Answer:
A spiral galaxy
Explanation:
Two strain gauges are mounted so that they sense axial strain on a steel member in uniaxial tension. The 120 Ω gauges form two legs of a Wheatstone bridge, and are mounted on opposite arms (e.g., arms 1 and 4). The gauge factor for each of the strain gauges is 2 and Em for this steel is 29 × 106 psi. For a bridge excitation voltage of 4 V and a bridge output voltage of 120 μV under load (i.e., Ei = 4 V and 0 E =120 μV ):
Answer:
a. The maximum strain is 60 * 10^-6
b. Resistance change = 0.014395 ohms
Explanation:
The complete question is as follows;
Two strain gauges are mounted so that they sense axial strain on a steel member in uniaxial tension. The 120 V gauges form two legs of a Wheatstone bridge, and are mounted on opposite arms (e.g., arms 1 and 4). The gauge factor for each of the strain gauges is 2 and E m for this steel is 29 × 106 psi. For a bridge excitation voltage of 4 V and a bridge output voltage of 120 μV under load (i.e., Ei = 4 V and 0 E =120 μV ):
(a) Estimate the maximum strain.
(b) What is the resistance change experienced by each gauge?
solution;
Please check attachment for complete solution and step by step explanation
What pushes against gravity in: a main sequence star, a white dwarf, a neutron star, and a black hole? electron degeneracy, neutron degeneracy, nothing, and heat pressure nothing in all cases. Gravity always wins. nothing, heat pressure, electron degeneracy, and neutron degeneracy heat pressure, neutron degeneracy, electron degeneracy, and nothing heat pressure, electron degeneracy, neutron degeneracy, and nothing
A main sequence star relies on heat pressure to counteract gravity, a white dwarf uses electron degeneracy pressure, a neutron star uses neutron degeneracy pressure, and in a black hole, gravity ultimately triumphs.
Explanation:In a main sequence star, it is the heat pressure, due to nuclear fusion, which pushes against gravity and maintains the star's stability. A white dwarf is stabilized against gravity by what is known as electron degeneracy pressure, a quantum mechanical effect that arises because no two electrons can be in the same place doing the same thing at the same time (the Pauli exclusion principle).
The neutron star is supported against gravity by neutron degeneracy pressure, the same quantum effect but explained by neutrons instead of electrons. However, if the mass of a star's core is high enough (more than about three times that of the Sun), no known force can stop it from collapsing into a black hole; gravity ultimately overcomes all other forces.
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Three liquids are at temperatures of 5 ◦C, 25◦C, and 34◦C, respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is 17◦C. Equal masses of the second and third are then mixed, and the equilibrium temperature is 30.7 ◦C. Find the equilibrium temperature when equal masses of the first and third are mixed. Answer in units of ◦C.
Answer:
m * c1 * (9 - 6) = m * c2 * (19 - 9) → mass m cancels
c2 = c1 * 3 / 10 = 0.3*c1
Explanation:
However, for the second mix, there should be nothing from the first mix's answer except for 0.3c1. Essentially, the left side of the second mix equation should look like: m *0.3c1 * (31.1-19). Then, you can properly solve the equation.
Consider the following descriptions of the vertical motion of an object subject only to the acceleration due to gravity. Begin with the acceleration equation a(t)equals=v primev′(t)equals=g, where gequals=minus−9.8 m divided by s squared9.8 m/s2. a. Find the velocity of the object for all relevant times. b. Find the position of the object for all relevant times. c. Find the time when the object reaches its highest point. What is the height? d. Find the time when the object strikes the ground. A softball is popped up vertically (from the ground) with a velocity of 30 m divided by s30 m/s.
Answer:
a) [tex]v = v_{o} - \left(9.8\,\frac{m}{s^{2}} \right)\cdot t[/tex], b) [tex]s = s_{o} + v_{o}\cdot t - \left(9.8\,\frac{m}{s^{2}} \right)\cdot t^{2}[/tex], c) [tex]t = \frac{v_{o}}{9.8\,\frac{m}{s^{2}} }[/tex], d) [tex]t = \frac{v_{o}}{2\cdot \left(9.8\,\frac{m}{s^{2}}\right)}+\frac{\sqrt{v_{o}^{2}+4\cdot s_{o}\cdot \left(9.8\,\frac{m}{s^{2}} \right)} }{2\cdot \left(9.8\,\frac{m}{s^{2}} \right)}[/tex]
Explanation:
a) The acceleration of the object is:
[tex]a = - 9.8\,\frac{m}{s^{2}}[/tex]
The velocity function is found by integration:
[tex]v = v_{o} - \left(9.8\,\frac{m}{s^{2}} \right)\cdot t[/tex]
b) The position function is found by integrating the velocity function:
[tex]s = s_{o} + v_{o}\cdot t - \left(9.8\,\frac{m}{s^{2}} \right)\cdot t^{2}[/tex]
c) The time when the object reaches its highest point ocurrs when speed is zero:
[tex]0\,m = v_{o} - \left(9.8\,\frac{m}{s^{2}} \right)\cdot t[/tex]
[tex]t = \frac{v_{o}}{9.8\,\frac{m}{s^{2}} }[/tex]
d) The time when the object hits the ground occurs when [tex]s = 0\,m[/tex]. The roots are found by solving the second-order polynomial:
[tex]t = \frac{-v_{o}\pm \sqrt{v_{o}^{2}+4\cdot s_{o}\cdot \left(9.8\,\frac{m}{s^{2}} \right)} }{2\cdot (-9.8\,\frac{m}{s^{2}} )}[/tex]
[tex]t = \frac{v_{o}}{2\cdot \left(9.8\,\frac{m}{s^{2}}\right)} \mp \frac{\sqrt{v_{o}^{2}+4\cdot s_{o}\cdot \left(9.8\,\frac{m}{s^{2}} \right)} }{2\cdot \left(9.8\,\frac{m}{s^{2}} \right)}[/tex]
Since time is a positive variable and [tex]v_{o} < \sqrt{v_{o}^{2}+4\cdot s_{o}\cdot \left(9.8\,\frac{m}{s^{2}} \right)}[/tex], the only possible solution is:
[tex]t = \frac{v_{o}}{2\cdot \left(9.8\,\frac{m}{s^{2}}\right)}+\frac{\sqrt{v_{o}^{2}+4\cdot s_{o}\cdot \left(9.8\,\frac{m}{s^{2}} \right)} }{2\cdot \left(9.8\,\frac{m}{s^{2}} \right)}[/tex]
The physics of vertical motion under gravity deals with velocity, position, and time. Velocity over time can be calculated by integrating the acceleration function, while position over time is calculated by integrating the velocity function. The height and time at the peak of motion, as well as when the object hits the ground, are determined by setting these equations to zero at appropriate points.
Explanation:Considering the physics of vertical motion under gravity, the concepts of velocity, position, and time are important. The acceleration needs to be considered for handling such scenarios, which in your case, has been described by the equation a(t)=v′(t)=g. The value of g=−9.8 m/s², where 'g' represents the acceleration due to gravity.
a. Velocity over time: Using the equation of acceleration a=v′=g, and because g is a constant, the velocity at any point can be determined by integrating the acceleration function. This gives v(t)=gt+c, where the constant c is the initial velocity. For a softball with an initial vertical velocity of 30 m/s, this becomes v(t)=-9.8*t+30.
b. Position over time: The position equation can be found by integrating the velocity function, that is: s(t)=½gt²+vt+s₀. The s₀ term is the initial height, which is 0 because the softball is popped up from the ground. Therefore, s(t)=-4.9*t²+30*t.
c. Time and height at the highest point: The motion reaches its highest point when the velocity is zero. So, set v(t) equal to zero and solve for t. Substituting this time into the position equation gives the maximum height.
d. Time when the softball strikes the ground: The softball will hit the ground when its position is again zero. Set s(t) equal to zero and solve for t to find this time.
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Suppose that you are visiting Champaign, IL and on July 10 you wake up early and note the rising azimuth of the Sun. How would the direction of the rising Sun change if you measured
Answer:
It would raise further south
Explanation:
To those of us who live on earth, the most important astronomical object by far is the sun. It provides light and warmth. Its motions through our sky cause day and night, the passage of the seasons, and earth's varied climates.
On any given day, the sun moves through our sky in the same way as a star. It rises somewhere along the eastern horizon and sets somewhere in the west. If you live at a mid-northern latitude (most of North America, Europe, Asia, and northern Africa), you always see the noon sun somewhere in the southern sky.
The sun's path through the rest of the sky is similarly farther north in June and farther south in December. In summary:
In late March and late September (at the "equinoxes"), the sun's path follows the celestial equator. It then rises directly east and sets directly west. The exact dates of the equinoxes vary from year to year but are always near March 20 and September 22.
After the March equinox, the sun's path gradually drifts northward. By the June solstice (usually June 21), the sun rises considerably north of due east and sets considerably north of due west. For mid-northern observers, the noon sun is still toward the south, but much higher in the sky than at the equinoxes.
After the June solstice, the sun's path gradually drifts southward. By the September equinox, its path is again along the celestial equator. The southward drift then continues until the December solstice (usually December 21), when the sun rises considerably south of due east and sets considerably south of due west.
Answer:
It would rise further south
Explanation:
To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^v→=(3.00×104m/s)j^. What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field B⃗ =(1.63T)i^+(0.980T)j^B→=(1.63T)i^+(0.980T)j^?
Answer:
The magnitude of the acceleration is [tex]a = 0.33 m/s^2[/tex]
The direction is [tex]- \r k[/tex] i.e the negative direction of the z-axis
Explanation:
From the question we are that
The mass of the particle [tex]m = 1.8*10^{-3} kg[/tex]
The charge on the particle is [tex]q = 1.22*10^{-8}C[/tex]
The velocity is [tex]\= v = (3.0*10^4 m/s ) j[/tex]
The the magnetic field is [tex]\= B = (1.63T)\r i + (0.980T) \r j[/tex]
The charge experienced a force which is mathematically represented as
[tex]F = q (\= v * \= B)[/tex]
Substituting value
[tex]F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \ ( 1.63 \r i + 0.980 \r j )T)[/tex]
[tex]= 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \ X \ \ \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \ \r j))[/tex]
[tex]= (-5.966*10^4 N) \r k[/tex]
Note :
[tex]i \ \ X \ \ j = k \\\\j \ \ X \ \ k = i\\\\k \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \ X \ \ j = -i\\\\i \ \ X \ \ k = - j\\[/tex]
Now force is also mathematically represented as
[tex]F = ma[/tex]
Making a the subject
[tex]a = \frac{F}{m}[/tex]
Substituting values
[tex]a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}[/tex]
[tex]= (-0.33m/s^2)\r k[/tex]
[tex]= 0.33m/s^2 * (- \r k)[/tex]
Which of the following correctly describes the result of a voltage difference?
A. Charges more randomly
B. Charges not to move
C. Charges flow from low voltage areas to high voltage areas.
D. Charges flow from high voltage areas to low voltage areas.
Answer:
D from high to low areas
A motorcycle starts from rest at and travels along a straight road with a constant acceleration of 2 6 / ft s until it reaches a speed of 50 / ft s . Afterwards it maintains this speed. Also, when t s 0 , a car located 6000 ft down the road is traveling toward the motorcycle at a constant speed of 30 / ft s . Determine the time and the distance traveled by the motorcycle when they pass each other.
Answer:
Time = 77.63 s
Distance = 3673.3 ft
Explanation:
Using equation of motion
v = u + at'
50 = 6 * t'
t' = 50 / 6
t' = 8.33 s
v² = u² + 2a(s - s•)
50² = 0² + 2 * 6 * (s - 0)
2500 = 12 * s
s' = 2500 / 12
s' = 208.3 ft
At t' = 8.33 the distance traveled by the car is
s'' = v• * t'
s'' = 30 * 8.33
s'' = 250 ft
Now, the distance between the motorcycle and car is
6000 - 208.33 - 250 = 5541.67
For motorcycle, when passing occurs,
s = v• * t, x = 50 * t''
For the car, when passing occurs,
s = v• * t, 5541.67 - x = 30 * t''
Solving both simultaneously, we have
5541.67 - [50 * t''] = 30 * t''
5541.67 = 30t''+ 50t''
5541.67 = 80t''
t''= 5541.67 / 80
t''= 69.3 s, substituting t'' = 69.3 back, we have
x = 50 * 69.3
x = 3465 ft
Therefore for motorcycle,
t = 69.3 + 8.33
t = 77.63 s
S = 3465 + 208.3
s = 3673.3 ft
Final Answer:
The total time [tex]\( t \)[/tex] for the motorcycle to pass the car is calculated to be [tex]\( 77.63 \)[/tex] seconds, and the distance traveled by the motorcycle when passing the car is approximately [tex]\( 3673.3 \)[/tex]feet.
Explanation:
The calculation begins by using the equation of motion ( v = u + at ), where (v ) represents final velocity, ( u ) is initial velocity, ( a ) is acceleration, and (t) is time. Substituting the given values, the time ( t' ) for the motorcycle to reach a speed of ( 50 ) ft/s with a constant acceleration of [tex]( 6\) ft/s\(^2\) )[/tex]is found to be approximately ( 8.33 ) seconds.
Next, the equation of motion [tex]\( v^2 = u^2 + 2a(s - s_0) \)[/tex] is used, where [tex]\( s \)[/tex] represents distance, [tex]\( s_0 \)[/tex] is initial distance, and other variables have the same meanings as before. Solving for [tex]\( s' \),[/tex] the distance traveled by the motorcycle during this time is approximately [tex]\( 208.3 \)[/tex] feet.
At [tex]\( t' = 8.33 \)[/tex] seconds, the distance traveled by the car is calculated to be [tex]\( 250 \)[/tex]feet.
The distance between the motorcycle and car when they pass each other is found to be approximately [tex]\( 5541.67 \)[/tex] feet.
To determine the time [tex]\( t'' \)[/tex] when the passing occurs, simultaneous equations for the motorcycle and car are set up using their respective distances. Solving these equations yields [tex]\( t'' = 69.3 \)[/tex] seconds. Substituting this value back, the distance traveled by the motorcycle when passing the car is approximately [tex]\( 3465 \)[/tex]feet.
A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 49° with respect to the normal to the surface. It passes into the second medium and refracts at an angle of θ2 = 69.5° with respect to the normal.
Required:
a. Write an expression for the index of refraction of the second material.
b. Numerically, what is this index?
c. Numerically, what is the light's velocity in medium 1, in meters per second?
d. Numerically, what is the light's velocity in medium 2, in meters per second?
Answer:
a) n2=(n1sin1)(sin2)
b) 1.18
c) 201081632.7m/s
d) 254237288.1m/s
Explanation:
a) We can calculate the index of refraction of the second material by using the Snell's law:
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]
[tex]n_2=\frac{n_1sin\theta_1}{sin\theta_2}[/tex]
b) By replacing in the equation of a) we obtain:
[tex]n_2=\frac{(1,47)sin49\°}{sin69.5\°}=1.18[/tex]
c) light velocity in the medium is given by:
[tex]v=\frac{c}{n_1}=\frac{3*10^{8}m/s}{1.47}=204081632.7\frac{m}{s}[/tex]
d)
[tex]v=\frac{c}{n_2}=\frac{3*10^{8}m/s}{1.18}=254237288.1\frac{m}{s}[/tex]
hope this helps!!
The index of refraction for the second material can be found using Snell's Law, which is given by: n₁sin(θ₁) = n₂sin(θ₂). The refractive index of the second material is approximately 1.19, The light's velocity in medium 1 is 2.04 x 10⁸ m/s. The light's velocity in medium 2 is 2.52 x 10⁸ m/s.
Using the provided angles and refractive index, we calculate the index of the second material to be approximately 1.19, with the corresponding light velocities as 2.04 x 10⁸ m/s in medium 1 and 2.52 x 10⁸ m/s in medium 2.
This question involves the concept of light refraction and the application of Snell's Law. Let's go through each part step-by-step.
The index of refraction for the second material can be found using Snell's Law, which is given by:
n₁sin(θ₁) = n₂sin(θ₂)where:
n₁ = 1.47 (refractive index of the first medium)θ₁ = 49° (angle of incidence)θ₂ = 69.5° (angle of refraction)Rearranging for n₂ gives:
n₂ = (n₁sin(θ₁))/sin(θ₂)Using the values:
n₂ = (1.47 * sin(49°)) / sin(69.5°)Let's calculate it step by step:
sin(49°) ≈ 0.7547sin(69.5°) ≈ 0.9336n₂ ≈ (1.47 * 0.7547) / 0.9336n₂ ≈ 1.1886So, the refractive index of the second material is approximately 1.19.
The velocity of light in a medium is given by:
v = c/nwhere c is the speed of light in vacuum (approximately 3.00 x 10⁸ m/s), and n is the refractive index of the medium. For medium 1:
v₁ = 3.00 x 10⁸ m/s / 1.47v₁ ≈ 2.04 x 10⁸ m/sSimilarly, for medium 2:
v₂ = 3.00 x 10⁸ m/s / 1.19v₂ ≈ 2.52 x 10⁸ m/sA quantity of gas with an initial volume of 5 cubic feet and a pressure of 1700 pounds per square foot expands to a volume of 9 cubic feet. Find the work done by the gas for the given volume and pressure. Round your answer to two decimal places. Assume the temperature of the gas in this process remain constant.
Answer:
Work done by the gas for the given volume and pressure [tex]= 4996.18[/tex] pounds foot
Explanation:
Given
Pressure applied [tex]= 1700[/tex] pounds per square foot
Initial Volume [tex]= 5[/tex] cubic feet
Final Volume [tex]= 9[/tex] cubic feet
As we know pressure is inversely proportional to V
[tex]P = \frac{k}{V}[/tex]
where k is the proportionality constant
V is the volume and
P is the pressure
Work done
[tex]\int\limits^{V_2}_{V_1} {P} \, dV[/tex]
[tex]\int\limits^{V_2}_{V_1} {\frac{k}{V} } \, dV\\= \int\limits^{V_2}_{V_1} {\frac{1700 * 5}{V} } \, dV\\= 8500* \int\limits^{V_2}_{V_1} {\frac{1}{V} } \, dV[/tex]
Integrating the above equation, we get-
[tex]8500 ln \frac{V_2}{V_1} \\8500 * 2.303 * \frac{9}{5} \\= 4996.18[/tex]
Work done by the gas for the given volume and pressure [tex]= 4996.18[/tex] pounds foot
A rod of mass M = 3.5 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 45 g, moving with speed v = 5.02 m/s, strikes the rod at angle θ = 42° from the normal at a distance D = 2/3 L, where L = 1.2 m, from the point of rotation and sticks to the rod after the collision.
What is the initial angular momentum of the ball, in kilogram meters squared per second, right before the collision relative to the pivot point of the rod?
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The initial angular momentum is [tex]L_i= 0.134 Kg .m^2/s[/tex]
Explanation:
From the question we are told that
The mass of the rod is [tex]M = 3.5 kg[/tex]
The mass of the ball is [tex]m =45g = \frac{45}{1000} = 0.045kg[/tex]
The speed is [tex]v = 5.02 m/s[/tex]
The angle the ball strikes the rod is [tex]\theta = 42^o[/tex]
The distance from the center of the rod is [tex]D = \frac{2}{3} L[/tex]
The length L is [tex]= 1.2m[/tex]
The initial angular momentum of the ball is mathematically represented as
[tex]L_i = m\ v\ D\ cos \theta[/tex]
Substituting the value
[tex]L_i = 45*10^{-3} * 5.02 * \frac{2}{3} * 1.2 * cos (42)[/tex]
[tex]= 0.134 Kg .m^2/s[/tex]
Following are the calculation of the initial angular momentum of the ball:
Given:
[tex]M = 3.5 \ kg \\\\m = 45\ g = 0.045\ kg\\\\ v = 5.02\ \frac{m}{s}\\\\ \theta = 42^{\circ}\\\\D = \frac{2}{3}\ L\\\\ L = 1.2\ m\\\\[/tex]
To find:
Li=?
Solution:
Using formula:
[tex]Li= m v D \cos \theta=mv\times \frac{2L}{3}\cos \theta\\\\[/tex]
[tex]=0.045 \times 5.02 \times \frac{2 \times 1.2}{3}\cos 42^{\circ}\\\\=0.2259 \times\frac{2 \times 1.2}{3}\cos 42^{\circ}\\\\=0.18072\times \cos 42^{\circ}\\\\=0.13430113286\\\\=13.43 \times 10^{-2} \ \frac{kg\ m^2}{s}\\\\[/tex]
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In a solar system far far away, the sun's intensity is 100 W/m2 for a planet located a distance R away. What is the sun's intensity for a planet located at a distance 5 R from the Sun
Answer:
The sun's intensity for a planet located at a distance 5 R from the Sun is 4 W/m²
Explanation:
The intensity of Sun located at a distance of R away is [tex]100\ W/m^2[/tex].
We need to find the sun's intensity for a planet located at a distance 5 R from the Sun. Intensity is given by :
[tex]I\propto \dfrac{1}{r^2}[/tex]
So,
[tex]\dfrac{I_1}{I_2}=(\dfrac{r_2}{r_1})^2[/tex]
Here, I₁ = 100 W/m² and r₁ = R and r₂ = 5R
[tex]I_2=\dfrac{I_1r_1^2}{r_2^2}[/tex]
[tex]I_2=\dfrac{100(R^2)}{(5R)^2}\\\\I_2=\dfrac{100}{25}\\\\I_2=4\ W/m^2[/tex]
So, the sun's intensity for a planet located at a distance 5 R from the Sun is 4 W/m².
The plates of a parallel-plate capacitor are maintained with a constant voltage by a battery as they are pushed together, without touching. How is the amount of charge on the plates affected during this process? Question 3 options: The amount of charge on the plates becomes zero. The amount of charge remains constant. The amount of charge on the plates increases during this process. The amount of charge on the plates decreases during this process.
Answer:
The amount of charge on the plates increases during this process
Explanation:
In order to maintain a constant voltage across the plates, pushing the plates closer will increase the charge on the plate
Two very large parallel metal plates, separated by 0.20 m, are connected across a 12-V source of potential. An electron is released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, how much kinetic energy (J) has the electron gained
Answer:
[tex]{\rm K} = 2.4\times 10^{-19}~J[/tex]
Explanation:
The electric field inside a parallel plate capacitor is
[tex]E = \frac{Q}{2\epsilon_0 A}[/tex]
where A is the area of one of the plates, and Q is the charge on the capacitor.
The electric force on the electron is
[tex]F = qE = \frac{qQ}{2\epsilon_0 A}[/tex]
where q is the charge of the electron.
By definition the capacitance of the capacitor is given by
[tex]C = \epsilon_0\frac{A}{d} = \frac{Q}{V}\\\frac{Q}{\epsilon_0 A} = \frac{V}{d} = \frac{12}{0.20} = 60[/tex]
Plugging this identity into the force equation above gives
[tex]F = \frac{qQ}{2\epsilon_0 A} = \frac{q}{2}(\frac{Q}{\epsilon_0 A}) = \frac{q}{2}60 = 30q[/tex]
The work done by this force is equal to change in kinetic energy.
W = Fx = (30q)(0.05) = 1.5q = K
The charge of the electron is [tex]1.6 \times 10^{-19}[/tex]
Therefore, the kinetic energy is [tex]2.4\times 10^{-19}[/tex]
a correlation coefficient represents what two things
Answer:
A correlation coefficient represents the following:
1- The direction of the relationship
2- The strength of the relationship
A correlation coefficient represents the relationship between two variables. It quantifies the strength and direction of the relationship.
Explanation:A correlation coefficient represents the relationship between two variables. It quantifies the strength and direction of the relationship. The correlation coefficient ranges from -1 to 1, where a value close to -1 indicates a strong negative correlation, a value close to 1 indicates a strong positive correlation, and a value close to 0 indicates no or weak correlation.
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A laser beam has diameter 1.30 mm . The beam exerts a force of 3.6×10−9 N on a totally reflecting surface.What is the amplitude of the electric field of the electromagnetic radiation in this beam?
Answer:
[tex]1.75\cdot 10^4 V/m[/tex]
Explanation:
For a totally reflecting surface, the radiation pressure is related to the intensity of radiation by
[tex]p=2\frac{I}{c}[/tex] (1)
where
I is the intensity of radiation
c is the speed of light
The radiation pressure is given by
[tex]p=\frac{F}{A}[/tex]
where in this case:
[tex]F=3.6\cdot 10^{-9}N[/tex] is the force exerted by the beam
A is the area on which the force is exerted
The beam has a diameter of d = 1.30 mm, so its area is
[tex]A=\pi (\frac{d}{2})^2=\pi (\frac{0.0013 m}{2})^2=1.33\cdot 10^{-6} m^2[/tex]
Now we can write eq(1) as
[tex]\frac{F}{A}=\frac{2I}{c}[/tex]
From which we find the intensity of radiation, I:
[tex]I=\frac{Fc}{2A}=\frac{(3.6\cdot 10^{-9})(3.0\cdot 10^8)}{2(1.33\cdot 10^{-6})}=4.06\cdot 10^5 W/m^2[/tex]
Now we can find the amplitude of the electric field using the equation
[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]
where
[tex]\epsilon_0=8.85\cdot 10^{-12}F/m[/tex] is the vacuum permittivity
E is the amplitude of the electric field
And solving for E,
[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(4.06\cdot 10^5)}{(3\cdot 10^8)(8.85\cdot 10^{-12})}}=1.75\cdot 10^4 V/m[/tex]
A radio transmitting station operating at a frequency of 115 MHzMHz has two identical antennas that radiate in phase. Antenna BB is 9.05 mm to the right of antenna AA. Consider point PP between the antennas and along the line connecting them, a horizontal distance xx to the right of antenna A For what values of x will constructive interference occur at point P?
Final answer:
Constructive interference at point P occurs when the path length difference between the two antennas is an integer multiple of the wavelength. The values of x where this occurs can be found using the equation path length difference = n * wavelength. Solving for n, we find that constructive interference occurs at x = 0.00906 m.
Explanation:
In order to have constructive interference at point P, the path length difference between the two antennas must be an integer multiple of the wavelength. Since the antennas are identical and radiating in phase, the path length difference is simply the distance between the two antennas, which is 9.05 mm.
The wavelength can be calculated using the formula: wavelength = speed of light / frequency. For a frequency of 115 MHz, the wavelength is approximately 2.6 m.
To find the values of x where constructive interference occurs, we can set up an equation: path length difference = n * wavelength, where n is an integer. So, 9.05 mm = n * 2.6 m.
Solving for n, we get:
n = (9.05 mm) / (2.6 m) = 0.00348
Therefore, constructive interference will occur at point P when x = n * wavelength. Substituting n = 0.00348 and wavelength = 2.6 m, we get:
x = (0.00348)(2.6 m) = 0.00906 m
This problem has been solved! See the answer A proton with a speed of 3.5x10^6 m/s is shot into a region between two plates that are separated by distance of 0.23 m. A magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. What must be the magnitude of the magnetic field so the proton just misses colliding with the opposite plate?
Answer:
The magnitude of the magnetic field 'B' is 0.16 Tesla.
Explanation:
The magnitude of the magnetic field 'B' can be determined by;
B = [tex]\frac{mV}{qR}[/tex]
where: m is the mass of proton, V is its speed , q is the charge of proton and R is the distance between the plates.
Given that: speed 'V' of the proton = 3.5 × [tex]10^{6} ms^{-2}[/tex], distance 'R' between the plates = 0.23m, the charge 'q' on proton = 1.9 × [tex]10^{-19}[/tex] C and mass of proton = 1.67 × [tex]10^{-27}[/tex]Kg.
Thus,
B = (1.67 × [tex]10^{-27}[/tex] × 3.5 ×[tex]10^{6}[/tex]) ÷ (1.6 × [tex]10^{-19}[/tex] × 0.23)
= [tex]\frac{5.845 * 10^{-21} }{3.68 * 10^{-20} }[/tex]
= 0.15883
B = 0.16 Tesla
The magnitude of the magnetic field 'B' is 0.16 Tesla.
Below is a schematic of a vapor power plant in which water steadily circulates through the four components. The water flows through the boiler and condenser at constant pressure and through the turbine and pump adiabatically. Kinetic and potential energy effects can be ignored. Process data follow: Process 4-1: constant pressure at 8 MPa from saturated liquid to saturated vapor Process 2-3: constant pressure at 8 kPa from x2
Complete Question:
the first attached file shows the schematic
Answer:
a. irreversible cycle
b. 0.182
c. 0.2646
Explanation:
From the saturated vapour pressure table
When the pressure is 1 MPa
Temperature, T₄₁ = 179.9 °C = 452.9 °k
When Pressure is 20 kPa
Temperature, T₃₂ = 60.06 °C = 333.06 °K
attached images 2, 3, 4 and 5 shows a comprehensive solution to the questions
A vapor power plant uses water as the working substance, which goes through a cycle of becoming vaporized and condensed in order to drive a turbine. The schematic provided includes four components: the boiler, condenser, turbine, and pump. Two specific processes mentioned include 4-1 and 2-3, both of which occur at constant pressures.
Explanation:A vapor power plant operates by using water as the working substance, which goes through a cycle of becoming vaporized, driving a turbine, and then condensing back into a liquid state. In this specific schematic, the water flows through four components: the boiler, condenser, turbine, and pump. The process data provided includes two specific processes: 4-1 which occurs at a constant pressure of 8 MPa, and 2-3 which also occurs at a constant pressure of 8 kPa.
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An initially stationary electron is accelerated by a uniform 640 N/C Electric Field. a) Calculate the kinetic energy of the electron after it has traveled 15 cm in a direction parallel to this field. b) Calculate the speed of the electron after it has traveled 15 cm in a direction parallel to this field.
Answer:
(a) 1.298 * 10^(-4) J
(b) 5.82 * 10^6 m/s
Explanation:
Parameters given:
Electric field, E = 640 N/C
Distance traveled by electron, r = 15 cm = 0.15 m
Mass of electron, m = 9.11 * 10^(-31) kg
Electric charge of electron, q = 1.602 * 10^(-19) C
(a) The kinetic energy of the electron in terms of Electric field is given as:
K = (q² * E² * r²) / 2m
Therefore, Kinetic energy, K, is:
K = [(1.602 * 10^(-19))² * 640² * 0.15²] / [2 * 9.11 * 10^(-31)]
K = {23651.981 * 10^(-38)} / [18.22 * 10^(-31)]
K = 1298.13 * 10^(-7) J = 1.298 * 10^(-4) J
(b) To find the final velocity of the electron, we have to first find the acceleration of the electron. This can be gotten by using the equations of force.
Force is generally given as:
F = ma
Electric force is given as:
F = qE
Therefore, equating both, we have:
ma = qE
a = (qE) / m
a = (1.602 * 10^(-19) * 640) / (9.11 * 10^(-31))
a = 112.54 * 10^(12) m/s² = 1.13 * 10^(14) m/s²
Using one of the equations of motion, we have that:
v² = u² + 2as
Since the electron started from rest, u = 0 m/s
Therefore:
v² = 2 * 1.13 * 10^(14) * 0.15
v² = 3.39 * 10^(13)
v = 5.82 * 10^6 m/s
The velocity of the electron after moving a distance of 15 cm is 5.82 * 10^6 m/s.
(a) The kinetic energy of the electron is 1.53×10⁻¹⁷ J
(b) The final speed of the electron is 5.8×10⁶ m/s
Kinetic energy and final speed:The electron is under motion due to the electrostatic force F produced by the electric field E = 640 N/C.
The equation of motion of the electron can be written as:
F = qE
⇒ ma = qE
where q = 1.6×10⁻¹⁹ C is the charge on the electron
m = 9.1×10⁻³¹ kg is the mass of the electron
and a is the acceleration.
So,
a = qE/m
a = 1.6×10⁻¹⁹×640 / 9.1×10⁻³¹
a = 1.12×10¹⁴ m/s²
The initial speed of the electron is zero, so u = 0
From the third equation of motion:
v² = u² + 2as
where v is the final speed,
and s is the distance traveled = 15cm = 0.15m
v² = 0 + 2× 1.12×10¹⁴×0.15
v = 5.8×10⁶ m/s
So the Kinetic energy of the electron is:
KE = ¹/₂mv²
KE = 0.5×9.1×10⁻³¹×(5.8×10⁶)²
KE = 1.53×10⁻¹⁷ J
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You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headlights that are 0.691 m apart. At what distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source? Take the wavelength of the light to be 539 nm and your pupil diameter as 5.11 mm. ______________km
To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,
[tex]\theta = \frac{1.22 \lambda }{D}[/tex]
Here,
D is diameter of the eye
[tex]D = \frac{1.22 (539nm)}{5.11 mm}[/tex]
[tex]D= 1.287*10^{-4}m[/tex]
The angle that relates the distance between the lights and the distance to the lamp is given by,
[tex]Sin\theta = \frac{d}{L}[/tex]
For small angle, [tex]sin\theta = \theta[/tex]
[tex]sin \theta = \frac{d}{L}[/tex]
Here,
d = Distance between lights
L = Distance from eye to lamp
For small angle [tex]sin \theta = \theta[/tex]
Therefore,
[tex]L = \frac{d}{sin\theta}[/tex]
[tex]L = \frac{0.691m}{1.287*10^{-4}}[/tex]
[tex]L = 5367m[/tex]
Therefore the distance is 5.367km.
You are working in a laser laboratory. Your current project involves suspending spherical glass beads in the Earth's gravitational field using a vertically directed laser beam. Today's experiment involves a black bead of radius r and density rho. Assuming that the radius of the laser beam is the same as that of the bead and that the beam is centered on the bead, determine the minimum laser power required to suspend this bead in equilibrium. (Use any variable or symbol stated above along with the following as necessary: g and c.)
Final answer:
The minimum laser power required to suspend the bead in equilibrium is (3c/8πr³ρg).
Explanation:
In order to suspend the bead in equilibrium, the radiation pressure from the laser beam must be equal to the gravitational force on the bead.
The radiation pressure is given by P = 2I/c, where P is the pressure, I is the laser intensity, and c is the speed of light. The gravitational force is given by F = (4/3)πr³ρg, where F is the force, r is the radius of the bead, ρ is the density, and g is the acceleration due to gravity.
Equating the radiation pressure and gravitational force, we have 2I/c = (4/3)πr³ρg. Rearranging this equation, we get I = (3c/8πr³ρg).
Therefore, the minimum laser power required to suspend the bead in equilibrium is (3c/8πr³ρg).
The lens-makers’ equation applies to a lens immersed in a liquid if n in the equation is replaced by n2/n1. Here n2 refers to the index of refraction of the lens material and n1 is that of the medium surrounding the lens. (a) A certain lens has focal length 79.0 cm in air and index of refraction 1.55. Find its focal length in water. (b) A certain mirror has focal length 79.0 cm in air. Find its focal length in water.
Answer:
a
The focal length of the lens in water is [tex]f_{water} = 262.68 cm[/tex]
b
The focal length of the mirror in water is [tex]f =79.0cm[/tex]
Explanation:
From the question we are told that
The index of refraction of the lens material = [tex]n_2[/tex]
The index of refraction of the medium surrounding the lens = [tex]n_1[/tex]
The lens maker's formula is mathematically represented as
[tex]\frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2} ][/tex]
Where [tex]f[/tex] is the focal length
[tex]n[/tex] is the index of refraction
[tex]R_1 and R_2[/tex] are the radius of curvature of sphere 1 and 2 of the lens
From the question When the lens in air we have
[tex]\frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2} ][/tex]
When immersed in liquid the formula becomes
[tex]\frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2} ][/tex]
The ratio of the focal length of the the two medium is mathematically evaluated as
[tex]\frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }[/tex]
From the question
[tex]f_{air }[/tex]= 79.0 cm
[tex]n_2 = 1.55[/tex]
and the refractive index of water(material surrounding the lens) has a constant value of [tex]n_1 = 1.33[/tex]
[tex]\frac{f_{water}}{79} = \frac{1.55- 1}{\frac{1.55}{1.44} -1}[/tex]
[tex]f_{water} = 262.68 cm[/tex]
b
The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.
A charge of +3.0 µC is distributed uniformly along the circumference of a circle with a radius of 20 cm. How much external energy is required to bring a charge of 25µC from infinity to the center of the circle?
Answer:
3.37x10^6J
Explanation:
The external energy is given as KQ1Q2/r
=(9*10^9)*(3*10^-3)*(25*10^-3)/(0.2)
=3.375 *10^6 J
The external energy required can be calculated using the electric potential energy formula. The external energy required to bring the charge of 25µC from infinity to the center of the circle is 337.5 joules.
Explanation:The external energy required to bring a charge of 25µC from infinity to the center of the circle can be calculated using the electric potential energy formula, which is given by:
U = k * (q1 * q2) / r
Where U is the electric potential energy, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges.
In this case, the charge at the center of the circle is 25µC, and the charge distributed uniformly along the circumference is 3.0µC. The radius of the circle is given as 20 cm. Plugging these values into the formula, we have:
U = (9 * 10^9 Nm^2/C^2) * (25 * 10^-6 C) * (3 * 10^-6 C) / (0.20 m) = 337.5 J
Therefore, the external energy required to bring the charge of 25µC from infinity to the center of the circle is 337.5 joules.
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A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like protons or electrons. You may have seen it used to make human hair stand on end or produce large sparks. One application of the Van de Graaff generator is to create x-rays by bombarding a hard metal target with the beam. Consider a beam of protons at 2.00 keV and a current of 5.05 mA produced by the generator. (a) What is the speed of the protons (in m/s)? b) How many protons are produced each second?
Answer:
v = 1.95*10^5 m/s
3.13 x 10^16 proton
Explanation:
Identify the unknown:
The speed of the protons
List the Knowns:
Energy of protons KE = 2 keV = 2*10^3 eV
Current produced by the generator: I= 5.05 mA = 5 x 10^-3 A
1 eV = 1.6 x 10-19 Joule
Mass of proton: m = 1.67 x 10^-27 kg
Charge of proton: q_p = 1.6 x 10^-19 C
Set Up the Problem:
Kinetic energy is given by:
KE= 1/2mv^2
v = √2KE/m
Solve the Problem:
v = √2 x 2*10^3 x 1.6 x 10^-19/ 1.67 x 10^-27
v = 1.95*10^5 m/s
b. Identify the unknown:
Number of protons produced each second
Set Up the Problem:
Current is given by:
I =ΔQ/Δt
So, the total charge in one second:
ΔQ =I*Δt = 5 x 10^-3 x 1 = 5 x 10^-3 C
Number of protons in this charge:
n = ΔQ/q_p
Solve the Problem:
n = 5 x 10^-3/1.6 x 10^-19
=3.13 x 10^16 proton
(6 points) A cylinder has a radius of 12 mm and a length of 2.5 m. It is made of steel with a thermal conductivity of LaTeX: 53.4 \frac{\mathrm{W}}{\mathrm{m K}}53.4 W m K. One end of the cylinder is held at a temperature of LaTeX: 361^{\circ}\text{C}361 ∘ C, the other at LaTeX: 105^{\circ}\text{C}105 ∘ C. In steady state how much power will flow through the cylinder in the form of heat?
Answer:
2.46 W
Explanation:
Thermal conductivity k = 53.4 W/m-K
Radius r = 12 mm = 12x10^-3 m
Lenght = 2.5 m
T1 = 361 °C
T2 = 105 ∘C
Area A = ¶r^2 = 3.142 x (12x10^-3)^2
= 0.00045 m^2
Power P = -AkdT/dx
Where dT = 361 - 105 = 256
dx = lenght of heat travel = 2.5 m
P = -0.00045 X 53.4 X (256/2.5)
= -0.024 X 102.4 = -2.46 W
The negative sign indicates that heat is lost from the metal.
A uniform ladder of length l rests against a smooth, vertical wall. If the mass of the ladder is m and the coefficient of static friction between the ladder and the ground is 0.40, find the minimum angle at which the ladder does not slip.
What If? What if a person begins to climb the ladder when the angle is 51°? Will the presence of a person on the ladder make it more or less likely to slip?
Answer:
51.3 Degrees
The presence of person above the center of mass of ladder will make the ladder more likely to slip.
Explanation:
Explanation in the attachment.
51.3 Degrees, The presence of person above the centre of mass of ladder will make the ladder more likely to slip.
What is uniform ladder?The middle of a uniform ladder is where the center of mass is located. A smooth wall signifies that there is no friction felt on the wall, whereas rough ground indicates that there is friction there.
How to use static equilibrium to determine the coefficient of friction between the bottom of the ladder and the ground is shown in the Ladder Problem from Static Equilibrium. The torque, the x-direction forces, and the y-direction forces are added together and set to zero.
It follows that the force exerted on the ladder as a result of the resultant moment of the forces must also equal zero due to a random point.
Thus, it is 51.3 Degrees.
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Enter the measured values of the angles of incidence and refraction below. Angle of incidence θi = 76.5 Correct: Your answer is correct. Your value is acceptable.° Angle of refraction θr = 76.5 Incorrect: Your answer is incorrect. Your value is too high.° Calculate the index of refraction using Snell's Law and the measured values of the angles of incidence and refraction. nacrylic = The accepted value of the index of refraction for acrylic is 1.49. What is the percent error between the accepted value and the experimental value of n? Hint Percent error = %
Answer:
Explanation:
Given that the inputted angle of incidence is accepted
Angle of incidence θi = 76.5°
But angle of refraction is not acceptable
Angle of refraction θr = 76.5°
We are told that the value is too high
Then, θr < 76.5°
We want to calculate index of refraction n?
The acceptable value of refraction index of acrylic is 1.49
So the true value is 1.49
So, let calculate the measure value
Refractive index is given as
n = Sin(i) / Sin(r)
Then,
n = Sin(76.5) / Sin(76.5)
Then, n = 1
Now, percentage error of the refractive index,
Percentage error is
%error= |true value—measure value| / true value × 100
The true value is 1.49
The measure value is 1
Then,
%error = ( |1.49—1| / 1.49 ) × 100
% error = ( 0.49 / 1.49 ) × 100
%error = 0.3289 × 100
%error = 32.89%
Answer:
Explanation:
32.89%
Wave motion is characterized by two velocities: the velocity with which the wave moves in the medium (e.g., air or a string) and the velocity of the medium (the air or the string itself). Consider a transverse wave traveling in a string. The mathematical form of the wave is y(x,t)=Asin(kx−ωt). a. Find the speed of propagation vp of this wave.b. Find the y velocity vy(x,t) of a point on the string as a function of x and t.
Answer with Explanation:
We are given that the mathematical form of wave
[tex]y(x,t)=Asin(kx-\omega t)[/tex]
a.We have to find the speed of propagation of this wave
In given mathematical form
k=Wave number
[tex]\omega[/tex]=Angular frequency
A=Amplitude
We know that
Speed of propagation of the wave=[tex]v_p=\frac{\omega}{k}[/tex]
b.Differentiate w.r.r t
[tex]v_y(x,t)=-A\omega cos(kx-\omega t)[/tex]
The velocity [tex]v_y(x,t)[/tex] of a point on the string as a function of x and t is given by
[tex]v_y(x,t)=-A\omega cos(kx-\omega t)[/tex]
(a) The speed of propagation of the wave is [tex]\frac{\omega}{k}[/tex]
(b) (b) The velocity of a point on the string as a function of x and t is [tex]v_y (x, t) = -A\omega cos(kx - \omega t)[/tex]
The given parameters;
[tex]y(x, t) = Asin(kx - \omega t)[/tex]
where;
t is the time of motiony is the displacement k is the wave number ω is the angular frequency A is the amplitude of the wave(a) The speed of propagation of the wave is calculated as follows;
[tex]v_p = \frac{\omega }{k}[/tex]
(b) The velocity of a point on the string as a function of x and t is calculated as follows;
[tex]v = \frac{dy}{dt} \\\\v_y (x, t) = -A\omega cos(kx - \omega t)[/tex]
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Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: A 2.50 kg stone thrown upward from the ground at 13.0 m/s returns to the ground in 4.50 s; the circumference of Mongo at the equator is 2.00×10^5km; and there is no appreciable atmosphere on Mongo. The starship commander, Captain Confusion, asks for the following information:
a. What is the mass of Mongo?
b. If the Aimless Wanderer goes into a circular orbit 30,000 km above the surface of Mongo, how many hours will it take the ship to complete one orbit?
Given Information:
Initial speed = v₁ = 13 m/s
time = t = 4.50 sec
Circumference of Mongo = C = 2.0×10⁵ km = 2.0×10⁸ m
Altitude = h = 30,000 km = 3×10⁷ m
Required Information:
a) mass of Mongo = M = ?
b) time in hours = t = ?
Answer:
a) mass of Mongo = M = 8.778×10²⁵ kg
b) time in hours = t = 11.08 h
Explanation:
We know from the equations of kinematics,
v₂ = v₁t - ½gt²
0 = 13*4.50 - ½g(4.50)²
58.5 = 10.125g
g = 58.5/10.125
g = 5.78 m/s²
Newton's law of gravitation is given by
M = gC²/4π²G
Where C is the circumference of the planet Mongo, G is the gravitational constant, g is the acceleration of planet of Mongo and M is the mass of planet Mongo.
M = 5.78*(2.0×10⁸)²/(4π²*6.672×10⁻¹¹)
M = 8.778×10²⁵ kg
Therefore, the mass of planet Mongo is 8.778×10²⁵ kg
b) From the Kepler's third law,
T = 2π*(R + h)^3/2/(G*M)^1/2
Where R = C/2π
T = 2π*(C/2π + h)^3/2/(G*M)^1/2
T = 2π*((2.0×10⁸/2π) + 3×10⁷)^3/2/(6.672×10⁻¹¹*8.778×10²⁵)^1/2
T = 39917.5 sec
Convert to hours
T = 39917.5/60*60
T = 11.08 hours
Therefore, it will take 11.08 hours for the ship to complete one orbit.
To determine the mass of Mongo, we can use the equation of motion for the stone thrown upward and calculate the gravitational acceleration on Mongo. The mass of Mongo can then be found using Newton's law of universal gravitation. To determine the time it takes for the Aimless Wanderer to complete one orbit around Mongo, we can use Kepler's third law of planetary motion.
Explanation:a. What is the mass of Mongo?
To determine the mass of Mongo, we can use the equation of motion for the stone thrown upward and calculate the gravitational acceleration on Mongo. The equation for the motion of the stone is given by:
h = ut + 0.5gt^2
Where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time. From the given information, we can find that the stone reaches a maximum height and returns to the ground in 4.50 s. Plugging in the values:
0.0 = (13.0 m/s)(4.50 s) + 0.5g(4.50 s)^2
Simplifying, we find:
g = 13.0 m/s / 4.50 s
Then, we can use Newton's law of universal gravitation to find the mass of Mongo:
F = G * (m1 * m2) / r^2
Where F is the gravitational force, G is the gravitational constant, m1 is the mass of the stone, m2 is the mass of Mongo, and r is the radius of Mongo. Since the stone is on the surface of Mongo, the radius is equal to the circumference divided by 2*pi:
r = 2.00×10^5 km / (2 * pi)
Substituting the known values, we can solve for m2:
m2 = (F * r^2) / (G * m1)
Calculating the gravitational force between the stone and Mongo using the known values and substituting them in the formula, the mass of Mongo is determined.
b. If the Aimless Wanderer goes into a circular orbit 30,000 km above the surface of Mongo, how many hours will it take the ship to complete one orbit?
To find the time it takes for the Aimless Wanderer to complete one orbit around Mongo, we need to use Kepler's third law of planetary motion. This law states that the square of the orbital period (T) is proportional to the cube of the orbital radius (r):
T^2 = (4 * pi^2 * r^3) / (G * m)
Where T is the orbital period, r is the orbital radius, G is the gravitational constant, and m is the mass of Mongo. Plugging in the known values, we can solve for T and convert it to hours.
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