The object initially had potential energy of 108,410 Joules when placed 6.5 meters higher. Since it maintains the same speed, this energy was used to overcome friction. Thus, the object lost 108,410 Joules to friction.
Explanation:To answer your question, we first need to calculate the initial potential energy of the object when it was 6.5 m higher. Potential energy (PE) is given by PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height. So, the initial potential energy is PE = (1700 kg)(9.8 m/s²)(6.5 m) = 108,410 Joules.
Since the object reaches the top of the loop with the same velocity in both scenarios, we can say that its kinetic energy remained the same. This implies all of the initial potential energy went into overcoming friction. Therefore, the object lost 108,410 Joules of energy to friction.
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The object lost 111,230 joules of energy to friction during its ascent.
Explanation:The question is asking how much energy, in joules, the object lost to friction if it starts 6.5 m higher than the previous part and still reaches the top of the loop with the same velocity. To calculate this, we need to find the change in gravitational potential energy, which is given by the formula: ΔPE = mgh.
The mass of the object is 1700 kg, and the change in height is 6.5 m. Substituting these values into the equation, we have:
ΔPE = (1700 kg)(9.8 m/s^2)(6.5 m) = 111,230 J.
Therefore, the object lost 111,230 joules of energy to friction during its ascent.
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A 32-kg child decides to make a raft out of empty 1.0-L soda bottles and duct tape. Neglecting the mass of the duct tape and plastic in the bottles, what minimum number of soda bottles will the child need to be able stay dry on the raft?
Answer:
32 bottles
Explanation:
If we create a free body diagram on the child we have his weight and the bouyant force
W-B=0
They must be equal to mantain equilibrium on the body and he can stay floating, this force is equivalent to the weight of water displaced
W=B=Ww
Mg=mg
32 kg=mass of water displaced
1 kilogram per liter (kg/L) is the density of water, this means that 32 Liters of water are displaced and since the bottles can retain 1 liter, the child needs 32 bottles
The child needs a minimum of 32 soda bottles to stay dry on the raft.
Explanation:To determine the minimum number of soda bottles needed for the child to stay dry on the raft, we need to consider the buoyant force exerted by the bottles. The buoyant force is equal to the weight of the displaced water. Since the child wants to stay dry, the buoyant force should be greater than or equal to the weight of the child.
The weight of the child can be calculated using the formula: weight = mass × gravity. Given the mass of the child is 32 kg, and the acceleration due to gravity is 9.8 m/s², we can find that the weight of the child is 32 kg × 9.8 m/s² = 313.6 N.
Next, we need to find the volume of one soda bottle. Since the empty soda bottles have a total volume of 1.0 L, we can assume that each bottle has a volume of 1.0 L ÷ the number of bottles needed. The mass of the water displaced by one bottle can be calculated using the formula: mass = density × volume. Given that the density of water is 1000 kg/m³, and 1 L = 0.001 m³, we can find that the mass of water displaced by one bottle is 1000 kg/m³ × 0.001 m³ = 1 kg.
To find the minimum number of bottles needed, we can set up the equation: buoyant force = weight of child. The buoyant force is equal to the mass of water displaced by one bottle × gravity × the number of bottles needed. Using the values we found earlier, we have: 1 kg × 9.8 m/s² × the number of bottles needed = 313.6 N. Solving for the number of bottles needed, we find that the minimum number of soda bottles the child needs is 313.6 N / (1 kg × 9.8 m/s²) = 32 bottles (rounded up to the nearest whole number).
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A physics student looks into a microscope and observes that small particles suspended in water are moving about in an irregular motion. Which of the following statements is the best explanation for this observation?
a. water molecules strike the particles giving them the same average kinetic energy as the water
b. the particles are carried by convection currents in the water
c. the small particles may be considered a fluid; and thus, move about randomly
d. the actual motion is regular, but the speeds of particles are too large to observe the regular motion
e. the particles are moving to be uniformly distributed throughout the volume of the water
Answer:
d. the actual motion is regular, but the speeds of particles are too large to observe the regular motion
Explanation:
The speeds of the particles are very large and comparatively the average free path is very small . Therefore time taken in covering the free path ( path between two consecutive collision with medium particles ) is very small . Hence the st line path covered by particles between two collision is less likely to be visible. Hence motion appears irregular or zig-zag.
As heat is added to water, is it possible for the temperature measured by a thermometer in the water to remain constant?
a. Maybe; it depends on the rate at which the heat is added.
b. No, adding heat will always change the temperature.
c. Yes, the water could be changing the phase.
d. Maybe; it depends on the initial water temperature.
Answer:
C. Yes, the water could be changing the phase.
Explanation:
Calculate the final speed (in m/s) of a 106 kg rugby player who is initially running at 7.75 m/s but collides head-on with a padded goalpost and experiences a backward force of 1.85 ✕ 10⁴ N for 7.50 ✕ 10⁻²s.
Answer:
Final velocity will be -5.33 m/sec
Explanation:
We have given mass of the rugby player m = 106 kg
Initial speed = 7.75 m/sec
Backward force [tex]F=1.85\times 10^4N[/tex]
Time is given as [tex]t=7.5\times 10^{-2}sec[/tex]
Impulse is given by impulse = force × time
So impulse [tex]=-1.85\times 10^4\times 7.5\times 10^{-2}=-1387.5N-s[/tex] ( as force is backward )
We know that impulse is given by change in momentum
So [tex]m(v_f-v_i)=-1387.5[/tex]
[tex]106\times (v_f-7.75)=-1387.5[/tex]
[tex]v_f=-5.33m/sec[/tex]
A projectile returns to its original height 4.08 s after being launched, during which time it travels 76.2 m horizontally. If air resistance can be neglected, what was the projectile's initial speed?
Answer:27.35 m/s
Explanation:
Given
Time of Flight of Projectile T=4.08 s
Range of Projectile =76.2 m
Time Of Flight of Projectile is given by
[tex]T=\frac{2u\sin \theta }{g}----------1[/tex]
where u=initial Velocity
[tex]\theta =[/tex]Launch angle
g=acceleration due to gravity
Range is given by [tex]R=\frac{u^2\sin 2\theta }{g}------2[/tex]
divide 1 and 2
[tex]\frac{R}{T}=\frac{u^2\sin 2\theta }{g}\times \frac{g}{2u\sin \theta }[/tex]
[tex]\frac{R}{T}=u\cos \theta ------3[/tex]
[tex]u\sin \theta =\frac{Tg}{2}------4[/tex]
squaring and adding 3 & 4 we get
[tex]u^2(\cos ^2\theta +\sin ^2\theta )=(\frac{R}{T})^2+(\frac{Tg}{2})^2[/tex]
[tex]u^2=(19.992)^2+(18.676)^2[/tex]
[tex]u=\sqrt{748.49}[/tex]
[tex]u=27.35 m/s[/tex]
The initial speed is the speed of the object at the beginning of the measurement or the starting speed.
The initial speed of the projectile is 27.35 m/s.
What is the initial speed?
The initial speed is the speed of the object at the beginning of the measurement or the starting speed.
Given information-
The time taken by projectile to returns its original height is 4.08 s.
The distance traveled by it is 76.2 m.
Air resistance can be neglected.
The time of flight of a projectile motion can be given as,
[tex]T=\dfrac{2u\sin \theta}{g}[/tex]
Let the above equation is equation 1.
Here, [tex]u[/tex] is the initial velocity, and [tex]\theta[/tex] is the angle of launch.
Rewrite the above equation as,
[tex]u\sin \theta =\dfrac{Tg}{2}[/tex]
Let the above equation is equation 2.
Now the range of the projectile motion can be given as,
[tex]R=\dfrac{u^2\sin (2\theta) }{g}\\R=\dfrac{u^22\cos \theta\sin \theta }{g}\\[/tex]
Divide this equation by the equation 1 as,
[tex]\dfrac{R}{T}=u\cos \theta[/tex]
Square and add the the above equation and equation 2 as,
[tex]u^2(\cos^2 \theta+\sin^2 \theta)=\dfrac{R}{T}+\dfrac{Tg}{2}\\u^2(1)=\dfrac{76.2}{4.08}+\dfrac{4.08\times9.81}{2}\\u=27.35\rm m/s[/tex]
Hence, the initial speed of the projectile is 27.35 m/s.
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A current I flows down a wire of radius a.
(a) If it is uniformly distributed over the surface, what is the surface current density K?
(b) If it is distributed in such a way that the volume current density is inversely.
Answer:
(a) [tex]K = \frac{I}{2\pi a}[/tex]
(b) [tex]J = \frac{I}{2\pi as}[/tex]
Explanation:
(a) The surface current density of a conductor is the current flowing per unit length of the conductor.
[tex]K = \frac{dI}{dL}[/tex]
Considering a wire, the current is uniformly distributed over the circumferenece of the wire.
[tex]dL = 2\pi r[/tex]
The radius of the wire = a
[tex]dL = 2\pi a[/tex]
The surface current density [tex]K = \frac{I}{2\pi a}[/tex]
(b) The current density is inversely proportional
[tex]J \alpha s^{-1}[/tex]
[tex]J = \frac{k}{s}[/tex] ......(1)
k is the constant of proportionality
[tex]I = \int\limits {J} \, dS[/tex]
[tex]I = J \int\limits \, dS[/tex] ........(2)
substituting (1) into (2)
[tex]I = \frac{k}{s} \int\limits\, dS[/tex]
[tex]I = k \int\limits^a_0 \frac{1}{s} {s} \, dS[/tex]
[tex]I = 2\pi k\int\limits\, dS[/tex]
[tex]I = 2\pi ka[/tex]
[tex]k = \frac{I}{2\pi a}[/tex]
substitute [tex]J = \frac{k}{s}[/tex]
[tex]J = \frac{I}{2\pi as}[/tex]
In a wire with current I and radius a, the surface current density K with uniform distribution is I/(2πaL). If volume current density varies inversely with the radius, the current is more dense at the center and less dense towards the surface, described by I/(πa²L).
Explanation:The subject here relates to the physical properties of the current flowing down a wire with a certain radius. (a) If the current I is uniformly distributed over the surface of the wire with a radius a, then the surface current density K can be given by the total current (I) divided by the surface area of the wire. In this context, the surface area of a cylindrical wire can be calculated by the formula 2π * a * L, where L is the length of the wire. Therefore, the surface current density K is I/(2πaL).
(b) If the current is distributed such that the volume current density J is inversely proportional to the radius, it implies that the current is more dense at the center of the wire and less dense as you approach the surface. The volume current density J can be described by the formula I/(πa²L). This complex distribution would likely require calculus to derive an exact relationship between current density and radius.
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If the atoms that share electrons have an unequal attraction for the electrons is called
If the atoms that share electrons have an unequal attraction for electrons, the bond is called a Polar covalent bond.
Explanation:A covalent chemical bond is formed in case of two different non-metals when one or more electron pairs are shared between bonding atoms. A difference in electronegativity of subsequent atoms of a covalent bond leads to formation of a small net charge around nucleus of each atom, pulling the shared electrons to one side of the bond, to the nucleus which has higher electronegativity.
HCl is an example of polar covalent bond and the HCl bond has Chlorine more electronegative. The bonding electrons are more close to Cl than H and hence Cl is partially negatively charged than H which has partial positive charge (HCl bond : [tex]H^{+} - Cl^{-}[/tex]). When electrons shared in a covalent bond have equal attraction, the bond is a Non-Polar covalent bond.
It's your birthday, and to celebrate you're going to make your first bungee jump. You stand on a bridge 110 m above a raging river and attach a 31-m-long bungee cord to your harness. A bungee cord, for practical purposes, is just a long spring, and this cord has a spring constant of 42 N/m. Assume that your mass is 80 kg. After a long hesitation, you dive off the bridge. How far are you above the water when the cord reaches its maximum elongation?
Answer:
h=20.66m
Explanation:
First we need the speed when the cord starts stretching:
[tex]V_2^2=V_o^2-2*g*\Delta h[/tex]
[tex]V_2^2=-2*10*(-31)[/tex]
[tex]V_2=24.9m/s[/tex] This will be our initial speed for a balance of energy.
By conservation of energy:
[tex]m*g*h+1/2*K*(h_o-l_o-h)^2-m*g*(h_o-l_o)-1/2*m*V_2^2=0[/tex]
Where
[tex]h[/tex] is your height at its maximum elongation
[tex]h_o[/tex] is the height of the bridge
[tex]l_o[/tex] is the length of the unstretched bungee cord
[tex]800h+21*(79-h)^2-63200-24800.4=0[/tex]
[tex]21h^2-2518h+43060.6=0[/tex] Solving for h:
[tex]h_1=20.66m[/tex] and [tex]h_2=99.24m[/tex] Since 99m is higher than the initial height of 79m, we discard that value.
So, the final height above water is 20.66m
Answer: using the conservation of potential energy stored in spring giving that at maximum amplitude velocity becomes zero.
Mgd= 1/2k(d-l)^2..... equation 1
M= 80kg=mass , g= 10m/s^2 =gravity, d=?=length of fully extended bungee rope, l=31m= length of bungee rope before extension, k=42N/m= spring constant
Simplifying equation above gives
2Mgh/k= d^2 - 2dl + l^2 ....eq 2
Substituting figures into the equ above gives
0 = d^2 - 100.1d +961 ...equ 3
Equ3 can be solved since it is a quadratic equation
d= (-b +or- square root (b^2 - 4ac))/2a ....equa4
Where a=1, b= -100.1, c= 961
Substituting figures into eequa4
d= 89.34m
So therefore the height above the river to me when bungee is fully extended is= 110 - 89.34
= 20.66
Explanation: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k.
Assuming that it performs simple harmonic motion.
The moon is a satellite that orbits the earth at a radius of 3.85 × 108 m. The mass of the earth is 5.98 × 1024 kg. What is the orbital speed of the moon?
Answer:
The orbital velocity of the moon is, V = 1018 m/s
Explanation:
Given data,
The radius of the moon's orbit, R = 3.85 x 10⁸ m
The mass of the Earth, M = 5.98 x 10²⁴ kg
The formula for orbital velocity is,
V = √(GM/R²)
Substituting the values,
V = √(6.673 x 10⁻¹¹ x 5.98 x 10²⁴ / 3.85 x 10⁸ )
= 1018 m/s
Hence, the orbital velocity of the moon is, V = 1018 m/s
The orbital speed of the Moon can be calculated using a specific formula that takes into account the mass of the Earth, the mass of the Moon, and the radius of the Moon's orbit.
Explanation:The orbital speed of an object in orbit around another object can be calculated using the formula:
v = √[G * (M+E) / r]
Where v is the orbital speed, G is the gravitational constant (approximately 6.67 × 10^-11 N.m^2/kg^2), M is the mass of the larger object (in this case, the Earth), E is the mass of the orbiting object (in this case, the Moon), and r is the radius of the orbit.
Substituting the given values into the formula, we can calculate the orbital speed of the Moon.
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A 5.20g bullet moving at 672 m/s strikes a 700g wooden block atrest on a frictionless surface. The bullet emerges, travelingin the same direction with its speed reduced to 428 m/s.
a. What is the resulting speed of the block?
b. What is the speed of the bullet-block center of mass?
Answer:
a) v=1.81 m/s; B) v=4.95 m/s
Explanation:
using momentum conservation
m1v1+m2v1=m1v2+m2v2
A)
5.2*672+700*0=5.2*428+700v2
The initial velocity of the block is 0, solving for v2
v2=1.81 m/s
B) Now both the bullet and the block travel together
m1v1+m2v1=(m1+m2)v2
5.2*672+700*0=(5.2+700)v
v=4.95 m/s
a. To find the resulting speed of the block after the bullet strikes it, we can use the law of conservation of momentum. b. The speed of the bullet-block center of mass can be found by calculating the weighted average of the speeds of the bullet and block.
Explanation:a. To find the resulting speed of the block, we can use the law of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.
The initial momentum of the bullet is given by m1 * v1, and the final momentum is given by (m1 + m2) * vf, where m1 is the mass of the bullet, v1 is the initial velocity of the bullet, m2 is the mass of the block, and vf is the final velocity of the bullet and block combined.
Using these values, we can solve for vf and then find the resulting speed of the block.
b. The speed of the bullet-block center of mass can be found by calculating the weighted average of the speeds of the bullet and block. Since the mass of the bullet is much smaller than the mass of the block, the center of mass will be closer to the speed of the block.
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Vector V⃗ 1 points along the z axis and has magnitude V1 = 76. Vector V⃗ 2 lies in the xz plane, has magnitude V2 = 60, and makes a -48 ∘ angle with the x axis (points below x axis).
What is the scalar product of the two vectors? Express answer using two significant figures.
Answer:
R= - 3388.74
Explanation:
Given that
V₁= 76 k ( in z-direction)
θ = 48°
V₂ = 60 cos48° i - 60 sin48° k
The dot product of two vector given as
We know that dot product of two vector is scalar and cross product of two vector is vector.
R= V₁ . V₂
We have to remember
i.i= j.j = k.k = 1
i.j = j.k = k.i = 0
Now
R= V₁ . V₂
R= (76 k ).( 60 cos48° i - 60 sin48° k)
R= 0 - 60 x 76 sin48°
R= - 3388.74
The scalar product of the vectors is - 3388.74
Given information:
V₁= 76k since it is in z-direction
Now vector V₂ makes an angle θ = 48° with x-axis so, it can be resolved as follows:
V₂ = 60 cos48°i - 60 sin48° k
Scalar Product:The scalar product of vectors is the product of the projection of one vector with the other vector.
The scalar product or the dot product of two vectors is given as
V= V₁ . V₂
The dot product of the x,y,and z direction components follow the below mentioned rule:
i.i= j.j = k.k = 1
i.j = j.k = k.i = 0
So, the required scalar product
V = V₁ . V₂
V = (76k ).(60cos48° i - 60sin48° k)
V = 0 -60 x 76sin48°
V = - 3388.74
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What would be your estimate of the age of the universe if you measured a value for Hubble's constant of H0 = 30 km/s/Mly ? You can assume that the expansion rate has remained unchanged during the history of the universe.
Answer:
The age of the universe would be 9.9 billion years
Explanation:
We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.
Thus, the time it has taken for the galaxies to reach their current separations is:
[tex]\displaystyle{t=D/v}[/tex]
and from Hubble's Law:
[tex]v =H_0D[/tex]
Therefore:
[tex]\displaystyle{t=D/v=D/(H_0\times D)=1/H_0}[/tex]
With the given value for the Hubble's constant we have:
[tex]H_0=(30\ km/s/Mly) \times (1 Mly/ 9.461 \times 10^{18} km) = 3.17\times 10^{-18}\ 1/s[/tex]
and thus,
[tex]t=1/H_0 = 1/(3.17\times 10^{-18} 1/s) = 0.315 \times 10^{18}\ s \approx 9988584474.8858\ years \approx 9.9\ billion\ years[/tex]
Explain how levitt and dubner’s argument effectively uses logical
Explanation:
Levitt and Dubner’s argument effectively uses logical, concrete evidence to arrive at conclusions about morality and cheating practices.the realities they present about sumo and the bagel business Their utilization of factual evidence in various illustrations shows morality and cheating practices are more common in high incentive situations where telling lies or an act of fooling people was rewarded awesomely. They use statistical evidence and different examples support the fact that Levitt and Dubner have arrived at a generalisation on moral grounds.
Answer:
Levitt and Dubner’s argument uses logic to present evidence to arrive at their final conclusions about morality and human cheating practices. The examples they used were of Sumo wrestler's in Japan and the bagel business of a self-employed man who provided bagel's and cream cheese to office worker's on the "honor system". Levitt and Dubner used factual evidence as well as some illustrations to present their claims. They concluded that although cheating exists when there is high incentive, most people are inherently honest when incentives are not a factor.
Explanation:
Why are unpaired electrons more significant than paired electrons in terms of magnetic properties
Answer:
Opposite spin neutralizes the magnetic fields.
Explanation:
The reason for the significance of the unpaired electrons with respect to the magnetic properties is because electrons have opposite spin and when the electrons are paired then as a result their opposite spins neutralizes the effect of their magnetic field thus resulting in no field effect.
Thus single electrons which are unpaired contributes to the magnetic properties of the material as compared to the paired electrons.
You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you pull the plates apart to a larger separation, do the following quantities increase, decrease or stay the same?i) Cii) Qiii) E between the platesiv) delta V
Answer:
i) C decreases
ii) Q remains constant
iii) E remains constant
iv) ΔV increases
Explanation:
i)
We know, capacitance is given by:
[tex]C=\frac{\epsilon_0.A}{d}[/tex]
[tex]\therefore C\propto \frac{1}{d}[/tex]
In this case as the distance between the plates increases the capacitance decreases while area and permittivity of free space remains constant.
ii)
As the amount of charge has nothing to do with the plate separation in case of an open circuit hence the charge Q remains constant.
iii)
Electric field between the plates is given as:
[tex]E=\frac{\sigma}{\epsilon_0}[/tex]
where:
charge density, [tex]\sigma=\frac{Q}{A}[/tex]
As we know that distance of plate separation cannot affect area of the plate. Charge Q and permittivity are also not affected by it, so E remains constant.
iv)
From the basic definition of voltage we know that it is the work done per unit charge to move it through a distance.Here we increase the distance so the work done per unit charge increases.i) Capacitance is decreases
ii) Charge Q remains constant
iii) Electric field E remains constant
iv) Change in potential ΔV is increases
Parallel-plate capacitor:The capacitance is computed as,
[tex]C=\frac{\epsilon A}{d}[/tex]
Where A is area of plates and d is distance between plates.
Following information is to be considered.
Given that the distance between the plates increases, the capacitance decreases while area and permittivity of free space remains constant.the amount of charge is independent on plate separation .Hence the charge Q remains constant.We know that distance of plate separation can not affect area of the plate. So that Charge Q and permittivity are also not affected by it. Thus, electric field E remains constant.Voltage is the work done per unit charge to move it through a distance.Here we increase the distance so the work done per unit charge increases.Learn more about the Parallel plate capacitor here:
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Two horizontal forces act on a 1.4 kg chopping block that can slide over a friction-less kitchen counter, which lies in an xy plane. One force is [tex]\vec{F}_1 = (3.9 N)\hat{i} + (3.3 N)\hat{j}[/tex]. Find the acceleration of the chopping block in unit-vector notation for each of the following second forces.[tex]a) \vec{F}_2= (-3.0N)\hat{i} + (-4.0N)\hat{j}\\b) \vec{F}_2= (-3.0N)\hat{i} + (4.0N)\hat{j}\\c) \vec{F}_2=(3.0N)\hat{i} + (-4.0N)\hat{j}[/tex]
Answer:
Part a)
[tex]a = (0.64\hat i - 0.5 \hat j)m/s^2[/tex]
Part b)
[tex]a = (0.64\hat i + 5.21 \hat j)m/s^2[/tex]
Part c)
[tex]a = (4.92\hat i - 0.5 \hat j)m/s^2[/tex]
Explanation:
As per Newton's II law we know that
F = ma
so we will have
[tex]a = \frac{F}{m}[/tex]
so we will have
[tex]a = \frac{F_1 + F_2}{m}[/tex]
Part a)
[tex]a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i - 4\hat j)}{1.4}[/tex]
[tex]a = \frac{0.9 \hat i - 0.7 \hat j}{1.4}[/tex]
[tex]a = (0.64\hat i - 0.5 \hat j)m/s^2[/tex]
Part b)
[tex]a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i + 4\hat j)}{1.4}[/tex]
[tex]a = \frac{0.9 \hat i + 7.3 \hat j}{1.4}[/tex]
[tex]a = (0.64\hat i + 5.21 \hat j)m/s^2[/tex]
Part c)
[tex]a = \frac{(3.9 \hat i + 3.3 \hat j) + (3\hat i - 4\hat j)}{1.4}[/tex]
[tex]a = \frac{6.9 \hat i - 0.7 \hat j}{1.4}[/tex]
[tex]a = (4.92\hat i - 0.5 \hat j)m/s^2[/tex]
Does the photoelectric effect support the wave theory of light? The particle theory of light?
Final answer:
The photoelectric effect supports the particle theory of light as it shows that light's ability to eject electrons depends on its frequency, indicative of light's particle-like behavior through photons.
Explanation:
The photoelectric effect does not support the wave theory of light, but rather supports the particle theory of light. According to classical wave theory, energy of light is related to its intensity or amplitude. However, the photoelectric effect demonstrates that the kinetic energy of ejected electrons from a metal surface depends on the light's frequency rather than its intensity.
This phenomenon can be explained by considering light to consist of particles called photons. Each photon carries a quantized amount of energy determined by the equation E = hv, where E is energy, h is Planck's constant, and v is frequency. If the frequency of light is above a certain threshold, it can dislodge electrons from the metal because the photons have sufficient energy, showing light's particle-like nature.
Calculate the work required to move a planet’s satellite of mass 1820 kg from a circular orbit of radius 2R to one of radius 3R, where 7.37×106 m is the radius of the planet. The mass of the planet is 7.51 × 1024 kg. Answer in units of J.
Answer:
The work required to move a planet's satellite is W = 2854.61 J
Explanation:
Given data,
The mass of the satellite, m = 1820 kg
The radius of the circular orbit, r = 2R
The radius of the planet, r = 5.37 x 10⁶ m
The mass of the planet, M = 7.5 x 10²⁴ kg
The formula for work done from the 2R to 3R is,
W = [tex]\int_{2R}^{3R}\frac{GMm}{r^{2}}dr[/tex]
W = GMm/3R - GMm/2R
W = (-0.17)GMm/R J
The negative sign indicates that the energy stored in the satellite as the potential energy.
Substituting the values
W = (-0.17) 6.673 x 10⁻¹¹ X 7.51 x 10²⁴ X 1820 / (7.37 x 10⁶)²
= -2854.61 J
Hence, the work required to move a planet's satellite is W = 2854.61 J
A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00 m and a total mass of 120 kg. The turntable is initially rotating at 3.00 rad>s about a vertical axis through its center. Suddenly, a 70.0-kg parachutist makes a soft landing on the turntable at a point near the outer edge. (a) Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.) (b) Compute the kinetic energy of the system before and after the parachutist lands. Why are these kinetic energies not equal
Answer:
a)1.385 rad/s
b) Before: 1080 J. After 498.46 J
Explanation:
The moments of inertia of the turn table, with the shape of uniform disk is:
[tex] I_1 = 0.5mr^2 = 0.5*120*2^2 = 240 kgm^2[/tex]
The angular momentum of the turn table before the impact is
[tex]A_1 = \omega_1I_1 = 3*240 = 720 radkgm^2[/tex]
The moments of inertia of the system after the impact is (treating the parachute man is a point particle)
[tex]I_2 = I_1 + Mr^2 = 240 + 70*2^2 = 240 + 280 = 520 kgm^2[/tex]
According to angular momentum conservation law:
[tex]A_1 = A_2[/tex]
[tex] 720 = \omega_2I_2[/tex]
[tex]\omega_2 = \frac{720}{I_2} = \frac{720}{520} = 1.385 rad/s[/tex]
(b) Before the impact:
[tex]K_1 = 0.5*I_1*\omega_1^2 = 0.5*240*3^2 = 1080 J[/tex]
After the impct
[tex]K_2 = 0.5*I_2*\omega_2^2 = 0.5*520*1.385^2 = 498.46 J[/tex]
The kinetic energies are not equal because the impact is causing the turn table to lose energy.
A box slides with uniform acceleration up an incline. The box has an initial speed of 9.0 m/s and rises vertically 2.60 m before coming to rest. If the angle of the incline is 30°, what is the coefficient of kinetic friction between the box and the incline?
Answer: 0.58
Explanation:
First we need to get the acceleration of the body using equation of motion
v²=u²-2as
v is the final velocity
u is the initial velocity
a is acceleration
s is the distance moved
0²=9²-2a(2.6)
-81=-5.2a
a=81/5.2
a= 15.6m/s²
Angle of inclination =30°
To get the coefficient of friction, we use the formula
Ff =nR
Ff is frictional force
n is coefficient of friction
R is normal reaction
n = Ff/R = Wsin30°/Wcos30°
n = tan30°
n = 0.58
A beam of electrons passes through a single slit, and a beam of protons passes through a second, but identical, slit. The electrons and the protons have the same speed. Which one of the following correctly describes the beam that experiences the greatest amount of diffraction?
(a) The electrons, because they have the smaller momentum and, hence, the smaller de Broglie wavelength
(b) The electrons, because they have the smaller momentum and, hence, the larger de Broglie wavelength
(c) The protons, because they have the smaller momentum and, hence, the smaller de Broglie wavelength
(d) The protons, because they have the larger momentum and, hence, the smaller de Broglie wavelength
(e) Both beams experience the same amount of diffraction, because the electrons and protons have the same de Broglie wavelength.
Answer:
(b) The electrons, because they have the smaller momentum and, hence, the larger de Broglie wavelength
Explanation:
de Broglie wavelength λ = h / m v
Since both electrons and protons have same velocity , momentum mv will be less for electrons because mass of electron is less .
for electron , momentum is less so . Therefore de Broglie wavelength λ will be more for electrons .
Amount of diffraction that is angle of diffraction is proportional to λ
Therefore electrons having greater de Broglie wavelength will show greater diffraction.
Calculate the energy that is required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C. The heat of fusion = 333 J/g, the heat of vaporization = 2256 J/g, and the specific heat capacities of ice = 2.06 J/gK and liquid water = 4.184 J/gK
Answer:
[tex]Q=35011.6\ J[/tex]
Explanation:
Given:
mass of ice, [tex]m=50\ g[/tex]initial temperature of ice, [tex]T_i=-30^{\circ}C[/tex]final temperature of liquid water, [tex]T_f=73^{\circ}C[/tex]heat of fusion of ice, [tex]L=333\ J.g^{-1}[/tex]specific heat capacity of ice, [tex]c_i=2.06\ J.g^{-1}[/tex]specific heat capacity of liquid water, [tex]c_w=4.184\ J.g^{-1}.K^{-1}[/tex]Now, total heat energy required get to the final state:
[tex]Q=m(c_i.\Delta T_i+L+c_w.\Delta T_w)[/tex]
where:
[tex]\Delta T_w=[/tex] change in temperature of water from 0 to 73 degree C
[tex]\Delta T_i=[/tex] change in temperature of ice from -30 to 0 degree C
[tex]\therefore Q=50(2.06\times 30+333+4.184\times 73)[/tex]
[tex]Q=35011.6\ J[/tex]
Answer:
The heat energy required is 350.1J
Explanation:
Given data
Mass of ice =50g---kg =50/1000= 0.05kg
Temperature of ice T1= - 30°c
Temperature of ice T2=0°c
Temperature of liquid T3=0°c
Temperature of water T4= 73°c
The heat of fusion = 333 J/g
heat of vaporization = 2256 J/g, and the specific heat capacities of ice = 2.06 J/gK
and liquid water = 4.184 J/gK
It will be a good idea to first understand the path this process will follow
Ice at - 30°c - - ice at 0°c
Ice at 0°c - - - - liquid at 0°c fusion
Liquid at 0°c -- - liquid at 73°c
First, you have to calculate the heat absorbed by ice going from -30 C to 0 C. Use the equation:
q = m c (T2-T1)
q= 0.5*2.06(0-(-30))
q= 30.9J
Then, calculate the heat required to melt that ice at 0C. Use the equation:
q = m *(heat of fusion)
q=0.5*333
q=166.5J
Then, calculate the heat required to raise the temperature of water from 0°C to 73°C. Use
q = m c (T4-T3)
q= 0.5*4.184(73-0)
q= 152.7J
Finally, we will sum up the heat required
Total heat energy required = 30.9+166.5+152.7= 350.1J
Adding or removing protons from an atom does what to the atom
Answer:
Changes the element of the atom.
Explanation:
The elements are classified by the number of protons they have in their nucleus, so if the number of protons is changed (added or removed), that atom will become one of a different element.
For example, hydrogen has only 1 proton in its nucleus, and helium has 2. So if a proton is added to a hydrogen atom, it becomes a helium atom, and consequently its atomic number, wich is determided by the protons in an element, will also change.
State in which all electrons are at their lowest possible energy level
Answer:
Ground state
Explanation:
At ground state all electrons are at the lowest energy level. At this level all the electrons, molecules or ions are said to be ground level. When electron get enough energy to jump they move to higher level. Any level higher than ground level is known as excited level. And energy of electron at excited state is higher than ground state. So the state at which all the electrons at their lowest possible energy level is the ground state.
A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of μs = 0.70 and μk = 0.50. The turntable very slowly speeds up.
Answer:
Angular speed will reach 6.833rad/s before the coin starts slipping
Explanation:
There is no question but I'll asume the common one: Calculate the speed of the turntable before the coin starts slipping.
With a sum of forces:
[tex]Ff = m*a[/tex]
[tex]Ff=m*V^2/R[/tex]
At this point, friction force is maximum, so:
[tex]\mu*N=m*V^2/R[/tex]
[tex]\mu*m*g=m*V^2/R[/tex]
Solving for V:
[tex]V=\sqrt{\mu*g*R}[/tex]
V=1.025 m/s
The angular speed of the turntable will be:
ω = V/R = 6.833 rad/s This is the maximum speed it can reach before the coin starts slipping.
A motorcyclist is traveling along a road and accelerates for 4.50s to pass another cyclist. The angular acceleration of each wheelis +6.70 rad/s^2, and, just after passing, the angular velocity ofeach is +74.5 rad/s, where the plus signs indicate counterclockwisedirections. What is the angular displacement of each wheel duringthis time?
a. +221 rad
b. +131 rad
c. +335 rad
d. +355 rad
e. +267 rad
Answer:
Angular displacement of the wheel, [tex]\theta=267.41\ rad[/tex]
Explanation:
It is given that,
Angular acceleration of the wheel, [tex]\alpha =6.7\ rad/s^2[/tex]
Final speed of the wheel, [tex]\omega_f=74.5\ rad/s[/tex]
Time taken, t = 4.5 s
Initially, it is required to find the initial angular velocity of the wheel. Using the first equation of rotational kinematics as :
[tex]\omega_f=\omega_o+\alpha t[/tex]
[tex]\omega_o[/tex] is the initial speed of the wheel
[tex]\omega_o=\omega_f-\alpha t[/tex]
[tex]\omega_o=74.5-6.7\times 4.5[/tex]
[tex]\omega_o=44.35\ rad/s[/tex]
Let [tex]\theta[/tex] is the angular displacement of each wheel during this time. Using the second equation of motion as :
[tex]\theta=\omega_o t+\dfrac{1}{2}\alpha t^2[/tex]
[tex]\theta=44.35\times 4.5+\dfrac{1}{2}\times 6.7\times (4.5)^2[/tex]
[tex]\theta=267.41\ rad[/tex]
So, the angular displacement of each wheel during this time is 267 radian.
The micturition reflex can be voluntarily controlled by the
Answer:
The micturition reflex can be voluntarily controlled by the relaxation of the external urethral sphincter.
The micturition reflex, or the process of urination, can be voluntarily controlled by the external urethral sphincter and the sacral micturition center. The external urethral sphincter is a skeletal muscle that we can consciously control, while the sacral micturition center is a group of neurons that normally act reflexively unless allowed for voluntary control by the brain. As the bladder fills, signals are sent through the sacral pelvic nerves which activate parasympathetic neurons to proceed with urination.
Explanation:The micturition reflex, which is another term for urination or voiding, can be voluntarily controlled by the external urethral sphincter and the sacral micturition center. The external urethral sphincter is a skeletal muscle that can be consciously controlled to maintain urinary continence, while the sacral micturition center, a group of neurons located in the sacral region of the spinal cord, acts reflexively unless its action is modified by higher brain centers for voluntary urination. In response to a filled bladder, these centers trigger the relaxation of both the internal and external urethral sphincters, the contraction of the detrusor muscle, and inhibit the somatic motor neurons for urination to occur.
Children learn to voluntarily control the urination process as they mature, thereby overriding the micturition reflex and delay voiding, a process known as potty training. Voluntary micturition requires (1) an intact spinal cord and (2) a functional pudendal nerve arising from the sacral micturition center. The sacral pelvic nerves play a crucial role in bladder control. Upon receiving signals of bladder stretch, they activate the parasympathetic neurons to release acetylcholine, which triggers detrusor muscle contraction and bladder emptying.
Learn more about Micturition Reflex here:https://brainly.com/question/36599868
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Nevine wants to improve her JavaScript program's efficiency and scalability by defining her own processes, or functions. Why are functions such an integral part of writing JavaScript code?
Because they perform specific tasks repeatedly throughout your program, as needed
Answer:
Because they perform specific tasks repeatedly throughout your program, as needed
Explanation:
Given the atomic radius of argon, 0.97 Å, and knowing that a sphere has a volume of 4πr3/3, calculate the fraction of space that Ar atoms occupy in a sample of argon at STP. Express your answer using two significant figures.
Answer:
1.0x10^-4
Explanation:
First, in order to do this, we need to calculate the volume of 1 simple atom of Ar. Using the formula of the volume of a sphere we have the following
Converting A to cm:
0.97 * 1x10^-8 = 9.7x10^-9 cm
Now the volume:
V = 4/3π(9.7x10^-9)³
V = 3.82x10^-24 cm³
We know that 1 cm³ is 1 mL, and 1 L is 1000 mL so:
V = 3.82x10^-24 mL / 1000 = 3.82x10^-27 L
Now, using avogadro's number, we should get the total volume of all atoms of Ar so:
3.82x10^-27 * 6.02x10^23 = 2.3x10^-3 L
Finally, at STP the volume of an ideal gas is 22.4 L so:
2.3x10^-3 / 22.4 = 1.03x10^-4
With two significant figure, it would be 1.0x10^-4
A baseball is thrown at an angle of 29 relative to the ground at a speed of 24.3 m/s. The ball is caught 51.0463 m from the thrower. The acceleration due to gravity is 9.81 m/s2. How long is it in the air? Answer in units of s.
Answer:
T = 2.4 s
Explanation:
given,
angle at which ball is thrown = 29°
speed relative to ground = 24.3 m/s
ball is caught at a distance = 51.0463
acceleration due to gravity = 9.8 m/s²
time for which ball was in the air = ?
now,
velocity of ball in x-direction
V_x = v cos θ
V_x = 24.3 x cos 29°
V_x = 21.25 m/s
velocity in y direction
V_y = v sin θ
V_y = 24.3 x sin 29°
V_y = 11.78 m/s
distance on the ground when ball will reach maximum height
x = 51.0463/2 = 25.52 m
at top most point velocity is equal to zero
time for which the ball was in air
v = u + a t
0 = 11.78 - 9.8 t
[tex]t = \dfrac{11.78}{9.8}[/tex]
t = 1.20
this time is taken to travel half distance
total time = 2 x 1.20
T = 2.4 s
time for which ball was in air is T = 2.4 s