/s
now we have
[tex]9.8 = \frac{97.22^2}{R}[/tex]
[tex]R = 964.5 m[/tex]
Radius of curvature of the flight path must be approximately 966.67 meters to create a condition of apparent zero weight inside the aircraft cabin.
The condition of weightlessness inside an aircraft can be simulated when the centripetal acceleration equals the acceleration due to gravity. According to the formula a = v^2 / r, where a is the centripetal acceleration, v is the velocity, and r is the radius of curvature, we need to rearrange the formula to solve for r: r = v^2 / g. Given that the aircraft is moving at 350 km/h, which is approximately 97.22 m/s (since 1 km/h = 0.27778 m/s), and using g as 9.8 m/s^2, we can calculate the necessary radius of curvature to create a condition of weightlessness. Therefore, the radius of curvature r is:
r = (97.22 m/s)^2 / 9.8 m/s^2
r = 966.67 m
Thus, the radius of curvature of the flight path must be approximately 966.67 meters to create a condition of apparent zero weight inside the aircraft cabin.
The atomic radius of main-group elements generally increases down a group because ________.
Answer:
The principal quantum number of the valence orbitals increases.
Explanation:
The atomic radius of main-group elements generally increases down a group because the principal quantum number of the valence orbitals increases.
The atomic radius of main-group elements increases down a group because the valence electron shell is getting larger and there is a larger principal quantum number, so the valence shell lies physically farther away from the nucleus.
Explanation:The atomic radius of main-group elements generally increases down a group because as you go down a column of the periodic table, the valence electron shell is getting larger and there is a larger principal quantum number, so the valence shell lies physically farther away from the nucleus. This trend can be summarized as follows:
As the atomic number increases down a group, there is an increase in the positive nuclear charge.However, there is also an increase in the number of occupied principal energy levels.Higher principal energy levels consist of orbitals which are larger in size than the orbitals from lower energy levels.The effect of the greater number of principal energy levels outweighs the increase in nuclear charge, and so atomic radius increases down a group.Suppose you were bungee jumping from a bridge while blowing a hand-held air horn. How would someone remaining on the bridge hear the pitch of the air horn as time increased?
a. The pitch would get progressively lower (i.e., smaller frequency)
b. The pitch would get progressively higher (i.e., larger frequency)
c. The pitch would remain the same (i.e., constant frequency)
Answer:
Answer to the question:
Explanation:
a. The pitch would get progressively lower (i.e., smaller frequency)
A person on the bridge would hear the pitch of an air horn during a bungee jump become lower as the jumper descends and higher as the jumper ascends, due to the Doppler effect.
Explanation:During a bungee jump, if someone is blowing an air horn, a person on the bridge would experience a change in the perceived pitch of the sound due to the Doppler effect. As you move away from the observer on the bridge, the pitch of the air horn would sound lower. This is because the frequency of the sound waves reaching the observer decreases as the distance between the air horn increases during the descent. Conversely, as the bungee jumper ascends and gets closer to the bridge, the frequency increases and the pitch sounds higher to the observer.
The correct answer to the question is:
a. The pitch would get progressively lower (i.e., smaller frequency) as the bungee jumper moves away from the bridge, and would get progressively higher (i.e., larger frequency) as the jumper ascends back towards the bridge.
Using the formula for work, Match the Force × Distance problems with their correct "Work Performed" answer. The first example has been completed for you.Force × Distance = Work Performed 10 newtons 6 meters =60 ng Click the item in the left column. Use the plus sign to move it up or the minus sign to move it down until it matches the correct entry in the right column. Lock your answer in place by clicking the square beside the item. (A check mark means it is locked.)5 ft.-lbs 40 pounds 100 feet 62 n-m 2.5 pounds 24 inches 200 ft.-lbs. 62 newtons 1 meter 40,000 ft.-lbs. 1 ton 20 foot 4000 ft.-lbs. 400 pounds 0.5 feet
Using the formula for work (Work = Force x Distance), force and distance pairs are matched to their corresponding 'Work Performed' measurements. This process involves converting all measurements to the metric system (newtons and meters) to calculate the work performed in newton-meters (n-m).
Explanation:The formula for work is Work = Force x Distance. Force is usually measured in newtons (N), and distance is measured in meters (m), so work is measured in newton-meters (n-m). One foot-pound (ft-lb) is equivalent to 1.35582 n-m and one pound is equivalent to 4.44822_newtons_ in the metric system. Therefore, the provided force and distance pairs can be matched to the correct 'Work Performed' answer as follows:
5 ft-lbs = 5 * 1.35582 = 6.7791 n-m (Approximately 7 n-m)40 pounds 100 feet = (40 * 4.44822_newtons_) * (100 * 0.3048_meters_) = 53096 n-m (Approximately 520 n-m)2.5 pounds 24 inches = (2.5 * 4.44822_newtons_) * (24 * 0.0254_meters_) = 6.8058 n-m (Approximately 7 n-m)62 newtons 1 meter = 62 * 1 = 62 n-m1 ton 20 foot = (2000 pounds * 4.44822_newtons_) * (20 * 0.3048_meters_) = 54680.4 n-m (Approximately 55000 n-m)400_pounds_ 0.5 feet = (400 * 4.44822_newtons_) * (0.5 * .3048_meters_) = 271.258 n-m (Approximately 270 n-m)Learn more about Workhttps://brainly.com/question/30763018
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A 2-m3 rigid tank initially contains air whose density is 1.18 kg/m3 . The tank is connected to a high-pressure supply line through a valve. The valve is opened, and air is allowed to enter the tank until the density in the tank rises to 5.30 kg/m3 . Determine the mass of air that has entered the tank?
Explanation:
It is known that density is the mass present in per unit volume.
Mathematically, Density = [tex]\frac{mass}{volume}[/tex]
Since, it is given that [tex]d_{1}[/tex] is 1.18 [tex]kg/m^{3}[/tex], [tex]d_{2}[/tex] is 5.30 [tex]kg/m^{3}[/tex], and volume is 2 [tex]m^{3}[/tex].
Therefore, mass of air that has entered will be [tex]m_{2}[/tex] - [tex]m_{1}[/tex] and it will be calculated as follows.
[tex]d_{2} - d_{1}[/tex] = [tex]\frac{m_{2} - m_{1}}{Volume}[/tex]
[tex]m_{2} - m_{1}[/tex] = [tex](5.30 kg/m^{3} - 1.18 kg/m^{3}) \times 2 m^{3}[/tex]
= 8.24 kg
Thus, we can conclude that mass of air that has entered the tank is 8.24 kg.
A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally away from and below the point of release. What is the speed of the ball just before it strikes the ground?
Answer:
v = 46.55 m/s
Explanation:
It is given that,
A ball is thrown horizontally from the top of a building 0.10 km high, d = 0.1 km = 100 m
The ball strikes the ground at a point 65 m horizontally away from and below the point of release, h = 65 m
At maximum height, velocity of the ball is 0. So, using the equation of motion as :
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
Here, a = g
[tex]100=0+\dfrac{1}{2}\times 9.8t^2[/tex]
[tex]t=4.51\ s[/tex]
Let [tex]v_x[/tex] is the horizontal velocity of the ball. It is calculated as :
[tex]v_x=\dfrac{65\ m}{4.51\ s}=14.41\ m/s[/tex]
Let [tex]v_y[/tex] is the final speed of the ball in y direction. It can be calculated as :
[tex]v_y^2+u_y^2=2as[/tex]
[tex]u_y=0[/tex]
[tex]v_y^2=2gd[/tex]
[tex]v_y^2=2\times 9.8\times 100[/tex]
[tex]v_y=44.27\ m/s[/tex]
Let v is the speed of the ball just before it strikes the ground. It is given by :
[tex]v=\sqrt{v_x^2+v_y^2}[/tex]
[tex]v=\sqrt{14.41^2+44.27^2}[/tex]
v = 46.55 m/s
So, the speed of the ball just before it strikes the ground is 46.55 m/s. Hence, this is the required solution.
The speed of the ball just before it strikes the ground is equal to 46.55 m/s.
Given the following data:
Horizontal distance = 65 metersHeight of building = 0.10 km = 100 metersWe know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 [tex]m/s^2[/tex].
To determine the speed of the ball just before it strikes the ground:
First of all, we would determine the time it took the ball to strike the ground by using the formula for maximum height.
[tex]H = \frac{1}{2} gt^2\\\\100 = \frac{1}{2} \times 9.8 \times t^2\\\\200 = 9.8t^2\\\\t^2 = \frac{200}{9.8} \\\\t^2=20.41\\\\t=\sqrt{20.41}[/tex]
Time, t = 4.52 seconds
Next, we would find the horizontal velocity:
[tex]Horizontal\;velocity = \frac{horizontal\;distance}{time} \\\\Horizontal\;velocity = \frac{65}{4.52}[/tex]
Horizontal velocity, V1 = 14.38 m/s
Also, we would find the velocity of the ball in the horizontal direction:
[tex]V_2^2 = U^2 + 2aS\\\\V_2^2 = 0^2 + 2(9.8)(100)\\\\V_2^2 = 1960\\\\V_2 = \sqrt{1960} \\\\V_2 = 44.27 \;m/s[/tex]
Now, we would calculate the speed of the ball just before it strikes the ground by finding the resultant speed:
[tex]V = \sqrt{V_1^2 + V_2^2} \\\\V = \sqrt{14.38^2 + 44.27^2}\\\\V = \sqrt{206.7844 + 1959.8329}\\\\V =\sqrt{2166.6173}[/tex]
Speed, V = 46.55 m/s
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Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.0 kg⋅m2 and for arms and legs in is 0.90 kg⋅m2 . If she starts out spinning at 5.2 rev/s , what is her angular speed (in rev/s) when her arms and one leg open outward?
Answer:
Her angular speed (in rev/s) when her arms and one leg open outward is [tex]1.56\frac{rev}{s}[/tex]
Explanation:
Initial moment of inertia when arms and legs in is [tex]I_i=0.90 kg.m^{2}[/tex]
Final moment of inertia when her arms and on leg open outward, [tex]I_f=3.0 kg.m^{2}[/tex]
Initial angular speed [tex]w_i=5.2\frac{rev}{s}[/tex]
Let the final angular speed be [tex]w_f[/tex]
Since external torque on her is zero so we can apply conservation of angular momentum
[tex]\therefore L_f=L_i[/tex]
=>[tex]I_fw_f=I_iw_i[/tex]
=>[tex]w_f=\frac{I_iw_i}{I_f}=\frac{0.9\times5.2 }{3.0}\frac{rev}{s}=1.56\frac{rev}{s}[/tex]
Thus her angular speed (in rev/s) when her arms and one leg open outward is [tex]1.56\frac{rev}{s}[/tex]
The movement of electrons is what type of energy? A. Kinetic energy B. Magnetic energy C. Charge energy D. Electrical energy
Answer:
The correct answer is kinetic energy. :) hope this helps!
Explanation:
A rectangular field is 300 meters long and 300 meters wide. What is the area of the field in square kilometers? Do not round your answer. Be sure to include the correct unit in your answer.
Answer:
Area of the rectangular field in kilometers is 0.09 [tex]km^2[/tex]
Explanation:
We know that 1 kilometers = 1000 meters
since we need to find the area in unit of kilometers
therefore converting length and width into kilometers
1000 meters = 1 kilometers
300 meters =[tex]\frac{300}{1000} = 0.3[/tex]
Likewise width = 0.3 km
Area = length x width
= 0.3km x 0.3 km
= 0.09 [tex]km^2[/tex]
In a follow-up experiment, two identical gurneys are placed side-by-side on a ramp with their wheels locked to eliminate spinning. Gurney 1 has a dummy placed on it to give it a total mass of 200 kg, while Gurney 2 is loaded with a dummy that makes it only 50 kg overall. If the ramp has a coefficient of friction of μs, which gurney is more likely to slide down the ramp?
Given the same ramp angle and coefficient of friction, the lighter gurney (50kg) is more likely to slide down the ramp since it requires less friction force to stay stationary as compared to the heavier gurney (200kg).
Explanation:The likelihood of the gurneys sliding down the slope depends upon the balance of forces on each. The force due to gravity on each gurney is mg sin θ, where m is mass, g is acceleration due to gravity, and θ is the angle of the ramp. The force of static friction on each gurney is μs N = μs mg cos θ, where μs is the coefficient of static friction, and N is the normal force, which equals mg cos θ.
In order for the gurneys not to slide, the friction force (μs mg cos θ) must be equal to or greater than the force due to gravity (mg sin θ). For a larger mass (like gurney 1 with the 200 kg dummy), the friction force is greater, so it is more likely to stay put. On the other hand, the smaller mass (gurney 2 with the 50 kg dummy) suffers less friction force due to its lesser weight. Therefore, as long as the angle of the ramp and the coefficient of friction are the same for both gurneys, gurney 2 (50 kg) is more likely to slide down the ramp.
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What area of the country is the main benefactor of the electricity generated by the hoover dam?
Answer:
Explanation:
Arizona Nevada and California. That includes some pretty big cities. Los Angeles, Las Vegas, San Diego to name 3.
The mass of a string is 5.5 × 10-3 kg, and it is stretched so that the tension in it is 230 N. A transverse wave traveling on this string has a frequency of 160 Hz and a wavelength of 0.66 m. What is the length of the string?
Answer:
The length of the string is 0.266 meters.
Explanation:
It is given that,
Mass of the string, [tex]m=5.5\times 10^{-3}\ kg[/tex]
Tension in the string, T = 230 N
Frequency of wave, f = 160 Hz
Wavelength of the wave, [tex]\lambda=0.66\ m[/tex]
We need to find the length of the string. Let l is the length of the string. The speed of a transverse wave is given by :
[tex]v=\sqrt{\dfrac{T}{M}}[/tex]
M is the mass per unit length, M = m/l
[tex]v=\sqrt{\dfrac{lT}{m}}[/tex]
[tex]l=\dfrac{v^2m}{T}[/tex]
The velocity of a wave is, [tex]v=\nu\times \lambda[/tex]
[tex]l=\dfrac{(\nu\times \lambda)^2m}{T}[/tex]
[tex]l=\dfrac{(160\ Hz\times 0.66\ m)^2\times 5.5\times 10^{-3}\ kg}{230\ N}[/tex]
l = 0.266 meters
So, the length of the string is 0.266 meters. Hence, this is the required solution.
Answer:
L = 0.275 m
Explanation:
velocity of transverse wave in a stretched string is given as
[tex]v =\sqrt \frac{T}{\mu}[/tex]
where T = tension = 230N
μ = linear density
[tex]μ = \frac[m}{L}[/tex]
where length L is in meters
Velocity = [tex]n\lambda[/tex]
so we have after equating both value of velocity
[tex]\sqrt \frac{T}{\mu} = n\lambda[/tex]
[tex]\frac{T}{\mu} =(n\lambda)^{2}[/tex]
μ = [tex]\frac{T}{(n\lambda)^{2}}[/tex]
μ = [tex] \frac{230}{(160*0.66)^{2}}[/tex]
μ = 0.020 kg/m
but μ = [tex]\frac[m}{L}[/tex]
so length of string is
L = [tex]\frac{5.5*10^{-3}}{0.020}[/tex]
L = 0.275 m
Atmospheric pressure varies from day to day. The level of a floating ship on a high-pressure day is (a) higher (b) lower, or (c) no different than on a low-pressure day.
Answer:
(c) no different than on a low-pressure day.
Explanation:
The force acting on the ship when it floats in water is the buoyant force. According to the Archimedes' principle: The magnitude of buoyant force acting on the body of the object is equal to the volume displaced by the object.
Thus, Buoyant forces are a volume phenomenon and is determined by the volume of the fluid displaced.
Whether it is a high pressure day or a low pressure day, the level of the floating ship is unaffected because the increased or decreased pressure at the all the points of the water and the ship and there will be no change in the volume of the water displaced by the ship.
The electronegativity of nonmetals is relatively __________ as compared to the electronegativity of metals.
We have that Metals are elements that mostly produce ions of Positive charges eg [tex]Na^{2+}[/tex]
While Non metals tends to have negatively charged ions as like [tex]O^-[/tex]
We have that The electro negativity of nonmetals is relatively High as compared to the electro negativity of metals which is Low
From the question we are told
The electro negativity of nonmetals is relatively __________ as compared to the electro negativity of metals.
Generally
Metals are elements that mostly produce ions of Positive charges eg [tex]Na^{2+}[/tex]
While Non metals tends to have negatively charged ions as like [tex]O^-[/tex]
Therefore
From the definition above
Metals are elements that mostly produce ions of Positive charges eg [tex]Na^{2+}[/tex]
While Non metals tends to have negatively charged ions as like [tex]O^-[/tex]
We have that The electro negativity of nonmetals is relatively High as compared to the electro negativity of metals which is Low
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A climber using bottled oxygen accidentally drops the oxygen bottle from an altitude of 4500 m. If the bottle fell straight down this entire distance, what is the velocity of the 3-kg bottle just prior to impact at sea level? (Note: ignore air resistance)
Answer:
300 m/s
Explanation:
As the cylinder drops off so its initial velocity is zero.
h = 4500 m, g = 10 m/s^2, u = 0
Use third equation of motion
v^2 = u^2 + 2 g h
v^2 = 0 + 2 x 10 x 4500
v^2 = 90000
v = 300 m /s
Final answer:
The velocity of the oxygen bottle just before impact at sea level, having fallen straight down from an altitude of 4500 m ignoring air resistance, is approximately 297.4 m/s.
Explanation:
To calculate the velocity of the oxygen bottle just before impact, we can use the principles of physics, specifically the conservation of energy or kinematics under the influence of gravity. Assuming air resistance is negligible, all potential energy (PE) of the bottle at the altitude of 4500 m will be converted into kinetic energy (KE) just before impact.
The potential energy (PE) at the height is given by PE = m*g*h, where m is mass, g is the acceleration due to gravity (approximately 9.81 m/s2), and h is the height from which it falls. The kinetic energy (KE) just before impact is given by KE = 0.5 * m * v2, where v is the velocity.
Setting PE equal to KE, we have m*g*h = 0.5 * m * v2. Solving for v (and noting that the mass cancels out), we find v = sqrt(2*g*h). Plugging in the values, we get v = sqrt(2*9.81*4500), which calculates to a velocity of approximately 297.4 m/s.
I'm spending 98 points please help!
Answer: C. 8.0 m west
Explanation: The arrows are going 15 m west and 7.0 m east. 7 meters of the west will cancel out because 15-7=8. Subtract the smaller number from the bigger number, which is west minus east. The answer will be 8.0 m west.
A car of mass m accelerates from speed v_1 to speed v_2 while going up a slope that makes an angle theta with the horizontal. The coefficient of static friction is mu_s, and the acceleration due to gravity is g. Find the total work W done on the car by the external forces.
Answer:
Work done by external force is given as
[tex]Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/tex]
Explanation:
As per work energy Theorem we can say that work done by all force on the car is equal to change in kinetic energy of the car
so we will have
[tex]Work_{external} + Work_{gravity} + Work_{friction} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/tex]
now we have
[tex]W_{gravity} = -mg(Lsin\theta)[/tex]
[tex]W_{friction} = -\mu mgcos(\theta) L[/tex]
so from above equation
[tex]Work_{external} - mgLsin\theta - \mu mgLcos(\theta) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/tex]
so from above equation work done by external force is given as
[tex]Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/tex]
A substance can absorb heat energy by the process of
Answer:
conduction,convection,radiation
Based on the processes of heat transfer, substance can absorb heat energy by the process of conduction, convection and radiation.
What is heat energy?Heat energy is the energy due to temperature difference between two bodies.
Heat energy always flow from hotter to colder bodies.
The processes of heat transfer are as follows:
conduction convection radiationTherefore, a substance can absorb heat energy by the process of conduction, convection and radiation.
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Two loudspeakers are located 2.59 m apart on an outdoor stage. A listener is 21.6 m from one and 22.7 m from the other. During the sound check, a signal generator drives the two speakers in phase with the same amplitude and frequency. The transmitted frequency is swept through the audible range (20 Hz to 20 kHz). (a) What is the lowest frequency fmin, one that gives minimum signal (destructive interference) at the listener's location
Answer:
Frequency [tex]f_{min,1}=155.90\ Hz[/tex]
Explanation:
Given data:
The distance between the speakers, d = 2.59 m
The distance between the listeners, ΔL = 22.7 - 21.6 = 1.1 m
Now, For a destructive interference, we know that
[tex]\frac{\Delta L}{\lambda}=0.5,1.5,2.5,.........[/tex]
where, λ = wavelength
thus,
frequency [tex]f_{min,n}=\frac{(n-0.5)v}{\Delta L}[/tex]
where,
v = speed of sound = 343 m/s
for n = 1
we get
frequency [tex]f_{min,1}=\frac{(1-0.5)\times 343}{1.1}[/tex]
or
Frequency [tex]f_{min,1}=155.90\ Hz[/tex]
A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 50.9 J and a maximum displacement from equilibrium of 0.204 m. (a) What is the spring constant? N/m (b) What is the kinetic energy of the system at the equilibrium point? J (c) If the maximum speed of the block is 3.45 m/s, what is its mass? kg (d) What is the speed of the block when its displacement is 0.160 m? m/s (e) Find the kinetic energy of the block at x = 0.160 m. J (f) Find the potential energy stored in the spring when x = 0.160 m. J (g) Suppose the same system is released from rest at x = 0.204 m on a rough surface so that it loses 12.2 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant? m
(a) 2446 N/m
When the spring is at its maximum displacement, the elastic potential energy of the system is equal to the total mechanical energy:
[tex]E=U=\frac{1}{2}kA^2[/tex]
where
U is the elastic potential energy
k is the spring constant
A is the maximum displacement (the amplitude)
Here we have
U = E = 50.9 J
A = 0.204 m
Substituting and solving the formula for k,
[tex]k=\frac{2E}{A^2}=\frac{2(50.9)}{(0.204)^2}=2446 N/m[/tex]
(b) 50.9 J
The total mechanical energy of the system at any time during the motion is given by:
E = K + U
where
K is the kinetic energy
U is the elastic potential energy
We know that the total mechanical energy is constant: E = 50.9 J
We also know that at the equilibrium point, the elastic potential energy is zero:
[tex]U=\frac{1}{2}kx^2=0[/tex] because x (the displacement) is zero
Therefore the kinetic energy at the equilibrium point is simply equal to the total mechanical energy:
[tex]K=E=50.9 J[/tex]
(c) 8.55 kg
The maximum speed of the block is v = 3.45 m/s, and it occurs when the kinetic energy is maximum, so when
K = 50.9 J (at the equilibrium position)
Kinetic energy can be written as
[tex]K=\frac{1}{2}mv^2[/tex]
where m is the mass
Solving the equation for m, we find the mass:
[tex]m=\frac{2K}{v^2}=\frac{2(50.9)}{(3.45)^2}=8.55 kg[/tex]
(d) 2.14 m/s
When the displacement is
x = 0.160 m
The elastic potential energy is
[tex]U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J[/tex]
So the kinetic energy is
[tex]K=E-U=50.9 J-31.3 J=19.6 J[/tex]
And so we can find the speed through the formula of the kinetic energy:
[tex]K=\frac{1}{2}mv^2 \rightarrow v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(19.6)}{8.55}}=2.14 m/s[/tex]
(e) 19.6 J
The elastic potential energy when the displacement is x = 0.160 m is given by
[tex]U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J[/tex]
And since the total mechanical energy E is constant:
E = 50.9 J
the kinetic energy of the block at this point is
[tex]K=E-U=50.9 J-31.3 J=19.6 J[/tex]
(f) 31.3 J
The elastic potential energy stored in the spring at any time is
[tex]U=\frac{1}{2}kx^2[/tex]
where
k = 2446 N/m is the spring constant
x is the displacement
Substituting
x = 0.160 m
we find the elastic potential energy:
[tex]U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J[/tex]
(g) x = 0
The postion at that instant is x = 0, since it is given that at that instant the system passes the equilibrium position, which is zero.
The response provides calculations for determining the spring constant, the kinetic energy at the equilibrium point, and the mass of the block. It explains the relevant formulas and step-by-step solutions for each part of the question.
Spring Constant Calculation: Using the formula for total mechanical energy, E = [tex]1/2kA^2[/tex], where A is the maximum displacement, the spring constant is calculated as k = [tex]2E / A^2[/tex]. Substituting the given values, k = 2 * 50.9 / [tex](0.204)^2[/tex] = 500 N/m.
Kinetic Energy at Equilibrium: At the equilibrium point, all energy is in the form of potential energy; hence, the kinetic energy is zero.
Mass Calculation: Using the formula for maximum speed vmax = sqrt(E / m), where E is the total mechanical energy and m is the mass, the mass is calculated as m = E / [tex]vmax^2[/tex]. Substituting the values, m = 50.9 / [tex](3.45)^2[/tex]= 4.82 kg.
The length that a hanging spring stretches varies directly with the weight placed at the end of the spring. If a weight of 11lb stretches a certain spring 7in., how far will the spring stretch if the weight is increased to 44lb ?
Final answer:
The spring will stretch approximately 28 inches when the weight is increased to 44lb.
Explanation:
The length that a hanging spring stretches varies directly with the weight placed at the end of the spring. To find out how far the spring will stretch when the weight is increased to 44lb, we can set up a proportion based on the given information:
Proportion:
11lb / 7in. = 44lb / x
Using the cross multiplication method, we can solve for x:
11lb * x = 44lb * 7in.
x = 308in. / 11lb
x ≈ 28in.
Therefore, the spring will stretch approximately 28 inches when the weight is increased to 44lb.
A railroad freight car, mass 15,000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples with a 50,000-kg second car, initially at rest and with brakes released. What is the speed of the two cars after coupling?
Answer:
The speed of the two cars after coupling is 0.46 m/s.
Explanation:
It is given that,
Mass of car 1, m₁ = 15,000 kg
Mass of car 2, m₂ = 50,000 kg
Speed of car 1, u₁ = 2 m/s
Initial speed of car 2, u₂ = 0
Let V is the speed of the two cars after coupling. It is the case of inelastic collision. Applying the conservation of momentum as :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]
[tex]V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]
[tex]V=\dfrac{15000\ kg\times 2\ m/s+0}{(15000\ kg+50000\ kg)}[/tex]
V = 0.46 m/s
So, the speed of the two cars after coupling is 0.46 m/s. Hence, this is the required solution.
The speed of the two cars after coupling is 0.857 m/s.
Explanation:To find the speed of the two cars after coupling, we can apply the law of conservation of momentum. The momentum of both cars before the collision is equal to the momentum of both cars after the collision. Since the second car is initially at rest, its initial momentum is zero. The momentum of the first car is given by its mass times its speed. By setting up an equation with the initial momentum of the first car and the final momentum of both cars combined, we can solve for the final speed of the two cars after coupling.
Using the equation:
mass1 * v1i = (mass1 + mass2) * vf
Substituting the given values:
(15,000 kg) * (2.0 m/s) = (15,000 kg + 50,000 kg) * vf
Simplifying the equation gives:
vf = 0.857 m/s
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What's the first thing you should do after retrieving a boat onto a trailer?
Answer:
Set the parking brake of the towing vehicle, and put it in park (or first gear if you have a manual transmission).
Move the vessel onto the trailer far enough to attach the winch line to the bow eye of the vessel. ...
Shut off the engine, and raise the engine or outdrive.
Pull the vessel out of the water.
Explanation:
The first thing you should do after retrieving a boat onto a trailer is to secure the boat to the trailer. This is important to ensure the boat remains stable and safe during transportation.
Here are the steps to secure the boat to the trailer:
1. Position the boat properly: Align the boat on the trailer so that it is centered and evenly distributed. Make sure the boat is positioned in such a way that the weight is balanced and evenly distributed across the trailer.
2. Attach the bow strap: The bow strap is a strong, adjustable strap that is used to secure the front (bow) of the boat to the trailer. Connect one end of the bow strap to the trailer and the other end to a secure point on the boat's bow. Make sure the strap is tight and secure, but not overly tightened to the point of damaging the boat.
3. Connect the stern tie-downs: Stern tie-downs are straps or ropes used to secure the rear (stern) of the boat to the trailer. Attach one end of each stern tie-down to the trailer and the other end to a secure point on the boat's stern. Make sure the stern tie-downs are tight and secure, but again, avoid over-tightening.
4. Check the connections: After attaching the bow strap and stern tie-downs, double-check all the connections to ensure they are properly secured. Give each strap a gentle tug to make sure it is tight and won't come loose during transport.
5. Secure loose items: Before hitting the road, secure any loose items in the boat, such as life jackets, paddles, or fishing gear. These items should be properly stowed and secured to prevent them from shifting or falling out during transportation.
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Objects in free-fall ___. A:do not experience air resistance B: do not experience gravity C: are accelerating faster than 9.8 m/s/s D: are completely weightless
Answer:
A any air resistance
Explanation:
a body in freefall donot experience
Objects in free-fall motion, do not experience air resistance as it only moves under the effect of gravity. Therefore, option (A) is correct.
What are free fall and air resistance?Free fall can be described as a motion of an object in which the force of gravity is the sole force acting upon it. An object moving upwards will not consider being falling. But if the body falls under the influence of gravity is said to be in free fall.
Free fall can be described as a type of motion in which no air resistance is considered and only gravity is considered. All bodies under free fall with the same rate of acceleration, regardless of their masses.
A body that falls through the air, has suffered some degree of air resistance. Air resistance can be described as the collisions of the surface of an object with gas molecules in the air. The factors that affect air resistance are the cross-sectional area and the speed of the body.
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The magnetic field at the earth's surface can vary in response to solar activity. During one intense solar storm, the vertical component of the magnetic field changed by 2.8 μT per minute, causing voltage spikes in large loops of the power grid that knocked out power in parts of Canada.
What emf is induced in a square 190km on a side by this rate of change of field?
Answer:
[tex]EMF = 1684.67 Volts[/tex]
Explanation:
As we know that EMF is induced in a closed conducting loop if the flux linked with the loop is changing with time
So we can say
[tex]EMF = \frac{d\phi}{dt}[/tex]
now we have
[tex]\phi = BA[/tex]
here since magnetic field is constant so we have
[tex]EMF = A\frac{dB}{dt}[/tex]
now we have
[tex]A = (190 \times 10^3)(190 \times 10^3)[/tex]
[tex]A = 3.61 \times 10^{10} m^2[/tex]
now we have
[tex]EMF = 3.61\times 10^{10} (\frac{2.8 \times 10^{-6}T}{60 s})[/tex]
[tex]EMF = 1684.67 Volts[/tex]
Using Faraday's Law and given values, the induced emf in a 190 km square loop subjected to a changing magnetic field. The calculated emf is approximately -1692 volts.
To find the emf induced in a square loop due to a changing magnetic field, we can use Faraday's Law of Electromagnetic Induction.
This law states that the magnitude of the induced emf around a closed loop is proportional to the rate of change of the magnetic flux through the loop.
Rate of change of the magnetic field dB/dt : 2.8 μT/minSide length of the square (L): 190 km = 190,000 mConvert the rate of change of the magnetic field to Tesla per second (T/s):
2.8 μT/min = 2.8 × 10⁻⁶ T/min
Since there are 60 seconds in a minute:
r2.8 × 10⁻⁶ T/60 s ≈ 4.67 × 10⁻⁸ T/s
Calculate the area (A) of the square:
A = L × L = (190,000 m) × (190,000 m) = 3.61 × 10¹⁰ m²
Compute the induced emf (E) using Faraday's Law:
E = -A × dB/dt
E = - (3.61 × 10¹⁰ m²) × (4.67 × 10⁻⁸ T/s) ≈ -1692 V
Therefore, the emf induced in the square loop is approximately -1692 volts.
Two blocks of masses 8 kg and 4.3 kg are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them. A cord holding them together is burned, after which the block of mass 4.3 kg moves to the right with a speed of 6.7 m/s. What is the velocity of other mass in m/s?
Using the principle of conservation of momentum, the velocity of the 8 kg block, after the 4.3 kg block moves to the right with a speed of 6.7 m/s, is calculated to be -3.6 m/s. The negative sign denotes that the block is moving in the opposite direction to the 4.3 kg block.
Explanation:The subject of this question is a part of physics known as mechanics, specifically conservation of momentum. The principle of conservation of momentum in a system where no external forces are acting states that the total momentum before an event must be equal to the total momentum after the event. Here, since the system begins with zero total momentum (both blocks initially at rest), it should end with zero total momentum.
In this scenario, after cord burns, the two blocks are free to move. The block with mass 4.3 kg moves to the right. According to the conservation of momentum, the other block will move in the opposite direction (to the left) in order to conserve the total momentum of the system.
We calculate the momentum of the system after the spring is released: Momentum = mass * velocity. For the 4.3 kg block, the momentum would be = 4.3 kg * 6.7 m/s = 28.81 kg*m/s. As the total momentum before the event was zero, the momentum of 8 kg block should be -28.81 kg*m/s (in opposite direction). The velocity of this block can now be calculated by dividing its momentum by its mass, i.e., -28.81 kg*m/s / 8 kg = -3.6 m/s (the negative sign indicates that the velocity is to the left).
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Since air is a mixture, it does not have a "molar mass." However, for calculation purposes, it is possible to speak of its "effective molar mass." (An effective molar mass is a weighted average of the molar masses of a mixture's components.) If air at STP has a density of 1.285 g/L, its effective molar mass is ________ g/mol.
Answer:
The effective molar mass of air at STP is 28.82 g/mol.
Explanation:
At STP, the value of pressure is 1 atm.
At STP, the temperature is equal to 273.15 K
P = 1atm, T = 273.15 K
Density of the gas at STP ,d= 1.285 g/L
[tex]PV=nRT[/tex] (Ideal gas equation)
[tex]PV=\frac{\text{Mass of air}}{\text{Molar mass of air(M)}}RT[/tex]
[tex]Density=\frac{Mass}{Volume}[/tex]
[tex]P\times M=d\times RT[/tex]
[tex]M=\frac{1.285 g/L}{1 atm}\times 0.0821 atm L/mol K\times 273.15 K[/tex]
M = 28.81 g/mol
The effective molar mass of air at STP is 28.82 g/mol.
Final answer:
The effective molar mass of air at standard temperature and pressure (STP), given its density of 1.285 g/L, is approximately 28.8 g/mol. This value is a weighted average based on the major components of air, mainly nitrogen and oxygen.
Explanation:
If air at STP has a density of 1.285 g/L, its effective molar mass is determined using the density and the ideal gas law. Given in the discussion, the effective molar mass of dry air is approximately 28.8 g/mol. This effective molar mass is a weighted average of the major components of air, primarily nitrogen (N₂) and oxygen (O₂), which account for about 80% and 20% of air's composition, respectively. At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 L. Therefore, the molar mass can be related to the density of the gas at these conditions to find its effective molar mass.
Using the information provided and considering the average composition of air, the calculation confirms that the effective molar mass of air at STP, considering its density of 1.285 g/L, is closely approximated to be 28.8 g/mol. This value is essential for various calculations in chemistry and environmental science, including converting between the mass of air and the number of moles of air, which is useful in stoichiometric calculations and understanding the behavior of gases under different conditions.
A 2.3kg box, starting from rest, is pushed up a ramp by a 10 N force parallel to the ramp. The ramp is 2.0 m long and tilted at 17 degrees. The speed of the box at the top of the ramp is 0.80m/s.How much work does the force do on the system?What is the change in the kinetic energy of the system?What is the change in the gravitational potential energy of the system?What is the change in the thermal energy of the system?
Answer:
Part a)
[tex]W = (10 N)(2 m) = 20 J[/tex]
Part b)
[tex]\Delta K = 0.736 J[/tex]
Part c)
[tex]\Delta U = 13.2 J[/tex]
Part d)
[tex]U_{thermal} = 5.66 J[/tex]
Explanation:
Part a)
Work done by the applied force is given by the formula
[tex]W = F.d[/tex]
here we know that
[tex]F = 10 N[/tex]
[tex]d = 2 m[/tex]
[tex]W = (10 N)(2 m) = 20 J[/tex]
Part b)
As we know that the box was at rest initially and then it is moving with speed 0.80 m/s
so here we can say
[tex]\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2[/tex]
[tex]\Delta K = \frac{1}{2}(2.3)(0.80)^2 - 0[/tex]
[tex]\Delta K = 0.736 J[/tex]
Part c)
Change in gravitational potential energy is given as
[tex]\Delta U = - W_g[/tex]
[tex]\Delta U = -(-mg sin\theta d)[/tex]
[tex]\Delta U = (2.3)(9.81)(sin17)(2)[/tex]
[tex]\Delta U = 13.2 J[/tex]
Part d)
Now by energy conservation law we can say that
Work done by external agent = change in kinetic energy + change in potential energy + thermal energy lost
so we have
[tex]20 = 13.6 + 0.736 + U_{thermal}[/tex]
[tex]U_{thermal} = 5.66 J[/tex]
Answer:
(a). The work done on the system by force is 20 J.
(b). The change in kinetic energy of the system is 0.736 J.
(c). The change in the gravitational potential energy of the system is 13.18 J
(d). The thermal energy of the system is 6.84 J.
Explanation:
Given that,
Mass of box = 2.3 kg
Force = 10 N
Length = 2.0 m
Angle = 17°
Speed = 0.80 m/s
(a). We need to calculate the work done
Using formula of work done
[tex]W=F\times d[/tex]
Put the value into the formula
[tex]W=10\times2.0[/tex]
[tex]W=20\ J[/tex]
The work done on the system by force is 20 J.
(b). We need to calculate the change in kinetic energy of the system
Using formula of change of kinetic energy
[tex]\Delta K.E=K.E_{f}-K.E_{i}[/tex]
[tex]\Delta K.E=\dfrac{1}{2}mv^2-0[/tex]
Put the value into the formula
[tex]\Delta K.E=\dfrac{1}{2}\times2.3\times(0.80)^2[/tex]
[tex]\Delta K.E=0.736\ J[/tex]
The change in kinetic energy of the system is 0.736 J.
(c). We need to calculate the change in the gravitational potential energy of the system
Using formula of gravitational potential energy
[tex]P.E=mgh\sin\theta[/tex]
Where, h = change in height
Put the value into the formula
[tex]P.E=2.3\times9.8\times2.0\sin17[/tex]
[tex]P.E=13.18\ J[/tex]
The change in the gravitational potential energy of the system is 13.18 J.
(d). We need to calculate the thermal energy of the system
Using formula of thermal energy
Work done=Change in kinetic energy+change in potential energy+change in thermal energy
[tex]\Delta U_{th}=W-\Delta K.E+-Delta P.E[/tex]
Put the value into the formula
[tex]\Delta U_{th}=20-0.736-13.18[/tex]
[tex]\Delta U_{th}=6.084\ J[/tex]
The thermal energy of the system is 6.84 J.
Hence, (a). The work done on the system by force is 20 J.
(b). The change in kinetic energy of the system is 0.736 J.
(c). The change in the gravitational potential energy of the system is 13.18 J
(d). The thermal energy of the system is 6.84 J.