Answer: entropy
Explanation: entropy is the degree of disorderliness or randomness in a system
Tracy (of mass 49 kg) and Tom (of mass 77 kg) are standing at rest in the center of the roller rink, facing each other, free to move. Tracy pushes off Tom with her hands and remains in contact with Tom’s hands, applying a constant force for 0.8 s. Tracy moves 0.4 m during this time. When she stops pushing off Tom, she moves at a constant speed. What is Tracy’s constant acceleration during her time of contact with Tom? Answer in units of m/s 2 .
Answer:
Tracy’s constant acceleration during her time of contact with Tom is 1.25 m/s².
Explanation:
Given that,
Mass of Tracy = 49 kg
Mass of Tom = 77 kg
Distance = 0.4 m
Time = 0.8 s
We need to calculate Tracy’s constant acceleration during her time of contact with Tom
Using equation of motion
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
Where, s = distance
a = acceleration
t = time
Put the value into the formula
[tex]0.4=0+\dfrac{1}{2}a\times(0.8)^2[/tex]
[tex]a=\dfrac{0.4\times2}{(0.8)^2}[/tex]
[tex]a=1.25\ m/s^2[/tex]
Hence, Tracy’s constant acceleration during her time of contact with Tom is 1.25 m/s².
Tracy’s constant acceleration during her time of contact with Tom is 1.25 m/s².
Calculation of the acceleration:
Here we applied the motion equation i.e. shown below:
[tex]S = ut + \frac{1}{2}at^2[/tex]
Here s = distance
a = acceleration
t = time
So,
[tex]0.4 = \frac{1}{2} a \times (0.8)^2\\\\a = \frac{0.4\times 2}{(0.8)^2}[/tex]
= 1.25 m/s²
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) Titan’s Evolving Atmosphere: Titan’s exopshere lies nearly 1400 km above its surface. What is the escape velocity from this altitude? What is the thermal speed of a hydrogen atom at the exospheric temperature of about 200 K?
Answer:
Explanation:
first we calculate the escape velocity using [tex]V_{esc} =\sqrt{\frac{2GM}{R} }[/tex]
Thermal velocity of an atom [tex]V_{therm} =\sqrt{\frac{2KT}{M} }[/tex]
Given [tex]G= 6.67*10^{-11} m^{3}/kgS^{2}\\\\Mx_{titan} = 1.35*10^{23}kg\\[/tex]
Resistance from the centr of the titan is:
R= R(titan) + height of exosphere = 2575 + 1400=3.975*10^6m
escape vel = [tex]V_{esc} =\sqrt{\frac{2GM}{R} }\\\\V_{esc} =\sqrt{\frac{2*6.67*10^{-11}*1.35*10^{23}}{3.975*10^{6}} }=2.129*10^{3}m/s[/tex]
Thermal velocity of an atom [tex]V_{therm} =\sqrt{\frac{2KT}{M} }\\\\V_{therm} =\sqrt{\frac{2*1.38*10^{-23}*200}{1.67*10^{-27}} }=1.8*10^{3}m/s[/tex]
The escape velocity from Titan's exosphere and the thermal speed of a hydrogen atom at the exospheric temperature of about 200 K can be calculated using specific formulas.
Explanation:The escape velocity from Titan's exosphere can be calculated using the formula:
V = sqrt((2GM)/r)
Where V is the escape velocity, G is the gravitational constant, M is the mass of Titan, and r is the distance from the center of Titan to the exosphere. The thermal speed of a hydrogen atom at the exospheric temperature can be calculated using the formula:
v = sqrt((3kT)/m)
Where v is the thermal speed, k is the Boltzmann constant, T is the temperature, and m is the mass of the hydrogen atom.
Given the information provided, the escape velocity from Titan's exosphere and the thermal speed of a hydrogen atom at the exospheric temperature of about 200 K can be determined.
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Neutron activation analysis for a sample of a rock revealed the presence of 131 53I, which has a half-life of 8.08 days . Assuming the isotope was freshly separated from its decay products, what is the mass of 131 53I in a sample emitting 1.00 mCi of radiation?
Answer:
The mass of [tex]_{53}^{131}I[/tex] is [tex]8.09\times10^{-9}\ g[/tex].
Explanation:
Given that,
Half life [tex]t_{\frac{1}{2}}=8.08\ days[/tex]
Sample emitting radiation = 1.00 mCi = [tex]3.7\times10^{7}\ dps[/tex]
We need to calculate the rate constant
Using formula of rate constant
[tex]\lambda=\dfrac{0.693}{t_{\frac{1}{2}}}[/tex]
[tex]\lambda=\dfrac{0.693}{8.08\times24\times60\times60}[/tex]
[tex]\lambda=9.92\times10^{-7}\ s^{-1}[/tex]
We need to calculate the numbers of atoms
Using formula of numbers of atoms
[tex]N_{0}=\dfrac{N}{\lambda}[/tex]
[tex]N_{0} =\dfrac{3.7\times10^{7}}{9.92\times10^{-7}}[/tex]
[tex]N_{0}=3.72\times10^{13}\ atoms[/tex]
We need to calculate the mass of [tex]_{53}^{131}I[/tex]
Using formula for mass
[tex]m=\dfrac{131\times3.72\times10^{13}}{6.023\times10^{23}}[/tex]
[tex]m=8.09\times10^{-9}\ g[/tex]
Hence, The mass of [tex]_{53}^{131}I[/tex] is [tex]8.09\times10^{-9}\ g[/tex].
The problem is related to the concept of radioactive decay, specifically of the isotope Iodine-131. The activity of the sample is calculated to be 37,000,000 decays/second. Through a series of calculations using formulas for decay constant, the number of atoms and finally the sample mass, we find the mass to be 9.3x10^-16 g/day.
Explanation:The problem given relates to nuclear physics and utilizes the concept of radioactive decay. It specifically involves the isotope Iodine-131 (131 53I), which decays with a half-life of 8.08 days.
Assuming the activity of the sample is 1 mCi, which is equivalent to 37,000,000 decays/second (since 1 Ci=3.7x10^10 decays/sec, 1 mCi = 1/1000 Ci), we can use the activity formula which is given by A = λN, where A is activity, λ is the decay constant, and N is the number of atoms in the sample. We first need to calculate λ, which we can find using the formula λ = Ln(2)/T(1/2), where T(1/2) is the half-life.
λ = Ln(2)/8.08 days = 0.086/day. Therefore, N = A/ λ = 37,000,000 / 0.086 = 4.3x10^8 atoms/day.
The remaining calculation is the mass of the sample. The atomic mass of Iodine-131 is approximately 131 g/mol. Using Avogadro's number (6.02x10^23 atoms/mol), we can calculate the mass: M = N * (131 g/mol) / (6.02x10^23 atoms/mol) = 9.3x10^-16 g/day.
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The speed of sound in air is around 345 m/s. A tuning fork vibrates at 610 Hz above the open end of the sound resonance tube. What is the wavelength (in cm) of the sound waves in the tube? Never include units with a numerical answer.
Answer:
[tex]\lambda=56.5cm[/tex]
Explanation:
Wavelength is calculated as:
[tex]\lambda=\frac{V}{f}[/tex]
Replacing the given values:
[tex]\lambda=\frac{345}{610}[/tex]
[tex]\lambda=0.565m[/tex]
Converting the result into cm:
[tex]\lambda=56.5cm[/tex]
A 350 kg mass, constrained to move only vertically, is supported by two springs, each having a spring constant of 250 kN/m. A periodic force with a maximum value of 100 N is applied to the mass with a frequency of 2.5 rad/s. Given a damping factor of 0.125, the amplitude of the vibration is
Answer:
Explanation:
Expression for amplitude of forced damped oscillation is as follows
A = [tex]\frac{F_0}{\sqrt{m(\omega^2-\omega_0^2)^2+b^2\omega^2} }[/tex]
where
ω₀ =[tex]\sqrt{\frac{k}{m} }[/tex]
ω₀ = [tex]\sqrt{\frac{500000}{350} }[/tex]
=37.8
b = .125 ,
ω = 2.5
m = 350
A = [tex]\frac{100}{\sqrt{350(37.8^2-2.5^2)^2+.125^2\times2.5^2} }[/tex]
A = 3.75 mm . Ans
Two balls have the same mass of 5.00 kg. Suppose that these two balls are attached to a rigid massless rod of length 2L, where L = 0.550 m.
One is attached at one end of the rod and the other at the middle of the rod.
If the rod is held by the open end and rotates in a circular motion with angular speed of 45.6 revolutions per second,
(a) What is the tension for the first half of the rod, i.e., between 0 and L if the pivot point is chosen as origin?
(b) What is the tension for the second half of the rod, i.e., between L and 2L if the pivot point is chosen as origin?
Answer:
Explanation:
Given
mass of balls [tex]m= 5 kg[/tex]
[tex]N=45.6 rev/s[/tex]
angular velocity [tex]\omega =2\pi N=286.55 rad/s[/tex]
Length of Rod [tex]2L=1.1 m[/tex]
Tension in the Second half of rod
[tex]T_2=m\omega ^2(2L)=2m\omega ^2L[/tex]
[tex]T_2=5\times (286.55)^2\times 1.1[/tex]
[tex]T_2=451.609 kN[/tex]
For First Part
[tex]T_1-T_2=m\omega ^2L[/tex]
[tex]T_1=T_2+m\omega ^2L[/tex]
[tex]T_1=3 m\omega ^2L[/tex]
[tex]T_1=3\times 5\times (286.55)^2\times 0.55[/tex]
[tex]T_1=677.41 kN[/tex]
Final answer:
The question deals with the tension in a rotating rod with attached masses. Tension in the first half of the rod is dictated by the centripetal force for one mass, while the second half must account for two masses. It's a problem in the field of Physics, specifically rotational dynamics at the college level.
Explanation:
The student's question revolves around the concepts of rotational motion and tension in a system consisting of a massless rod with balls of equal mass attached at different points. Given the angular speed and the positions of the masses, we are to find the tension in two parts of the rod during rotation.
The tension for the first half of the rod can be calculated by considering the centripetal force required to keep the ball rotating in a circle of radius L, the middle of the rod. Since the ball at L is the only mass in the first segment, we only need to consider its centripetal force requirement.
For the second segment of the rod, from L to 2L, the tension must accommodate the centripetal force for both masses, one at the middle and the other at the end. The ball at the end experiences more tension because it is further from the pivot point and thus has a larger radius of rotation.
Note that actual equations and calculations are not provided here, as the question seems to request conceptual explanations rather than specific numerical solutions.
Find the radius R of the orbit of a geosynchronous satellite that circles the Earth. (Note that R is measured from the center of the Earth, not the surface of the Earth.) Use the following values if needed in this problem: The universal gravitational constant G is 6.67
The radius R of a geosynchronous satellite's orbit is 42.2 x 10⁶ meters, or approximately 6.63 times the radius of the Earth.
The radius R of the orbit of a geosynchronous satellite that circles the Earth can be computed using the principles of circular motion and the law of universal gravitation, considering a geostationary satellite takes 23 hours 56 minutes and 4 seconds to complete one orbit. The gravitational force provides the necessary centripetal force to keep the satellite in orbit. Given the mass of the Earth (me = 5.98 x 10²⁴kg), the mean radius of the earth (R2 = 6.37 x 10⁶ m), and the universal gravitational constant (G = 6.67 x 10⁻¹¹ N m²kg⁻²), the radius R is already provided as 42.2 x 10⁶ m. This is the distance from the center of the Earth to the satellite. When expressed in terms of Earth's radii, this is approximately 6.63 Earth radii, since the radius of the Earth is 6.37 x 10⁶ m.
Because you were desperate for a gift for your mom, you have a picture of yourself imbedded in the center of a cubic block of Lucite (n = 3/2) with sides of length 8cm. As you turn the block you notice the location of the picture seems to change. You know the picture is physically in the exact center of the block. How far from one of the faces of the block does the picture appear to be, that is what is the image distance of the picture from the face you are looking through?
Select One of the Following:
(a) 1.5cm
(b) 2.7cm
(c) 6.0cm
(d) 9.0cm
Answer:
correct option is (b) 2.7cm
Explanation:
given data
n = 3/2 = 1.5
sides of length = 8 cm
to find out
How far from one of the faces of the block that what is the image distance of the picture from the face
solution
we get here distance that is
distance d = [tex]\frac{d_o}{n}[/tex] ......................1
put here value as do is [tex]\frac{8}{2}[/tex] = 4 cm
so
distance d = [tex]\frac{d_o}{n}[/tex]
distance d = [tex]\frac{4}{1.5}[/tex]
distance d = 2.6666666 = 2.67
so correct option is (b) 2.7cm
As the image distance of the picture from the face is 2.7 cm, the correct answer is (b) 2.7cm.
The problem is related to optics, specifically the refraction of light through a medium with a given refractive index, here Lucite, which has a refractive index of n = 1.5.
The picture is physically at the center of the block. Since the block has sides of length 8 cm, the center is 4 cm from any face.
To find the apparent distance due to refraction, we use the formula for the apparent depth (d') from the real depth (d) and the refractive index (n) given as:
[tex]d' = \frac{d}{n}[/tex]
Substituting the given values, we get:
d' = [tex]\frac{4 cm}{1.5}[/tex] ≈ 2.67 cm.
Rounding to the closest option, the apparent distance is 2.7 cm from one of the faces of the block.
A sled plus passenger with total mass m = 52.3 kg is pulled a distance d = 21.8 m across a horizontal, snow-packed surface for which the coefficient of kinetic friction with the sled is μk = 0.13. The pulling force is constant and makes an angle of φ = 36.7 degrees above horizontal. The sled moves at constant velocity.a) Write an expression for the work done by the pulling force in terms of m, g (acceleration due to gravity), φ, μk, and d.b) What is the work done by the pulling force, in joules?c) Write an expression for the work done on the sled by friction in terms of m, g (acceleration due to gravity), φ, μk, and d.d) What is the work done on the sled by friction, in joules?
Answer:
Explanation:
Since the sled plus passenger moves with constant velocity , force applied will be equal to frictional force. Let the force applied be F
a ) Frictional force = μ R = F cosφ
R = mg - F sinφ
μ(mg - F sinφ) = F cosφ
μmg = F (μsinφ+cosφ)
F = μmg / (μsinφ+cosφ)
Work done
= F cosφ x d
= μmg x cosφ x d / (μsinφ+cosφ)
b )Work done
= 0.13 x 52.3 x 9.8 cos36.7 x 21.8 / ( 0.13 sin36.7 +cos36.7)
= 1164.61 / .87946
1324.23 J
c ) work done on the sled by friction
= - (work done by force)
= - μmg x cosφ x d / (μsinφ+cosφ)
d ) work done on the sled by friction
= - 1324.23 J
The work done by the pulling force is given by the formula W = F * d * cos(φ), and in this case, the force of friction can be calculated as Ff = μk * m * g * cos(φ).
Explanation:The work done by the pulling force is given by the formula W = F * d * cos(φ), where F is the magnitude of the pulling force, d is the distance, and φ is the angle above the horizontal. In this case, the force of friction can be calculated as Ff = μk * m * g * cos(φ). Therefore, the work done by the pulling force can be written as W = Ff * d.
Using the given values, we can substitute in the values to calculate the work done by the pulling force, which is W = μk * m * g * cos(φ) * d.
The work done on the sled by friction can be calculated by multiplying the force of friction by the distance, since the force is acting in the opposite direction. Therefore, the work done on the sled by friction is given by Wfr = -Ff * d. Substituting the given values, we can calculate the work done by friction, which is Wfr = -μk * m * g * cos(φ) * d.
Inserting given values in this formula, we can calculate the work done by friction as Wf = - (0.13) * (52.3 kg) * (9.8 m/s²) * (21.8 m) = -1475.24 Joules.
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You have two springs. One has a greater spring constant than theother. You also have two objects, one with a greater mass than theother. Which object should be attached to which spring, so that theresulting spring-object system has the greatest possible period ofoscillation?
The object with the smaller mass should be attached to the springwith the greater spring constant.
The object with the smaller mass should be attached to the springwith the smaller spring constant.
The object with the greater mass should be attached to the springwith the greater spring constant.
The object with the greater mass should be attached to the springwith the smaller spring constant.
The object with the greater mass should be attached to the spring with the smaller spring constant, so that the resulting spring-object system has the greatest possible period of oscillation.
Answer: Option D
Explanation:
According to the simple harmonic motions, from physics, it gives a relation between deformation force and the deflection. The more deflection results in more time period of oscillation.
F = - k x
where ‘k’ is the spring constant, and ‘F’ is the deformation force.
So, deflection is directly proportionate to forces, and inversely proportionate to its spring constant. Hence, we can derive that the force must be maximum, and hence weight must be maximum, with the spring constant lesser. Then, the deflection will be high. So, time period increases.
Our two intrepid relacar drivers are named Pam and Ned. We use these names to make it easy to remember: measurements made by Pam are primed (x', t') and those made by Ned are not primed (x, t). v is the velocity of Pam (the other frame of reference) as measured by Ned. What is the interpretation of v'?
This is a problem based on the logic and interpretation of the variables. From the measured data taken
what is collected by the two individuals is expressed as,
- NED reference system: (x, t)
- PAM reference system: (x ', t')
From the reference system we know that ν is the speed of PAM (the other reference system) as a measurement by NED.
Then ν' is the speed of NED (from the other system of the reference) as a measurement by PAM.
In the context of Special Relativity, v' represents the relative velocity of Ned as observed from Pam's reference frame. This velocity takes into account the motion of both Ned and Pam and can be computed using the principles of Conservation of Momentum.
Explanation:The symbol v' can be interpreted as the velocity of Ned as measured from Pam's frame of reference. This concept forms the basis of Special Relativity as proposed by Einstein, where observations are taken from different frames of reference, in this case, Pam and Ned. Relative velocity, like v', is the velocity of an object (in this case, Ned) as observed from a particular frame of reference (in this case, Pam's), and it can change depending on who is making the observation. It takes into account the motion of both Ned and Pam and describes how fast Ned is moving and in what direction from Pam's perspective.
For example, if Pam and Ned were moving towards each other, the relative velocity v' would be the sum of their individual speeds as observed from Pam's perspective. If they were moving in the same direction, v' would be the difference in their individual speeds. Also, Pam's velocity in Ned's frame can be defined as -v'. This negative sign indicates they are moving in opposite directions.
To compute v', you would need to understand the principle of Conservation of Momentum, represented by equations defining momentum conservation in the x and y directions. Using this principle, you can solve for V.
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A coil 3.95 cm radius, containing 520 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×10−2 T/s )t+( 3.45×10−5 T/s4 )t4. The coil is connected to a 560-Ω resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil.
a-)Find the magnitude of the induced emf in the coil as a function of time.
Find the magnitude of the induced emf in the coil as a function of time.
E= 1.07×10−2 V +( 1.05×10−4 V/s3 )t3
E= 3.36×10−2 V +( 8.26×10−5 V/s3 )t3
E= 3.36×10−2 V +( 3.30×10−4 V/s3 )t3
E= 1.07×10−2 V +( 3.30×10−4 V/s3 )t3
b-)What is the current in the resistor at time t0 = 5.25 s ?
Final answer:
The magnitude of the induced emf in the coil as a function of time is given by the equation E = 3.36×10^(-2) V + (3.30×10^(-4) V/s^3)t^3. The current in the resistor at time t0 = 5.25 s can be calculated using Ohm's law by dividing the induced emf in the coil by the resistance.
Explanation:
In order to find the magnitude of the induced emf in the coil as a function of time, we can use Faraday's law of electromagnetic induction. Faraday's law states that the induced emf in a coil is equal to the rate of change of magnetic flux through the coil. The magnetic flux through the coil can be determined by multiplying the magnetic field by the area of the coil. Therefore, the induced emf in the coil is given by the equation E = -N * dΦ/dt, where E is the induced emf, N is the number of turns in the coil, and dΦ/dt is the rate of change of magnetic flux.
In this case, the magnetic field varies with time according to B = (1.20×10^(-2) T/s)t + (3.45×10^(-5) T/s^4)t^4. The magnetic field is perpendicular to the plane of the coil, so the area of the coil through which the magnetic flux passes is given by A = πr^2, where r is the radius of the coil.
Substituting these values into the equation for the induced emf, we get:
E = -N * (A * dB/dt)
E = -N * (πr^2 * dB/dt)
E = -520 * (π(0.0395)^2 * ((1.20×10^(-2) T/s) + (3.45×10^(-5) T/s^4)(4t^3)))
Simplifying this equation gives the magnitude of the induced emf in the coil as a function of time: E = 3.36×10^(-2) V + (3.30×10^(-4) V/s^3)t^3.
To find the current in the resistor at time t0 = 5.25 s, we can use Ohm's law, which states that the current through a resistor is equal to the voltage across the resistor divided by the resistance. The voltage across the resistor is equal to the induced emf in the coil. Therefore, the current in the resistor is given by the equation I = E/R, where I is the current, E is the induced emf, and R is the resistance.
Substituting the values given in the question into this equation, we get:
I = (3.36×10^(-2) V + (3.30×10^(-4) V/s^3)(5.25 s)^3) / 560 Ω
Calculating this gives the current in the resistor at time t0 = 5.25 s.
Four point charges are placed at the corners of a square. Each charge has the identical value +Q. The length of the diagonal of the square is 2a.
What is the electric potential at the center of the square?
Answer:[tex]V_{net}=4\frac{kQ}{a}[/tex]
Explanation:
Given
charge on each Particle is Q
Length of diagonal of the square is 2a
therefore distance between center and each charge is [tex]\frac{2a}{2}=a[/tex]
Electric Potential of charged Particle is given by
For First Charge
[tex]V_1=\frac{kQ}{a}[/tex]
[tex]V_2=\frac{kQ}{a} [/tex]
[tex]V_3=\frac{kQ}{a}[/tex]
[tex]V_4=\frac{kQ}{a}[/tex]
total Electric Potential At center is given by
[tex]V_{net}=V_1+V_2+V_3+V_4[/tex]
[tex]V_{net}=4\times \frac{kQ}{a}[/tex]
[tex]V_{net}=4\frac{kQ}{a}[/tex]
(a) How far is the center of mass of the Earth-Moon system from the center of the Earth? (Appendix C gives the masses of the Earth and the Moon and the distance between the two.) m
(b) Express the answer to (a) as a fraction of the Earth's radius.'
Answer:
a) [tex]r_{cm} = 4672 km[/tex]
b)r_cm = 0.733 R_{earth}
Explanation:
Location of center of mass of Earth- moon system is
[tex]r_{cm}= \frac{M_{earth}r_{earth}+M_{moon}r_{moon}}{M_{moon}+M{earth}}[/tex]
[tex]r_{cm}= \frac{6\times10^{24}(0)+7.34\times10^{22}384000\times1000}{6\times10^{24}+7.34\times10^{22}}[/tex]
=4.672×10^(6) m
a) [tex]r_{cm} = 4672 km[/tex]
b) [tex]\frac{r_{cm}}{R{earth}} = \frac{4672}{6371} = 0.733[/tex]
therefore
r_cm = 0.733 R_{earth}
The center of mass of the Earth-Moon system is the point around which the two bodies orbit and it is located within Earth but not at the center. The distance from the Earth's center to this point can be calculated using the masses of the Moon and Earth, and the distance between them. This distance expressed as a fraction of the Earth's radius is approximately 0.73.
Explanation:The center of mass of the Earth-Moon system is the point about which the two bodies orbit, and it's located inside the Earth, not at the Earth's center. The distance from the Earth's center to the system's center of mass can be calculated using the formula: d = D*(m/(M+m)), where D is the distance between the Earth and the Moon, m is the Moon's mass, M is the Earth's mass.
To express the previous value as a fraction of the Earth's radius, you must divide it by the radius of the Earth. For instance, let's assume the distance turned out to be 4671 km and the Earth's radius is 6371 km. The resulting fraction would be 4671/6371 which simplifies to approximately 0.73.
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Two concrete spans of a 250-m-long bridge are placed end to end so that no room is allowed for expansion. If a temperature increase of 20.0C occurs, what is the height y to which the spans rise when they buckle?
Answer:
y = 2.74 m
Explanation:
The linear thermal expansion processes are described by the expression
ΔL = α L ΔT
Where α the thermal dilation constant for concrete is 12 10⁻⁶ºC⁻¹, ΔL is the length variation and ΔT the temperature variation in this case 20ªc
If the bridge is 250 m long and is covered by two sections each of them must be L = 125 m, let's calculate the variation in length
ΔL = 12 10⁻⁶ 125 20
ΔL = 3.0 10⁻² m
Let's use trigonometry to find the height
The hypotenuse Lf = 125 + 0.03 = 125.03 m
Adjacent leg L₀ = 125 m
cos θ = L₀ / Lf
θ = cos⁻¹ (L₀ / Lf)
θ = cos⁻¹ (125 / 125.03)
θ = 1,255º
We calculate the height
tan 1,255 = y / x
y = x tan 1,255
y = 125 tan 1,255
y = 2.74 m
When a bridge is subjected to a temperature increase, the concrete spans rise or buckle due to thermal expansion. The height to which they rise can be calculated using the formula for change in length. In this case, the height is 0.42 m.
Explanation:When a bridge is subjected to a temperature increase, it expands due to thermal expansion. In this case, the two concrete spans of the bridge are placed end to end without allowing any room for expansion. Therefore, when the temperature increases by 20.0°C, the spans rise or buckle. The height to which they rise can be calculated using the formula:
change in length = coefficient of linear expansion x original length x change in temperature
In this case, the change in length is 0.84 m. To calculate the height y to which the spans rise, divide this change in length by the number of spans, which is 2. Therefore, the height is 0.42 m.
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Determine the ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy (1/2mv2) when a particle has a speed of (a) 2.71 x 10-3c. and (b) 0.855c.
Final answer:
The ratio of relativistic kinetic energy to nonrelativistic kinetic energy depends on the speed of the particle. For speeds much less than the speed of light, the ratio is close to 1, but for speeds approaching the speed of light, the relativistic effect becomes significant and the ratio increases.
Explanation:
The ratio of relativistic kinetic energy to nonrelativistic kinetic energy can be calculated using the formula for relativistic kinetic energy KErel = mc2((1/(1 - v2/c2)1/2) - 1), where m is the particle mass, v is its velocity, and c is the speed of light. The nonrelativistic kinetic energy is given by KEnr = (1/2)mv2. Firstly, to determine the ratios at given speeds, we calculate the relativistic factor γ, which is (1/(1 - v2/c2)1/2), for each speed
For part (a) with v = 2.71 x 10-3c, the relativistic factor is close to 1, indicating that the speeds are nonrelativistic, and the ratio KErel/KEnr would be close to 1. For part (b), at v = 0.855c, the relativistic factor is significantly greater than 1, and the relativistic kinetic energy will be much larger than the classical value, yielding a ratio much greater than 1.
A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is 6.68 × 10-3 Wb. What is the flux that passes through the circular loop?
Answer:
0.0085 Wb
Explanation:
a = Side of square
r = Radius of circle
[tex]\phi_s[/tex] = Magnetic flux through square loop = [tex]6.68\times 10^{-3}\ Wb[/tex]
Magnetic flux is given by
[tex]\phi=BA[/tex]
For square
[tex]\phi_{s}=Ba^2[/tex]
The length of the square will be equal to the circumference of the circle
[tex]4a=2\pi r\\\Rightarrow r=\frac{2a}{\pi}[/tex]
For circle
[tex]\phi_{c}=B\pi r^2\\\Rightarrow \phi_{c}=B\pi \left(\frac{2a}{\pi}\right)^2\\\Rightarrow \phi_c=Ba^2\frac{4}{\pi}\\\Rightarrow \phi_c=\phi_s\frac{4}{\pi}\\\Rightarrow \phi_c=6.68\times 10^{-3}\frac{4}{\pi}\\\Rightarrow \phi_c=0.0085\ Wb[/tex]
The flux that passes through the circular loop is 0.0085 Wb
The magnetic flux that passes through the circular loop is 0.00848 Wb.
What is the magnetic flux?The magnetic flux is defined as the measurement of the total magnetic field which passes through a given area.
Given that the magnetic flux that passes through the square loop is 6.68 × 10-3 Wb.
Let's consider that r is the radius of the circle and a is the side of the square.
The magnetic flux through the square loop is given below.
[tex]\phi_s = Ba^2[/tex]
Where B is the magnetic field.
The length of the square will be equal to the circumference of the circle
[tex]4a = 2\pi r[/tex]
[tex]r = \dfrac {2a}{\pi}[/tex]
The magnetic flux through the circular loop is given below.
[tex]\phi = BA[/tex]
Where A is is the cross-sectional area of the circular loop.
[tex]\phi = B \pi r^2[/tex]
[tex]\phi = B \pi (\dfrac {2a}{\pi})^2[/tex]
[tex]\phi = B\pi \times \dfrac {4a^2}{\pi^2}[/tex]
[tex]\phi = Ba^2 \dfrac {4}{\pi}[/tex]
[tex]\phi = \phi_s \dfrac {4}{3.14}[/tex]
[tex]\phi = 6.68 \times 10^{-3} \times 1.27[/tex]
[tex]\phi = 0.00848 \;\rm Wb[/tex]
Hence we can conclude that the magnetic flux that passes through the circular loop is 0.00848 Wb.
To know more about the magnetic flux, follow the link given below.
https://brainly.com/question/15359941.
An Atwood machine consists of a mass of 3.5 kg connected by a light string to a mass of 6.0 kg over a frictionless pulley with a moment of inertia of 0.0352 kg m2 and a radius of 12.5 cm. If the system is released from rest, what is the speed of the masses after they have moved through 1.25 m if the string does not slip on the pulley?
Please note: the professor has told us that the correct answer is 2.3 m/s. how does one arrive at this answer?
Answer:
[tex]v=2.28m/s[/tex]
Explanation:
For the first mass we have [tex]m_1=3.5kg[/tex], and for the second [tex]m_2=6.0kg[/tex]. The pulley has a moment of intertia [tex]I_p=0.0352kgm^2[/tex] and a radius [tex]r_p=0.125m[/tex].
We solve this with conservation of energy. The initial and final states in this case, where no mechanical energy is lost, must comply that:
[tex]K_i+U_i=K_f+U_f[/tex]
Where K is the kinetic energy and U the gravitational potential energy.
We can write this as:
[tex]K_f+U_f-(K_i+U_i)=(K_f-K_i)+(U_f-U_i)=0J[/tex]
Initially we depart from rest so [tex]K_i=0J[/tex], while in the final state we will have both masses moving at velocity v and the tangential velocity of the pully will be also v since it's all connected by the string, so we have:
[tex]K_f=\frac{m_1v^2}{2}+\frac{m_2v^2}{2}+\frac{I_p\omega_p^2}{2}=(m_1+m_2+\frac{I_p}{r_p^2})\frac{v^2}{2}[/tex]
where we have used the rotational kinetic energy formula and that [tex]v=r\omega[/tex]
For the gravitational potential energy part we will have:
[tex]U_f-U_i=m_1gh_{1f}+m_2gh_{2f}-(m_1gh_{1i}+m_2gh_{2i})=m_1g(h_{1f}-h_{1i})+m_2g(h_{2f}-h_{2i})[/tex]
We don't know the final and initial heights of the masses, but since the heavier, [tex]m_2[/tex], will go down and the lighter, [tex]m_1[/tex], up, both by the same magnitude h=1.25m (since they are connected) we know that [tex]h_{1f}-h_{1i}=h[/tex] and [tex]h_{2f}-h_{2i}=-h[/tex], so we can write:
[tex]U_f-U_i=m_1gh-m_2gh=gh(m_1-m_2)[/tex]
Putting all together we have:
[tex](K_f-K_i)+(U_f-U_i)=(m_1+m_2+\frac{I_p}{r_p^2})\frac{v^2}{2}+gh(m_1-m_2)=0J[/tex]
Which means:
[tex]v=\sqrt{\frac{2gh(m_2-m_1)}{m_1+m_2+\frac{I_p}{r_p^2}}}=\sqrt{\frac{2(9.8m/s^2)(1.25m)(6.0kg-3.5kg)}{3.5kg+6.0kg+\frac{0.0352kgm^2}{(0.125m)^2}}}=2.28m/s[/tex]
Final answer:
To solve this problem, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system (kinetic energy + potential energy) remains constant if no external forces are acting on the system.
Explanation:
Let's denote the initial position of the masses as position 1 and the final position as position 2. At position 1, the 3.5 kg mass is at a height h above the ground, and the 6.0 kg mass is at a height 1.25 m - h above the ground. At position 2, both masses are at a height 1.25 m above the ground.
The potential energy (PE) of the system at position 1 is:
PE1 = m1 * g * h + m2 * g * (1.25 m - h)
The potential energy (PE) of the system at position 2 is:
PE2 = (m1 + m2) * g * 1.25 m
Since the system is released from rest, the initial kinetic energy (KE) of the system is zero. The final kinetic energy (KE) of the system is:
KE2 = (1/2) * (m1 + m2) * v^2
where v is the speed of the masses after they have moved through 1.25 m.
According to the principle of conservation of mechanical energy, the total mechanical energy at position 1 is equal to the total mechanical energy at position 2:
PE1 + KE1 = PE2 + KE2
Substituting the expressions for PE1, PE2, and KE2, we get:
m1 * g * h + m2 * g * (1.25 m - h) = (m1 + m2) * g * 1.25 m + (1/2) * (m1 + m2) * v^2
Solving for v^2, we get:
v^2 = 2 * g * (m1 * h + m2 * (1.25 m - h) - (m1 + m2) * 1.25 m)
Substituting the given values for g, m1, m2, and h, we get:
v^2 = 2 * 9.8 m/s^2 * (3.5 kg * 1.25 m + 6.0 kg * (1.25 m - 1.25 m) - (3.5 kg + 6.0 kg) * 1.25 m)
Solving for v, we get:
v ≈ 2.3 m/s
So, the speed of the masses after they have moved through 1.25 m is approximately 2.3 m/s.
A horizontal water jet strikes a stationary vertical plate at a rate of 5 kg/s with a velocity of 35 km/hr. Assume that the water stream moves in the vertical direction after the strike. The force, in N, needed to prevent the plate from moving horizontally is _____ .
Answer:
48.6 N
Explanation:
rate of mass per second, dm/dt = 5 kg/s
Velocity, v = 35 km/hr = 9.72 m/s
Force acting on the plate
F = v x dm/dt
F = 9.72 x 5 = 48.6 N
Thus, the force acting on the plate is 48.6 N.
To prevent the plate from moving horizontally, a force of 48.6 N is required. The initial horizontal momentum of the water is 48.6 kg*m/s.
To calculate the force needed to prevent the plate from moving horizontally when a horizontal water jet strikes a vertical plate, we first need to determine the change in momentum of the water. The water strikes the plate at a rate of 5 kg/s with a velocity of 35 km/hr. Let's convert the velocity to meters per second:
35 km/hr = (35 * 1000) / 3600 = 9.72 m/s.
Next, we calculate the initial horizontal momentum of the water jet:
Initial momentum = mass flow rate * velocity = 5 kg/s * 9.72 m/s = 48.6 kg*m/s.
The water stream moves in the vertical direction after the strike, so its horizontal momentum is reduced to zero. The change in momentum is:
Change in momentum = Initial momentum - Final momentum = 48.6 kg*m/s - 0 kg*m/s = 48.6 kg*m/s.
We use the change in momentum (impulse) to find the force since force is the rate of change of momentum:
Force = Change in momentum / Time
The mass flow rate gives the change in momentum per second, which means the force can be directly calculated as:
Force = 48.6 N
Thus the force required to prevent the plate from moving horizontally, a force of 48.6 N
A quantity of an ideal gas is kept in a rigid container of constant volume. If the gas is originally at a temperature of 28 °C, at what temperature (in °C) will the pressure of the gas triple from its base value?
Answer:
[tex]T_2=630^{\circ}C[/tex]'
Explanation:
Original temperature of the gas, [tex]T_1=28^{\circ}C=301\ K[/tex]
From the ideal gas equation,
[tex]P_1V_1=nRT_1[/tex]
Since,
[tex]P_2=3P_1[/tex]
[tex]nRT_2=3(nRT_1)[/tex]
[tex]T_2=3T_1[/tex]
[tex]T_2=3\times 301[/tex]
[tex]T_2=903\ K[/tex]
or
[tex]T_2=630^{\circ}C[/tex]
So, the new temperature of the gas is 630 degree Celsius. Hence, this is the required solution.
The space shuttle, in circular orbit around the Earth, collides with a small asteroid which ends up in the shuttle's storage bay. For this collision,
a. neither momentum nor kinetic energy is conserved.
b. only momentum is conserved.
c. both momentum and kinetic energy are conserved.
d. only kinetic energy is conserved.
Answer:
b
Explanation:
The space shuttle, in circular orbit around the Earth, collides with a small asteroid which ends up in the shuttle's storage bay.
This form of collision is called inelastic collision. And inelastic collision momentum is conserved but the kinetic energy is not conserved. Hence the correct option is b. only momentum is conserved.
Answer:
b
Explanation:
This form of collision is called inelastic collision.
A long, thin solenoid has 700 turns per meter and radius 2.50 cm. The current in the solenoid is increasing at a uniform rate of 36.0 A/s . a. What is the magnitude of the induced electric field at a point near the center of the solenoid? b. What is the magnitude of the induced electric field at a point 0.500 cm from the axis of the solenoid?
To solve this problem it is necessary to apply the concepts related to the magnetic field in a solenoid.
By definition we know that the magnetic field is,
[tex]B = \mu_0 n I[/tex]
[tex]\frac{dB}{dt} = \frac{d}{dt} \mu_0 NI[/tex]
[tex]\frac{dB}{dt} = \mu_0 n\frac{dI}{dt}[/tex]
At the same tome we know that the induced voltage is defined as
[tex]\epsilon = \frac{\Phi}{dt}[/tex]
[tex]\epsilon = A \frac{dB}{dt}[/tex]
Replacing
[tex]\epsilon = A \mu_0 n\frac{dI}{dt}[/tex]
[tex]\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}[/tex]
PART A) Substituting with our values we have that
[tex]\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}\\\epsilon = \frac{(4\pi*10^{-7})700(0)(36)}{2}\\\epsilon = 0V/m\\[/tex]
Therefore there is not induced electric field at the center of solenoid.
PART B) Replacing the radius for 0.5cm
[tex]\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}\\\epsilon = \frac{(4\pi*10^{-7})700(0.5*10^{-2})(36)}{2}\\\epsilon = 7.9168*10^{-5}V/m[/tex]
Therefore the magnitude of the induced electric field at a point 0.5cm is [tex]7.9168*10^{-5}V/m[/tex]
The magnitude of the induced emf near the center of the solenoid is 0 V/m.
The magnitude of the induced electric field at a point 0.500 cm from the axis of the solenoid is 8 x 10⁻⁵ V/m.
The given parameters;
number of turns of the solenoid, N = 700 turns/mradius of the wire, r = 2.5 cmcurrent in the solenoid, I = 36 A/sThe magnitude of the induced emf near the center of the solenoid is calculated as follows;
[tex]B = \mu_0 nI\\\\\frac{dB}{dt} = \mu_0 n \frac{dI}{dt} \\\\\frac{dB}{dt} = (4\pi \times 10^{-7} \times 700 \times 36) = 0.032 \ T/s[/tex]
[tex]E = \frac{r}{2} (\frac{dB}{dt} )\\\\\ E = \frac{0}{2} (0.032)\\\\ E = 0 \ V/m[/tex]
The magnitude of the induced electric field at a point 0.500 cm from the axis of the solenoid is calculated as follows;
[tex]E = \frac{r}{2} (\frac{dB}{dr} )\\\\ E= \frac{0.5\times 10^{-2} }{2}( 0.032)\\\\E = 8\times 10^{-5} \ V/m[/tex]
Learn more here:https://brainly.com/question/15399686
In the year 1178, five monks at Canterbury Cathedral in England observed what appeared to be an asteroid colliding with the moon, causing a red glow in and around it. It is hypothesized that this event created the crater Giordano Bruno, which is right on the edge of the area we can usually see from Earth. How long after the asteroid hit the Moon, which is 3.84 x 105 km away, would the light first arrive on Earth in seconds?
To solve this problem it is necessary to apply the kinematic equations of description of the movement in which it is understood that the velocity is the travel of a particle in a fraction of time, that is to say
[tex]v = \frac{x}{t}[/tex]
Where,
x = Displacement
t = time
In our case the speed is equivalent to that of the Light, and the distance is necessary to reach the moon by the asteroid.
[tex]v = 3.8*10^8m/s[/tex]
[tex]x = 3.84*10^5km = 3.84*10^8m[/tex]
Re-arrange to find t,
[tex]t = \frac{x}{V}[/tex]
[tex]t = \frac{3.84*10^8}{3.8*10^8}[/tex]
[tex]t = 1.28s[/tex]
Therefore will take 1.28 s for the light arrive on Earth.
The light from the asteroid collision on the Moon would take approximately 1.28 seconds to reach Earth.
Explanation:The speed of light in a vacuum is approximately 3 x 10^8 meters per second. Since the Moon is 3.84 x 10^5 km away from Earth, we can calculate the time it takes for light to travel from the Moon to Earth using the formula:
Time = Distance / Speed
Converting the distance from km to meters, we get 3.84 x 10^8 meters. Plugging in the values:
Time = (3.84 x 10^8 meters) / (3 x 10^8 meters per second) = 1.28 seconds
So, it would take approximately 1.28 seconds for the light from the asteroid collision on the Moon to reach Earth.
A satellite explodes in outer space, far from any other body, sending thousands of pieces in all directions. Is the linear momentum of the satellite before the explosion less than, equal to, or greater than the total linear momentum of all the pieces after the explosion?
The answer to this problem can be given through energy conservation as well as Newton's first law.
Newton's first law states that as long as there is no force exerted on a body, its movement will be constant or at rest. In this way the amount of linear movement of the satellite before the explosion will be equal to the sum of the movement generated in the pieces after the explosion. In the absence of external forces but the preservation of the internal force as the only one to act, these will not change the total momentum of the system.
A hoop, a disk, and a solid sphere each have mass 1.4 kg and diameter 16 cm. Starting from rest, all three objects roll down a 7° slope. If the slope is 3 m long and all bodies roll without slipping, find the speed of each at the bottom.
I know I have to use rotational kinetic energy and translational kinetic energy to get the answer but im not sure how.
The answers are Hoop=1.89 m/s disk=2.18 m/s and sphere=2.26 m/s
Answer:
The velocities of hoop , disk and sphere are 1.89 m/s , 2.18 m/s , 2.26 m/s.
Explanation:
Lets find the speed of any general body of mass 'm' , moment of inertia 'I' , radius 'r'.
Let 'v' be the speed and 'ω' be the angular velocity of the body at bottom of the slope.
Since there is no external force acting on the system (Eventhough friction is acting at the point of contact of the body and slope , it does no work as the point of contact is always at rest and not moving) , we can conserve energy for this system.
Initially the body is at rest and at a vertical height 'h' from the ground.
Here , h=3sin(7°)
Initial energy = mgh.
Finally on reaching bottom h=0 but the body has both rotational and translational kinetic energy.
∴ Final energy = [tex]\dfrac{1 }{2}[/tex]I[tex]ω^{2}[/tex] + [tex]\dfrac{1 }{2}[/tex]m[tex]v^{2}[/tex].
Since the body is rolling without slipping.
v=rω
and
Initial Energy = Final Energy
mgh = [tex]\dfrac{1 }{2}[/tex]I[tex]ω^{2}[/tex] + [tex]\dfrac{1 }{2}[/tex]m[tex]v^{2}[/tex]
∴ mgh = [tex]\dfrac{1 }{2}[/tex]I[tex]\dfrac{v^{2} }{r^{2} }[/tex] + [tex]\dfrac{1 }{2}[/tex]m[tex]v^{2}[/tex]
∴ v = [tex]\sqrt{\frac{2mgh}{\frac{I}{r^{2} }+m } }[/tex]
For a hoop ,
I = m[tex]r^{2}[/tex]
Substituting above value of I in the expression of v.
We get,
v = [tex]\sqrt{gh}[/tex] = [tex]\sqrt{9.81×3sin(7°) }[/tex] = 1.89 m/s
Similarly for disk,
I = [tex]\dfrac{1}{2}[/tex]m[tex]r^{2}[/tex]
We get,
v = [tex]\sqrt{\frac{4gh}{3} }[/tex] = 2.18 m/s
For solid sphere ,
I = [tex]\dfrac{2}{5}[/tex]m[tex]r^{2}[/tex]
v = [tex]\sqrt{\frac{10gh}{7} }[/tex] = 2.26 m/s.
An airplane is flying at Mach 1.48 at an altitude of 7,800.00 meters, where the speed of sound is v = 311.83 m/s. How far away from a stationary observer will the plane be when the observer hears the sonic boom? (Enter the total distance in m.)
Answer:
The plane will be 11545.46 m far when the observer hears the sonic boom
Explanation:
Step 1: Data given
Altitude of the plane = 7800 meters
speed of sound = 311.83 m/s
Step 2:
The mach number M = vs/v
This means v/vs = 1/M
Half- angle = ∅
sin∅= v/vs
∅ = sin^-1 (v/vs)
∅ = sin^-1 (1/M)
∅ = sin^-1(1/1.48)
∅= 42.5 °
tan ∅ = h/x
⇒ with h= the altitude of the plane = 7800 meter
⇒ with x = the horizontal distance moved by the plane
x = h/tan ∅
x = 7800 / tan 42.5
x = 8512.2 meters
d = the distance between the observer and the plane when the observer hears the sonic boom is:
d = √(8512.2² + 7800²)
d = 11545.46 m
The plane will be 11545.46 m far when the observer hears the sonic boom
A spacecraft is placed in a circular orbit around a planet with mass 6.4 x 1023 kg. The spacecraft orbits at a height of 4.5 x 107 m above the planet’s surface. What additional information is needed to calculate the speed of the spacecraft in the orbit?
Answer:
It is necessary to know the radius of the planet.
Explanation:
The speed of the spacecraft can be found by means of the equation of the Universal law of gravity:
[tex]F = G \frac{M.m}{r^{2}}[/tex] (1)
Where F is the gravitational force, G is gravitational constant, M is the mass of the planet, m is the mass of the spacecraft and r is the orbital radius of the spacecraft.
Equation 1 can be express in terms of the speed by using Newton's second law and the equation for centripetal acceleration:
[tex]F = ma[/tex] (2)
Replacing equation 2 in equation 1 it is gotten:
[tex]ma = G \frac{M.m}{r^{2}}[/tex] (3)
the centripetal acceleration is defined as:
[tex]a = \frac{v^{2}}{r}[/tex] (4)
Replacing equation 4 in equation 3 it is gotten:
[tex]m\frac{v^{2}}{r} = G \frac{M.m}{r^{2}}[/tex] (5)
Then, v can be isolated from equation 5:
[tex]mv^{2} = G \frac{M.m}{r}[/tex]
[tex]v^{2} = G \frac{M.m}{rm}[/tex]
[tex]v^{2} = G \frac{M}{r}[/tex]
[tex]v = \sqrt{\frac{GM}{r}}[/tex]
However, the orbital radius of the spacecraft is obtained by the sum of the radius of the planet and the height of the spacecraft above the surface of the planet (r = R+h)
[tex]v = \sqrt{\frac{GM}{R+h}}[/tex] (6)
Hence, by equation 6 it can be concluded that it is necessary to know the radius of the planet in order to calculate the speed of the spacecraft.
On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on the moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 14 m/s at an angle 20 ∘ above the horizontal. For how much more time was the ball in flight?
Answer:
4.86 seconds
Explanation:
Velocity of projection, u = 14 m/s
angle of projection, θ = 20°
Formula for the time of flight
[tex]T=\frac{2uSin\theta }{g}[/tex]
For earth
Te = (2 x 14 x Sin 20) / 9.8
Te = 0.98 s
For moon
g' = g/6 = 1.64 m/s^2
Tm = ( 2 x 14 x Sin 20) / 1.64
Tm = 5.84 seconds
Tm - Te = 5.84 - 0.98 = 4.86 s
So, it takes 4.86 s more time of flight on moon than the earth.
The golf ball hit by Alan Shepard was in flight approximately 4.87 seconds longer on the moon than on Earth.
To analyze the flight duration of a projectile under different gravitational conditions.
1. Initial Parameters:
Initial velocity of the golf ball, [tex]v_0 = 14 \, \text{m/s}[/tex]
Launch angle, [tex]\theta = 20^{\circ}[/tex]
Acceleration due to gravity on Earth, [tex]g = 9.81 \, \text{m/s}^2[/tex]
Acceleration due to gravity on the moon, [tex]g_{moon} = \frac{g}{6} = 1.635 \, \text{m/s}^2[/tex]
2. Determine the Vertical Component of Initial Velocity:
The vertical component of the initial velocity can be calculated using the formula:
[tex]v_{0y} = v_0 \sin(\theta)[/tex]
Substituting the known values:
[tex]v_{0y} = 14 \, \text{m/s} \cdot \sin(20^{\circ}) \approx 4.78 \, \text{m/s}[/tex]
3. Calculate Time of Flight on Earth:
The time of flight [tex]T[/tex] for a projectile launched and landing on a flat surface can be calculated using the formula:
[tex]T = \frac{2 v_{0y}}{g}[/tex]
For Earth:
[tex]T_{Earth} = \frac{2 \cdot 4.78 \, \text{m/s}}{9.81 \, \text{m/s}^2} \approx 0.975 \, \text{s}[/tex]
4. Calculate Time of Flight on the Moon:
Using the same formula but with the moon's gravity:
[tex]T_{Moon} = \frac{2 v_{0y}}{g_{moon}}[/tex]
Substituting the values:
[tex]T_{Moon} = \frac{2 \cdot 4.78 \, \text{m/s}}{1.635 \, \text{m/s}^2} \approx 5.84 \, \text{s}[/tex]
5. Calculate the Difference in Flight Time:
Finally, to find how much more time the ball was in flight on the moon compared to Earth:
[tex]\Delta T = T_{Moon} - T_{Earth}[/tex]
[tex]\Delta T = 5.84 \, \text{s} - 0.975 \, \text{s} \approx 4.87 \, \text{s}[/tex]
Which one of the following statements is false?
(a) The orbits in the Bohr model have precise sizes, whereas in the quantum mechanical picture of the hydrogen atom they do not.
(b) In the absence of external magnetic fields, both the Bohr model and quantum mechanics predict the same total energy for the electron in the hydrogen atom.
(c) The spin angular momentum of the electron plays a role in both the Bohr model and the quantum mechanical picture of the hydrogen atom.
(d) The magnitude of the orbital angular momentum cannot be zero in the Bohr model, but it can be zero in the quantum mechanical picture of the hydrogen atom.
Answer:
d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero
Explanation:
Affirmations
a) true. The orbits are accurate in the Bohr model and probabilistic in quantum mechanics
b) True. If both give the same results and use the same quantum number (n)
c) True. If in angular momentum it is quantized, in the Bohr model too but it does not justify it
d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero
An astronaut drops a marble on the surface of Mars and observes that it takes 1.02 s for the marble to fall 2.00 m. She also knows that the radius of Mars is 3.39 × 106 m and that G = 6.67 x 10-11 N·m2/kg2. From this information, she can conclude that the mass of Mars is
Answer:
The mass of mars is [tex]6.61\times10^{23}\ kg[/tex]
Explanation:
Given that,
Time = 1.02 s
Height = 2.00 m
Radius of mars [tex]r= 3.39\times10^{6}\ m[/tex]
We need to calculate the acceleration due to gravity on mars
Using equation of motion
[tex]h = ut+\dfrac{1}{2}gt^2[/tex]
[tex]h = \dfrac{1}{2}gt^2[/tex]
[tex]g=\dfrac{2h}{t^2}[/tex]
Where, u = initial velocity
t = time
h = height
Put the value into the formula
[tex]g=\dfrac{2\times2.00}{(1.02)^2}[/tex]
[tex]g=3.84\ m/s^2[/tex]
We need to calculate the mass of mars
Using formula of gravity
[tex]g=\dfrac{Gm}{r^2}[/tex]
Put the value into the formula
[tex]3.84=\dfrac{6.67\times10^{-11}\times m}{(3.39\times10^{6})^2}[/tex]
[tex]m=\dfrac{3.84\times(3.39\times10^{6})^2}{6.67\times10^{-11}}[/tex]
[tex]m=6.61\times10^{23}\ kg[/tex]
Hence, The mass of mars is [tex]6.61\times10^{23}\ kg[/tex]
In order to find the mass of Mars, we can use a physics formula for gravity. The given values, such as the drop distance, time it took, the radius of Mars, and the universal gravitational constant, can be substituted into this formula. This, then, gives a theoretical value for the mass of Mars.
Explanation:To calculate the mass of Mars, we use the formula to find gravity, which is defined as g = GM/r², where G is the universal gravitational constant. We can set this equal to something else: 2d / t², where d is the drop distance and t is the time it took for the marble to drop. We then can isolate M (the mass of Mars) on one side.
By substituting the given values into the formula, we get: M = (g * r²) / G = ((2 * 2.00 m) / (1.02 s)² * (3.39 × 10⁶ m)²) / (6.67 x 10⁻¹¹ N·m²/kg²).
This Netwonian calculation gives the mass of Mars, theoretically, based on the observed falling time of the marble.
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