Answer:
Intensity Of MRI signal is related as square of [tex]B_{0}[/tex]
or
Intensity is related as [tex]B_{0}^{2}[/tex].
Explanation:
MRI simply means Magnetic Resonance Imaging. In this process by the resonance of magnetic field the frequency is produced which is used to image the body parts (Mainly the brain).
The signal Strength is the rate of change of Magnetization (M) with respect to the time. For the calculation of Intensity we use Bloch equation which is given by
[tex]\frac{dH}{dt} = M \times \gamma B[/tex]
On further solving
[tex]\frac{dH}{dt} =\mu B \times \gamma B \\\frac{dH}{dt} = \mu \times \gamma B^{2}[/tex]
So, Intensity is related as [tex]B_{0}^{2}[/tex].
Which molecule of the gpcr-adenylyl cyclase signal transduction system phosphorylates proteins to create the cellular response?
Answer:
Protein kinase A
Explanation:
Protein kinase A is also called Cyclic AMP- dependent protein kinase or A kinase. It is an enzyme that enhance protein covalently using the phosphate group. The function of this enzyme is that it helps to end the effect of different hormones working via the Cyclic AMP signalling pathway. Protein kinase A can be found in the cytoplasm which phosphorylate proteins.
Protein kinase A helps the cell in regulating sugar, glycogen and lipids metabolism level.
Answer:
The answer is Protein Kinase A.
Refer below for the explanation.
Explanation:
In cell science, protein kinase A is a group of compounds whose movement is reliant on cell levels of cyclic AMP. PKA is otherwise called cAMP-subordinate protein kinase. Protein kinase A has a few capacities in the cell, including guideline of glycogen, sugar, and lipid digestion.
TheAurora Borealisis a night display in the Northern latitudes caused by ionizing radiation interacting with the Earth's magnetic field and the upper atmosphere. The distinctive green color is caused by the interaction of the radiation with oxygen and has a frequency of 5.38 x 1014units. What is the wavelength of this light?
Answer:
The answer is
The wavelength of the light is 557.2 x 10⁻⁻⁹nm
Explanation:
The wavelength of a wave is the distance between adjacent troughs and crests while the frequency of a wave is the number of completed cycles that pass a given per unit time
Electromagnetic radiation, such as light is usually described n terms of its frequency and wavelength
The equation relating the three quantities of speed of light, frequency and wavelength is as follows
Speed of light, c = Frequency of the light wave, ν × Wavelength of the light, λ
That is c = ν × λ
Where c = 299792458 m/s, v = 5.38 x 10¹⁴ Hz
Therefore the wavelength = [tex]\frac{c}{v}[/tex] = 5.572 x 10⁻⁷ m 557.2 x 10⁻⁻⁹nm
A spotlight on a boat is y = 2.2 m above the water, and the light strikes the water at a point that is x = 8.5 m horizontally displaced from the spotlight (see the drawing). The depth of the water is 4.0 m. Determine the distance d, which locates the point where the light strikes the bottom.
Answer:
The answer to the question is
The distance d, which locates the point where the light strikes the bottom is 29.345 m from the spotlight.
Explanation:
To solve the question we note that Snell's law states that
The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction
n₁sinθ₁ = n₂sinθ₂
y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and
θ₁ = 14.511 °
n₁ = 1.0003 = refractive index of air
n₂ = 1.33 = refractive index of water
Therefore sinθ₂ = [tex]\frac{n_1sin\theta_1}{n_2}[/tex] = [tex]\frac{1.003*0.251}{1.33}[/tex] = 0.1885 and θ₂ = 10.86 °
Since the water depth is 4.0 m we have tanθ₂ = [tex]\frac{4}{x_2}[/tex] or x₂ = [tex]\frac{4}{tan\theta_2 }[/tex] =[tex]\frac{4}{tan(10.86)}[/tex] = 20.845 m
d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.
Match the name of each gas law to the properties it compares. Part AA) Temperature and volumeB) Pressure and temperatureC) Pressure and volumePart B1. Boyle's law2. Charles's law 3. Gay-Lussac's law
Answer:
A. CHARLES' LAW
B. GAY-LUSSAC'S LAW
C. BOYLE'S LAW
Explanation:
THE QUESTION SEEKS ANSWER BY MATCHING THE PROPERTIES COMPARED BY A GAS LAW TO THE GAS LAW IN QUESTION.
A. CHARLES' LAW COMPARES TEMPERATURE AND VOLUME.
IT ESTABLISHED THE FACT THAT VOLUME AND TEMPERATURE ARE DIRECTLY PROPORTIONAL AT A FIXED PRESSURE
B. GAY-LUSSAC'S LAW COMPARE PRESSURE AND TEMPERATURE.
IT ESTABLISHED THE FACT THAT PRESSURE AND TEMPERATURE ARE DIRECTLY PROPORTIONAL FOR A GIVEN VOLUME OF GAS
C. BOYLE'S LAW COMPARES PRESSURE AND VOLUME
IT ESTABLISHED THAT BOTH ARE INVERSELY PROPORTIONAL AT A GIVEN TEMPERATURE
Answer:
1) D
2) B
3) E
Explanation: Edge 2020
Is the magnitude of the acceleration of the center of mass of spool A greater than, less than, or equal to the magnitude of the acceleration of the center of mass of spool B?
Answer:
The acceleration of the centre of mass of spool A is equal to the magnitude of the acceleration of the centre of mass of spool B.
Explanation:
From the image attached, the description from the complete question shows that the two spools are of equal masses (same weight due to same acceleration due to gravity), have the same inextensible wire with negligible mass is attached to both of them over a frictionless pulley; meaning that the tension in the wire is the same on both ends.
And for the acceleration of both spools, we mention the net force.
The net force acting on a body accelerates the body in the same direction as that in which the resultant is applied.
For this system, the net force on either spool is exactly the same in magnitude because the net force is a difference between the only two forces acting on the spools; the tension in the wire and their similar respective weights.
With the net force and mass, for each spool equal, from
ΣF = ma, we get that a = ΣF/m
Meaning that the acceleration of the identical spools is equal also.
Hope this Helps!
Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m. A positive point charge q3 is located at the origin. a) What must the value of q3 be for the net force on thispoint charge to have a magnitude 4.40 μN ?
Answer:
q₃=5.3nC
Explanation:
First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:
[tex]F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\[/tex]
Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:
[tex]F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC[/tex]
In words, the value of q₃ must be 5.3nC.
What is the frequency of highly energetic ultraviolet radiation that has a wavelength of 129 nm? The speed of light is 3 × 108 m/s. Answer in units of Hz.
Answer:
[tex]2.33*10^{15}Hz[/tex]
Explanation:
The relationship between velocity v, frequency f and wavelength for electromagnetic waves is given by;
[tex]v=\lambda f..............(1)[/tex]
Given;
[tex]v=3*10^8m/s\\\lambda=129nm=129*10^{-9}m\\f=?[/tex]
We make substitutions into equation (1) as follows;
[tex]f=\frac{v}{\lambda}\\f=\frac{3*10^8}{129*10^{-9}}[/tex]
Explanation:
Below is an attachment containing the solution.
You are on an airplane that is landing. The plane in front of your plane blows a tire. The pilot of your plane is advised to abort the landing, so he pulls up, moving in a semicircular upward-bending path. The path has a radius of 450m with a radial acceleration of 17m/s2. What is the plane’s speed?
Answer:
[tex]v=87.46m/s[/tex]
Explanation:
Objects moving in circular path would be have either centripetal or centrifugal force.The force is either to center or away from center. When the object is moving along the circular path the centripetal force is
[tex]F=\frac{mv^{2}}{r}[/tex]
Here m is mass, v is velocity and r is radius of circular path
The acceleration is given by:
[tex]a_{r}=\frac{v^{2}}{r}[/tex]
The point of interest is lowest point on circle.The acceleration of plane at this position point up.The speed of plane from radial acceleration equation is:
[tex]v=\sqrt{a.r}\\[/tex]
Substitute 17 m/s² for a and 450m for r
So
[tex]v=\sqrt{17m/s^{2}*450m }\\ v=87.46m/s[/tex]
A diver makes 2.5 revolutions on the way from a 9.3-m-high platform to the water. Assuming zero initial vertical velocity, find the diver's average angular velocity during a dive.
Answer:
11.4 rad/s
Explanation:
The motion of the diver is a free-fall motion, so its center of mass falls down with constant acceleration of
[tex]g=9.8 m/s^2[/tex] towards the water
Therefore, we can use the following suvat equation:
[tex]s=ut-\frac{1}{2}gt^2[/tex]
where:
s = 9.3 m is the vertical displacement of the diver
u = 0 is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex]
And t is the total time of flight. Solving for t,
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(9.3)}{9.8}}=1.38 s[/tex]
So, the diver takes 1.38 s to reach the water.
During this time, the diver makes 2.5 revolutions; since 1 revolution is equal to an angle of [tex]2\pi[/tex] radians, then the total angular displacement is
[tex]\theta=2.5\cdot 2\pi =15.7 rad[/tex]
Therefore, the average angular velocity of the diver is the ratio between the total angular displacement and the time taken:
[tex]\omega=\frac{\theta}{t}=\frac{15.7}{1.38}=11.4 rad/s[/tex]
For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser pointer (λ=680nm) through the slit onto a screen 5.5 m away, a diffraction pattern appears. The bright band in the center of the pattern is 8.0 cm wide. What is the width of the slit?
Answer:
width of slit(a)≅ 0.1mm
Explanation:
Wave length of laser pointer =λ = 685 nm
Distance between screen and slit = L = 5.5 m
Width of bright band = W=8.0cm=0.08m
width of slit=a
recall the formula;
W=(2λL)/a
a=2λL/W
a=(2 *685*10⁻⁹*5.5m)/0.08m
a=7535*10⁻⁹/0.08
a=94187.5 *10⁻⁹
a=0.0000941875m
a=0.0941875mm
a≅0.1mm
Answer:
The wide of silt is a=93.5×10⁻⁶m
Explanation:
Given data
Wavelength λ=680 nm
Length L=5.5 m
Width w=8.0 cm
To find
Width of slit a
Solution
A single slit of width a has a bright central maximum of width
ω=2λL/a
a=2λL/ω
Substitute the given values
[tex]a=\frac{2*(680*10^{-9}m)5.5m}{0.08m} \\a=93.5*10^{-6}m[/tex]
The wide of silt is a=93.5×10⁻⁶m
When a mass of 24 g is attached to a certain spring, it makes 21 complete vibrations in 3.3 s. What is the spring constant of the spring? Answer in units of N/m.
Answer: 3.889N/m
Explanation:
f=(1/2)*√k/m
f=n/t
f= 21/3.3=6.4Hz
6.4*2=√k/m
12.73=√k/m
12.73^2=k/m
162.0529=k/m
Since m=24g
Convert g to kg
m=24/1000
m=0.024kg
K=162.0529*0.024
K=3.889N/m
Final answer:
The spring constant of the spring is approximately 24 N/m.
Explanation:
When a 24-gram mass is attached to a spring, causing it to make 21 complete vibrations in 3.3 seconds, we are looking at a problem that involves simple harmonic motion and can use the properties of a mass-spring system to determine the spring constant (k).
The formula to calculate the frequency (f) is:
f = number of vibrations / time
f = 21 / 3.3 s
f = 6.36363636 Hz
The frequency is related to the spring constant (k) and mass (m) as follows:
f = (1 / (2π)) * (√(k / m))
Given the mass (m) and frequency (f), we can solve for k:
k = (2πf)² * m
k = (2 * π * 6.36363636 Hz)² *0.024 kg
k ≈ 24 N/m
Therefore, the spring constant of the spring is approximately 24 N/m.
Your roommate is working on his bicycle and has the bike upside down. He spins the 54.0 cm diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. A. What is the pebble's speed? B. What is the period of the pebble stone?
Answer:
If the diameter is 54cm, then the radius is half that, the radius is:
r = 54cm/2 = 27cm
Then the perimeter of the wheel is p = 2*pi*27cm
this would mean that the pebble travels the distance:
d = 3*2*pi*27cm in one second.
Then the velocity of the pebble is:
speed = 3*2*3.1416*27cm/s = 509cm/s
Now, we know that the pebble does 3 cycles in a second, so does each cycle in 1/3 seconds, so the period would be the time that it needs to do one cycle, that we already find that is equal to 1/3 seconds.
Final answer:
The pebble stuck in the bicycle wheel tread has a speed of 5.091 m/s, and the period of its motion is 0.333 seconds.
Explanation:
To solve for the pebble's speed and the period of its motion, we will perform calculations based on the given wheel diameter and the frequency of the pebble's motion. The wheel's diameter is 54.0 cm, which gives us a radius (r) of 27.0 cm or 0.27 meters.
The speed (v) of the pebble on the tread of the wheel can be found using the formula for the circumference (C) of the wheel and multiplying by the frequency (f) of the pebble's motion:
C = 2πr
v = C × f
Plugging in the values we get:
C = 2 × π × 0.27 m = 1.697 m
f = 3 rev/s (since it goes by three times every second)
v = 1.697 m × 3 rev/s = 5.091 m/s
The period (T) is the inverse of frequency:
T = 1 / f
T = 1 / 3 rev/s = 0.333 s
Therefore, the pebble's speed is 5.091 meters per second, and the period of the pebble stone is 0.333 seconds.
Two identical loudspeakers separated by distance d emit 200Hz sound waves along the x-axis. As you walk along the axis, away from the speakers, you don't hear anything even though both speakers are on.What are the three lowest possible values for d? Assume a sound speed of 340m/s.
Answer:
So three lowest possible values of d ar 0.85m, 2.55m and 4.25m
Explanation:
The wavelength of wave:
λ=v/f
λ=340/200=1.7m
The destructive interference condition is:
Δd=(m+1/2)λ where m=0,1,2,3........
For minimum destructive interference ,the value of m is equal to zero
Δd=(0+1/2)×1.7
Δd=0.85m
For m=1
Δd=(1+1/2)×1.7
Δd=2.55m
For m=2
Δd=(2+1/2)×1.7
Δd=4.25m
So three lowest possible values of d ar 0.85m, 2.55m and 4.25m
The interference of sound waves is causing silence at your location. The three lowest possible values for the distance between two speakers creating this effect, given the given frequency of 200Hz and sound speed of 340m/s, are 0.85m, 2.55m and 4.25m.
Explanation:This is a problem related to sound interference; that is when two or more sound waves overlap, the resultant sound you hear is different than when you hear the sound waves individually.
You are not hearing anything because the sounds from the two speakers are arriving out of phase at your location, causing destructive interference.
At destructive interference, the path difference is (n + 1/2)λ. For the first three lowest values for d, n = 0, 1, 2. The speed of sound v = λf, so λ = v / f. Substituting these values in we get d = (n + 1/2) * v/f.
Now, let's plug in the given values: f=200Hz, v=340m/s.
d for n=0: d = (0 + 1/2) * 340 / 200 = 0.85 md for n=1: d = (1 + 1/2) * 340 / 200 = 2.55 md for n=2: d = (2 + 1/2) * 340 / 200 = 4.25 mLearn more about Sound Interference here:
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When the compressor of an air conditioner starts up, it draws a current of 50 A. If the start-up time is 0.60 s, then the amount of charge passing through a cross-sectional area of the circuit during this time is:
Answer:
30 C
Explanation:
Given:
Current flowing in the circuit (I) = 50 A
Start-up time (t) = 0.60 s
Now, we know that, charge drawn in through a cross sectional area of the circuit is given as:
[tex]q=It[/tex]
Where, 'q' is the amount of charge drawn, 'I' is the current and 't' is the start-up time.
Now, plug in 50 A for 'I', 0.60 s for 't' and solve for 'q'. This gives,
[tex]q=50\ A\times 0.60\ s\\\\q=30\ C[/tex]
Therefore, the amount of charge drawn in the circuit at the start-up of the compressor of an air conditioner is 30 C.
A point charge Q moves on the x-axis in the positive direction with a speed of 160 m/s. A point P is on the y-axis at y = +20 mm. The magnetic field produced at the point P, as the charge moves through the origin, is equal to -0.6 μT k^k^. What is the charge Q? (μ 0 = 4π × 10-7 T · m/A)
Answer:
The charge on the particle = -0.00075 C = -0.75 mC = -750 μC
Explanation:
The solution to this question is presented in the attached image to this answer.
The Biot Savart's formula for calculating magnetic field due to moving point charge is used in this calculation.
Hope this Helps!!!
The formula v = √ 2.3 r models the maximum safe speed, v , in miles per hour, at which a car can travel on a curved road with radius of curvature r , in feet. A highway crew measures the radius of curvature at an exit ramp on a highway as 370 feet. What is the maximum safe speed? FThe formula v = √ 2.3 r models the maximum safe speed, v , in miles per hour, at which a car can travel on a curved road with radius of curvature r , in feet. A highway crew measures the radius of curvature at an exit ramp on a highway as 370 feet. What is the maximum safe speed? F
The maximum safe speed on a curved road can be calculated using the formula v = √(2.3r), where v represents speed in miles per hour and r is the radius of curvature in feet. By substituting the given radius of curvature into the formula, we can find the maximum safe speed. In this case, the maximum safe speed is approximately 29.18 miles per hour.
Explanation:The formula v = √(2.3r) models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r, in feet. To find the maximum safe speed at a given radius of curvature, we can substitute the value of r into the formula and solve for v. In this case, the radius of curvature is given as 370 feet, so we substitute r = 370 into the formula:
v = √(2.3 * 370)
Simplifying the expression, we get:
v = √(851)
Taking the square root, we find that the maximum safe speed is approximately v ≈ 29.18 miles per hour.
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1. Impulse is the product of force and distance. True or false?
2. The force necessary to lift an object is g. True or false?
3. A joule is a newton times a second. True or false?
4. Work is done to lift an object in the classroom. True or false?
5. Kinetic energy is the energy of position. True or false?
6. Stopping distance is doubled, if speed is quadrupled.True or false?
Answer:
Explanation:
1. Impulse, I = F.t
The statement impulse is the product of Force and distance is false.
2. F = m g
Force necessary to lift the object depends on the mass of the object.
statement 2 is false.
3. Joule is equal to Newton times meter.
Statement 3 is false.
4. Work done to lift an object is correct statement.
Statement 4 is true.
5. Kinetic energy of an object is due to motion.
Statement 5 is false.
6. Stopping distance is directly proportional to the square of velocity.
If velocity is doubled, stopping distance is quadrupled.
Statement 6 is false.
The correct answer is 1.False; 2.False; 3.False; 4.True; 5.False; 6.False.
1. Impulse is the product of force and time, not force and distance. Therefore, the statement is false.
2. The force necessary to lift an object is not simply g, but weight, which is calculated as mass times the gravitational acceleration g. So, this statement is false.
3. A joule is the unit of work or energy, equivalent to one newton meter (N·m), not a newton times a second. Thus, the statement is false.
4. Work is indeed done to lift an object in the classroom since work is the force applied over a distance in the direction of the force. Therefore, this statement is true.
5. Kinetic energy is the energy of motion, not the energy of position. Thus, this statement is false.
6. If the stopping distance is doubled when speed is quadrupled, it overlooks the fact that stopping distance increases with the square of the speed, meaning a quadrupling of speed typically increases the stopping distance by a factor of sixteen, not two. So, this statement is false.
The circuit to the right consists of a battery ( V 0 = 64.5 V) (V0=64.5 V) and five resistors ( R 1 = 711 (R1=711 Ω, R 2 = 182 R2=182 Ω, R 3 = 663 R3=663 Ω, R 4 = 534 R4=534 Ω, and R 5 = 265 R5=265 Ω). Find the current passing through each of the specified points. -g
To find the Current in a circuit passing through each specified point in the circuit, we can use Ohm's law and the known resistances and voltages.
To find the current passing through each specified point in the circuit, we can use Ohm's law and the known resistances and voltages.
For example, to find the current passing through resistor R₂, we first need to find the voltage applied to it. This can be done by subtracting the voltage drop across resistor R₁ from the battery voltage. Once we have the voltage, we can use Ohm's law to find the current through R₂.
We can repeat this process for the other specified points in the circuit using the appropriate resistances and voltages.
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The ______ clause in an if statement allows a program to make one choice if the condition is true, and anothere choice if the condition is false.
Answer: The IF clause.
Explanation: This is the IF clause; you can use it as:
IF (something = true) then "something happens"
else "other thing happens"
Some example of it can be, suppose that your program reads a number X that the user inputs, then you can do:
If ( X > 5) then
print: "the number X is bigger than five"
Else
print: "the number X is smaller than five"
Where, of course, the statements depend on the language used, but the "if" clause works almost the same in every language.
A 75-hp compressor in a facility that operates at full load for 2500 h a year is powered by an electric motor that has an efficiency of 93 percent. If the unit cost of electricity is $0.11/kWh, the annual electricity cost of this compressor is
The annual electricity cost of the compressor is $15,386.25.
Explanation:To calculate the annual electricity cost of the compressor, we first need to find the energy consumption of the motor. We can use the formula:
Energy Consumption = Power x Time
Given that the compressor operates at full load for 2500 hours a year and has a power of 75 hp, we need to convert the power to watts:
1 hp = 746 watt
So, the power of the motor is 75 x 746 = 55,950 watts.
Now, we can calculate the energy consumption:
Energy Consumption = 55,950 watts x 2500 hours
Next, we need to convert the energy consumption to kilowatt-hours (kWh):
1 kWh = 1000 watt-hours
So, the energy consumption in kWh is 55,950 watts x 2500 hours / 1000 = 139,875 kWh.
Finally, we can calculate the annual electricity cost by multiplying the energy consumption by the unit cost of electricity:
Annual Electricity Cost = 139,875 kWh x $0.11/kWh
Therefore, the annual electricity cost of this compressor is $15,386.25.
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Suppose we placed a positive charge Q on the Moon and an equal positive charge Q on the Earth. What value of Q would be needed to neutralize the gravitational attraction of the Moon and the Earth?
Answer: q=5.70 x 10^13 C
Explanation:
gravitational attraction = electrostatic repulsion GMm/d^2 = kQ^2/d^2 as you can see the d^2 cancel out. that is why lunar distance is irrelevant. G is the universal gravitational constant = 6.67 x 10^-11 m^3 / kgs^2 M is earth's mass = 5.972 × 10^24 kg m is moon's mass = 7.342×10^22 kg Q is charge on earth and moon. k is coulomb's constant = 9 x10^9 N m^2 /C^2 On solving equation for Q. Q = sqrt (GMm/k) = sqrt ( 6.67 x 10^-11 x 5.972 x 10^24 * 7.342×10^22 / 9 x10^9) = 5.70 x 10^13 C
To neutralize the gravitational attraction between the Moon and the Earth using equal charges, calculate the total gravitational force and equate it to the electrostatic force of the charges. q=5.70 x 10¹³ C
gravitational attraction = electrostatic repulsion
GMm/d² = kQ²/d² as you can see the d² cancel out. that is why lunar distance is irrelevant. G is the universal gravitational constant = 6.67 x 10⁻¹¹ m³ / kgs² M is earth's mass = 5.972 × 10²⁴ kg m is moon's mass = 7.342×10²² kg Q is charge on earth and moon. k is coulomb's constant = 9 x10⁹ N m² /C² On solving equation for Q. Q = √ (GMm/k) = √ ( 6.67 x 10⁻¹¹x 5.972 x 10²⁴ × 7.342×10²² / 9 x10⁹) = 5.70 x 10¹³ C
An object with mass 0.900kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00m to the right (the x - direction) to stretch the spring, and released. What is the speed of the object when it is 0.50m to the right of the x
Answer:
7.85 m/s
Explanation:
We are given that
Mass of object=m=0.900 kg
[tex]F(x)=\alpha x-\beta x^2[/tex]
[tex]\alpha=60 N/m[/tex]
[tex]\beta=18N/m^2[/tex]
[tex]F(x)=-60x-18x^2[/tex]
U=0 when x=0
Potential energy=[tex]-\int F(x)dx[/tex]
Substitute the values
[tex]U(x)=-\int (-60x-18x^2)dx[/tex]
[tex]U(x)=60(\frac{x^2}{2})+18(\frac{x^3}{3})+C[/tex]
Using the formula
[tex]\int x^n dx=\frac{x^{n+1}}{n+1}+C[/tex]
Substitute x=0
[tex]U(0)=C\implies C=0[/tex]
[tex]U(x)=30x^2+6x^3[/tex]
[tex]x_1=0.5,x_2=1[/tex]
[tex]v_2=0[/tex]
Using law of conservation energy
[tex]\frac{1}{2}mv^2_1+U(x_1)=\frac{1}{2}mv^2_2+U(x_2)[/tex]
Substitute the values
[tex]\frac{1}{2}(0.9)v^2_1+30(0.5)^2+6(0.5)^3=0+30(1)^2+6(1)^3[/tex]
[tex]\frac{1}{2}(0.9)v^2_1+8.25=36[/tex]
[tex]\frac{1}{2}(0.9)v^2_1=36-8.25=27.75[/tex]
[tex]v^2_1=\frac{27.75\times 2}{0.9}[/tex]
[tex]v_1=\sqrt{\frac{27.75\times 2}{0.9}}[/tex]
[tex]v_1=7.85 m/s[/tex]
A small space telescope at the end of a tether line of length L moves at linear speed v about a central space station. What will be the linear speed of the telescope if the length of the line is changed to x*L ? x = 2.8; v = 2 m/s?
Answer:
v' = 0.714 m/s
Explanation:
Solution:
- Assuming no external torque is acting on the system then the angular momentum is conserved for the system.
- The initial momentum angular Mi and final angular momentum Mf are as follows:
Mi = Mf
m*L*v = m*x*L*v'
Where,
m : mass of the telescope
L : Length of teether line
v: Initial speed
v' : Changed speed.
- Then we have:
L*v = x*L*v'
v' = v / x
v' = 2 / 2.8
v' = 0.714 m/s
Answer:
The answer to the question is
The linear speed of the telescope will be 5.6 m/s if the length of the line is changed to x*L where x = 2.8; and initial velocity v = 2 m/s
Explanation:
Speed = v₁ = ωL = 2 m/s
When the line is changed to x*L where x = 2.8 the linear speed will be
v₂ = 2.8 × L× ω = 2.8× 2 = 5.6 m/s
The linear speed varies with the angular speed following the relation v/r =ω where
ω = angular speed
v = linear speed and
r = radius of the path of travel of the object at the vertex
In a cyclic process, a gas performs 123 J of work on its surroundings per cycle. What amount of heat, if any, transfers into or out of the gas per cycle?
123 J transfers out of the gas
123 J transfers into the gas
246 J transfers into the gas
0 J (no heat transfers)
Answer:
123 J transfer into the gas
Explanation:
Here we know that 123 J work is done by the gas on its surrounding
So here gas is doing work against external forces
Now for cyclic process we know that
[tex]\Delta U = 0[/tex]
so from 1st law of thermodynamics we have
[tex]dQ = W + \Delta U[/tex]
[tex]dQ = W[/tex]
so work done is same as the heat supplied to the system
So correct answer is
123 J transfer into the gas
If the volume of a sample of gas is reduced at constant temperature, the average velocity of the molecules _______, the average force of an individual collision _________, and the average number of collisions with the wall, per unit area, per second_______.
When the volume of gas is reduced at constant temperature, the average velocity of the molecules, the average force of an individual collision, and the average number of collisions with the wall all increase.
When the volume of a sample of gas is reduced at constant temperature, the average velocity of the molecules increases, the average force of an individual collision between molecules and the container walls increases, and the average number of collisions with the wall, per unit area, per second increases.
When the volume of a gas is reduced at constant temperature, the gas molecules’ average velocity remains unchanged, the force of each collision increases, and collisions with the wall per unit area per second become more frequent.
If the volume of a sample of gas is reduced at constant temperature, according to Boyle's law, the average velocity of the molecules remains the same, the average force of an individual collision increases, and the average number of collisions with the wall, per unit area, per second increases.
At a constant temperature, the kinetic energy of the gas molecules, as well as their average velocity, does not change. This is in accordance with the kinetic theory of gases. However, as the volume decreases, the gas molecules have less space to move around, which means they will encounter the container walls more frequently. Hence, the number of collisions with the wall per unit area per second increases. Since the same amount of molecules exert force over a smaller area of the container wall (because the volume is reduced), the average force per collision also increases.
It's important to clarify that while the average velocity does not change, the frequency of collisions leads to the increased pressure we observe in a confined volume of gas. Consequently, the gas exerts a greater pressure on the walls of the container.
Calculate the acceleration of the object from 3 seconds to 7 seconds (1pt). Show your work (1pt) and make sure to include the correct units (1pt)! Speed at 3 seconds - 0 m/s Speed at 7 seconds - 8 m/s
Final answer:
The acceleration of the object from 3 seconds to 7 seconds is 2 m/s².
Explanation:
To calculate the acceleration of the object, you can use the formula:
acceleration = (final velocity - initial velocity) / time
Using the given information, the final velocity is 8 m/s, the initial velocity is 0 m/s, and the time is 7 seconds - 3 seconds = 4 seconds.
Substituting the values into the formula:
acceleration = (8 m/s - 0 m/s) / 4 s = 2 m/s²
Therefore, the acceleration of the object from 3 seconds to 7 seconds is 2 m/s².
The acceleration of the object from 3 seconds to 7 seconds is 2 m/s².
To calculate the acceleration of the object from 3 seconds to 7 seconds, we use the formula:
Acceleration = (Final Velocity - Initial Velocity) / Time Interval
Given:
Initial velocity ([tex]V_{i}[/tex]) at 3 seconds = 0 m/sFinal velocity ([tex]V_{f}[/tex]) at 7 seconds = 8 m/sTime interval (Δt) = 7 seconds - 3 seconds = 4 secondsSo, the calculation is:
Acceleration = (8 m/s - 0 m/s) / 4 s = 2 m/s²
The acceleration of the object from 3 seconds to 7 seconds is 2 m/s².
Technician A says that a heavy engine load results in high intake manifold vacuum and a high MAP sensor signal voltage. Technician B says that a manifold absolute pressure (MAP) sensor uses a perfect vacuum (zero absolute pressure) in the sensor to determine the pressure. Who is right?
Answer:
Technician B
Explanation:
A 2.00-nF parallel-plate capacitor is charged to an initial potential
difference AV}: 100 V and is then isolated. The dielectric material
between the plates is mica, with a dielectric constant of 5.00. (a) How
much work is required to withdraw the mica sheet? (b) What is the
potential difference across the capacitor after the mica is withdrawn?
Answer:
The answer to the question is;
(a) The amount of work required to withdraw the mica sheet is 4.0×10⁻⁵ J
(b) The potential difference across the capacitor after the mica is withdrawn is 500 V.
Explanation:
To solve the question, we note that
The Capacitance of the capacitor is proportional to the dielectric constant, that is removal of the dielectric material will reduce the capacitance of the capacitor by a factor of the dielectric constant as follows
Energy stored in a capacitor is given by
[tex]E = \frac{1}{2} CV^{2}[/tex]
Where:
C = Capacitance of the capacitor = 2.00 nF
V = Voltage = 100 V
Therefore before the dielectric material is removed, we have
E = [tex]\frac{1}{2}[/tex]×2.00 × 10⁻⁹×100² = 1.0×10⁻⁵ J
Asfter the dielectric material is removed we have
E = 5× [tex]\frac{1}{2}[/tex]×2.00 × 10⁻⁹×100² = 5.0×10⁻⁵ J
Since energy is neither created nor destroyed, we have
Change in energy of system = 5.0×10⁻⁵ J - 1.0×10⁻⁵ J = 4.0×10⁻⁵ J
Work done on the system to remove the mica sheet = energy change of the system = 4.0×10⁻⁵ J.
(b) From the equation [tex]C = \frac{Q}{V}[/tex]
Where
Q = Charge of the capacitor
V= Voltage
C = Capacitance of the capacitor, we have
Q, the charge of the capacitor is constant
and the final capacitance = [tex]\frac{Initial .capacitance}{5} = \frac{C_1}{5}[/tex]
Therefore we have
V₁ = [tex]\frac{Q}{C_1}[/tex]
V₂ = [tex]\frac{Q}{\frac{C_1}{5} }[/tex] = [tex]{5} *\frac{Q}{C_1}[/tex] = 5·V₁
The velocity increases by a factor of 5 and
V₂ = 5×V₁ = 5 × 100 V = 500 V.
Tech A says motor action occurs through the interaction of the magnetic fields of the field coils and the armature, which causes a rotational force to act on the armature, creating the turning motion. Tech B says two magnetic fields are required for motor action: one in the casing and the other in the rotating armature. Who is correct
Answer:
Both A and B
Explanation:
The interaction of magnetic fields and armature results into a rotational force of the armature hence turning motion. It's important to note that you will always need two magnetic fields in order to experience the force since one magnetic field is at the rotating armature and another at the casing. Considering the arguments of these two technicians, both of them are correct in their arguments.
Two blocks with masses M1 and M2 hang one under the other. For this problem, take the positive direction to be upward, and use g for the magnitude of the free-fall acceleration. A. Find T2, the tension in the lower rope. B. Find T1, the tension in the upper rope.C. Find T2, the tension in the lower rope. D Find T1, the tension in the upper rope.
Answer:
A. [tex]T_2=M_2g[/tex]
B. [tex]T_2=(M_1+M_2)g[/tex]
Explanation:
Since the only forces acting on the blocks are the tensions and the weights (both in the vertical direction), and the system has acceleration zero, we can write the equilibrium equations for M₁ and M₂ as:
[tex]T_1-T_2-M_1g=0\\ \\T_2-M_2g=0[/tex]
From the second equation, we get:
[tex]T_2=M_2g[/tex],
Which is the answer to the part A.
Next, we substitute this result in the first equation and obtain:
[tex]T_1-M_2g-M_1g=0\\ \\T_1=(M_1+M_2)g[/tex],
Which is the answer to the part B.
Final answer:
To find the tensions in ropes supporting hanging blocks, one must draw free-body diagrams for each mass, and apply Newton's Second Law. The tension in the lower rope equals the weight of the lower mass or this weight adjusted for acceleration. The tension in the upper rope is the sum of the weight of the upper mass and the tension in the lower rope.
Explanation:
The student's question revolves around the concept of tension in ropes in a physics context, where two blocks are hanging one under the other. The physics involved here includes Newton's second law and the interplay of gravitational and tensional forces in a system. To find the tensions T₁ and T₂, we need to draw free-body diagrams for each block and apply Newton's Second Law, F = ma, where F is the net force, m is the mass and a is the acceleration of the block.
For block M₂, which hangs at the bottom, the tension T₂ is the force counteracting the gravitational pull downward. Hence, assuming the system is at equilibrium or moving at constant speed (acceleration = 0), T₂ would equal the weight of M₂, given by the formula T₂ = M₂ × g.
For block M₁, which is above M₂, the tension T₁ is slightly more complicated to calculate as it must support its own weight as well as the tension from the lower rope T₂. Therefore, the tension T₁ in the upper rope would be T₁= (M₁ + M₂) × g again assuming no net acceleration.
If the entire system is accelerating, we must incorporate the net force necessary to accelerate both masses. This changes the calculations respectively:
For the lower block (M₂), T₂ = M₂ × (g + a) if accelerating upward, or T₂ = M₂ × (g - a) if accelerating downward.
For the upper block (M₁), T₁ = (M₁ × g) + T₂ because T₁ has to support both the weight of M₁ and the tension from the lower rope.