Answer:
Circuit breaker
Explanation:
Circuit breaker is the devise designed to protect the circuit from over current by opening the circuit automatically. Breaker can also be off manually by toggle switch. Earlier fuses were used but circuit breakers have replaced them. Fuse and circuit breakers operates differently. in case of overloading fuses blown off and opens the circuit while circuit breaker opens the circuit automatically without being blown off.
Final answer:
A circuit breaker is a safety device in a circuit that opens the circuit when an overcurrent occurs. It is faster than fuses and can be reset.
Explanation:
A circuit breaker is a device designed to open and close a circuit by non-automatic means and to open the circuit automatically on a predetermined overcurrent without damage to itself when properly applied within its rating. It acts as a safety device that switches off an appliance if the current in the circuit is too strong.
Circuit breakers are rated for a maximum current and can be reset, reacting much faster than fuses. They consist of components like bimetallic strips that respond to heat to break the electrical connection in the circuit.
The filament in an incandescent light bulb is made from tungsten. The resistivity of tungsten = 5.6e-8 Ω*m. The radius of the tungsten wire is 0.045 mm. If the bulb is plugged into a 120 V outlet and is to draw a current of 1.24 A, how long must the wire be?
Answer:
11m
Explanation:
Given:
Resistivity ρ = 5.6e-8 Ωm
Radius r = 0.045 mm [tex]=\frac{0.045}{1000}[/tex] = 4.5 x 10⁻⁵ m
Voltage V = 120V
Current I = 1.24A
From Ohm's law, [tex]R = \frac{V}{I}[/tex]
[tex]R = \frac{120}{1.24}[/tex]
R = 96.77 Ω
Resistivity = (Resistance × Area)/ length
ρ = (RA)/L
Therefore, the length of a wire is given by;
L = (RA)/ρ
Calculating the area A of the wire;
A = πr²
A = π × (4.5 x 10⁻⁵)²
A = 6.36 x 10⁻⁹ m²
Substituting area of the wire A = 6.36 x 10⁻⁹ m² into the equation of the length of wire
L = (96.77 × 6.36×10⁻⁹ ) / 5.6×10⁻⁸
L = 10.9977m
L = 11m (approximately)
Which fossil fuel is produced as a by-product that occurs when bacteria decompose organic material under anaerobic conditions?
Bio-gas is the naturally produced fossil fuel, a by-product when bacteria decompose organic material under anaerobic conditions.
Explanation:Organic matter particularly waste material is broken down by bacteria through fermentation in an environmental condition without any presence of oxygen. This process of decomposition leads to formation of bio-gas with "carbon dioxide and methane" in a 2:3 ratio.
The above biological process is termed as bio-digestion or anaerobic digestion. Methane is flammable and thus bio-gas can be used as "energy source", a waste-to-energy transformation. The remaining decomposed matter is ideal as manure for plants due to its rich nutrient level.
A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with an angular speed of 4.0 rad/s. neglecting friction and air resistance, find (a) the rod’s kinetic energy at its lowest position, (b) and how far above that position the center of mass rises.
Final answer:
The rod's kinetic energy at its lowest position is 1.26 J, calculated using the formula for rotational kinetic energy. The center of mass of the rod rises to a height of approximately 30.6 cm above the lowest point, using the principle of conservation of mechanical energy to equate potential and kinetic energies.
Explanation:
A thin rod acting as a pendulum is an example of physical principles related to rotational motion and conservation of energy. To find the rod's kinetic energy at its lowest position, we use the formula for rotational kinetic energy, K = (1/2)Iω2, where 'I' is the moment of inertia and 'ω' is the angular speed. For a rod rotating about one end, the moment of inertia, 'I', is given by (1/3)ml2, with 'm' being the mass of the rod and 'l' its length.
We can find the kinetic energy of the rod at its lowest position, where the angular speed is 4.0 rad/s: K = (1/2)Iω2 = (1/2)(1/3)(0.42 kg)(0.75 m)2(4.0 rad/s)2. After calculation, this gives K = 1.26 J (Joules).
To determine how far above the lowest position the center of mass rises, we have to consider the conservation of mechanical energy. At the highest point, all kinetic energy is converted into potential energy (assuming no energy losses), so we have mgh = K, where 'g' is the acceleration due to gravity, 'h' is the height gained, and 'm' is the mass of the rod. Therefore, h = K / (mg). Plugging in the values, we get h = 1.26 J / (0.42kg * 9.81 m/s2) = 0.306 m, or approximately 30.6 cm above the lowest position.
According to the principle of conservation of momentum, A. the amount of momentum of all the objects in the universe is constant. B. in an open system, the total amount of momentum of the objects is conserved. C. in a closed system, the speed of any two colliding objects will remain constant. D. in a closed system, an object's momentum before a collision will equal its momentum after the collision.
The principle of conservation of momentum states that in a closed system, an object's momentum before a collision will equal its momentum after the collision.
Explanation:The principle of conservation of momentum states that in a closed system, the total momentum of the objects before a collision will be equal to the total momentum of the objects after the collision. This means that in a closed system, an object's momentum before a collision will equal its momentum after the collision.
An unknown mass is hung from a very light spring with a spring constant of 13.2 N/m. Neglect the mass of the spring. If the period of oscillation of the mass-spring system is 2.60 s, then the unknown mass is kg.
Amplitude is the maximum displacement of the mass from the _______.
end point equilibrium point top point vibrational point
A 15.5 kg mass vibrates in simple harmonic motion with a frequency of 9.73 Hz. It has a maximum displacement from equilibrium of +14.8 cm at time, t = 0.00 s. The displacement from equilibrium of the mass at time, t = 1.25 s is cm.
A ball of mass 1.55 kg is hung from a spring which stretches a distance of 0.3800 m. The spring constant of the spring (a scalar quantity) is N/m
A mass m on a spring with a spring constant k is in simple harmonic motion with a period T. If the same mass is hung from a spring with a spring constant of 3k, the period of the motion will be _______.
increased by a factor of the square root of 3 decreased by a factor of the square root of 3 decreased by a factor of 3 increased by a factor of 3
Answer:
Explanation:
Time period of oscillation of spring mass system
T = 2π[tex]\sqrt{\frac{m}{k} }[/tex]
m = [tex]\frac{T^2\times k}{4\pi^2}[/tex]
= [tex]\frac{(2.6)^2\times 13.2}{4\pi^2}[/tex]
= 2.26 kg
Amplitude is the maximum displacement of the mass from the _equilibrium point .
maximum displacement from equilibrium of +14.8 cm at time, t = 0.00 s
x(t) = 14.8cosωt
ω = 2πn = 2π x 9.73 = 61.1 rad /s
x(t) = 14.8cos61.1 t
when t = 1.25
x(t) = 14.8cos61.1 x 1.25
= 14.8cos61.1 x 1.25
= 14.8cos76.38
= 8.22 cm
A ball of mass 1.55 kg is hung from a spring which stretches a distance of 0.3800 m
spring constant
k = mg / x
= 1.55 x 9.8 / .38
= 40 N / m
T = 2π[tex]\sqrt{\frac{m}{k} }[/tex]
so if mass is hung from a spring with a spring constant of 3k
T decreases by a factor of the square root of 3
For the first question, the unknown mass is approximately 0.71 kg. For the second question, the term 'equilibrium point' completes the statement. For the third question, the displacement at time 1.25 seconds is -14.2 cm. For the fourth question, the spring constant k is approximately 40.0 N/m. For the fifth question, the period of the motion will be decreased by a factor of the square root of 3.
Explanation:The unknown mass hung from a spring can be calculated using the formula for the period of a mass-spring system, T = 2π√(m / k), where T is the period, m is the mass, and k is the spring constant. Rearranging for m gives us m = (T²k) / (4π²). Given that T = 2.60 seconds and k = 13.2 N/m, we find m = (2.60^2 * 13.2) / (4*π^2), which gives a mass of approximately 0.71 kg.
The amplitude is the maximum displacement of the mass from the equilibrium point.
The displacement of a harmonically oscillating mass can be found using the equation: x(t) = A*cos(2πft + φ), where A is the amplitude, f is the frequency, and φ is the phase angle. Given the information we have a displacement at 1.25 seconds of -14.2 cm.
The spring constant can be calculated using the equation k = mg / x, where m is the mass, g is gravity (9.81 m/s²), and x is the distance stretched. Given that the mass is 1.55 kg and the distance stretched is 0.3800 m, we find that the spring constant k is approximately 40.0 N/m.
If the same mass is hung from a spring with a spring constant of 3k, the period of the motion will be decreased by a factor of the square root of 3.
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What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then released from rest, and whose period is 0.66 s?
Assume that the displacement at the start of the motion is positive.
What will be its displacement after 2.3 s ?
Answer:
[tex]0.836cm[/tex] in the positive direction.
Explanation:
The equation that describes the motion of this mass-spring system is given by;
[tex]y=Asin\omega t...................(1)[/tex]
Where A is the amplitude, which defined as the maximum displacement from the equilibrium position for a body in simple harmonic motion.
[tex]\omega[/tex] is the angular velocity measured in [tex]rads^{-1[/tex], this is the angle turned through per unit time.
[tex]y[/tex] is the displacement along the axis of the amplitude, and [tex]t[/tex] is any instant of time in the motion.
Given; A = 8.8cm = 0.088m
The angular velocity is given by the following relationship also;
[tex]\omega=2\pi/T[/tex]
Where T is the period, which is defined as the time taken for a body in simple harmonic motion to make one complete oscillation.
Given; T=0.66s
Therefore;
[tex]\omega=2\pi/0.66\\\omega=3.03\pi rads^{-1[/tex]
Substituting all values into equation (1) we obtain the following;
[tex]y=0.088sin3.03\pi t................(2)[/tex]
Equation (2) is the equation describing the motion of the mass on the spring.
At an instant of time t = 2.3s, the displacement is therefore given as follows;
[tex]y=0.088sin[3.03\pi(2.3)]\\y=0.088sin6.97\pi\\[/tex]
By conversion, [tex]6.97\pi rad=6.97*180^o=1254.55^o[/tex]
Therefore
[tex]y=0.088sin1254.54=0.00836m\\[/tex]
[tex]y=0.00836m=0.836cm[/tex]
A weather balloon is designed to expand to a maximum radius of 21 m at its working altitude, where the air pressure is 0.030 atm and the temperature is 200 K. If the balloon is filled at atmospheric pressure and 323 K, what is its radius at liftoff?
Answer:
7.65 m
Explanation:
[tex]P_1[/tex] = Initial pressure = 0.03 atm
[tex]P_2[/tex] = Final pressure = 1 atm
[tex]r_1[/tex] = Inital radius = 21 m
[tex]V_1[/tex] = Intial volume of gas = [tex]\frac{4}{3}\pi r_1^3[/tex]
[tex]V_2[/tex] = Final volume of gas = [tex]\frac{4}{3}\pi r_2^3[/tex]
[tex]T_1[/tex] = Initial temperature = 200 K
[tex]T_2[/tex] = Final temperature = 323 K
From ideal gas law we have
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\Rightarrow \frac{P_1\frac{4}{3}\pi r_1^3}{T_1}=\frac{P_2\frac{4}{3}\pi r_2^3}{T_2}\\\Rightarrow \frac{P_1 r_1^3}{T_1}=\frac{P_2r_2^3}{T_2}\\\Rightarrow r_2=\frac{P_1r_1^3T_2}{T_1P_2}\\\Rightarrow r_2=\left(\frac{0.03\times 21^3\times 323}{200\times 1}\right)^{\frac{1}{3}}\\\Rightarrow r_2=7.65\ m[/tex]
The radius at liftoff is 7.65 m
The initial radius of the weather balloon at liftoff can be calculated using the ideal gas law by comparing the conditions at the ground and at the balloon's working altitude.
Explanation:The subject of this question deals with the physics principle of the behavior of gases, namely the ideal gas law. The ideal gas law links the pressure, temperature and volume of a gas, represented by the equation PV=nRT. Here, 'P' denotes pressure, 'V' is volume, 'n' is the number of moles, 'R' is the universal gas constant, and 'T' is temperature.
In the case of the weather balloon, when it is filled at atmospheric pressure and 323 K on the ground, and then expands to a maximum radius of 21m at a working altitude where the air pressure is 0.030 atm and the temperature is 200K, the principle of the ideal gas law can be applied. Assuming the same quantity of gas (n) we have P1V1/T1=P2V2/T2 where indices 1 and 2 refer to conditions at lift off and at altitude respectively.
We know that the volume of the balloon (V) is related to its radius (r) by the equation for volume of a sphere V=(4/3)*pi*r^3. So we can substitute the volumes in above equation with their expressions in terms of radius and solve for r1 (radius at liftoff).
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When output is below potential and the policy rate has hit the floor of zero, the resulting fall in inflation leads to ________ real interest rates, which ________ output further, which causes inflation to fall further.
Answer:
high real rate of interest, depressing
Explanation:
If output is less than potential and the rate of policy has fall to the ground of zero, the subsequent decline in inflation results in high real rate of interest which further depressing output, causing inflation to even further decline.
If the actual output is less than potential output, demand or supply has fallen, resulting in job and output declines
Which type of intermolecular attractive force operates between all molecules?
Answer:
Hydrogen bonding
Ionic bonding
Non- induced dipole forces
Ion dipole forces
Van-der wall forces
Explanation:
Intermolecular forces are the forces which operates at molecular level they may be attractive or repulsive in nature. But the given question ask only about attractive molecular forces. These are of basically five types:-
1. Hydrogen bonding- It is the force of attraction between hydrogen atom and an electro-negative atom.
2 .Ionic bonding- It is the force of attraction between cationic and anionic pair.
3.Non - Induced dipole forces- It is the dipole- dipole interaction between permanent electromagnetic materials.
4.Ion-dipole forces- These bonds are stronger than hydrogen bond and also stronger than dipole-dipole bond.
Van-der walls forces- It is the force of attraction between uncharged molecules.
Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg · m2. The father exerts a force on the merry-go-round perpendicular to its 1.50 m radius to achieve a torque of 375 N · m.(a) Calculate the rotational kinetic energy (in J) in the merry-go-round plus child when they have an angular velocity of 29.2 rpm. J(b) Using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest. revolutions(c) Again, using energy considerations, calculate the force (in N) the father must exert to stop the merry-go-round in seven revolutions. N
In this part, we discuss the expected decrease in angular acceleration when the child is on the merry-go-round due to the greater moment of inertia. We calculate the child's moment of inertia by approximating the child as a point mass at a specific distance from the axis.
Explanation:b. We expect the angular acceleration for the system to be less in this part because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia I, we first find the child's moment of inertia Ic by approximating the child as a point mass at a distance of 1.25 m from the axis. Then
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Final answer:
The rotational kinetic energy of the merry-go-round plus child at 29.2 rpm is 394.5834 J. The father must push for approximately 0.167 revolutions to reach this velocity from rest. A force of -8.975 N (in the opposite direction of motion) must be exerted to stop the merry-go-round in seven revolutions.
Explanation:
The problem addresses the concepts of angular velocity, rotational kinetic energy, and the application of work-energy principles within a physics context, specifically relating to a merry-go-round system.
(a) To calculate the rotational kinetic energy (KErot) of the merry-go-round plus child, we use the formula KErot = (1/2)Iω^2 where I is the moment of inertia and ω is the angular velocity in radians per second. First, convert the angular velocity from rpm to rad/s using the conversion factor 1 rpm = (2π/60) rad/s.
KErot = (1/2)(84.4 kg · m^2)(29.2 rpm × 2π/60 rad/s)^2
= (1/2)(84.4)(3.058 rad/s)^2
= (1/2)(84.4)(9.351)
= (1/2)(789.1668) J
= 394.5834 J
(b) The number of revolutions needed to achieve this angular velocity from rest can be found using the work-energy principle, where work done (W) by the father's force equals the rotational kinetic energy of the merry-go-round system.
W = τr(2πN)
394.5834 J = 375 N · m × (2πN)
N = 394.5834 J / (375 N · m × 2π)
N = 0.167 revolutions
(c) The force to stop the merry-go-round in seven revolutions can be calculated using the concept that the work done by the father's force (W) to stop the merry-go-round equals the negative change in the rotational kinetic energy. Thus, W = -KErot, and the rotational kinetic energy is initially computed in part (a). This gives:
W = -394.5834 J
Force (F) = W / (2π×7)
F = -394.5834 J / (2π × 7)
F = -8.975 N (negative sign indicates that the force is applied in the opposite direction of motion)
At a certain harbor, the tides cause the ocean surface to rise and fall a distance d (from highest level to lowest level) in simple harmonic motion, with a period of 12.5 h. How long does it take for the water to fall a distance 0.250d from its highest level?Number__________Unit____________.
Answer:
2.08335 hours
Explanation:
T = Time period = 12.5 h
Angular frequency is given by
[tex]\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{12.5}\\\Rightarrow \omega=0.50265\ rad/h[/tex]
The distance moved from highest to lowest level is given by
[tex]d=2x_{m}\\\Rightarrow x_m=\frac{d}{2}\\\Rightarrow x_m=0.5d[/tex]
At the ocean surface [tex]x_m=0.25d[/tex]
[tex]x_0=x_mcos(\omega t+\phi)[/tex]
[tex]\phi[/tex] = Phase constant = 0 as clock is started at [tex]x_0=x_m[/tex]
[tex]0.25d=0.5dcos(0.50265t)\\\Rightarrow cos(0.50265t)=\frac{0.25}{0.5}\\\Rightarrow 0.50265t=cos^{-1}0.5\\\Rightarrow t=\frac{1.047}{0.50265}\\\Rightarrow t=2.08335\ h[/tex]
The time taken for the water to fall the distance is 2.08335 hours
2. Which of the following is an example of a passive immunization? A. Antibodies against rabies given to someone who was bitten by a potentially rabid dog B. Heat-killed flu antigen grown in chicken eggs C. Live viral antigen given on a sugar cube to protect against polio D. Both A and B. E. Both B and C.
Answer:
A. Antibodies against rabies given to someone who was bitten by a potentially rabid dog
Explanation:
passive immunization is the immunity or protection against certain infectious diseases by injecting antigens or antibodies to the blood stream or certain antibodies can be gotten through colostrum or breast milk.
for option A. Antibodies against rabies given to someone who was bitten by a potentially rabid dog. the immunity against rabies infection is administer to the person
For option B:is only telling us the type of infection but not stating the immunity administered
option c can be fairly correct, because its another method of immunization but its not passive immunization
The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for there to be only one photon per second per square meter? Assume that the photons spread out spherically. The area of a sphere is 4????????2.
Answer:
The distance from the radio station is 0.28 light years away.
Solution:
As per the question:
Distance, d = 4 ly
Frequency of the radio station, f = 854 kHz = [tex]854\times 10^{3}\ Hz[/tex]
Power, P = 50 kW = [tex]50\times 10^{3}\ W[/tex]
[tex]I_{p} = 1\ photon/s/m^{2}[/tex]
Now,
From the relation:
P = nhf
where
n = no. of photons/second
h = Planck's constant
f = frequency
Now,
[tex]n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s[/tex]
Area of the sphere, A = [tex]4\pi r^{2}[/tex]
Now,
Suppose the distance from the radio station be 'r' from where the intensity of the photon is [tex]1\ photon/s/m^{2}[/tex]
[tex]I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}[/tex]
[tex]1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}[/tex]
[tex]r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m[/tex]
Now,
We know that:
1 ly = [tex]9.4607\times 10^{15}\ m[/tex]
Thus
[tex]r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly[/tex]
Two parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plates is 2,000 N/C. If the voltage is doubled and the distance between the plates is reduced to 1/5 the original distance, the magnitude of the new electric field is
A) 800 N/C
B) 1600 N/C
C) 2400 N/C
D) 5000 N/C
E) 20000 N/C
Answer:
E) E₂= 20000 N/C
Explanation:
The electric field between two parallel conductive plates is thus calculated
E= V/d
Where:
E: Electric field (N/C)
V: voltage (V)
d : distance between the plates (m)
Problem development
E₁= V₁/d₁ = 2000 N/C
E₂= V₂/d₂
[tex]E_{2} = \frac{2V_{1} }{\frac{d_{1} }{5} }[/tex]
[tex]E_{2} = 10(\frac{V_{1} }{d_{1} })[/tex]
E₂= 10* (2000)
E₂= 20000 N/C
Final answer:
When the voltage between two parallel conducting plates is doubled and the distance between them is reduced to 1/5 of the original distance, the magnitude of the new electric field is 20,000 N/C.
Explanation:
The question relates to the electric field between two parallel conducting plates connected to a voltage source. The original electric field's magnitude is given as 2,000 N/C, and we're asked to find the new electric field magnitude when the voltage is doubled and the distance between the plates is reduced to 1/5 of the original distance. The electric field (E) between two parallel plates is given by the equation E = V/d, where V is the potential difference (voltage) between the plates, and d is the distance between them. When the voltage is doubled, V becomes 2V, and when the distance is reduced to 1/5, d becomes d/5. Thus, the new electric field E' = (2V)/(d/5) = 10(V/d) = 10E. So, the new electric field strength is 10 * 2,000 N/C = 20,000 N/C, which corresponds to option E).
Technician A says that magnetic lines of force can be seen by placing iron filings on a piece of paper and then holding them over a magnet. Technician B says that the effects of magnetic lines of force can be seen using a compass. Which technician is correct?
Answer:
Both A and B are correct
Explanation:
Magnetic line of force can be seen by placing iron fillings on a piece of paper and then hold a magnet over them, the iron fillings align themselves in the magnetic field of the magnetic along the line of forces thus showing the loops of the magnetic line of forces.Also, we know that unlike poles attract where as like poles repel each other. This fact can be proved when we hang a bar magnet it aligns itself in the North-South direction because of the the magnetic field due to the alignment of the geographical North-South poles and the poles of the bar magnet and the compass works the same way.Thus both the technicians are correct.
Answer:
Technician A is correct
Magnetic lines of force can be seen by placing iron filings on a piece of paper and then holding them over a magnet.
Explanation:
As we know that the magnetic field lines are always originated out from the North pole of magnet and then terminate at south pole of the magnet
Here we know that when magnetic material aligns itself in the direction of external magnetic field
So here when we put the iron filings on the cardboard and then place a magnet near it then all the iron filing will align itself in the direction of the magnetic field
Now if we place a compass near the magnet then it will aligns itself in the tangential direction of the magnetic field at the same position
So here we can say that Technician A is correct
A molecule with a seesaw molecular geometry has a bond angle of
Answer
A molecule with seesaw molecular geometry has a bond angle of 90° and 120°.
Seesaw is a type of molecular geometry where there are four bonds attached to a central atom.
Bond is observed in the shape of playground seesaw that is why it is known as a seesaw bond.
Four bonds attached to the central bond formed results as tetrahedral or square planar geometry.
In see-saw, shape maximizes the bond angle of the lone pair and the other atoms in the molecule.
lone pair in the equatorial position offer 120 and 90-degree bond angles whereas bond angle will be 90-degree bond angles if placed at axial position.
A molecule with seesaw molecular geometry has bond angles that are less than 120 degrees for equatorial positions and less than 90 degrees for axial positions, due to lone pair repulsion. Hence the correct option is 1.
Bond Angles in Seesaw Molecular Geometry
A molecule with a seesaw molecular geometry has four nuclei and one lone pair of electrons. This molecular structure is based on a trigonal bipyramidal geometry but with one equatorial position occupied by a lone pair. As a result, the bond angles are adjusted due to electron pair repulsion.
The bond angles for a seesaw geometry are:
Less than 120 degrees for equatorial bonds, as the lone pair causes slight compression of these angles.Less than 90 degrees for axial bonds due to the same repulsive effects from the lone pair.Approximately 173 degrees between the two axial positions, slightly less than the ideal 180 degrees due to lone pair repulsion.The correct question is:
A molecule with a seesaw molecular geometry has a bond angle of
1. <120 for equatorial bonds and <90 for axial bonds.
2.180
3. <90
4. 120 for equatorial bonds and 90 for axial bonds.
5.120
Listed following are several objects in the solar system. Rank these objects from left to right based on the distance from the Sun at which they are presumed to have formed, from nearest to farthest.
a typical asteroid in the asteroid belt, a typical oort cloud object, a typical kuiper belt object
Although Oort cloud comets are now located far beyond the Kuiper belt, they are thought to have formed in the region of the jovian planets. They were then "kicked out" to their current orbits by gravitational encounters with the jovian planets.
The objects in the solar system can be ranked from nearest to farthest based on their formation distance from the Sun: Mercury, Venus, Earth, Mars, and Jupiter.
Explanation:The objects in the solar system can be ranked based on their distance from the Sun as follows:
Mercury: This planet is the closest to the Sun and is presumed to have formed at its current location.Venus: Venus is the second planet from the Sun and also formed relatively close to it.Earth: Our home planet is the third planet from the Sun and formed within the habitable zone.Mars: Mars is the fourth planet from the Sun and formed farther away than Earth.Jupiter: As the largest planet in the solar system, Jupiter formed even farther away from the Sun than Mars.Therefore, the ranking from nearest to farthest would be Mercury, Venus, Earth, Mars, and Jupiter.
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A type of cuckoo clock keeps time by having a mass bouncing on a spring, usually something cute like a cherub in a chair. What force constant (in N/m) is needed to produce a period of 0.660 s for a 0.0100 kg mass?
Answer:0.906 N/m
Explanation:
Given
time period [tex]T=0.66 s[/tex]
mass [tex]m=0.01 kg[/tex]
System can be considered as spring mass system
Time Period of spring mass system is given by
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
squaring
[tex]k=\frac{4\pi ^2m}{T^2}[/tex]
[tex]k=\frac{4\times \pi ^2\times 0.01}{0.66^2}[/tex]
[tex]k=0.906 N/m[/tex]
An aircraft is loaded 110 pounds over maximum certificated gross weight. If fuel (gasoline) is drained to bring the aircraft weight within limits, how much fuel should be drained?
Answer:
242 lts de fuel
Explanation:
We know density of fuel (is variable) but we can say that is close to 1 Kg/lts.
If we have 110 pounds over maximun certificated gross weight, we need to get rid of that 110 pounds or 110 * 2.2 = 242 Kgs
Now by rule of three
If 1 lt weight 1 kg.
x 242 kgs
x = 242 lts de fuel
To bring the aircraft within the maximum gross weight limits, approximately 18.33 gallons of fuel need to be drained, calculated by dividing the excess weight (110 pounds) by the weight of fuel per gallon (6 pounds/gallon).
To calculate how much fuel needs to be drained to bring an aircraft weight within the maximum certificated gross weight limits, we need to know two pieces of information: the weight by which the aircraft is over its maximum gross weight and the weight of the fuel that needs to be drained.
In this scenario, the aircraft is 110 pounds over the maximum certificated gross weight. In general, aviation gasoline (avgas) weighs about 6 pounds per gallon. To solve this problem, we will divide the excess weight by the weight of the fuel per gallon to find out how many gallons need to be drained.
So, the calculation is as follows:
Excess weight: 110 pounds.
Weight of fuel per gallon: 6 pounds/gallon.
Number of gallons to be drained = Excess weight / Weight of fuel per gallon.
Number of gallons to be drained = 110 pounds / 6 pounds/gallon.
Number of gallons to be drained = 18.33 gallons (approximately).
Therefore, to bring the aircraft back within weight limits, approximately 18.33 gallons of fuel need to be drained.
) The sprocket assembly on a bicycle consists of a chain and two sprockets, one on the pedal and the other on the rear wheel. If the sprocket on the pedal is 6 inches in diameter, the sprocket on the rear wheel is 4 inches in diameter, and the rear wheel is 26 inches in diameter, how fast is the bicycle traveling in miles per hour when the cyclist is pedaling at the rate of 1.1 revolutions per second? Round your answer to the nearest tenth mph.
Answer:
7.8 mph
Explanation:
Rate of cycling = 1.1 rev/s
Rear wheel diameter = 26 inches
Diameter of sprocket on pedal = 6 inches
Diameter of sprocket on rear wheel = 4 inches
Circumference of rear wheel = [tex]\pi d=26\pi[/tex]
Speed would be
[tex]\text{Rate of cycling}\times \frac{\text{Diameter of sprocket on pedal}}{\text{Diameter of sprocket on rear wheel}}\times{\text{Circumference of rear wheel}}\\ =1.1\times \frac{6}{4}\times 26\pi\\ =134.77432\ inches/s[/tex]
Converting to mph
[tex]1\ inch/s=\frac{1}{63360}\times 3600\ mph[/tex]
[tex]134.77432\ inches/s=134.77432\times \frac{1}{63360}\times 3600\ mph=7.65763\ mph[/tex]
The Speed of the bicycle is 7.8 mph
To determine the bicycle speed, calculate the rear wheel's circumference, use the gear ratio to find the rear wheel's rotation rate, and then convert the distance traveled per second into miles per hour. The resulting speed is 17.3 mph when pedaling at 1.1 revolutions per second.
Explanation:The student's question involves calculating the speed of a bicycle based on the diameters of the sprockets on the pedal and rear wheel, the diameter of the rear wheel, and the pedaling rate in revolutions per second. To find the speed in miles per hour, we first calculate the circumference of the rear wheel, which gives us the distance traveled per revolution. We then use the gear ratio to find the effective revolutions of the rear wheel based on the pedaling speed. The total distance traveled per second is then converted to miles per hour.
The diameter of the rear wheel is 26 inches, so its circumference is π(26 inches) ≈ 81.68 inches. The gear ratio is the diameter of the pedal sprocket (6 inches) over the diameter of the rear wheel sprocket (4 inches), giving us a 1.5:1 gear ratio. For every pedal revolution, the rear wheel rotates 1.5 times. Pedaling at 1.1 revolutions per second, the rear wheel rotates at 1.5 * 1.1 = 1.65 revolutions per second. The distance traveled per second is then 81.68 inches * 1.65 ≈ 134.77 inches per second.
To convert inches per second to miles per hour, we use the conversion factors 12 inches per foot and 5280 feet per mile, and 3600 seconds per hour. Thus, the bike speed is (134.77 inches/second) * (1 foot/12 inches) * (1 mile/5280 feet) * (3600 seconds/hour) ≈ 17.26 miles per hour. Rounded to the nearest tenth, the bicycle is traveling at 17.3 mph.
An engine does 19900 J of work and rejects 5300 J of heat into a cold reservoir whose temperature is 285 K. What would be the smallest possible temperature of the hot reservoir?
Answer:
[tex]T_H = 1335\ J[/tex]
Explanation:
Given,
Work done by engine = 19900 J
rejected energy of heat = 5300 J
temperature = 285 K
efficiency of the engine
[tex]\eta = \dfrac{W}{Q_H}[/tex]
where
[tex]W = Q_H - Q_C[/tex]
[tex]19900 = Q_H - 5300[/tex]
[tex]Q_H = 25200\ J[/tex]
efficiency in terms of temperature
[tex]\eta = \dfrac{T_H-T_C}{T_H}[/tex]
from the above two equation
[tex]\dfrac{Q_H-Q_C}{Q_H}= \dfrac{T_H-T_C}{T_H}[/tex]
[tex]1-\dfrac{Q_C}{Q_H}= 1- \dfrac{T_C}{T_H}[/tex]
[tex]\dfrac{Q_C}{Q_H}= \dfrac{T_C}{T_H}[/tex]
[tex]T_H=T_C\ \dfrac{Q_H}{Q_C}[/tex]
[tex]T_H=285\times \dfrac{25200}{5300}[/tex]
[tex]T_H = 1335\ J[/tex]
The smallest possible temperature of the hot reservoir, given the work done by the engine, heat rejected into the cold reservoir, and temperature of the cold reservoir, is approximately 1357.14 Kelvin.
Explanation:The problem is based on the Carnot heat engine, a theoretical model used for understanding the thermodynamics of heat conversion into work. The efficiency of Carnot's engine depends on the temperatures of the hot and cold reservoirs (Th and Tc respectively). The efficiency, represented as e, is calculated as e = (Th-Tc) / Th.
Before we do the calculation, it's important to note that temperatures must be expressed in Kelvin for this formula to be accurate. Tc, the temperature of the cold reservoir, is given as 285 K. The work done by the engine, represented as W, is 19900 J, and the heat rejected by the engine into the cold reservoir, represented as Qc, is 5300J. The heat absorbed by the engine from the hot reservoir, represented as Qh, can be calculated using the first law of thermodynamics, which states Qh = W + Qc. Therefore, Qh = 19900 J + 5300 J = 25200 J.
Now, to find Th, we rearrange the efficiency formula to Th = Tc / (1 - e), where e is now W / Qh = 19900 J / 25200 J ≈ 0.79. Substituting Tc = 285 K and e ≈ 0.79 into the formula, Th = 285 K / (1 - 0.79) ≈ 1357.14 K. This is the smallest possible temperature of the hot reservoir for the given conditions.
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A particle of mass 4.0 kg is constrained to move along the x-axis under a single force F(x) = −cx3 , where c = 8.0 N/m3 . The particle’s speed at A, where xA = 1.0 m, is 6.0 m/s. What is its speed at B, where xB = −2.0 m?
Answer:4.58 m/s
Explanation:
Given
mass of Particle [tex]m=4 kg[/tex]
[tex]F=-cx^3[/tex]
[tex]a=\frac{F}{m}[/tex]
[tex]a=-\frac{cx^3}{m}[/tex]
[tex]a=-\frac{8x^3}{4}[/tex]
[tex]a=-2x^3[/tex]
[tex]v\frac{\mathrm{d} v}{\mathrm{d} x}=-2x^3[/tex]
[tex]vdv=-2x^3dx[/tex]
integrating
[tex]\int_{6}^{v_b}vdv=\int_{1}^{-2}-2x^3dx[/tex]
[tex]\frac{v_b^2-6^2}{2}=-\frac{1}{2}\left [ \left ( -2\right )^4-\left ( 1\right )^4\right ][/tex]
[tex]\frac{v_b^2-36}{2}=-0.5\times 15[/tex]
[tex]v_b^2=36-15[/tex]
[tex]v_b=\sqrt{21}[/tex]
[tex]v_b=4.58 m/s[/tex]
A particle of mass 4.0 kg is constrained to move along the x-axis under a single force F(x) = −cx³ The speed of the particle at point B is 5.3 m/s.
The potential energy at any point x is given by:
F(x) = -cx³
U(x) = -∫F(x) dx
U(x)= ∫cx³dx
U(x) = c/4 × x⁴ + C
The total mechanical energy E remains constant:
E = K + U
E_A = K(A) + U(A)
E_A = (1/2)mv(A)²+ c÷4 × x(A)⁴ + C
At point B, the potential energy U(B) is:
U(B) = c/4 × x(B)⁴ + C
The total mechanical energy at point B is:
E(B) = K(B) + U(B)
Since the total mechanical energy is conserved, we have:
E(A) = E(B)
(1/2)mv(A)² + c÷4 × x(A)⁴ + C = (1/2)mv(B)² + c÷4 × x(B)⁴ + C
(1/2)(4.0)(6.0)² + ((8.0) ÷ 4) × (1.0)⁴ = (1/2)(4.0)v(B)² + ((8.0) ÷ 4) × (-2.0)⁴
72.0 + 2.0= 2.0 × v(B)² + 16.0
56.0 J = 2.0× v(B)²
v(B)² = 28.0
v(B) = ±5.3 m/s
Since speed cannot be negative, the speed of the particle at point B is 5.3 m/s.
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A small spinning asteroid is in a circular orbit around a star, much like the earth's motion around our sun. The asteroid has a surface area of 7.70 m2. The total power it absorbs from the star is 3800 W.
Assuming the surface is an ideal absorber and radiator, calculate the equilibrium temperature of the asteroid (in K)
Answer:
Temperature will be 305 K
Explanation:
We have given The asteroid has a surface area [tex]A=7.70m^2[/tex]
Power absorbed P = 3800 watt
Boltzmann constant [tex]\sigma =5.67\times 10^{-8}Wm/K^4[/tex]
According to Boltzmann rule power radiated is given by
[tex]P=\sigma AT^4[/tex]
[tex]3800=5.67\times 10^{-8}\times 7.70\times T^4[/tex]
[tex]T^4=87.0381\times 10^8[/tex]
[tex]T=305K[/tex]
So temperature will be 305 K
In the experiment, you will determine the wavelength of a laser by shining the laser beam on a diffraction grating, a set of regularly spaced lines. If the screen is 10 m away, the spots are 5.4 cm apart, and the lines of the grating are separated by 0.1 mm,
what is (in nm)?
Answer:
The wavelength is 539 nm.
Explanation:
Given that,
Distance of screen L= 10 m
Width s= 5.4 cm
Grating separation d= 0.1 mm
Suppose the pattern is displayed on a screen a distance L from the grating and the spots are separated by s. we need to find the wavelength.
The angle between the central and first maximum is given as
[tex]\tan\theta=\dfrac{s}{L}[/tex]
[tex]\theta=\tan^{-1}\dfrac{s}{L}[/tex]
[tex]\theta=\tan^{-1}\dfrac{5.4\times10^{-2}}{10}[/tex]
[tex]\theta=0.309[/tex]
We need to calculate the wavelength
The condition for maximum of a diffraction grating is
[tex]d\sin\theta=m\lambda[/tex]
[tex]\lambda=\dfrac{d\sin\theta}{m}[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{0.1\times10^{-3}\times\sin0.309}{1}[/tex]
[tex]\lambda=539\times10^{-9}\ m[/tex]
[tex]\lambda=539\ nm[/tex]
Hence, The wavelength is 539 nm.
The wavelength of a laser can be determined using the diffraction grating equation. Calculate the θ value with θ = tan⁻¹(y/L), then substitute the known values into the equation λ = (d sin θ) / m. Convert the resultant wavelength to nanometers.
Explanation:Your question relates to the physics of light and diffraction grating. We can find the wavelength of the laser by using the diffraction grating equation. The equation is: d sin θ = mλ
Where:
d is the grating spacing (distance between the lines represented in meters)θ is the angle to the m'th order maximumm is the order of maximum (in your case, it's likely 1)λ is the wavelength we are trying to findIn your case, we need to calculate θ first: θ = tan⁻¹(y/L) where y is the distance between the spots (5.4cm) and L is the distance from the grating to the screen (10m).
Substitute the known values and solve for λ: λ = (d sin θ) / m
Remember, the resultant wavelength should be converted to nanometers (1m = 1 x 10⁹ nm).
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Calculate the angular velocity of the earth in its orbit around the sun.
Answer:
[tex]0.0172rad/day=1.99x10^{-7}rad/second[/tex]
Explanation:
The definition of angular velocity is as follows:
[tex]\omega=2\pi f[/tex]
where [tex]\omega[/tex] is the angular velocity, and [tex]f[/tex] is the frequency.
Frequency can also be represented as:
[tex]f=\frac{1}{T}[/tex]
where [tex]T[/tex] is the period, (the time it takes to conclude a cycle)
with this, the angular velocity is:
[tex]\omega=\frac{2\pi}{T}[/tex]
The period T of rotation around the sun 365 days, thus, the angular velocity:
[tex]\omega=\frac{2\pi}{365days}=0.0172rad/day[/tex]
if we want the angular velocity in rad/second, we need to convert the 365 days to seconds:
Firt conveting to hous
[tex]365days(\frac{24hours}{1day} )=8760hours[/tex]
then to minutes
[tex]8760hours(\frac{60minutes}{1hour} )=525,600minutes[/tex]
and finally to seconds
[tex]525,600minutes(\frac{60seconds}{1minute})=31,536x10^3seconds[/tex]
thus, angular velocity in rad/second is:
[tex]\omega=\frac{2\pi}{31,536x10^3seconds}=1.99x10^{-7}rad/second[/tex]
A very loud train whistle has an acoustic power output of 100 W. If the sound energy spreads out spherically, what is the intensity level in dB at a distance of 100 meters from the train?
The intensity level in dB at a distance of 100 meters from the train is 98.46 dB.
Explanation:To calculate the intensity level in dB at a distance of 100 meters from the train, we need to first calculate the intensity of sound at that distance.
Since the sound energy spreads out spherically, the intensity decreases with the square of the distance from the source.
Using the formula for intensity, I = P/4πr², where P is the acoustic power output and r is the distance from the source, we can calculate the intensity at 100 meters:
I = 100W / (4π * (100m)²) = 0.0796 W/m².
Now, we can use the formula for the intensity level in dB, β = 10log(I/I₀), where I₀ is the reference intensity (10⁻¹² W/m²).
Substituting the values, β = 10log(0.0796 / 10⁻¹²) = 98.46 dB.
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The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope. The sleds move on an icy surface. The dogs exert a 240-N force on the rope attached to the front sled. The front rope pulls horizontally. Find the acceleration of the sleds.
Answer:
Acceleration, [tex]a=1.2\ m/s^2[/tex]
Explanation:
Given that,
The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope, m = 100 kg
Force exerted by the doges on the rope attached to the front sled, F = 240 N
To find,
The acceleration of the sleds.
Solution,
Let a is the acceleration of the sleds. The product of mass and acceleration is called force. Its expression is given by :
F = ma
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{240\ N}{2\times 100\ kg}[/tex] (m = 2m)
[tex]a=1.2\ m/s^2[/tex]
So, the acceleration of the sleds is [tex]1.2\ m/s^2[/tex].
A gas, behaving ideally, fills a fixed volume container at a pressure P₁ and at a temperature T₁. The temperature of the container is changed to T₂. Using Avogadro’s, Charles’, or Boyle’s law, develop an expression that would solve for the new pressure P₂.a. V₁/T₁=V₂/T₂b. P₁*V₁=P₂*V₂c. P₁/T₁=P₂/T₂d. P₁*T₁=P₂*T₂
Answer:
c. P₁/T₁=P₂/T₂
Explanation:
neither Avogadro’s, Charles’, or Boyle’s law formula can be used, since some parameters like volume is not given,
to find P₂, given P₁, T₁, and T₂ we will therefore use Gay-lussac's law.
gay lussacs law state that, provided volume is kept constant, pressure is directly proportional to temperature.
the volume volume is said to be filled, i.e its is kept constants when temperature is change
Suppose you are asked to find the amount of time t, in seconds, it takes for the turntable to reach its final rotational speed. Which of the following equations could you use to directly solve for the numerical value of t?
A) θ=θ0+ω0t+(1/2)αt2
B) ω=ω0+αt
C) ω2=ω02+2α(θ−θ0)
D) More information is needed before t can be found.
Answer:
option (D)
Explanation:
Here initial rotation speed is given, final rotation speed is given and asking for time.
If we use
A) θ=θ0+ω0t+(1/2)αt2
For this equation, we don't have any information about the value of angular displacement and angular acceleration, so it is not useful.
B) ω=ω0+αt
For this equation, we don't have any information about angular acceleration, so it is not useful.
C) ω2=ω02+2α(θ−θ0)
In this equation, time is not included, so it is not useful.
D) So, more information is needed.
Thus, option (D) is true.
The equation to directly solve for time t in this context is ω = ω₀ + αt, as it includes all necessary variables except for t, making it easy to solve algebraically. Hence, option(b) is correct.
The correct equation to use in this problem is ω = ω₀ + αt.
By rearranging this equation, you can directly find the value of t using given angular velocities and acceleration. This is applicable in analyzing rotational motion in physics.
To solve for the time t, we can rearrange the equation algebraically:
[tex]\\t = \frac{\omega - \omega_0}{\alpha}\\[/tex]
This equation directly relates
the final angular velocity (ω), initial angular velocity (ω₀), and angular acceleration (α) to time t.Given the initial and final angular velocities and the angular acceleration, you can plug these values in to find t.
The electric motor of a model train accelerates the train from rest to 0.740 m/s in 30.0 ms. The total mass of the train is 560 g. Find the average power delivered to the train during its acceleration.
Answer:
= 5.1 W
Explanation:
time (t) = 30 ms = 0.03 s
mass (m) = 560 g = 0.56 kg
initial velocity (U) = 0 m/s
final velocity (V) = 0.74 m/s
power = \frac{work done}{t} = \frac{f x d}{t} = f x v = m x a x v
m x a x v = m x \frac{V-U}{t} x \frac{V + U}{2}
m x \frac{V-U}{t} x \frac{V + U}{2} = 0.56 x \frac{0.74 - 0}{0.03} x \frac{0.74+0}{2}
= 5.1 W