The orbital speed of stars in a binary system can be calculated using Kepler's laws and principles of circular motion, given the combined mass and orbital period. In a binary system, both stars orbit around their common center of mass, and their speed indicates how quickly they complete their orbit.
Explanation:The problem involves calculating the orbital speed of two stars in a binary system. For this, we will apply the principles of Kepler's laws of planetary motion and principles of circular motion. We are given that the two stars M1 and M2 are of equal mass and their total mass is 6.95 solar masses.
The semi-major axis of the orbit, 'D', can be found using the formula D³ = (M₁ + M₂)P². Plugging in the given mass and orbital period, we can find D. Then, we can calculate the speed, 'v', based on the principle of circular motion: v = 2πD / P.
An important point in a binary star system is that both stars move around their common center of mass. In case of equal masses, it is at the exact midpoint of the line that separates them. Remember that the orbital speed of the stars is a measure of how fast they travel in their orbit, completing it in the given period of 2.20 days.
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The total magnification of a specimen being viewed with a 10X ocular lens and a 40X objective lens is _____.
Answer:
total magnification = 400 X
Explanation:
given data
ocular lens = 10 X
objective lens = 40 X
to find out
total magnification
solution
we know that total magnification is express as
total magnification = Objective magnification × ocular magnification .................1
put here value we get
total magnification = 10 × 40
total magnification = 400 X
The total magnification for a microscope with a 10X ocular lens and a 40X objective lens is 400X.
Explanation:When using a microscope, the total magnification can be found by multiplying the magnification of the ocular lens (also known as the eyepiece) by the magnification of the objective lens. Thus, in your case, the total magnification of the specimen being viewed with a 10X ocular lens and a 40X objective lens would be 400X. This is because 10 times 40 equals 400. Therefore, you are observing the specimen at 400 times its actual size.
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Two balls collide elastically the balls have masses of 3.4 and 5.7, if the 3.4 ball is moving at 10.5 m/s toward the other ball and the 5.7 kg ball is moving with a speed of 6.7 what are the final speeds of the balls?
Answer: The final speed of the balls will be 8.12m/s
Explanation:
For least collision of bodies, both momentum and kinetic energy is conserved.
According to Law of conservation of energy,
The sum of momentum of the objects before collision is equal to the sum of their momentum after collision. Note that the objects will move with a common velocity after impact.
Let m1 and m2 be the masses of the objects
u1 and u2 be their velocities before impact
v be their common velocity after impact
Since momentum is mass × velocity
Mathematically,
m1u1 + m2u2 = (m1+m2)v
Given m1= 3.4kg m2 = 5.7kg u1 = 10.5m/s u2 = 6.7m/s v = ?
Substituting this values in the formula, we have
3.4(10.5)+5.7(6.7) = (3.4+5.7)v
35.7+38.19 = 9.1v
73.89 = 9.1v
V = 73.89/9.1
V= 8.12m/s
As a cat pounces on a mouse, its muscles consume 10 units of potential energy (which the cat previously gained from the food it consumed). However, the pounce itself only required 4 units of kinetic energy. How many units of energy were dissipated as heat?
The cat used 10 units of potential energy to pounce but only 4 units were converted meaningfully into kinetic energy of the pounce. The remaining 6 units were dissipated as heat.
Explanation:When the cat pounces on a mouse, the energy needed for this action is less than the total potential energy consumed. The cat's body uses 10 units of potential energy, but the pounce only required 4 units of kinetic energy. The difference between these two values signifies the amount of energy dissipated as heat. Therefore, we can perform the subtraction: 10 units (total potential energy) - 4 units (kinetic energy used) = 6 units. Thus, "6 units of energy were dissipated as heat."
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When two charge producers with different surface materials are rubbed together to create a charge imbalance. True or False
Answer:
True
Explanation:
Answer:
The answer is TRUE
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"The work done on an ideal gas system in an isothermal process is -400 J. What is the change in internal (thermal) energy of the gas?"
Answer:
zero
Explanation:
Work done = - 400 J
In an isothermal process, the temperature remains constant. So, according to the first law of thermodynamics
dQ = dU + dW
As the temperature remains constant and the change in internal energy is the function of change in temperature, here change is temperature is zero so the change in internal energy is also zero.
Final answer:
The change in internal energy of an ideal gas during an isothermal process is 0 J. This is because in an isothermal process for an ideal gas, any work done by the gas is invariably compensated by an equivalent amount of heat absorbed, resulting in no net change in internal energy.
Explanation:
The question pertains to the change in internal energy of an ideal gas during an isothermal process in the context of thermodynamics, specifically according to the first law of thermodynamics. The first law states that the change in internal energy of a system (Triangle U) is equal to the heat added to the system (Q) minus the work done by the system (W). In an isothermal process, the temperature remains constant, and if the process is performed on an ideal gas, the internal energy also remains constant since it depends only on temperature for an ideal gas. Therefore, the change in internal energy ( triangle U) is zero.
In this specific scenario where the work done on the system is -400 J (-W), which means that the system has done 400 J of work on the surroundings. Since no heat transfer information is given, we can infer that in an isothermal process, whatever work is done by the gas is compensated by the heat received from the surroundings, keeping the internal energy unchanged. Thus, the change in internal energy of the gas is 0 J, as the energy spent as work would have been absorbed as heat from the surroundings.
What is the magnitude of the net force ona(n) 97 kg driver operating a dragster as it accelerates horizontally along a straight line from rest to 40 m/s in 5 s?
Answer:
Force acting on the driver will be 776 N
Explanation:
We have given mass of the driver m = 97 kg
It starts from rest so initial velocity u = 0 m /sec
And reaches to a velocity of 40 m /sec in 5 sec
So final velocity v = 40 m /sec
And time taken t = 5 sec
From first equation of motion v = u +at
So [tex]40=0+a\times 5[/tex]
[tex]a=8m/sec^2[/tex]
Now we have to find the force acting the driver
From newtons law we know that F = ma
So force F = 97×8 = 776 N
So force acting on the driver will be 776 N
A truck is hoisted a certain distance in a garage and therefore has potential energy with respect to the floor. If it were lifted twice as high, it would have ____ as much potential energy.
Answer:
If the truck were lifted twice as high, it would have twice as much potential energy
Explanation:
Potential energy = mgh
where;
m is mass of the truck (kg)
g is acceleration due to gravity (m/s²)
h is the height above the floor.
Initial Potential Energy, E₁ = mgh₁
When the truck was lifted twice as high, New height (h₂) = 2h₁
New potential Energy, E₂ =mgh₂ = mg(2h₁)
E₂ = 2mgh₁
Recall, E₁ = mgh₁, then substitute in E₁ into E₂
E₂ = 2(mgh₁)
E₂ = 2E₁
Therefore, If the truck were lifted twice as high, it would have twice as much potential energy
how do you describe and determine the direction of the magnetic field produced by an electric current?
Answer:
the magnet or electric current produces a magnetic field
Explanation:
this magnetic field can be visualized as a pattern of a circular field lines surrounding a wire. hall probes can determine the magnitude of the field
hope it helps you
The direction of a magnetic field produced by an electric current in a straight wire can be determined using the right-hand rule. The Biot-Savart law is used to calculate the magnetic field strength at a distance from the wire.
Explanation:To describe and determine the direction of the magnetic field produced by an electric current, one commonly uses the right-hand rule. For a straight, current-carrying wire, point your right-hand thumb in the direction of the current; then, the direction in which your fingers curl around the wire represents the direction of the magnetic field lines, which form concentric circles around the wire.
For example, if the current flows from right to left in a wire, and you are looking at the wire end-on from the left end, your thumb would point to the left, indicating the direction of the current. Your fingers would naturally curl in a counterclockwise direction, suggesting the magnetic field direction at that point.
For calculating the magnetic field strength, B, produced by a long straight wire carrying current I, the Biot-Savart law can be used. It states that the strength of the magnetic field is directly proportional to the magnitude of the current and inversely proportional to the distance from the wire (assuming the wire is very long compared to the distance).
Charge is flowing through a conductor at the rate of 420 C/min. If 742 J. of electrical energy are converted to heat in 30 s., what is the potential drop across the conductor?
Answer:
3.53 V
Explanation:
Electric charge: The is the rate of flow of electric charge along a conductor.
The S.I unit of electric charge is C.
Mathematically it is expressed as,
Q = It ............................ Equation 1
Where Q = electric charge, I = current, t = time.
I = Q/t.......................... Equation 2
From the question, charge flows through the conductor at the rate of 420 C/mim
Which means in 1 min, 420 C of charge flows through the conductor.
Hence,
Q = 420 C, t = 1 min = 60 seconds
Substitute into equation 2
I = 420/60
I =7 A
Also
P = VI......................... Equation 3
Where P = power, V = potential drop, I = current.
V = P/I................... Equation 4
Note: Power = Energy/time
From the question, P = 742/30 = 24.733 W. and I = 7 A.
Substitute these values into equation 4
V = 24.733/7
V = 3.53 V
Hence the potential drop across the conductor = 3.53 V
The potential drop across the conductor is 3.54 V
The rate of flow of electric charge along a conductor per unit time is known as current.
Mathematically it is expressed as,
As we know that
Power, [tex]P = VI[/tex]
Where P is power, V is potential drop and I is current.
Also,
It is given that, energy = 742 J and time = 30s
Power [tex]=\frac{742}{30} =24.74watt[/tex]
Substituting the value of power in equation [tex]P=VI[/tex]
[tex]V=\frac{P}{I} =\frac{24.74}{7}=3.54V[/tex]
Thus, the potential drop across the conductor is 3.54 V
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A certain compact disc (CD) contains 783.216 megabytes of digital information. Each byte consists of exactly 8 bits. When played, a CD player reads the CD's information at a constant rate of 1.1 megabits per second. How many minutes does it take the player to read the entire CD?
Answer:
Time takes the player to read the entire CD is 94.9 minutes.
Explanation:
compact disc (CD) contains 783.216 megabytes of digital information.
Each byte consists of exactly 8 bits.
constant rate of 1.1 megabits per second
How many minutes does it take the player to read the entire CD?
1 Megabites = 1000000 bites
783.216 megabytes = 783216000 bites = 783216000 x 8 = 6265728000 bits
1.1 megabits/s = 1100000 bits/s
thus
t = 6265728000 bits/1100000 bits/s
= 5696 s
= 94.9 min
A cannon shoots a ball at an angle θ above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find the ball's position as a function of time. (Use axes with x measured horizontally and y vertically.) (b) Let r(t) denote the ball's distance from the cannon. What is the largest possible value of θ if r(t) is to increase throughout the ball's flight?
The question involves the application of Newton's law to a ball's position over time, launched from a cannon at an angle. The x and y-axis positions can be calculated using horizontal and vertical principles of motion respectively. For the ball's distance (r) to increase throughout the flight, the launch angle needs to be 90°.
Explanation:The subject of this question relates to the application of Newton's second law in two dimensions, specifically in cases of projectile motion. The position of the ball fired from a cannon is determined by the initial velocity of the ball, the launch angle θ, and the acceleration due to gravity.
The x-component of the ball's position can be calculated using horizontal motion principles. Here, the x position x(t) = V₀cos(θ)t, where V₀ is the initial speed of the ball, cos(θ) is the cosine of the launch angle and t is the time.The y-component of the ball's position can be calculated using vertical motion principles. In this case, y(t) = V₀sin(θ)t - 0.5gt², where sin(θ) is the sine of the launch angle, g is the acceleration due to gravity, and t is the time. For r(t) to increase throughout the ball's flight, the launch angle θ needs to be 90°. This is because the maximum range or distance is attained when objects are projected at an angle of 45°. But the ball's range will start to decrease after this angle, while the height (r(t)) will continue to increase until 90°.Learn more about Projectile Motion here:
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What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and approaching it?
Answer:
The frequency is 302.05 Hz.
Explanation:
Given that,
Speed = 18.0 m/s
Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .
We need to calculate the frequency
Using formula of frequency
[tex]f'=f(\dfrac{v+v_{p}}{v-v_{s}})[/tex]
Where, f = frequency
v = speed of sound
[tex]v_{p}[/tex] = speed of passenger
[tex]v_{s}[/tex] = speed of source
Put the value into the formula
[tex]f'=262\times(\dfrac{344+18}{344-30})[/tex]
[tex]f'=302.05\ Hz[/tex]
Hence, The frequency is 302.05 Hz.
A Sense of Proportion: The Sun is roughly 100 times the diameter of Earth. If you built a model solar system with a golf ball (diameter of 1.68 inches) for Earth, how big would the model Sun be?
Explanation:
Diameter of Sun = 100 x Diameter of Earth.
We are creating a model where Earth is same as golf ball,
Diameter of Earth in model = Diameter of golf ball
Diameter of Earth in model = 1.68 inches
Diameter of Sun in model = 100 x Diameter of Earth in model
Diameter of Sun in model = 100 x 1.68
Diameter of Sun in model = 168 inches = 14 feet
Diameter of Sun in model = 168 inches = 14 feet = 4.27 m
How long (in s) will it take an 875 kg car with a useful power output of 42.0 hp (1 hp = 746 W) to reach a speed of 17.0 m/s, neglecting friction? (Assume the car starts from rest.)
Answer:
The time taken by the car with a useful power output is 4.03 seconds.
Explanation:
Given that,
Mass of the car, m = 875 kg
Initial speed of the car, u = 0
Final speed of the car, v = 17 m/s
Power of the car, P = 42 hp = 31332 W
The power output of the car is given by the work done per unit time. It is given by :
[tex]P=\dfrac{W}{t}[/tex]
Initial kinetic energy of the car will be 0 as it is at rest.
Final kinetic energy of the car is :
[tex]K=\dfrac{1}{2}mv^2[/tex]
[tex]K=\dfrac{1}{2}\times 875\times (17)^2[/tex]
K = 126437.5
As per the work energy theorem, the change in kinetic energy of the object is equal to the work done.
So,
[tex]P=\dfrac{W}{t}[/tex]
[tex]t=\dfrac{W}{P}[/tex]
[tex]t=\dfrac{126437.5}{31332}[/tex]
t = 4.03 seconds
So, the time taken by the car with a useful power output is 4.03 seconds. Hence, this is the required solution.
For every factor of 10 difference in Keq' (e.g., going from 102 to 10, or from 10–3 to 10–4), what is the difference in standard free energy change, ΔG°' (kJ/mol)?
Answer:
The standard free energy is the difference between that of products and reactants. The value of G has a greater influence on the reaction been considered.
Explanation:
The Gibb's free energy also referred to as the gibb's function represented with letter G. it is the amount of useful work obtained from a system at constant temperature and pressure. The standard gibb's free energy on the other hand is a state function represented as Delta-G, as it depends on the initial and final states of the system. The spontaneity of a reaction is explained by the standard gibb's free energy.
If Delta-G = -ve ( the reaction is spontaneous)if Delta -G = +ve ( the reaction is non-spontaneous)if Delta-G = 0 ( the reaction is at equilibrium)Use this hints for any reaction involving the Gibb's free, Enthalpy and entropy.
For each factor of 10 change in Keq, there is a corresponding change of approximately 5.7 kJ/mol in standard free energy change, ΔG°', at standard biochemical conditions (298.15 K).
Explanation:For every factor of 10 difference in the equilibrium constant, Keq, there is a corresponding change in the standard free energy change, ΔG°', as calculated using the relationship between ΔG°' and Keq:
ΔG°' = -RT ln Keq, where R is the universal gas constant and T is the temperature in Kelvin.
Given that R = 8.314 J/mol·K and at standard biochemical conditions T = 298.15 K, a tenfold change in Keq results in a change of (8.314 J/mol·K x 298.15 K x ln(10)), which equates to approximately 5.7 kJ/mol. Therefore, for example, increasing Keq from 102 to 103 would decrease ΔG°' by 5.7 kJ/mol, making the reaction more product-favored under standard conditions. Conversely, decreasing Keq from 10-3 to 10-4 would increase ΔG°' by 5.7 kJ/mol, making the reaction less product-favored under standard conditions.
If a ball end mill cutter's stepover is kept at a fixed value but the diameter of the tool is increased, the scallop height increases as a consequence.Select one:a. Trueb. False
Answer: FALSE
Explanation:Ball End Mills are hemispherical tip used to cut and and shape/ milling rounded objects, such as the metal bearing grooves, slotting and pocketing.
Ball End Mill also called Ball nose end mills, the scallop height will not increase as a result of the increase in diameter. The diameter is the distance between one end of the round tool to another when taken from the middle.
You throw a ball horizontally from the top of a building, with a speed of 3 m/s. In this problem, you can neglect the force of air resistance.
After 2 Seconds, what is the balls horizontal velocity?
A. Impossible to tell
B.Less than 3 m/s
C. 3 m/s
D. 0
E.More than 3 m/s
Answer:
Option C. 3 m/s
Explanation:
From the question it is stated that we can neglect the force of air resistance, therefore there is no force acting on the ball horizontally. From Newton's first law of motion an object will remain at rest or in uniform motion unless acted upon by an external force. thus the ball velocity will remain 3m/s.
A 45.0 g hard-boiled egg moves on the end of a spring with force constant 25.0 N/m. Its initial displacement 0.500 m. A damping force Fx= −bvx acts on the egg, and the amplitude of the motion decreases to 0.100 m in a time of 5.00 s. Calculate the magnitude of the damping constant b. Express the magnitude of the damping coefficient numerically in kilograms per second, to three significant figures.
Answer:
0.02896 kg/s
Explanation:
[tex]A_1[/tex] = Initial displacement = 0.5 m
[tex]A21[/tex] = Final displacement = 0.1 m
t = Time taken = 0.5 s
m = Mass of object = 45 g
Displacement is given by
[tex]x=Ae^{-\dfrac{b}{2m}t}cos(\omega t+\phi)[/tex]
At maximum displacement
[tex]cos(\omega t+\phi)=1[/tex]
[tex]\\\Rightarrow A_2=A_1e^{-\dfrac{b}{2m}t}\\\Rightarrow \dfrac{A_1}{A_2}=e^{\dfrac{b}{2m}t}\\\Rightarrow ln\dfrac{A_1}{A_2}=\dfrac{b}{2m}t\\\Rightarrow b=\dfrac{2m}{t}\times ln\dfrac{A_1}{A_2}\\\Rightarrow b=\dfrac{2\times 0.045}{5}\times ln\dfrac{0.5}{0.1}\\\Rightarrow b=0.02896\ kg/s[/tex]
The magnitude of the damping coefficient is 0.02896 kg/s
Final answer:
To find the damping constant b for a damped harmonic motion scenario with a given change in amplitude over time, you employ the decay of amplitude formula in damped harmonic motion, then rearrange and solve for b.
Explanation:
To calculate the magnitude of the damping constant b for a 45.0 g hard-boiled egg moving at the end of a spring with a force constant of 25.0 N/m given a decrease in amplitude from 0.500 m to 0.100 m over 5.00 seconds, we apply the principle of damped harmonic motion. The damping force is given by Fx= −bvx, where b is the damping coefficient we wish to find. From the information provided, it's understood that this is a case of exponentially decaying amplitude in harmonic motion.
The equation that relates the exponential decay of amplitude in damped harmonic motion is A(t) = A0e−(bt/2m), where A0 is the initial amplitude, A(t) is the amplitude at time t, and m is the mass of the object. By re-arranging this equation to solve for b, and substituting the given values A0=0.500 m, A(t)=0.100 m, t=5.00 s, and m=0.045 kg (since 45.0 g=0.045 kg), we can calculate b.
The rearranged equation for b would be b = 2m[ln(A0/A(t))]/t. Substituting the values gives b = 2(0.045)[ln(0.500/0.100)]/5.00, which after calculation yields the magnitude of b.
A very light cotton tape is wrapped around the outside surface of a uniform cylinder of mass M and radius R. The free end of the tape is attached to the ceiling. The cylinder is released form rest and as it descends it unravels from the tape without slipping. The moment of inertia of the cylinder about its center is I = 1/2 MR2
a. On the diagram above show all the forces applied on the cylinder. b. Find the acceleration of the center of the cylinder when it moves down. c. Find the tension force in the tape.
Answer:
Explanation:
Check attachment for answer.
The acceleration of the center of the cylinder and the tension force in the tape is mathematically given as
a=2g/3
T=ug/3
What are the forces applied on the cylinder, the acceleration of the center of the cylinder when it moves down, and the tension force in the tape?Generally, the equation for the Liner motion is mathematically given as
Mg-T=Ma
Therefore
Mg.R=(0.5MR^2+MR^2)\alpha
[tex]\alpha=2g/3R[/tex]
b)
Where
a=\alpha R
Hence
a=2g/3
c)
For the tension T
ug-T=Ma
T=u(g-a)
T=ug/3
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The ability of water molecules to form hydrogen bonds with other water molecules and water's ability to dissolve substances that have charges or partial charges are __________.
Answer:
The ability of water molecules to form hydrogen bonds with other water molecules and water's ability to dissolve substances that have charges or partial charges are due to water's partial charges.
Explanation:
The partial negative charge on oxygen and partial positive charge on hydrogen enables them to make hydrogen bond and also makes it to dissolve the the other substances having partial charges.
During heavy rain, a section of a mountainside measuring 2.3 km horizontally (perpendicular to the slope), 0.75 km up along the slope, and 1.4 m deep slips into a valley in a mud slide. Assume that the mud ends up uniformly distributed over a surface area of the valley measuring 0.52 km x 0.52 km and that the mass of a cubic meter of mud is 1900 kg. What is the mass of the mud sitting above a 5.5 m2 area of the valley floor?
Answer:
93328 kg
Explanation:
[tex]\rho[/tex] = Density of mud = 1900 kg/m³
Volume of the cuboid is given by
[tex]V=2300\times 750\times 1.4\\\Rightarrow V=2415000\ m^3[/tex]
Volume is also given by
[tex]V=Al\\\Rightarrow l=\dfrac{V}{A}\\\Rightarrow l=\dfrac{2415000}{520\times 520}\\\Rightarrow l=8.9312\ m[/tex]
The length of the cuboid is 0.89312 m
For the small volume
[tex]V_a=al\\\Rightarrow V_a=5.5\times 8.9312\\\Rightarrow V_a=49.12\ m^3[/tex]
Mass is given by
[tex]m=\rho V_a\\\Rightarrow m=1900\times 49.12\\\Rightarrow m=93328\ kg[/tex]
The mass of the mud is 93328 kg
____ is the process of sending data over a signal by varying either its amplitude, frequency, or phase.
Answer:
Modulation
Explanation:
Modulation is the the process whereby a waveform has one or more of it's properties varied. Amplitude, frequency and phase are all properties of a waveform which can be varied hence Modulation is the process of sending data over a signal by varying either its amplitude, frequency or phase.
Assume that you land at an airport with the altimeter set to 29.92 instead of the current setting of 30.00. What will the altimeter read if the field elevation is 2,000 feet MSL?
When the altimeter is set to 29.92 inHg, it will read 4,000 feet MSL at a field elevation of 2,000 feet MSL.
To determine the altimeter reading when the field elevation is 2,000 feet MSL (Mean Sea Level) and the altimeter is set to 29.92 inches of mercury (inHg), we need to consider the difference in pressure settings.
The altimeter measures altitude based on the atmospheric pressure. When the altimeter setting (also known as the barometric pressure setting) is changed, it adjusts the reference point for sea level pressure. The standard atmospheric pressure at sea level is approximately 29.92 inHg.
Given:
Field elevation = 2,000 feet MSL
Altimeter setting = 29.92 inHg
We need to find the altimeter reading when the altimeter is set to 29.92 inHg.
To calculate the altimeter reading, we need to determine the difference in pressure between the current pressure (altimeter setting) and the pressure at the field elevation (2,000 feet above sea level).
The standard atmospheric pressure decreases as we go higher in altitude. A typical lapse rate for pressure is around 1 inHg per 1,000 feet of altitude gain. Since the field elevation is 2,000 feet above sea level, the difference in pressure would be approximately 2 inHg.
Therefore, the corrected altimeter reading would be 2,000 feet + 2,000 feet (for the 2 inHg difference) = 4,000 feet MSL.
Hence, when the altimeter is set to 29.92 inHg, it will read 4,000 feet MSL at a field elevation of 2,000 feet MSL.
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Radiometric dating of a magnetic anomaly stripe of rock that is 225 km away from the mid-ocean ridge axis gives an age of 9 million years. Assuming a constant rate, seafloor spreading in this area occurs at a rate of __________.
A. 5 cm per year
B. 1,012.5 km per year
C. 20,000 cm per year
D. 50 km per year
The rate of seafloor spreading in this area, based on the given radiometric dating results and calculated assuming a constant rate, is 25 km/million years.
Explanation:
To calculate the rate of seafloor spreading, we divide the distance that the seafloor has spread by the amount of time it took. Given the radiometric dating results, we know that this distance (225 km) was covered in 9 million years.
So, the calculation would be: 225 km / 9 million years = 0.025 km/year. Or, to convert this figure to kilometers per million years, you'd multiply by 1 million, giving you a seafloor spreading rate of 25 km/million years.
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A thread holds a 1.5-kg and a 4.50-kg cart together. After the thread is burned, a compressed spring pushes the carts apart, giving the 1.5 kg cart a speed of 26 cm/s to the left. What is the velocity of the 4.5-kg cart? a. 10 cm/s b. 4.56 cm/s c. 8.67 cm/s d. 7.3 cm/s
Answer:
c. 8.67 cm/s
Explanation:
From the law of conservation of momentum,
Total momentum before the thread was burned = Total momentum after was burned
mu + m'u' = mv + m'v'...................... Equation
Where m = mass of the heavier cart, m' = mass of the lighter cart, u = initial velocity of the bigger cart, u' = initial velocity of the smaller cart, v = final velocity of the bigger cart, v' = final velocity of the smaller cart.
Note: Both cart where momentarily at rest, as such u = u' = 0. i.e the total momentum before the thread was burn = 0
And assuming the left is positive,
We can rewrite equation 1 as
mv + m'v' = 0............................................ Equation 2
Given: m = 4.5 kg, m' = 1.5 kg, v' = 26 cm/s
Substitute into equation 2,
4.5v + 1.5(26) = 0
4.5v + 39 = 0
4.5v = -39
v = -39/4.5
v = -8.67 cm/s.
Note: v is negative because it moves to right.
Hence the velocity of the 4.5 kg cart = 8.67 cm/s.
The right option is c. 8.67 cm/s
The velocity of the 4.5 kg cart after the thread is burned is 10 cm/s.
Explanation:When the thread is burned, the compressed spring pushes the carts apart. Since the 1.5 kg cart moves to the left with a speed of 26 cm/s, the 4.5 kg cart would also move to the left but with a slower speed. To determine the velocity of the 4.5 kg cart, we can use the principle of conservation of momentum:
Momentum before = Momentum after
(Mass of 1.5 kg cart) x (Initial velocity) + (Mass of 4.5 kg cart) x (Initial velocity) = (Mass of 1.5 kg cart + Mass of 4.5 kg cart) x (Final velocity)
Solving for the final velocity of the 4.5 kg cart:
(1.5 kg x 26 cm/s + 4.5 kg x 0 cm/s) / (1.5 kg + 4.5 kg) = (6.9 kg) x (Final velocity)
Final velocity = 10 cm/s
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Two canoes are touching and at rest on a lake. The occupants push away from each other in opposite directions, giving canoe 1 a speed of 0.56 m/s and canoe 2 a speed of 0.45 m/s .
If the mass of canoe 1 is 320 kg , what is the mass of canoe 2?
Answer:
398.22 kg
Explanation:
[tex]m_1[/tex] = Mass of canoe 1 = 320 kg
[tex]m_2[/tex] = Mass of canoe 2
[tex]v_1[/tex] = Velocity of canoe 1 = 0.56 m/s
[tex]v_2[/tex] = Velocity of canoe 2 = 0.45 m/s
In this system the linear momentum is conserved
[tex]m_1v_1=m_2v_2\\\Rightarrow m_2=\dfrac{m_1v_1}{v_2}\\\Rightarrow m_2=\dfrac{320\times 0.56}{0.45}\\\Rightarrow m_2=398.22\ kg[/tex]
The mass of the second canoe is 398.22 kg
Answer:
398.22 kg
Explanation:
mass of I canoe, m1 = 320 kg
initial velocity of both the canoe = 0
final velocity of I canoe, v1 = 0.56 m/s
final velocity of II canoe, v2 = - 0.45 m/s
Let the mass of second canoe is m2.
Use conservation of momentum
momentum before push = momentum after push
0 + 0 = m1 x v1 + m2 x v2
0 = 320 x 0.56 - m2 x 0.45
m2 = 398.22 kg
Thus, the mass of second canoe is 398.22 kg.
A bakery measured the mass of a whole cake as 0.870 kg. A customer bought one slice of the cake and measured the mass of the slice as 0.1151 kg. What is the mass of the remaining cake, without the one slice?
Answer:
0.7549kg
Explanation:
The mass of the slice + mass of the remaining cake = total mass of cake.
mass of remaining cake = total mass of cake - the mass of the slice
total mass=0.870kg
mass of slice = 0.1151kg
mass of remaining cake = 0.870 - 0.1151
mass of remaining cake=0.7549kg
At time t in seconds a particles distance s(t) in micrometers from a point is given by s(t)=e^t+2-sin(t) what is the average velocity of the particle from t=1 to t=5
Answer:
[tex]v_{avg}=29.499\ m.s^{-1}[/tex]
Explanation:
Given:
distance as a function of time is:
[tex]s=e^t+2-sin\ t[/tex]
Average velocity of the particle from time t=1s to t=5s can be given by the total distance covered divided by the total time consumed.
Position at t=1s:
[tex]s_1=e^1+2-sin\ 1[/tex]
[tex]s_1=3.8768\ m[/tex]
Position at t=5s:
[tex]s_5=e^5+2-sin\ 5[/tex]
[tex]s_5=151.3721\ m[/tex]
Therefore the distance covered during this time is:
[tex]\Delta s=s_5-s_1[/tex]
[tex]\Delta s=151.3721-3.8768[/tex]
[tex]\Delta s =147.4952\ m[/tex]
Now the average speed for the duration:
[tex]v_{avg}=\frac{\Delta s}{\Delta t}[/tex]
[tex]v_{avg}=\frac{147.4952}{5}[/tex]
[tex]v_{avg}=29.499\ m.s^{-1}[/tex]
Radio Station: What is the wavelength of a radio station photon from an AM radio station that broadcast at 1000 kilo-hertz? Remember that λ = c/f, where λ is the wavelength, f is the frequency and c is the speed of light (c = 300000 km/s)
Answer: The wavelength of a radio station photon from an AM radio station that broadcast at 1000 kilo-hertz is 3000 m.
Explanation:
To calculate the wavelength of light, we use the equation:
[tex]\lambda=\frac{c}{\nu}[/tex]
where,
= wavelength of the light = ?
c = speed of light = 300000 km/s = [tex]3\times 10^8m/s[/tex] (1km=1000m)
[tex]\nu[/tex] = frequency of light = 100kHz = 100000Hz or (1kHz=1000Hz)
[tex]\lambda=\frac{3\times 10^8m/s}{100000s^{-1}}[/tex]
[tex]\lambda=3000m[/tex]
Thus the wavelength of a radio station photon from an AM radio station that broadcast at 1000 kilo-hertz is 3000 m.
A 1.25 in. by 3 in. rectangular steel bar is used as a diagonal tension member in a bridge truss. the diagonal member is 20 ft long, and its modulus of elasticity is 30,000,000 psi. if the strain in the diagonal member is measured as 0.001200 in./in., determine:
Answer:
axial stress in the diagonal bar =36,000 psi
Explanation:
Assuming we have to find axial stress
Given:
width of steel bar: 1.25 in.
height of the steel bar: 3 in
Length of the diagonal member = 20ft
modulus of elasticity E= 30,000,000 psi
strain in the diagonal member ε = 0.001200 in/in
Therefore, axial stress in the diagonal bar σ = E×ε
= 30,000,000 psi× 0.001200 in/in =36,000 psi
The stress in the bridge truss diagonal tension member is computed to be 36000 psi, and the elongation of the member under this stress is calculated to be approximately 0.384 inches.
Explanation:The subject of the question is in the field of Engineering, particularly dealing with the stress and strain effects in a steel structural member of a bridge.
The stress in the member can be calculated using stress and strain relationship along with Young's modulus, given by the formula Stress = Strain x Young's Modulus. Given that strain is 0.001200 in./in. and Young's Modulus (E) is 30,000,000 psi, stress (σ) in the member is σ = E x Strain = 30,000,000 psi x 0.001200 in./in. = 36000 psi.
The steel bar's cross-sectional area can also be calculated as Area = Width x Depth = 1.25 in. x 3 in. = 3.75 in^2. The change in length (∆L) of the member under this stress can be calculated using the formula ∆L = (Stress x Original length) / (Young's Modulus x Area) = (36000 psi x 20 ft) / (30,000,000 psi x 3.75 in^2) = 0.032 ft or 0.384 in.
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