The magnitude of the torque on the current loop is 0.0055 Nm
Explanation:
Given data,
We have the formula,
T= u x B
Where u= i x A
T= i×A×B ×sin(30)
T=0.46×0.52² ×0.900×0.5
T=0.0055 Nm
The magnitude of the torque on the current loop is 0.0055 Nm
An electron moves with a speed of 8.0 × 10^{6} m/s along the +x-axis. It enters a region where there is a magnetic field of 2.5 T, directed at an angle of 60° to the +x-axis and lying in the xy-plane. What is the magnitude of the magnetic force of the electron?
Answer:
The magnitude of the magnetic force of the electron is 2.77 x 10⁻¹² N
Explanation:
Given;
speed of the electron, v = 8.0 × 10⁶ m/s
magnetic field strength, B = 2.5 T
angle of inclination of the field, θ = 60°
The magnetic force experienced by the electron in the magnetic field is given as;
F = qvBsinθ
where;
q is charge of electron = 1.6 x 10⁻¹⁹ C
B is strength of magnetic field
v is speed of the electron
Substitute the given values and solve for F
F = (1.6 x 10⁻¹⁹)( 8.0 × 10⁶)(2.5)sin60
F = 2.77 x 10⁻¹² N
Thus, the magnitude of the magnetic force of the electron is 2.77 x 10⁻¹² N
The magnitude of the magnetic force on the electron is approximately 2.06 x 10^-13 N.
Explanation:In this scenario, the magnitude of the magnetic force on the electron can be calculated using the equation F = qvBsinθ, where q is the charge of the electron (+1.6 x 10-19 C), v is its velocity (8.0 x 106 m/s), B is the magnetic field (2.5 T), and θ is the angle between the velocity and the magnetic field (60°). Plugging in the values, we get:
F = (1.6 x 10-19 C)(8.0 x 106 m/s)(2.5 T)sin(60°)
Using the trigonometric identity sin(60°) = sqrt(3)/2, we can simplify the equation and calculate the magnitude of the magnetic force to be approximately 2.06 x 10-13 N.
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The speed of sound in air is 320 ms-1 and in water it is 1600 ms-1. It takes 2.5 s for sound to reach a certain distance from the source placed in air. a. Find the distance. b. How much time will it take for sound to travel the same distance when the source is in water?
Answer:
Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.
Explanation:
Given:
Speed of sound in air = 320 m/s
Speed of sound in water = 1600 m/s
Time taken to reach certain distance in air = 2.5 sec
a.
We have to find the distance traveled by sound in air.
Distance = Product of speed and time.
⇒ [tex]Distance = Speed\times time\ taken[/tex]
⇒ [tex]Distance = 320\times 2.5[/tex]
⇒ [tex]Distance = 800[/tex] meters.
b.
Now we have to find how much time the sound will take to travel in water.
⇒ Time = Ratio of distance and speed
⇒ [tex]Time =\frac{distance}{speed}[/tex]
⇒ [tex]Time =\frac{800}{1600}[/tex] ...distance = 800 m and speed = 1600 m/s
⇒ [tex]Time =\frac{1}{2}[/tex]
⇒ [tex]Time =0.5[/tex] seconds.
Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.
Pesticides and cancer drugs are both created by:
selective breeding
hybridization
recombinant DNA
cloning
Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyond Mars when furthest from the Sun (aphelion), but with an average distance (semi-major axis) of 1 AU. How long will it take to complete an orbit and where will it spend most of its time
Answer:
16.63min
Explanation:
The question is about the period of the comet in its orbit.
To find the period you can use one of the Kepler's law:
[tex]T^2=\frac{4\pi}{GM}r^3[/tex]
T: period
G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2
r: average distance = 1UA = 1.5*10^11m
M: mass of the sun = 1.99*10^30 kg
By replacing you obtain:
[tex]T=\sqrt{\frac{4\pi}{GM}r^3}=\sqrt{\frac{4\pi^2}{(6.67*10^{-11}Nm^2/kg^2)(1.99*10^{30}kg)}(1.496*10^8m)^3}\\\\T=997.9s\approx16.63min[/tex]
the comet takes around 16.63min
A dog has a mass of 60kg and an acceleration of 2m/s/s. What is the force of the dog?
Answer:
120 N
Explanation:
F=ma therefore 60kg times 2m/s^2 is 120 N
A source of emf is connected by wires to a resistor, and electrons flow in the circuit. The wire diameter is the same throughout the circuit. Compared to the drift speed of the electrons before entering the source of emf, the drift speed of the electrons after leaving the source of emf is __________.A. faster. B. slower. C. the same. D. either A or B depending on circumstances. E. any of A, B, or C depending on circumstances.
Answer:
C. the same.
Explanation:
The drift speed of electrons in a circuit is the same all through the circuit.
The drift speed of the electrons remains the same before and after they pass through the source of emf because the current is constant throughout the circuit assuming a steady state situation. The emf source does not affect the drift speed despite functioning like a pump that maintains the potential difference.
Explanation:The drift speed of the electrons after leaving the source of emf is the same as before entering it. This is so because, in a steady-state situation where the current is constant throughout a circuit, the rate at which electrons enter any given section of the circuit must equal the rate at which they exit. The drift speed is the average speed of the electrons as they move through the conductor, and it remains constant throughout the entirety of the circuit assuming the diameter of the wire is consistent.
It's important to understand that the emf source acts like a pump, helping move the electrons through the circuit and maintaining a potential difference. However, this doesn't result in an increase in the electron drift speed, as the overall velocity of the electron stays steady. Even in the presence of an electric field, the phenomenon known as drift velocity remains unaffected due to the near-random movements of the free electrons caused by their collisions with atoms and other free electrons.
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What are the similarities and differences between novae and supernovae?
Answer:
Explanation:
A novae in astronomy means an explosion in the white dwarf star which had tapped enough gas from a companion star,hence it releases an incredible amount of energy which is Over a million times brighter than it normal stars.
A super novae on the other hand is a cosmic explosion that can be a billion times brighter than the normal.
From this one can see that a perculiar similarity between a novae and super novae is that both generate huge explosion and bright Ness, and a major difference is super novae release huge amount of brightness and energy more than the novae
How much time would it take to exert 700W of power while doing 700J of work?
Answer:
The answer to your question is time = 1 s
Explanation:
Data
time = ?
Power = 700 W
Work = 700 J
Power is defined as the work done per unit of time.
Formula
Power = Work / Time
-Solve for time
Time = Work/Power
-Substitution
Time = 700 / 700
-Result
Time = 1 s
Conclusion
In 1 second and with 700 J there will be produced 700 watts
Consider a loop of wire whose plane is horizontal and that carries a current in the clockwise direction when viewed from above. If we were to represent the current loop as a bar magnet or magnetic dipole, in what direction would the north pole be pointing?
Answer:
Downwards into the plane
Explanation:
Solution:-
- This is a conceptual application of hand rule. We will place our palm fingers open vertical to a plane surface. Then curl our fingers in and naturally point the thumb.
- The direction of curl of fingers denotes the direction of of current flow in the coil. Which in our case is "clockwise direction". We will orient/invert our right hand palm in such a way that we curl our fingers in clockwise fashion. Then stick the thumb out to give us the direction of magnetic field or North pole end. In our case the the thumb points downwards into the plane denoting that the magnetic field within the loop is also acting downwards into the plane.
- The bar magnet would be placed in such a way that North pole is pointing downward into the plane in the direction of magnetic field and end up at south pole pointing up out of the plane.
when all else remains the same, what effect would decreasing the focal have known a convex lens
Answer:
when all else remains the same, what effect would decreasing the focal have known a convex lens
Explanation:
It would cause the lens to produce only real images
( Hope This Will Help U Out!!)
1. What will happen to the brightness of the light bulb if the switch in this circuit is suddenly closed?
Answer:
There is no closed-loop path for the current to flow through the circuit.When the switch is closed,the light bulb operates since the current flows through the circuit.The bulb glows at its full brightness since its receives its full 120 volts and has the design current flow.
Explanation:
The voltage V in a circuit that satisfies the law V = IR is slowly dropping as the battery wears out. At the same time, the resistance R is increasing as the resistor heats up. Use the equation dV dt = ∂V ∂I dI dt + ∂V ∂R dR dt to find how the current I is changing at the instant when R = 600 ohms, I = 0.04 amps, dR/dt = 0.5 ohms/sec, and dV /dt = −0.01 volts/sec. Hint: We need to find the rate of change of I, with respect to time. Find the partial derivatives of V with respect to I and R, then substitute into the equation for dV /dt.
Answer:
[tex]-0.5\times 10^{-4} A/s[/tex]
Explanation:
We are given that
[tex]\frac{dV}{dt}=-0.01 V/s[/tex]
R=600 ohms
I=0.04 A
[tex]\frac{dR}{dt}=0.5ohm/s[/tex]
[tex]V=IR[/tex]
[tex]\frac{dV}{dt}=\frac{\partial V}{dI}\frac{dI}{dt}+\frac{\partial V}{dR}\frac{dR}{dt}[/tex]
[tex]\frac{dV}{dt}=R\frac{dI}{dt}+I\frac{dR}{dt}[/tex]
Substitute the values
[tex]-0.01=600\times \frac{dI}{dt}+0.04\times 0.5[/tex]
[tex]-0.01-0.04\times 0.5=600\frac{dI}{dt}[/tex]
[tex]-0.03=600\frac{dI}{dt}[/tex]
[tex]\frac{dI}{dt}=\frac{-0.03}{600}=-0.5\times 10^{-4}A/s[/tex]
Final answer:
The student's question concerns finding the rate of change of current in a circuit as described by Ohm's law. By finding partial derivatives of the voltage with respect to current and resistance and applying the provided rates, you solve for the rate of change of current.
Explanation:
The student is asking how the current I is changing over time in a circuit with a given resistance R, current I, and rates of change dR/dt and dV/dt according to Ohm's law. To solve this, we need to find the partial derivatives of voltage V with respect to current I and resistance R, and then use the given differential equation dV dt = ∂V ∂I dI dt + ∂V ∂R dR dt.
Using Ohm's law, we can express the voltage as V = IR. The partial derivative of V with respect to I is R, and with respect to R is I. Substituting these derivatives and the given values into the differential equation, we have: -0.01 = 600 dI dt + 0.04 * 0.5. Solving for dI dt, we get the rate at which the current is changing over time, which is the answer the student is looking for.
A man driving a car traveling at 30m/s slams on the brakes and accelerates in the negative direction at 4.75 m/s2, how far does the car travel before it stops?
Answer:
94.73 meters.
Explanation:
Using SUVAT.
s = ?
u = 30
v = 0
a = -4.75
t = ?
[tex]v^{2}=u^2+2as\\0=30^2+2(-4.75)s\\\\900=9.5s\\s = 94.73[/tex]
The car will travel approximately 94.74 meters before it stops.
To answer this question, we can use the equations of motion.
When the car stops, its final velocity will be 0 m/s.
We can use the equation v^2 = u^2 + 2as,
where v is the final velocity,
u is the initial velocity,
a is the acceleration, and
s is the distance.
Plugging in the values, we have (0)^2 = (30)^2 + 2(-4.75)s.
Solving for s, we get s = 900/9.5, which is approximately 94.74 meters.
Therefore, the car will travel approximately 94.74 meters before it stops.
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An inventor claims to have developed a heat pump that produces 200 kW of heating for a 293 K heated zone whilst only using 75 kW of power and a heat source at 273 K. Using the max. theoretical efficiency of this device, justifythe validity of this claim. [5]
Answer:
Answer is 14.65
Refer below.
Explanation:
Refer to the picture for brief explanation.
A helium balloon containing 5m³ of gas at a pressure of 30kPa is released into the air. Assuming that the temperature is constant calculate the volume the balloon would have when the pressure inside the balloon has fallen to 5kPa. Answer in m³.
Answer:
30 m³
Explanation:
Parameters given:
Initial volume of helium, V1 = 5 m³
Initial pressure in balloon, P1 = 30 kPa
Final pressure, P2 = 5 kPa
To find the volume of the balloon at that volume, we apply Boyle's law.
It states that at constant temperature, the pressure of a gas is inversely proportional to the volume of the gas.
Mathematically:
P = k / V
Where k = constant of proportionality
This implies that:
P * V = k
This means that if the pressure or volume of the gas changes at the same temperature, the product of the pressure and volume would be the same:
Hence:
P1 * V1 = P2 * V2
Hence, to find the final volume:
30 * 5 = 5 * V2
=> V2 = (30 * 5) / 5
V2 = 30 m³
The volume of the gas when the pressure is 5 kPa is 30 m³.
An ordinary egg can be approximated as a 5.5-cm diameter sphere. The egg is initially at a uniform temperature of 8°C and is dropped into boiling water at 97°C. Taking the properties of the egg to be r 1020 kg/m3 and cp 3.32kJ/kg · °C, determine (a) how much heat is transferred to the egg by the time the average temperature of the egg rises to70°C and (b) the amount of entropy generation associated with this heat transfer process.
Answer:
a) [tex]Q_{in} = 13.742\,kW[/tex], b) [tex]\Delta S = 370.15\,\frac{kJ}{K}[/tex]
Explanation:
a) The heat transfered to the egg is computed by the First Law of Thermodynamics:
[tex]Q_{in} +U_{sys,1} - U_{sys,2} = 0[/tex]
[tex]Q_{in} = U_{sys,2} - U_{sys,1}[/tex]
[tex]Q_{in} = \rho_{egg}\cdot \left(\frac{4\pi}{3}\cdot r^{3}\right)\cdot c \cdot (T_{2}-T_{1})[/tex]
[tex]Q_{in} = \left(1020\,\frac{kg}{m^{3}}\right)\cdot \left(\frac{4\pi}{3}\right)\cdot (0.025\,m)^{3}\cdot \left(3.32\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (70\,^{\textdegree}C - 8\,^{\textdegree}C)[/tex]
[tex]Q_{in} = 13.742\,kW[/tex]
b) The amount of entropy generation is determined by the Second Law of Thermodynamics:
[tex]\Delta S = \frac{Q_{in}}{T_{in}}[/tex]
[tex]\Delta S = \frac{13.742\,kJ}{370.15\,K}[/tex]
[tex]\Delta S = 370.15\,\frac{kJ}{K}[/tex]
The heat transferred to the egg is approximately 18.24 kJ. The entropy generation associated with this heat transfer is about 5.68 kJ/K.
To determine how much heat is transferred to the egg, we use the following formula for heat transfer in a substance:
Q = m × cp × (T_final - T_initial)
Given data:
Initial temperature, T_initial: 8°CFinal temperature, T_final: 70°CDensity, ρ: 1020 kg/m³Specific heat, cp: 3.32 kJ/kg·°CDiameter of the egg: 5.5 cm (radius is 2.75 cm or 0.0275 m)The volume of the egg as a sphere is:
V = (4/3) × π × r³ = (4/3) × π × (0.0275)³ ≈ 8.72 × 10⁻⁵ m³
The mass of the egg is then:
m = ρ × V = 1020 kg/m³ × 8.72 × 10⁻⁵ m³ ≈ 0.0889 kg
Now, calculate the heat transferred:
Q = m × cp × (T_final - T_initial) = 0.0889 kg × 3.32 kJ/kg·°C × (70°C - 8°C) ≈ 18.24 kJ
(b) Calculating Entropy Generation:
Using the formula for entropy change, we have:
ΔS = m × cp × ln(T_final/T_initial)
Temperatures in Kelvin:
T_initial = 8°C + 273.15 = 281.15 KT_final = 70°C + 273.15 = 343.15 KThen:
ΔS = 0.0889 kg × 3.32 kJ/kg·°C × ln(343.15 / 281.15) ≈ 5.68 kJ/K
Hence, the amount of entropy generated during the process is approximately 5.68 kJ/K.
If a wave has a speed of 100 m/s and a wavelength of 20 meters, what is the frequency?
Answer:
Frequency 5hz
Explanation:
Final answer:
To calculate the frequency of a wave with a speed of 100 m/s and a wavelength of 20 meters, divide the speed by the wavelength, resulting in a frequency of 5 Hz.
Explanation:
To find the frequency of a wave when you know its speed and wavelength, you can use the formula: speed = (wavelength) x (frequency). Rearranging this formula to solve for frequency gives us frequency = speed / wavelength. In this case, with a wave speed of 100 m/s and a wavelength of 20 meters, the calculation would be frequency = 100 m/s / 20 m.
Therefore, the frequency of the wave is 5 Hz. This means the wave completes 5 cycles per second.
A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force constant of 8.08 N/m. When the cannon is fired, the ball moves 15.8 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.031 0 N on the ball.With what speed does the projectile leave the barrel of the cannon?
Answer:
Speed will be equal to 1.40 m/sec
Explanation:
Mass of the rubber ball m = 5.24 kg = 0.00524 kg
Spring is compressed by 5.01 cm
So x = 5.01 cm = 0.0501 m
Spring constant k = 8.08 N/m
Frictional force f = 0.031 N
Distance moved by ball d = 15.8 cm = 0.158 m
Energy gained by spring
[tex]KE=\frac{1}{2}kx^2=\frac{1}{2}\times 8.08\times 0.0501^2=0.0101J[/tex]
Energy lost due to friction
[tex]W=Fd=0.031\times 0.158=0.0048J[/tex]
So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J
This energy will be kinetic energy
[tex]\frac{1}{2}mv^2=0.0052[/tex]
[tex]\frac{1}{2}\times 0.00524\times v^2=0.0052[/tex]
v = 1.40 m/sec
The toy cannon's projectile leaves the barrel at approximately 1.42 m/s.
To find the speed at which the 5.24-g soft rubber ball leaves the barrel of the toy cannon, we follow these steps:
1.) Determine the Potential Energy in the Compressed Spring:
The potential energy stored in the spring when it is compressed can be calculated using the formula:
Potential Energy = (1/2) * k * x²where
k is the spring constant (8.08 N/m) and x is the compression distance (5.01 cm = 0.0501 m). Potential Energy = (1/2) * 8.08 N/m * (0.0501 m)²
Potential Energy = 0.01017504 J
2.) Calculate the Work Done by Friction:
Friction force is constant at 0.0310 N over a distance of 15.8 cm (0.158 m):
Work = Friction Force * DistanceWork = 0.0310 N * 0.158 mWork = 0.004898 J3.) Find the Net Energy Available:
Net Energy = Potential Energy - Work done by friction:Net Energy = 0.01017504 J - 0.004898 JNet Energy = 0.00527704 J4.) Calculate the Speed of the Ball:
Using the energy conservation principle where Net Energy is converted into kinetic energy:
Kinetic Energy = (1/2) * m * v²The ball's mass (m) = 5.24 g = 0.00524 kg.Solving for velocity (v):
0.00527704 J = (1/2) * 0.00524 kg * v²Thus, the speed at which the ball leaves the barrel of the cannon is approximately 1.42 m/s.
The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity as measured in W/m2 changes by a multiplicative factor. The number of decibels increases by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I is
β=10log(II0)dB,
where I0 is a reference intensity. For sound waves, I0 is taken to be 10−12W/m2. Note that log refers to the logarithm to the base 10.
Part A) What is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity (i.e., I=10I0)?
Part B) What is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity (i.e. I=100I0)?
Express the sound intensity numerically to the nearest integer.
β = dB
One often needs to compute the change in decibels corresponding to a change in the physical intensity measured in units of power per unit area. Take m to be the factor of increase of the physical intensity (i.e., I=mI0).
Part C) Calculate the change in decibels ( Δβ2, Δβ4, and Δβ8) corresponding to m=2, m=4, and m=8.
Give your answers, separated by commas, to the nearest integer--this will give an accuracy of 20%, which is good enough for sound.
Answer:
a) 10 dB, b) 20dB, c) 10², 10⁴, 10⁸
Explanation:
The logarithmic scale has a great advantage when measuring magnitudes of a large number of scales, since it converts these values to linear, allowing easier viewing.
Part A
Let's look for decibels for an intensity I = 10 Io
We calculate
β = 10 log (10Io / Io)
β = 10 dB
Part b
Let's find the intensity for I = 100 Io
We calculate
β = 10 log (100Io / Io)
β = 10 log 100
β = 10 2
β = 20 db
Part c
Δβ2 corresponds to an intensity change of 10² Io, therefore it corresponds to an intensity increase of 10²
Δβ4 corresponds to a change in intensity of 10⁴Io
Δβ8 is an intensity change of 10⁸ Io
The sound intensity level in decibels can be calculated using the formula β = 10log(I/I_0). For a sound wave with intensity 10 times the reference intensity, the sound intensity level is 10 dB. For a sound wave with intensity 100 times the reference intensity, the sound intensity level is 20 dB.
Explanation:Part A: To find the sound intensity level when the intensity is 10 times the reference intensity, we use the formula β = 10log(I/I_0). Plugging in the values, β = 10log(10I_0/I_0) = 10log(10) = 10 x 1 = 10 dB.
Part B: Similarly, when the intensity is 100 times the reference intensity, we get β = 10log(100I_0/I_0) = 10log(100) = 10 x 2 = 20 dB.
Part C: To calculate the change in decibels for different intensities (m), we use the formula Δβ = 10log(m). For m = 2, Δβ2 = 10log(2) = 10 x 0.3 = 3 dB. For m = 4, Δβ4 = 10log(4) = 10 x 0.6 = 6 dB. And for m = 8, Δβ8 = 10log(8) = 10 x 0.9 = 9 dB.
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what is the difference between condensation and conduction
Answer:
Convection is the heat transfer due to the bulk movement of molecules within fluids such as gases and liquids, including molten rock. Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds. These clouds may produce precipitation, which is the primary route for water to return to the Earth's surface within the water cycle.
Explanation:
Their a difference....... A huge One
Condensation is the transition from gas to liquid, whereas conduction is a form of heat transfer through direct contact between objects.
The difference between condensation and conduction lies in their definition and the processes they describe. Condensation is a phase change in which a substance transitions from a gaseous state to a liquid state, often as a result of cooling. It is not directly related to heat transfer but is a thermodynamic process. On the other hand, conduction is a mode of heat transfer where thermal energy is transmitted through stationary matter due to physical contact, as when heat from an electric burner moves to the bottom of a pan.
Examples of conduction include the warming of a glass when holding it in a warm hand or the transfer of heat from a stove's burner to a pan. Convection is another mode of heat transfer involving the movement of fluids, which plays an important role in weather systems and heating systems. In contrast, conduction relies on direct contact between objects for heat to be transferred.
Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much less than the masses of the stars.
In System A , Planet A of mass Mp orbits Star A of mass Ms in a circular orbit of radius R .
In System B , Planet B of mass 4Mp orbits Star B of mass Ms in a circular orbit of radius R .
In System C , Planet C of mass Mp orbits Star C of mass 4Ms in a circular orbit of radius R .
(a) The gravitational force exerted on Planet A by Star A has a magnitude of F0 . Determine the magnitudes of the gravitational forces exerted in System B and System C .
___ Magnitude of gravitational force exerted on Planet B by Star B
___ Magnitude of gravitational force exerted on Planet C by Star C
(b) How do the tangential speeds of planets B and C compare to that of Planet A ? In a clear, coherent paragraph-length response that may also contain equations and/or drawings, provide claims about
why the tangential speed of Planet B is either greater than, less than, or the same as that of Planet A , and
why the tangential speed of Planet C is either greater than, less than, or the same as that of Planet A .
a) 4F0
b) Speed of planet B is the same as speed of planet A
Speed of planet C is twice the speed of planet A
Explanation:
a)
The magnitude of the gravitational force between two objects is given by the formula
[tex]F=G\frac{m_1 m_2}{r^2}[/tex]
where
G is the gravitational constant
m1, m2 are the masses of the 2 objects
r is the separation between the objects
For the system planet A - Star A, we have:
[tex]m_1=M_p\\m_2 = M_s\\r=R[/tex]
So the force is
[tex]F_A=G\frac{M_p M_s}{R^2}=F_0[/tex]
For the system planet B - Star B, we have:
[tex]m_1 = 4 M_p\\m_2 = M_s\\r=R[/tex]
So the force is
[tex]F=G\frac{4M_p M_s}{R^2}=4F_0[/tex]
So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.
For the system planet C - Star C, we have:
[tex]m_1 = M_p\\m_2 = 4M_s\\r=R[/tex]
So the force is
[tex]F=G\frac{M_p (4M_s)}{R^2}=4F_0[/tex]
So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.
b)
The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:
[tex]G\frac{mM}{r^2}=m\frac{v^2}{r}[/tex]
where
m is the mass of the planet
M is the mass of the star
v is the tangential speed
We can re-arrange the equation solving for v, and we find an expression for the speed:
[tex]v=\sqrt{\frac{GM}{r}}[/tex]
For System A,
[tex]M=M_s\\r=R[/tex]
So the tangential speed is
[tex]v_A=\sqrt{\frac{GM_s}{R}}[/tex]
For system B,
[tex]M=M_s\\r=R[/tex]
So the tangential speed is
[tex]v_B=\sqrt{\frac{GM_s}{R}}=v_A[/tex]
So, the speed of planet B is the same as planet A.
For system C,
[tex]M=4M_s\\r=R[/tex]
So the tangential speed is
[tex]v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A[/tex]
So, the speed of planet C is twice the speed of planet A.
donna is attempting to make an electromagnet to pick up old nails in and around her driveway. which of these materals would be best to use the core in order to make the strongest electromagnet to pick up the most nails
Answer:
D. Iron nail
Explanation:
A horizontal wooden beam sags a bit when supported at its ends. In between the top and bottom surfaces is a region of
Answer: neither tension nor compression
Explanation:
The tension in the form of load or forces
And also compression in the form of weight or load since the sagging is negligible as interpreted from the question which say that - horizontal wooden beam sags a bit when supported at its ends.
Describe how the properties of water contribute to two of the following. Transpiration thermoregulation in endotherms plasma membrane structure
Answer:
Explanation:
The contributions of water properties in Thermoregulation in endotherms and in the plasma membrane structure.
A) Thermoregulation in endotherms: Water contributes to thermoregulation by its property of High specific heat. Water's high specific heat acts as a heat buffer, it is much harder to cool down/heat up, therefore, it will make warmer than the cold environment around it or colder than the hot environment around it. Thermoregulation works in humansas well. Recall Homeostasis which is the body's way of regulating our temperature.
B) Plasma membrane structure:
The property of water, polarity, contributes to the plasma membrane because it creates the arrangement of phospholipids making it a semi-permeable membrane.
Recall that the Plasma Membrane has a hydrophilic end on the outside that is charged and likes water and a hydrophobic end on the inside that has no charge and doesn't like water.
The membrane that separates the interior of the cell from the external environment is known as the plasma membrane, sometimes known as the cell membrane, and it is present in all cells. A cell wall is affixed to the plasma membrane on the exterior of bacterial and plant cells.
What is transpiration?Like all other living things, plants need an excretory system to remove extra water from their bodies. Transpiration is the term for this process of removing extra water from the body of the plant. Typically, it is the evaporation of water from the leaf surface.
Now, according to the question :
(A) Thermoregulation in endotherms :
Water helps regulate body temperature since it has a high specific heat. Due to its high specific heat, which acts as a heat buffer, water will either make an area warmer than the surrounding cold environment or colder than the surrounding hot environment. Humans are capable of thermoregulation as well. Remember that the body uses homeostasis to control our body temperature.
(B) Plasma membrane structure:
Because it forms the arrangement of phospholipids that makes the plasma membrane a semi-permeable membrane, water's polarity contributes to the plasma membrane.
Remember that the Plasma Membrane has an exterior hydrophilic end that is charged and prefers water, and an interior hydrophobic end that is uncharged and dislikes water.
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A wheel 1.70 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 3.60 rad/s2. The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3° with the horizontal at this time. At t = 2.00 s. What is the tangential speed, total acceleration, and angular position of point P.
Final answer:
To calculate the tangential speed, total acceleration, and angular position of point P on the wheel at t = 2.00 s, we can use the formulas for tangential speed, total acceleration, and angular position. By substituting the given values into these equations, we can find the required values.
Explanation:
To calculate the tangential speed of a point on the wheel, we can use the formula:
Tangential Speed = Angular Velocity x Radius
In this case, the angular velocity is given by the equation:
Angular Velocity = Initial Angular Velocity + (Angular Acceleration x Time)
Substituting the given values and solving the equations, we can find the tangential speed, which is the speed of the point P on the rim at time t = 2.00 s.
To find the total acceleration, we can use the formula:
Total Acceleration = Tangential Acceleration + Radial Acceleration
The tangential acceleration can be calculated using the equation:
Tangential Acceleration = Angular Acceleration x Radius
The radial acceleration can be calculated using the equation:
Radial Acceleration = (Angular Velocity x Angular Velocity) x Radius
By substituting the given values into these equations, we can find the total acceleration at time t = 2.00 s.
To find the angular position of point P at time t = 2.00 s, we can use the equation:
Angular Position = Initial Angular Position + (Initial Angular Velocity x Time) + (0.5 x Angular Acceleration x Time x Time)
Substituting the given values, we can find the angular position of point P at t = 2.00 s.
Technician A says a pull to the right during braking could be caused by a defective metering valve. Technician B says a pull to the left could be caused by a defective proportioning valve. Which technician is correct
Answer:
The correct option is;
Neither A nor B
Explanation:
The common cause of a vehicle pulling to the right or to the left during braking is due to a contaminated breaking surface or a faulty caliper. Other causes include tire size variation or worn out suspension components.
Other causes include;
Leaking break fluid
Piston frozen in wheel cylinder or caliper
Adjuster of rear break
Tire fault.
As soon as it is observed that the vehicle pulls on one side when braking, the vehicle should should be taking for checks by a qualified mechanic as soon as possible.
Final answer:
Both Technician A and B are incorrect, as a pull during braking is more likely due to uneven brake pad wear, sticking calipers, or collapsed brake lines rather than a defective metering or proportioning valve.
Explanation:
The question relates to whether a pull to the right during braking could be caused by a defective metering valve, and whether a pull to the left could be caused by a defective proportioning valve. Neither of these assertions is typically accurate.
A metering valve is designed to ensure that the rear brakes engage just after the front brakes, to prevent rear wheel lockup during early brake application. A proportioning valve adjusts the pressure between the front and rear brakes to prevent the rear wheels from locking up during heavy braking. Neither valve is designed to balance the braking force from side to side.
A pull to one side during braking is more likely to be caused by uneven brake pad wear, sticking calipers, or a collapsed brake line, rather than issues with the metering or proportioning valves. Therefore, both Technician A and Technician B are incorrect regarding the cause of a pull during braking.
How do the magnetic poles of earth serve to protect life on earth?
Answer:
The block solar winds and other harmful things
Explanation:
The Earth's magnetic field serves to deflect most of the solar wind, whose charged particles would otherwise strip away the ozone layer that protects the Earth from harmful ultraviolet radiation.
Answer:
Explanation:The Earth's magnetic field serves to deflect most of the solar wind, whose charged particles would otherwise strip away the ozone layer that protects the Earth from harmful ultraviolet radiation. One stripping mechanism is for gas to be caught in bubbles of magnetic field, which are ripped off by solar winds.
A British bicycle light company advertises flashing bicycle lights that require no batteries and produce no resistance to riding. A magnet attached to a spoke on the bicycle tire moves past a generator coil on the bicycle frame, inducing an emf that causes a light to flash. The magnet and coil never touch. Does this lighting system really produce no resistance to riding? Justify your answer.
Answer:
Lenz's law is responsible for the resistance to advance due to the change in magnetic flux
Explanation:
This system produces electricity by Faraday's law,
E = - d / dt B.A
Where the electromotive force is generated by the change of the field strength maintained due to the circular movement of the wheel.
As this field changes when Lenz's law increases or decreases, it creates a field that opposes this change, this field is applied to the bicycle and is in the opposite direction to movement, so it is an effect that creates resistance to the advance of the bike.
In summary Lenz's law is responsible for the resistance to advance due to the change in magnetic flux
Recently, astronomers have observed stars and other objects that orbit the center of the Milky Way Galaxy farther out than our Sun, but move around faster than we do. How do astronomers think such an observation can be explained?
Answer:
That scenario can be explained by the idea of the contribution of dark matter on that point.
Explanation:
It can be explained through the idea of dark matter, this one was born to explain why stars (or any object) that were farther for the supermassive black hole in the center of the Milky Way galaxy didn't decrease it rotational velocity as it was expected according to equation 1.
[tex]v = \sqrt{\frac{G M}{r}}[/tex] (1)
Where v is the rotational velocity, G is the gravitational constant, M is the mass of the supermassive black hole, and r is the orbital radius.
Notice, that If the distance increases the orbital speed decreases (inversely proportional).
Final answer:
Astronomers explain the high velocities of stars orbiting the Milky Way's center outside our Sun's orbit by the gravitational influence of dark matter. This halo of invisible matter adds mass to the galaxy, affecting the orbital speeds in accordance with Kepler's third law.
Explanation:
Astronomers have observed that stars and other objects orbiting the center of the Milky Way Galaxy farther out than our Sun move around faster than we do, which is an unexpected observation according to Kepler's third law. Typically, as per Kepler's third law, we would expect objects that are farther from the center of mass to orbit more slowly. However, the presence of invisible matter, which we now understand as dark matter, affects the gravitational pull and thus the orbital velocities of these outer objects, leading them to move at unexpectedly high speeds.
The Milky Way is surrounded by a halo of this invisible matter, which does not emit or reflect light, making it undetectable through traditional means. Instead, its presence is inferred from its gravitational effects on the motions of stars and gas in the galaxy. The observations mentioned, where objects beyond the luminous part of the Milky Way are moving faster than expected, provide evidence for the existence of dark matter, as this additional mass can account for the higher orbital velocities.
A student waiting on a platform for an incoming train hears a train whistle blow some place down the track but she is not certain from which direction it came. If the student notices that the frequency of the train whistle slowly decreases. What can the student conclude about the speed and direction of the train blowing the whistle
Answer:
Considering the two cases of when the train is moving towards and away from the stationary observer. The observed frequency of sound waves from the train can decrease when
1) The speed of the train is slowly reducing and its direction is towards the stationary observer.
2) The speed of the train is slowly increasing and its direction is away from the observer for the sound's observed frequency to keep decreasing.
Explanation:
This phenomenon is due to Doppler's Effect.
Doppler's Effect explains how relative frequency of a sound source varies with the velocity of the source or the observer.
Generally, the mathematical expression for Doppler's Effect is given below
f' = f [(v + v₀)/(v - vₛ)]
where
f' = observed frequency
f = actual frequency
v = velocity of sound waves
v₀ = velocity of observer
vₛ = velocity of sound source
When the train is moving towards the stationary observer,
f' = observed frequency of the sound wave = f₁
f = actual frequency of the sound wave = f
v = velocity of sound waves = v
v₀ = velocity of observer = 0 m/s
vₛ = velocity of sound source = vₛ
f' = f [(v + v₀)/(v - vₛ)]
f₁ = f [(v + 0)/(v - vₛ)]
f₁ = fv/(v - vₛ) (eqn 1)
Observed frequency is obviously higher than the frequency of the train.
But if the train keeps reducing its speed (vₛ) towards from the stationary observer, the observed frequency will decrease.
When the train is moving away from the stationary observer,
f' = observed frequency of the sound wave = f₂
f = actual frequency of the sound wave = f
v = velocity of sound waves = v
v₀ = velocity of observer = 0 m/s
vₛ = velocity of sound source = - vₛ (train is moving away from the observer, hence, the negative sign)
f' = f [(v + v₀)/(v - vₛ)]
f₂ = f [(v + 0)/(v - (-vₛ)]
f₂ = fv/(v + vₛ) (eqn 2)
From the expression, it is clear that the observed frequency is smaller than the frequency of the sound waves and it keeps decreasing as the speed of the train (vₛ) increases as the train moves away from the stationary observer.
Hope this Helps!!!
Answer:
The student can conclude that the speed of the train is increasing while the direction of the train is away from her or him
Explanation:
Here we have the Doppler effect given by
For an approaching wave
[tex]F_{obs} =\left [ \frac{v}{v - v_{source}} \right ] f_{source[/tex]
For a receding wave we have
[tex]F_{obs} =\left [ \frac{v}{v + v_{source}} \right ] f_{source[/tex]
Therefore, comparing the denominators of the two equation, it is seen that the frequency heard by the observer is decreases for a receding wave.
That is the student can conclude that the train is moving away from her or him or that the speed of the train is increasing while the direction of the train is away from her or him.