Answer:
The coefficient of static friction is : 0.36397
Explanation:
When we have a box on a ramp of angle [tex]\alpha[/tex] , and the box is not sliding because of friction, one analyses the acting forces in a coordinate system system with an axis parallel to the incline.
In such system, the force of gravity acting down the incline is the product of the box's weight times the sine of the angle:
[tex]F_g=W\,*\, sin(\alpha)\\F_g=m*g\,*\, sin(\alpha)[/tex]
Recall as well that component of the box's weight that contributes to the Normal N (component perpendicular to the ramp) is given by:
[tex]N=m*g*cos(\alpha)[/tex]
and the force of static friction (f) is given as the static coefficient of friction ([tex]\mu[/tex]) times the normal N:
[tex]f=\mu *m*g*cos(\alpha)[/tex]
When the box starts to move, we have that the force of static friction equals this component of the gravity force along the ramp:
[tex]f=F_g\\\mu *\,m*g*cos(\alpha)=m*g\,*\, sin(\alpha)[/tex]
Now we use this last equation to solve for the coefficient of static friction, recalling that the angle at which the box starts moving is 20 degrees:
[tex]\mu *\,m*g*cos(\alpha)=m*g\,*\, sin(\alpha)\\\mu=\frac{sin(\alpha)}{cos(\alpha)}\\\mu = tan(\alpha)\\\mu=tan(20^o)\\\mu=0.36397[/tex]
The coefficient of static friction (μ) for the box on the 20-degree incline can be found using the tangent of the angle, resulting in μ ≈ 0.3640.
To find the value of the coefficient of static friction (μ) when the box begins to slide at a 20-degree angle, we first need to understand the relationship between the normal force (Fn), the weight of the box (W), and the angle of the incline (θ).
Since the box is sliding at an angle of 20 degrees, we can use the component of the gravitational force perpendicular to the ramp to find Fn. The normal force is given by Fn = Wcos(θ), where W = mg and g is the acceleration due to gravity (9.8 m/s²). In this case, Fn = (12 pounds)(cos(20°)) × 4.44822, converting pounds to Newtons.
At the point of sliding, the static friction force is at its maximum, and it equals the component of the weight of the box parallel to the ramp, which is Wsin(θ). Thus, Fstatic_max = Wsin(θ), and since Fstatic_max = μFn, we can solve for μ.
μ = Fstatic_max / Fn = (Wsin(θ)) / (Wcos(θ)) = tan(θ). Substituting the given values, we get μ = tan(20°) ≈ 0.3640, which is the coefficient of static friction.
At the surface of Venus the average temperature is a balmy 460∘C due to the greenhouse effect (global warming!), the pressure is 92 Earth-atmospheres, and the acceleration due to gravity is 0.894 gEarthgEarth. The atmosphere is nearly all CO2 (molar mass 44.0g/m) and the temperature remains remarkably constant. We shall assume that the temperature does not change at all with altitude.
a. What is the atmospheric pressure of 2.00 km above the surface of Venus? Express your answer in Earth-atmospheres.
b. What is the atmospheric pressure of 2.00 km above the surface of Venus? Express your answer in Venus-atmospheres.
c. What is the root-mean-square speed of the CO2 molecules at the surface of Venus and at an altitude of 2.00 km?
Answer:
a) 86 atm
b) 86 atm
c) 645 m/s
Explanation:
See attachment for calculations on how i arrived at the answer
The atmospheric pressure of 2.00 km above the surface of Venus is 86 atm and the root-mean-square speed of the CO2 molecules at the surface of Venus and at an altitude of 2.00 km is 645 m/sec.
Given :
At the surface of Venus, the average temperature is a balmy 460∘C due to the greenhouse effect (global warming!), the pressure is 92 Earth-atmospheres, and the acceleration due to gravity is 0.894 g.The atmosphere is nearly all CO2 (molar mass 44.0g/m) and the temperature remains remarkably constant.a) and b) In order to determine the atmospheric pressure of 2.00 km above the surface of Venus using the formula given below:
[tex]\rm P = P_0\times L^{\frac{Mg}{\rho T}}[/tex] --- (1)
The value of the expression [tex]\rm Mg/\rho T[/tex] is given below:
[tex]\rm \dfrac{Mg}{\rho T} = \dfrac{44\times 10^{-3}\times 9.8\times 10^3}{8.314\times 733}[/tex]
[tex]\rm \dfrac{Mg}{\rho T}=0.02076[/tex]
Now, substitute the values of the known terms in the expression (1).
[tex]\rm P=92\times L^{0.02076}[/tex]
P = 86 atm
c) The root-mean-square speed of the CO2 molecules at the surface of Venus and at an altitude of 2.00 km can be calculated as given below:
[tex]\rm V_{rms}=\sqrt{\dfrac{3RT}{M}}[/tex]
Now, substitute the values of the known terms in the above formula.
[tex]\rm V_{rms}=\sqrt{\dfrac{3\times 8.314\times 733}{44\times 10^{-3}}}[/tex]
Simplify the above expression.
[tex]\rm V_{rms} = 645\;m/sec[/tex]
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A traffic controller at the airport watches a commercial plane circling above at a distance of 24.0 km as the pilot waits for clearance to land. If the Moon subtends an angle of 9.89 10-3 radians at the controller's location, what is the distance the jet travels as its nose moves across the diameter of the Moon?
Answer:
The distance travelled is [tex]S = 237.36 \ m[/tex]
Explanation:
From the question we are told that
The distance of the airplane from the ground is [tex]L = 24.0\ km[/tex]
The angle subtended by the moon is [tex]\theta = 9.89*10^{-3} \ radians[/tex]
The distance traveled can be mathematically represented as
[tex]S = R \theta[/tex]
Substituting values
[tex]S = 24 *10^3 * 9.89*10^{-3}[/tex]
[tex]S = 237.36 \ m[/tex]
The number of protons in the nucleus of an atom determines the species of the atom, i.e., the element to which the atom belongs. An atom has the same number of protons and neutrons. But the electron number cannot be used instead because
A. electrons are not within the nucleus
B. electrons are negatively charged
C. electrons can be removed from or added to an atom
D.electrons are lighter than protons
The number of protons in the nucleus of an atom determines the element that the atom belongs to.
Explanation:The subject of this question is Chemistry. The number of protons in the nucleus of an atom determines the species or element to which the atom belongs. This is because each element has a unique number of protons, known as the atomic number. The other options are not correct because:
Electrons are not within the nucleus: Electrons are found in electron shells outside the nucleus of an atom.Electrons are negatively charged: While electrons are negatively charged, this does not affect their use in determining the species of an atom.Electrons can be removed from or added to an atom: While electrons can be added or removed from an atom, their number does not define the species of the atom.Electrons are lighter than protons: Although electrons are lighter than protons, this is not the reason why their number cannot be used to determine the species of an atom.Learn more about protons in the nucleus here:https://brainly.com/question/30654172
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9) A bug of mass 0.020 kg is at rest on the edge of a solid cylindrical disk (M = 0.10 kg, R = 0.10 m) rotating in a horizontal plane around the vertical axis through its center. The disk is rotating at 10.0 rad/s. The bug crawls to the center of the disk. (a) What is the new angular velocity of the disk? (b) What is the change in the kinetic energy of the system? (c) What is the cause of the increase and decrease of kinetic energy?
The bug moving toward the center of the disk changes the moment of inertia and the angular velocity, resulting in a change in the rotational kinetic energy of the system. The principles of conservation of angular momentum and calculations of kinetic energy apply here.
Explanation:This problem involves the principle of conservation of angular momentum. Initially, when the bug is at the edge of the disk, the system's angular momentum (L) is conserved. The initial angular momentum (L_initial) is the sum of the angular momenta of the bug and the disk. It can be calculated as L_initial = I_bug(ω) + I_disk(ω) where I_bug = m*R², I_disk = (1/2)*M*R², m is the mass of the bug, M is the mass of the disk, R is the radius of the disk, and ω is the initial angular velocity.
When the bug moves to the center of the disk, the angular velocity will change but the total angular momentum will still remain conserved as L_final = I_bug(ω') + I_disk(ω') where ω' is the final angular velocity and I_bug is now 0 because the bug is at the center. From the conservation of angular momentum, we can solve for ω'.
The change in kinetic energy can be calculated from the difference between the initial kinetic energy and the final kinetic energy of the system. An increase or decrease in kinetic energy in this scenario is due to the displacement of the bug from the edge to the center of the disk, which changes the moment of inertia of the system and thus, the rotational kinetic energy.
Learn more about Conservation of Angular Momentum here:https://brainly.com/question/1597483
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Final answer:
The explanation covers the new angular velocity of the disk, the change in kinetic energy, and the causes for the increase and decrease in kinetic energy.
Explanation:
A: The new angular velocity of the disk can be calculated using the principle of conservation of angular momentum. The bug crawling towards the center decreases the moment of inertia, resulting in an increase in angular velocity.
B: The change in kinetic energy of the system can be determined by calculating the initial and final kinetic energies and finding their difference.
C: The increase in kinetic energy is due to the bug moving towards the center, reducing the moment of inertia. The decrease in kinetic energy occurs due to the redistribution of mass towards the center of rotation.
Which two elements have the same number of valence electrons?
Elements in the same group on the periodic table, like alkali metals lithium and sodium or alkaline earth metals beryllium and magnesium, have the same number of valence electrons. This similarity is a key factor in their chemical reactivity and the formation of covalent bonds, evidenced through electron dot diagrams.
Explanation:Elements that have the same number of valence electrons are located in the same group or column on the periodic table. For instance, the alkali metals such as lithium (Li) and sodium (Na) each possess one valence electron. Similarly, the alkaline earth metals like beryllium (Be) and magnesium (Mg) each have two valence electrons.
Another example can be found in the halogens group, as elements such as fluorine (F) and chlorine (Cl) each exhibit seven valence electrons. These elements have not only the same valence electron count but also exhibit similar chemical properties due to this shared characteristic. It's important to note that the number of valence electrons contributes to an element's ability to bond with others through the loss, gain, or sharing of these electrons.
Understanding that elements in the same group on the periodic table share the same number of valence electrons can greatly aid in predicting their chemical behavior and reactivity. Electron dot diagrams provide a visual representation of the valence electron distribution in each element and are particularly useful for visualizing how elements bond covalently. For example, elements in the first group have a single dot representing the single valence electron they possess.
A parallel-plate capacitor of capacitance 20 µF is fully charged by a battery of 12 V. The battery is then disconnected. A dielectric slab of K = 4 is slipped between the two plates of the capacitor:
(a) Find the change in potential energy of the capacitor.
(b) Does the potential energy increase or decrease? Explain
Answer:
Explanation:
capacitance = 20 x 10⁻⁶ F .
potential V = 12 V
charge = CV
= 20 x 10⁻⁶ x 12
Q = 240 x 10⁻⁶ C
energy = Q² / 2C
= (240 x 10⁻⁶ )² / 2 x 20 x 10⁻⁶
= 1440 x 10⁻⁶ J
b )
In this case charge will remain the same but capacity will be increased 4 times
new capacity C = 4 x 20 x 10⁻⁶
= 80 x 10⁻⁶
energy = Q² / 2C
= (240 x 10⁻⁶ )² / 2 x 4 x 20 x 10⁻⁶
= 360 x 10⁻⁶ J .
potential energy will decrease from 1440 x 10⁻⁶ J to 360 x 10⁻⁶ J
A ring is attached at the center of the underside of a trampoline. A sneaky teenager crawls under the trampoline and uses the ring to pull the trampoline slowly down while his 67-kg mother is sleeping on it. When he releases the trampoline, she is launched upward. As she passes through the position at which she was before her son stretched the trampoline, her speed is 3.0m/s.
How much elastic potential energy did the son add to the trampoline by pulling it down? Assume the interaction is nondissipative.
Answer:
E = 301.5 J
Explanation:
We have,
Mass of mother, m = 67 kg
Here, a sneaky teenager crawls under the trampoline and uses the ring to pull the trampoline slowly down. As she passes through the position at which she was before her son stretched the trampoline, her speed is 3 m/s.
It is required to find the elastic potential energy the son add to the trampoline by pulling it down. It is based on the conservation of energy.
The elastic potential energy of the mother = the elastic potential energy the son add to the trampoline.
[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2[/tex]
So, the elastic potential energy is :
[tex]E=\dfrac{1}{2}mv^2\\\\E=\dfrac{1}{2}\times 67\times 3^2\\\\E=301.5\ J[/tex]
So, the elastic potential energy of 301.5 J the son add to the trampoline by pulling it down.
You are spinning a rock, of mass 0.75 kg, at the end of a string of length 0.86m in a vertical circle in uniform circular motion. What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path? If the string's maximum tension is 45 N, what is the maximum speed the rock can have so that the string does not break? At what point in the vertical circle does this maximum value occur?
Answer:
v (minimum speed) = 2.90 m/sec.
[tex]\\ \\ maximum speed (v)= 6.57 m/sec.\\[/tex]
Maximum value of speed will occur at lowest point of vertical circle.
Explanation:
a) What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path?
Using the force balance expression at the top of the circle,
Gravitational Force + Tension force = Centrifugal force
[tex]m*g + T = m*v^2/R[/tex]
Given that : T = 0
R = length of string = 0.86 m
mass of the spinning rock = 0.75 kg
[tex]v = \sqrt{g*R}[/tex]
[tex]v = \sqrt{9.81*0.86}[/tex]
v (minimum speed) = 2.90 m/sec.
b) what is the maximum speed the rock can have so that the string does not break?
Here the force balance at bottom of circle is represented by the illustration:
[tex]T = m*g + m*v^2/R[/tex]
Given that:
maximum tension T = 45 N
maximum speed v = ??
mass m = 0.75 kg
∴
[tex]45 - 0.75*9.81 = 0.75*\frac{v^2}{0.86} \\\\v^2 = 0.86*(45 - 0.75*9.81)/0.75 \\ v = \sqrt{0.86*(45 - 0.75*9.81)/0.75\\ maximum speed (v)= 6.57 m/sec.\\[/tex]
c)
At what point in the vertical circle does this maximum value occur?
Maximum value of speed will occur at lowest point of vertical circle.
This is so because at the lowest point; the tension in string will be maximum.
Certain neutron stars (extremely dense stars) are believed to be rotating at about 0.83 rev/s. If such a star has a radius of 40 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation? Number Enter your answer in accordance to the question statement Units Choose the answer from the menu in accordance to the question statement
Answer:
The mass of the star would be [tex]M = 2.644*10^{24} \ kg[/tex]
Explanation:
From the question we are told that
The angular speed is [tex]w = 0.83\ rev/s = 0.83 * 2 \pi = 1.66 rad/s[/tex]
The radius of the star is [tex]r = 40km = 40 *1000 = 40 * 10^{3} m[/tex]
Generally the minimum mass of the start is mathematically evaluated as
[tex]M = \frac{r^3 w^2}{G}[/tex]
Where is the gravitational constant with a values of [tex]G = 6.67*10^{-11} N \cdot m^2 /kg[/tex]
[tex]M = \frac{(40*10^3)^3 * 1.66^2}{6.67*10^{-11}}[/tex]
[tex]M = 2.644*10^{24} \ kg[/tex]
A large diameter closed top tank is filled with a depth of 3 meters of a fluid (density 1200 kg/m3). A small pipe leading from the bottom of the tank must carry the fluid to a height of 5 meters (2 meters above the top level of the tank fluid level). a) What pressure must be maintained in the space above the fill line in the tank to provide an exit speed of 5 m/sec for the fluid out of the pipe? (the pressure at the exhaust end of the pipe is 1 atmosphere or about 1 x 105 pascal) 15 pt If the pipe diameter is 2 cm, what will be the initial volume rate of flow (m3 /sec) 5pt
Answer:
1.57×10^-3 m^3/s
Explanation:
Please see the attached filw for a detailed and step by step solution of the given problem.
The attached file explicitly solves it.
a cross section of three parallel wires each carrying a current of 20 A. The current in wire B is out of the paper, while that in wires A and C are into the paper. If the distance R = 15.0 mm,
what is the magnitude of the force on a 200 cm length of wire C?
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The magnitude of force is [tex]F_R} = 5.33 \ m N[/tex]
Explanation:
From the question we are told that
The current in wire A , B and C are [tex]I _a = I_b =I_c = I= 20 A[/tex]
The distance is [tex]R = 15.0mm = \frac{15}{1000} = 0.015m[/tex]
The length of wire C is [tex]L_c = 200cm = \frac{200}{100} = 2 m[/tex]
Generally the force exerted per unit length that is acting in between two current carrying conductors can be mathematically represented as
[tex]\frac{F}{L} = \frac{\mu_o }{2 \pi} \cdot \frac{I_1 I_2 }{R}[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with a constant value of
[tex]\mu_o = 4 \pi * 10^{-7} N/A^2[/tex]
When the current in the wire are of the same direction the force is positive and when they are in opposite direction the force is negative
Considering force between A and C
[tex]F_{A -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_a I_c }{2R} * L[/tex]
Considering force between B and C
[tex]F_{B -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_b I_c }{R} * L[/tex]
The resultant force is
[tex]F_R = F_{B -C} - F_{A -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_a I_c }{2R} * L - \frac{\mu_o }{2 \pi} \cdot \frac{I_b I_c }{R} * L[/tex]
[tex]F_R} = \frac{\mu_o }{2 \pi} \cdot \frac{I * I }{R} * L * ({1 - \frac{1}{2} })[/tex]
Substituting values
[tex]F_R} = \frac{4 \pi * 10^{-7}}{2 * 3.142} \cdot \frac{ 20*20 }{0.015} * 2 ({\frac{1}{2} })[/tex]
[tex]F_R} = 5.33 *10^{-3} N[/tex]
[tex]F_R} = 5.33 \ m N[/tex]
Assume that a certain location on the Earth reflects 33.0% of the incident sunlight from its clouds and surface. (a) Given that the intensity of solar radiation at the top of the atmosphere is 1368 W/m2, find the radiation pressure on the Earth, in pascals, at the location where the Sun is straight overhead. Pa (b) State how this quantity compares with normal atmospheric pressure at the Earth's surface, which is 101 kPa. Patm Prad
Answer:
a) The radiation pressure on the Earth is 6.065x10⁻⁶Pa
b) The atmospheric pressure is 1.665x10¹⁰ greater than the pressure exerted by the radiation.
Explanation:
Given:
I = intensity of solar radiation = 1368 W/m²
Earth reflects 33%, therefore Earth absorbs 67%
P = pressure = 101 kPa = 1.01x10⁵Pa
c = speed of light = 3x10⁸m/s
Questions:
a) Find the radiation pressure on the Earth, in pascals, at the location where the Sun is straight overhead, P = ?
b) State how this quantity compares with normal atmospheric pressure at the Earth's surface
a) To solve, it is necessary to calculate the pressure exerted by both the reflected light and the light that is absorbed, in this way:
The pressure exerted by the reflected light:
[tex]P_{1} =\frac{2*0.33*I}{c} =\frac{2*0.33*1368}{3x10^{8} } =3.01x10^{-6} Pa[/tex]
The pressure exerted by the absorbed light:
[tex]P_{2} =\frac{0.67*I}{c} =\frac{0.67*1368}{3x10^{8} } =3.055x10^{-6} Pa[/tex]
The radiation pressure on the Earth:
Pt = P₁ + P₂ = 3.01x10⁻⁶ + 3.055x10⁻⁶ = 6.065x10⁻⁶Pa
b) Comparing with normal atmospheric pressure
[tex]Ratio=\frac{P_{atm} }{P_{t} } =\frac{1.01x10^{5} }{6.065x10^{-6} } =1.665x10^{10}[/tex]
According to this result, the atmospheric pressure is 1.665x10¹⁰ greater than the pressure exerted by the radiation.
Which part of the electromagnetic Spectrum is nearest to X-rays?
microwaves
infrared light
gamma rays
radio waves
Answer:gamma rays
Explanation:
Out of all these spectrum listed in The question , gamma rays is closest to x-ray in The electromagnetic spectrum
What is heat?
A. A measure of the movement of molecules inside an object
B. The transfer of energy from a hot object to a cold object
C. The force exerted on an area by an object
D. A measure of mass per unit volume of an object
Answer:
B. The transfer of energy from a hot object to a cold object
Explanation:
Answer:
The answer is option B
Explanation:
To apply Problem-Solving Strategy 21.1 Faraday's law. A closely wound rectangular coil of 80 turns has dimensions 25.0 cm by 40.0 cm. The plane of the coil is rotated from a position in which it makes an angle of 37.0 degrees with a magnetic field of 1.10 T to a position perpendicular to the field. The rotation takes 0.0600 s. What is the average emf E induced in the coil
Answer:
58.37 V
Explanation:
Given that
Number of winding on coil, N = 80 turns
Area of the coil, A = 25 cm * 40 cm = 0.25 m * 0.4 m
Magnitude of magnetic force, B = 1.1 T
Time of rotation, t = 0.06 s
See the attachment for calculations
Answer:
ℰ[tex]Ф_{av}[/tex]=58.37V
Explanation:
Given:
Number of winding on coil 'N' = 80 turns
Area 'A' = 25 cm x 40 cm = 0.25 m x 0.4 m =>0.10m²
Magnitude of magnetic force 'B' = 1.1 T
Time ' t' = 0.06 s
The average magnitude of the induced emf is given by:
ℰ[tex]Ф_{av}[/tex]=N|ΔФB/Δt| => N |Ф[tex]Ф_{B,f}[/tex] - Ф[tex]Ф_{B,i}[/tex]| /Δt
The flux through the coil is Ф[tex]Ф_{B[/tex] = BA cos∅
ℰ[tex]Ф_{av}[/tex]=NBA |cos(∅[tex]Ф_{f}[/tex]) - cos(∅[tex]Ф_{i}[/tex])| /Δt
As the initial angle is ∅= 97-37 =>53° and the final angle is ∅=0°
ℰ[tex]Ф_{av}[/tex]=[tex]\frac{(80)(1.1)(0.1)|cos(0)-cos(53)|}{0.06}[/tex]
ℰ[tex]Ф_{av}[/tex]= [tex]\frac{(8.8)|1-0.002|}{0.06}[/tex]
ℰ[tex]Ф_{av}[/tex]=58.37V
A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)=2.30mmcos[(6.98rad/m)x+(742rad/s)t]. Being more practical-minded, you measure the rope to have a length of 1.35 mm and a mass of 3.38 grams. Assume that the ends of the rope are held fixed and that there is both this traveling wave and the reflected wave traveling in the opposite direction.
What is the wave function y(x,t) for the standing wave that is produced?
Answer:
Explanation:
y(x,t)=2.30mmcos[(6.98rad/m)x+(742rad/s)t]
The angular velocity ω = 742 rad /s
wave function k = 6.98 rad /m
Amplitude A = 2.3 mm
y(x,t) = A cos( ωt + kx )
equation of wave reflected wave
y(x,t) = A cos( ωt - kx )
resultant standing wave
= y₁ +y₂
= A cos( ωt + kx ) +A cos( ωt - kx )
y(x,t) = 2 A cosωt cos kx
= 2 x 2.3 cos 742t .cos6.98x
= 4.6 mm . cos 742t .cos6.98x
Consider three starships that pass by an observer on Earth. Starship A is traveling at speed v=c/3v=c/3 relative to Earth and has a clock placed aboard. Starship B is traveling at speed v=c/3v=c/3 relative to Earth and in the same direction as Starship A. Starship C is traveling at speed v=c/3v=c/3 relative to Earth, but in the opposite direction as Starship A. In which reference frame can a time interval be measured that equals the time interval measured by the clock aboard Starship A?
An observer on Starship B will measure the same time interval as that on Starship A, because they are both travelling at the same speed and in the same direction relative to Earth. Starship C's measurement would differ due to its opposite direction of travel relative to A and B.
Explanation:To determine which reference frame measures a time interval equal to the time interval measured by the clock aboard Starship A, we need to consider the principles of the theory of special relativity. Since Starship B is traveling at the same speed and in the same direction as Starship A relative to Earth, the time dilation effect will be the same for both starships. Therefore, an observer on Starship B will measure the same time interval on their clock as that measured on Starship A's clock.
In contrast, Starship C is moving at the same speed but in the opposite direction relative to Earth, and its relative velocity to Starships A and B is not zero. As a result, the time interval measured on Starship C will not match the time interval measured on Starship A's clock.
So, the time interval that equals the time interval measured by the clock aboard Starship A can be measured in the reference frame of Starship B, which is moving at the same speed and in the same direction as Starship A relative to Earth.
There are two possibilities for final stage of extremely massive stars. The first is a
neutron star and the second is a
Answer: Black hole.
Explanation:
As the massive star "compacts" under its own gravity, it triggers a massive supernova, after this point the remains of the star can become a neutron star, which is a very compact star made primarily, as the name says, of neutrons. The other possibility is a black hole, which is a finite region of space wherein it's interior there is a big concentration of mass, which creates a gravitational field strong enough that there is no particle that can escape it.
If a light bulb is missing or broken in a parallel circuit, will the other light bulb right ?
The light pattern shown is a result from a beam of light being passed through a single slit. The pattern is created from constructive and destructive interference. This phenomenon is due to
A) the wave nature of light.
B) the particle nature of light.
C) the resonance effect of light.
D) the photoelectric effect of light.
Answer:
A
Explanation:
PART ONE
A steel railroad track has a length of 28 m
when the temperature is 2◦C.
What is the increase in the length of the
rail on a hot day when the temperature is
35 ◦C? The linear expansion coefficient of
steel is 11 × 10^−6(◦C)^−1
.
Answer in units of m
PART TWO
Suppose the ends of the rail are rigidly
clamped at 2◦C to prevent expansion.
Calculate the thermal stress in the rail if
its temperature is raised to 35 ◦C. Young’s
modulus for steel is 20 × 10^10 N/m^2
Answer in units of N/m^2
Answer:[tex]\Delta L=0.0101\ m[/tex]
Explanation:
Given
Length of track [tex]L_o=28\ m[/tex] when
[tex]T_o=2^{\circ}C[/tex]
Coefficient of linear expansion [tex]\alpha =11\times 10^{-6}\ ^{\circ}C^{-1}[/tex]
When Temperature rises to [tex]T=35^{\circ}C[/tex]
[tex]\Delta T=35-2=33^{\circ}C[/tex]
and we know length expand on increasing temperature
[tex]L=L_o[1+\alpha \Delta T][/tex]
[tex]L-L_o=L_o\alpha \Delta T[/tex]
[tex]\Delta L=28\times 11\times 10^{-6}\times (33)[/tex]
[tex]\Delta L=0.0101=10.164\ mm[/tex]
(b)When rails are clamped thermal stress induced
we know [tex]E=\frac{stress}{strain}[/tex]
[tex]Stress=E\times strain[/tex]
[tex]Stress=20\times 10^{10}\times \frac{\Delta L}{L_o}[/tex]
[tex]Stress=20\times 10^{10}\times \frac{0.0101}{28}[/tex]
[tex]Stress=72.14\ MPa[/tex]
[tex]Stress=72.14\times 10^{6}\ N/m^2[/tex]
Sandra is riding a bicycle at 10.0 meters/second. She slows down to 2.0 meters/second in 10 seconds.
Answer:
Acceleration= final velocity-initial velocity/time taken
so
a= 10-2/10
a=-8/10
a= -.8 meters per second
Answer: 8.0 = 8 meters/seconds.
Explanation:
Started at= 10.0 meters/seconds.
Slows down= 2.0
10-2=8
Sandra is currently riding her bicycle at 8.0 meters/seconds.
*Mark brainlist* please :))
What should Jaime do to increase the number of energy storage molecules that producers can make? You may choose more than one answer.
Decrease the amount of carbon dioxide in the ecosystem. Increase the amount of carbon dioxide in the ecosystem. Decrease the amount of sunlight in the ecosystem. Increase the amount of sunlight in the ecosystem.
Answer: Increase the amount of carbon dioxide in the ecosystem.
Explanation:
The producers are the organisms which can make their own food by conducting the process of photosynthesis. The carbon dioxide, water are the reactants and carbohydrates and oxygen are the products of photosynthesis. The entire process takes place in the presence of sunlight energy.
The carbon dioxide is the chief component for this reaction and the components of carbon are produced in the form of carbohydrates. The carbohydrates are the source of energy that are utilized by the producers for cellular metabolism and other functions. These get stored in the form of storage molecules in producers. Thus to increase the no. of storage molecules that producers can make Jaime need to increase the concentration of carbon dioxide in the ecosystem.
You find an unmarked blue laser on your way to physics class. When you get to class you realize you can determine the wavelength of the laser by doing a double slit experiment. Shining the laser through a double slit with a slit separation of 0.329 mm on the wall 2.20 m away the first bright fringe is 2.90 mm from the center of the pattern. What is the wavelength
Explanation:
We have,
Slit separation, d = 0.329 mm
Distance between slit and screen, D = 2.20 m
The first bright fringe is formed at a distance off 2.9 mm from the center of the pattern. We need to find the wavelength.
For double slit experiment, the fringe width is given by :
[tex]\beta=\dfrac{D\lambda}{d}[/tex]
[tex]\lambda[/tex] is wavelength
[tex]\lambda=\dfrac{\beta d}{D}\\\\\lambda=\dfrac{2.9\times 10^{-3}\times 0.329\times 10^{-3}}{2.2}\\\\\lambda=4.33\times 10^{-7}\ m\\\\\lambda=433\ nm[/tex]
So, wavelength is 433 nm.
Which of the following do not make their own energy through nuclear fusion? Select all that apply.
A. giant star
B. protostar
C. dwarf star
D. black hole
E. neutron star
F. main sequence star
8. A tuning fork vibrating with a sonometer wire 20cm long produces 5 beats per second. Given
that the beat frequency does not change if the length of the wire is changed to 21cm, calculate
the frequency of the tuning fork and frequency of sonometer wire when changed to 21cm.
Answer:
Explanation:
length of wire is proportional to frquency of sound produced by it.
n₁ /n₂ = l₂ / l₁
= 21 / 20
If n be the frequency of tuning fork
n₁ = n + 5 ; n₂ = n - 5 ( no of beat / s = frequency difference )
n + 5 / n - 5 = 21 / 20
20n + 100 = 21n - 105
n = 205 Hz
frequency of tuning fork = 205
the frequency of the tuning fork and frequency of sonometer wire when changed to 21cm. = n - 5 = 200 Hz .
Given the beat frequency and the inverse relationship between the length of the sonometer wire and frequency, we determine the tuning fork frequency. Beat frequency remains constant despite the change in wire length. Calculation involves solving system of linear equations with fundamentals of wave physics.
To find the frequency of the tuning fork and the frequency of the sonometer wire for both lengths, we utilize the concept of beat frequency. The beat frequency is given by the difference in frequency between the tuning fork and the sonometer wire.
Let’s denote the frequency of the tuning fork as ft and the frequency of the sonometer wire for 20 cm as f1. Given that the beat frequency is 5 Hz, we have:
|ft - f1| = 5 Hz
When the length of the wire is changed to 21 cm, assume the new frequency of the sonometer wire to be f2. Beat frequency remains unchanged:
|ft - f2| = 5 Hz
The frequency of a vibrating string is inversely proportional to its length:
f1 * L1 = f2 * L2
Given L1 = 20 cm and L2 = 21 cm,
f1 * 20 = f2 * 21 which simplifies to:
f2 = (20/21) * f1
Since |ft - f1| = 5 and |ft - f2| = 5, solving these equations requires knowing the possible frequencies of the sonometer.
Determine f1 and f2 using the equations:
Solve both scenarios for ft + 5 = f1 and ft - 5 = f2:
If ft= f1 ± 5 Hz,
Scenario 1: for f1 = (n/20), f2 = (20/21) * (n/20) = n/21, verify both.
The instructions for an electric lawn mower suggest that an A gauge extension cord ( cross sectional area = 4.2 x 10-7 m2 ) should only be used for distances up to 35 m. The resistivity of copper (used in the extension cord) is 1.72 x 10-8 .m at 20oC and the temperature coefficient of resistivity of copper is 0.004041 (oC)-1. What is the resistance of a A type extension cord of length 35 m at 40oC
Answer:
1.43 Ω
Explanation:
Given that
Cross sectional area of the cord, A = 4.2*10^-7 m²
Distance meant to be used, L = 35 m
Resistivity of copper, p = 1.72*10^-8 Ωm
Temperature of copper, t = 20° C
Temperature coefficient of copper, t' = 0.004041 °C^-1
To solve this, we would use the resistance and resistivity formula
R = pL / A, on substituting
R = (1.72*10^-8 * 35) / 4.2*10^-7
R = 6.02*10^-7 / 4.2*10^-7
R = 1.43 ohm
Therefore, the resistance of the extension cord is 1.43 ohm
Tamsen and Vera imagine visiting another planet, planet X, whose gravitational acceleration, gX, is different from that of Earth's. They envision a pendulum, whose period on Earth is 2.243 s, that is set in motion on planet X, and the period is measured to be 2.000 s. What is the ratio of gX/gEarth? Neglect any effects caused by air resistance.
Answer:
1.27
Explanation:
Period of a pendulum T is
T = 2¶(l/g)^0.5
Where g is acceleration due to gravity
l is lenght of pendulum
For earth, T = 2.243 s
2.243 = 2 x 3.142 x (l/9.81)^0.5
0.36 = (l^0.5)/3.13
1.13 = l^0.5
l = 1.28 m
For planet X of the same lenght
Period T = 2.00 s
2 = 2 x 3.142 (1.28/gx)^0.5
0.32 = 1.13 / gx^0.5
3.53 = gx^0.5
gx^0.5 = 3.53
gx = 12.46 m/s^2
gx/gearth = 12.46/9.81 = 1.27
The ratio of the gravitational acceleration on planet X to that on Earth is 1.26, calculated using the periods of a pendulum on both planets and the relationship between period and gravitational acceleration.
Explanation:The period of a simple pendulum is given by the formula T = 2*pi*sqrt(L/g) where T is the period of the pendulum, L is the length of the pendulum, and g is the acceleration due to gravity. To find the ratio gX/gEarth, we can use the formula T_earth/T_x = sqrt(g_x/g_earth) (let's denote T_earth as the period on earth and T_x as the period on planet X).
Since we know the periods, we can substitute these values into the formula: 2.243s/2.000s = sqrt(g_x/g_earth), which gives us 1.1215 = sqrt(g_x/g_earth). To find the ratio g_x/g_earth, square both sides, giving (1.1215)² = g_x/g_earth. Thus, g_x/g_earth = 1.26.
Learn more about Gravitational Acceleration here:https://brainly.com/question/30429868
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A 0.468 g sample of pentane, C5H12, was burned in a bomb calorimeter. The temperature of the calorimeter and 1.00 kg of water in it rose from 20.45 °C to 23.65 °C. The heat capacity of the calorimeter by itself is 2.21 kJ/°C and the specific heat capacity of water is 4.184 J/g.°C What is the heat of combustion per mole of pentane?
Answer:
E = 3,154.37 KJ
The heat of combustion per mole of pentane is 3,154.37 KJ
Explanation:
Given;
Change in temperature of system ∆T = 23.65-20.45 = 3.2 °C
Mass of water m1 = 1 kg
Specific heat capacity of water C1 = 4.184J/g°C = 4184J/kg °C
Heat capacity of calorimeter mC2 = 2.21 kJ/°C = 2210J/°C
Heat gained by both calorimeter and water is;
H = (m1C1 + mC2)∆T
Substituting the values;
H = (1×4184 + 2210)×3.2
H = 20460.8 J
Mass of pentane burned = 0.468 g
Molecular mass of pentane = 72.15g
If 0.468g of pentane releases 20460.8 J of heat,
72.15g of pentane will release;
E = (72.15/0.468) × 20460.8 J
E = 3154373.333333J
E = 3,154.37 KJ
The heat of combustion per mole of pentane is 3,154.37 KJ
An 80-eV electron impinges upon a potential barrier 100 eV high and 0.20 nm thick. What is the probability the electron will tunnel through the barrier? (1 eV = 1.60 × 10-19 J, m proton = 1.67 × 10-27 kg, = 1.055 × 10-34 J · s, h = 6.626 × 10-34 J · s) a. 7.7 × 10-10 % b. 0.11% c. 1.1% d. 0.011% e. 1.1 × 10-4 %
Answer:
d) 0.011%
Explanation:
The probability for tunneling the barrier is given by the following formula:
[tex]P=exp(-2d\sqrt{\frac{2m_e(U_o-E)}{\hbar ^2} }\ )[/tex] ( 1 )
me: mass of the electron
Uo: energy of the barrier
E: energy of the electron
d: thickness of the barrier
By replacing the values of the parameters in (1), you obtain:
[tex]P=exp(-2(0.20*10^{-9}m)\sqrt{\frac{2(9.11*10^{-31}kg)(100eV-80eV)(1.60*10^{-19}J)}{(1.055*10^{-34}Js)^2}})\\\\P=e^{-9.15}=1.08*10^{-4}\approx0.011\%[/tex]
hence, the probability is 0.011% (answer d)