When a glass rod is rubbed with a silk cloth, electrons are removed from the rod, resulting in a positive charge.
Explanation:When an object, such as a glass rod, is rubbed with a silk cloth, it can acquire a positive charge. In this case, electrons are removed from the object, leaving behind a net positive charge. The process of rubbing causes the cloth to transfer some of its excess electrons to the glass rod, resulting in the glass rod gaining a positive charge.
A raft of mass 199 kg carries two swimmers of mass 52 kg and 70 kg. The raft is initially floating at rest. The two swimmers simultaneously dive off opposite ends of the raft, each with a horizontal velocity of 4 m/s. With what velocity the raft start to move
To solve this problem we will apply the concept related to the conservation of the Momentum. We will then start considering that the amount of initial momentum must be equal to the amount of final momentum. Considering that all the objects at the initial moment have the same initial velocity (Zero, since they start from rest) the final moment will be equivalent to the multiplication of the mass of each object by the velocity of each object, so
Initial Momentum = Final Momentum
[tex](m_B+m_1+m_2)v_i = m_1v_1+m_2v_2+m_Bv_B[/tex]
Here,
[tex]m_B[/tex] = mass of Raft
[tex]m_1[/tex] = Mass of swimmers 1
[tex]m_2[/tex] = Mass of swimmers 2
[tex]v_i[/tex] = Initial velocity (of the three objects)
[tex]v_B[/tex] = Velocity of Raft
Replacing,
[tex](199+52+70)*0 = (52)(4)+(70)(-4)+199v_B[/tex]
Solving for [tex]v_B[/tex]
[tex]vB = \frac{72}{199}[/tex]
[tex]v_B = 0.3618m/s[/tex]
Therefore the velocity the rarft start to move is 0.3618m/s
Charge q1 is placed a distance r0 from charge q2 . What happens to the magnitude of the force on q1 due to q2 if the distance between them is reduced to r0/4 ?
What is the electrostatic force between and electron and a proton separated by 0.1 mm?
Answer:
The electrostatic force between and electron and a proton is [tex]F=2.30\times 10^{-20}\ N[/tex]
Explanation:
It is given that, charge [tex]q_1[/tex] is placed at a distance [tex]r_o[/tex] from charge [tex]q_2[/tex]. The force acting between charges is given by :
[tex]F=\dfrac{kq_1q_2}{r_o^2}[/tex]
We need to find the force if the distance between them is reduced to [tex]r_o/4[/tex]. It is given by :
[tex]F'=\dfrac{kq_1q_2}{(r_o/4)^2}[/tex]
[tex]F'=16\times \dfrac{kq_1q_2}{r_o^2}[/tex]
[tex]F'=16\times F[/tex]
So, if the the distance between them is reduced to [tex]r_o/4[/tex], the new force becomes 16 times of the previous force.
The electrostatic force between and electron and a proton separated by 0.1 mm or [tex]10^{-4}\ m[/tex] is :
[tex]F=\dfrac{kq_1q_2}{r_o^2}[/tex]
[tex]F=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(10^{-4})^2}[/tex]
[tex]F=2.30\times 10^{-20}\ N[/tex]
So, the electrostatic force between and electron and a proton is [tex]F=2.30\times 10^{-20}\ N[/tex]. Hence, this is the required solution.
When the distance between two charges is reduced, the magnitude of the force on one charge due to the other increases. This is because the electrostatic force is inversely proportional to the square of the distance between the charges.
Explanation:When the distance between two charges, q1 and q2, is reduced from r0 to r0/4, the magnitude of the force on q1 due to q2 increases. This is because the electrostatic force is inversely proportional to the square of the distance between the charges.
For example, if we have charges q1 = 2 C and q2 = 4 C, and the original distance r0 = 2 m, the force between them would be F = k × (q1 × q2) / r0^2 = k × (2 × 4) / (2^2) = k × 4, where k is the Coulomb's constant.
If we reduce the distance to r0/4 = 0.5 m, the force would become F' = k × q1 × q2) / (r0/4)^2 = k × 4) / (0.5^2) = k ×64, which is 16 times greater than the original force.
If the clock runs slow and loses 15 s per day, how should you adjust the length of the pendulum?
Answer:
L= 1 m, ΔL = 0.0074 m
Explanation:
A clock is a simple pendulum with angular velocity
w = √ g / L
Angular velocity is related to frequency and period.
w = 2π f = 2π / T
We replace
2π / T = √ g / L
T = 2π √L / g
We will use the value of g = 9.8 m / s², the initial length of the pendulum, in general it is 1 m (L = 1m)
With this length the average time period is
T = 2π √1 / 9.8
T = 2.0 s
They indicate that the error accumulated in a day is 15 s, let's use a rule of proportions to find the error is a swing
t = 1 day (24h / 1day) (3600s / 1h) = 86400 s
e= Δt = 15 (2/86400) = 3.5 104 s
The time the clock measures is
T ’= To - e
T’= 2.0 -0.00035
T’= 1.99965 s
Let's look for the length of the pendulum to challenge time (t ’)
L’= T’² g / 4π²
L’= 1.99965 2 9.8 / 4π²
L ’= 0.9926 m
Therefore the amount that should adjust the length is
ΔL = L - L’
ΔL = 1.00 - 0.9926
ΔL = 0.0074 m
In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The front end crumples so much that the truck is 0.62 m shorter than before.
(a) What is the average speed of the truck during the collision (that is, during the interval between the first contact with the wall and coming to a stop)?
(b) About how long does the collision last? (That is, how long is the interval between first contact with the wall and coming to a stop?)
(c) What is the magnitude of the average force exerted by the wall on the truck during the collision?
(d) It is interesting to compare this force to the weight of the truck. Calculate the ratio of the force of the wall to the gravitational force mg on the truck. This large ratio shows why a collision is so damaging.
(e) What approximations were necessary in making this analysis? (Select all that apply.)
a. Neglect the horizontal component of the force of the road on the truck tires.
b. Assume a nearly constant force exerted by the wall, so the speed changes at a nearly constant rate.
c. The deceleration of the truck is approximately equal to g.
Answer:
a) v_average = 11 m / s, b) t = 0.0627 s
, c) F = 7.37 10⁵ N
, d) F / W = 35.8
Explanation:
a) truck speed can be found with kinematics
v² = v₀² - 2 a x
The fine speed zeroes them
a = v₀² / 2x
a = 22²/2 0.69
a = 350.72 m / s²
The average speed is
v_average = (v + v₀) / 2
v_average = (22 + 0) / 2
v_average = 11 m / s
b) The average time
v = v₀ - a t
t = v₀ / a
t = 22 / 350.72
t = 0.0627 s
c) The force can be found with Newton's second law
F = m a
F = 2100 350.72
F = 7.37 10⁵ N
.d) the ratio of this force to weight
F / W = 7.37 10⁵ / (2100 9.8)
F / W = 35.8
.e) Several approaches will be made:
- the resistance of air and tires is neglected
- It is despised that the force is not constant in time
- Depreciation of materials deformation during the crash
An object thrown straight up a distance ymaxymax. After tt seconds, it falls back and is caught again, just as it reaches the height from which it was thrown. Claudia says its average velocity was zero, and Hossein says its average speed was 2ymaxt2ymaxt . What would you say to help them out?
Answer:
Avg.velocity=(Δy/ Δt) =(net displacement/ total time for journey)
Δy = 0
Δt = t
so avg. velocity = 0/t =0
Avg. speed =(total distance traveled/ total time for journey)
total distance = up +down = Ymax+Ymax= 2 Ymax
total time = t
avg. speed = 2 Ymax/t
Explanation:
Since there is no net displacement from the original position,velocity is zero. Claudia is right!
while it covered some distance in time t so its speed is not as qouted by Hossien
In a liquid with a density of 1400 kg/m3 , longitudinal waves with a frequency of 370 Hz are found to have a wavelength of 8.40 m . Calculate the bulk modulus of the liquid.
Answer:
Bulk modulus = 1.35 × [tex]10^{10}[/tex] Pa
Explanation:
given data
density = 1400 kg/m³
frequency = 370 Hz
wavelength = 8.40 m
solution
we get here bulk modulus of the liquid that is
we know Bulk Modulus = [tex]v^2*\rho[/tex] ...............
here [tex]\rho[/tex] is density i.e 1400 kg/m³
and v is = frequency × wavelength
v = 370 × 8.40 = 3108 m/s
so here bulk modulus will be as
Bulk modulus = 3108² × 1400
Bulk modulus = 1.35 × [tex]10^{10}[/tex] Pa
With what tension must a rope with length 2.30 mm and mass 0.105 kgkg be stretched for transverse waves of frequency 36.0 HzHz to have a wavelength of 0.790 mm?
Answer:
Tension must a rope be stretched is 36.8828 N
Explanation:
Wave speed in term of wavelength:
[tex]v=f. \lambda[/tex]
Where:
f is the frequency
[tex]\lambda[/tex] is the wavelength
Now:
[tex]v=36*0.790\\v=28.44 m/s[/tex]
Wave speed in term of tension force and mass per unit length
[tex]v=\sqrt{\frac{F}{Mass\ per\ unit\ length}}[/tex] Eq (1)
Where:
F is the tension force
[tex]Mass\ per\ unit\ length=\frac{0.105}{2.30} \\Mass\ per\ unit\ length=0.0456 Kg/m\\[/tex]
Since [tex]v[/tex] is calculated above.On rearranging Eq (1) we will get:
[tex]F=v^2 *Mass\ per\ unit\ length\\F=(28.44)^2*0.0456\\F=36.8828 N[/tex]
Tension must a rope be stretched is 36.8828 N
Determine a valid way of finding the wire’s diameter if you know the resistivity of the material, \rho , and can measure the current flowing through the wire, I, and the voltage drop between two points of the wire, V, a known distance apart, L. find the diameter of the wire in terms of \rho L V and I.
How will you find the measurement error (i.e., uncertainty) in the diameter of the wire?
Answer:
To find the diameter of the wire, when the following are given:
Resistivity of the material (Rho), Current flowing in the conductor, I, Potential difference across the conductor ends, V, and length of the wire/conductor, L.
Using the ohm's law,
Resistance R = (rho*L)/A
R = V/I.
Crossectional area of the wire A = π*square of radius
Radius = sqrt(A/π)
Diameter = Radius/2 = [sqrt(A/π)]
Making A the subject of the formular
A = (rho* L* I)V.
From the result of A, Diameter can be determined using
Diameter = [sqrt(A/π)]/2. π is a constant with the value 22/7
Explanation:
Error and uncertainty can be measured varying the value of the parameters used and calculating different values of the diameters. Compare the values using standard deviation
The diameter of the wire can be derived from the resistivity, length, current, and voltage using Ohm's Law and the equation for the resistance of a wire. To find the uncertainty in the diameter, use error propagation techniques considering the uncertainties of the resistivity, length, and (voltage/current).
Explanation:The diameter of the wire can be found using Ohm's Law and the formula for the resistance of a wire. The guiding formula is R = V/I (Ohm’s Law), where R is the resistance, V is the voltage, and I is the current.
The formula for resistance of a wire is R = ρL/A, where A (cross-sectional area) is π(d/2)^2 because the wire is cylindrical. Equating these two equations gives V/I = ρL/π(d/2)^2. Solving this for d (diameter) gives d = sqrt((4ρL(V/I))/π). To calculate the measurement error or uncertainty in the diameter of the wire, one would need to use error propagation techniques.
If the uncertainties in ρ, L, V/I are δρ, δL and δ(V/I) respectively, then the uncertainty in d, δd is given by δd = 1/2 [((4δρρ)/(ρL(V/I)))^2+ ((4δL/L)/(ρL(V/I)))^2+ ((4δ(V/I)/(V/I))/(ρL(V/I)))^2]^1/2.
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How long does it take for a rotating object to speed up from 15.0 sto 33.3 rad/s if it has an angular acceleration of 3.45rad/s2?
a) 9.57 s
b) 5.30 s
c) 63.1 s
d) 4.35 s
Answer:
The time taken by the rotating object to speed up from 15.0 s to 33.3 rad/s is 5.30 seconds.
Explanation:
It is given that,
Initial angular speed of the object, [tex]\omega_o=15\ rad/s[/tex]
Final angular speed of the object, [tex]\omega_f=33.3\ rad/s[/tex]
Angular acceleration of the object, [tex]\alpha =3.45\ rad/s^2[/tex]
Angular acceleration of an object is object is defined as the change in angular velocity per unit time. It is given by :
[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}[/tex]
[tex]t =\dfrac{\omega_f-\omega_i}{\alpha}[/tex]
[tex]t =\dfrac{33.3-15}{3.45}[/tex]
t = 5.30 seconds
So, the time taken by the rotating object to speed up from 15.0 s to 33.3 rad/s is 5.30 seconds. Hence, this is the required solution.
The time taken will be "5.30 seconds".
According to the question,
Initial angular speed,
[tex]\omega_i = 15 \ rad/s[/tex]Final angular speed,
[tex]\omega_f = 33.3 \ rad/s[/tex]Angular acceleration,
[tex]\alpha = 3.45 \ rad/s^2[/tex]As we know,
→ [tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
or,
→ [tex]t = \frac{\omega_f -\omega_i}{\alpha}[/tex]
By substituting the values, we get
[tex]= \frac{33.3-15}{3.45}[/tex]
[tex]= \frac{18.3}{3.45}[/tex]
[tex]= 5.30 \ seconds[/tex]
Thus the above answer i.e., "option b" is right.
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Each of the following equations was given by a student during an examination. Do a dimensional analysis of each equation and explain why the equation can't be correct.
(a) 1 2 mv2 = 1 2 mv02 + mgh
(b) v = v0 + at2
(c) ma = v2
a) The equation 1/2 mv² = 1/2 mv0² + mgh is not dimensionally consistent; b) The equation v = v0 + at² is not dimensionally consistent; c) The equation ma = v² is not dimensionally consistent.
(a) The left-hand side of the equation has dimensions of energy (M L² T⁻²), while the right-hand side has dimensions of energy plus length (M L² T⁻² + ML).
This is because the term mgh represents the potential energy gained by an object of mass m when it is lifted to a height h above the ground. Therefore, the dimensions of the two sides of the equation do not match, and the equation cannot be correct.
(b)The left-hand side of the equation has dimensions of velocity (L T⁻¹), while the right-hand side has dimensions of velocity plus time squared (L T⁻¹ + T²).
This is because the term at² represents the change in velocity over time due to acceleration. Therefore, the dimensions of the two sides of the equation do not match, and the equation cannot be correct.
(c) The left-hand side of the equation has dimensions of force (M L T⁻²), while the right-hand side has dimensions of velocity squared (L² T⁻²).
This is because the term v² represents the square of the velocity of an object. Therefore, the dimensions of the two sides of the equation do not match, and the equation cannot be correct.
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The dimensional analysis shows that equation (a) is correct while equations (b) and (c) are incorrect. Equation (b) is missing a time component, while equation (c) incorrectly equates force to velocity squared.
Explanation:Let's analyze each equation in terms of their dimensions:
(a) 1/2 mv² = 1/2 m(v0)² + mgh: This equation is actually dimensionally correct. It represents the conservation of mechanical energy in Physics, where 1/2 mv² is the kinetic energy and mgh is the potential energy.(b) v = v₀ + at²: This equation is not dimensionally correct. In Physics, the equation of motion is v = v₀ + at, where v is the final velocity, v0 is the initial velocity, a is acceleration and t is time.(c) ma = v²: This equation is also not dimensionally correct. Force (ma) has dimensions of [M][L][T]⁻² while velocity squared (v)² has dimensions of [L]²[T]⁻².By comparing dimensions on both sides of an equation, we can spot if the equations are incorrect or not.
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An ideal spring has spring constant ks (to distinguish it from the electrostatic constant k) and equilibrium length l. Then, you glue two identical negative point charges to the ends of the spring and observe that the equilibrium length doubles. Determine the amount of charge on each end of the spring
Answer:
q = square root (4KsL³/k)
The force of extension of the spring is equal to the force of repulsion between the two like charges. Two like charges(positive or negative) would always repel each other and two unlike charges would always attract each other. This electric force between the charges is what is responsible for the stretching of the spring. The electric force causes the spring to increase in length from L to 2L. Equating these forces, that is the electric force between the charges and the elastic force of the spring and rearranging the variables gives the expression to obtain q.
Explanation:
See the attachment below for full solution.
By equating the spring force with the electrostatic repulsion force at the new equilibrium length, we find that the charge on each end of the spring is [tex]q = 2\sqrt(ks ke l^3)[/tex]. This derivation uses Hooke's Law and Coulomb's Law for calculation.
To solve this problem, we need to understand the balance between the spring force and the electrostatic repulsion force. We start by looking at Hooke's Law for the spring force:
[tex]F_spr = -k_s (x - l)[/tex]
Where k_s is the spring constant, x is the stretched length (which is 2l in this case, since the equilibrium length doubles), and l is the original equilibrium length. The spring force is given by:
[tex]F_spr = -k_s (2l - l) \\= -k_s l[/tex]
The electrostatic repulsion force between two identical charges q separated by a distance 2l is given by Coulomb's Law:
[tex]F_e = \frac{k_e q^2}{(2l)^2} \\= \frac{k_e q^2}{4l^2}[/tex]
At equilibrium, the magnitudes of these two forces are equal:
[tex]k_s l = \frac{k_e q^2}{4l^2}[/tex]
Rearranging to solve for q:
[tex]q^2 = 4k_s k_e l^3[/tex]
[tex]q = \sqrt{4k_s k_e l^3} \\= 2\sqrt{k_s k_e l^3}[/tex]
So, each end of the spring must have a charge of[tex]q = 2\sqrt{k_s k_e l^3}.[/tex]
describe one new piece of information that you learned about one of the featured scientist
Answer:
Development in science:
There a number of individuals in the world around us who has improved the field of science and made it easy for us to know this universe more easily then ever. This universe is still unexplored and there are going to be more then billions of stars which are still not studied and has an immense amount of information regarding the universe.
Stephen Hawking: The scientist who made it easy for each of us to know the universe and more over the cosmos in a more detailed form, as before this no one knew about the different patterns of entities and there properties.
A genius who was handicapped:
He was a genius in cosmology, mathematics, and other subjects of science as he was diagnosed by the amyotrophic lateral sclerosis(ALS), which made it unable for him to live a normal life.But, he communicated through a computer which detected his nerve signals and shared his thoughts about any thing or subject.
Why him?
Stephen hawking was just tremendous in making it more understandable about the black holes, time worms, and space etc.As it was known to very small number of people before he took the stage.
A car, on a straight road, is stopped at a traffic light. When the light turns to green the car accelerates with a constant acceleration. It reaches a speed of 19.1 m/s (68.8 km/h) in a distance of 101 m. Calculate the acceleration of the car.
The car accelerates from 0 m/s to 19.1 m/s in 101 m, with an acceleration of approximately 1.81 m/s².
Given that,
The car is stopped at a traffic light on a straight road.
When the light turns green, the car accelerates.
The car reaches a speed of 19.1 m/s (68.8 km/h).
The distance covered during acceleration is 101 m
To calculate the acceleration of the car,
Use the kinematic equation:
v² = u² + 2as
Where,
v is the final velocity (19.1 m/s in this case),
u is the initial velocity (0 m/s since the car is stopped),
a is the acceleration that we're trying to find,
s is the distance covered (101 m).
Plugging in the known values, we have:
19.1² = 0² + 2a(101)
Simplifying further:
365.81 = 202a
Now, let's solve for a:
a = 365.81 / 202
Calculating that, we find:
a ≈ 1.81 m/s²
Hence,
The acceleration of the car is approximately 1.81 m/s².
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The acceleration of the car is 1.85 m/s^2.
Explanation:To calculate the acceleration of the car, we can use the equation:
v^2 = u^2 + 2as
Where:
v is the final velocityu is the initial velocity (0 m/s in this case)a is the accelerations is the displacementPlugging in the given values, we have:
19.1^2 = 0 + 2a(101)
Simplifying the equation gives us:
a = (19.1^2) / (2 * 101) = 1.85 m/s^2
Therefore, the acceleration of the car is 1.85 m/s^2.
Calculate the average linear momentum of a particle described by the following wavefunctions: (a) eikx, (b) cos kx, (c) e−ax2 , where in each one x ranges from −[infinity] to +[infinity].
Answer:
a) p=0, b) p=0, c) p= ∞
Explanation:
In quantum mechanics the moment operator is given by
p = - i h’ d φ / dx
h’= h / 2π
We apply this equation to the given wave functions
a) φ = [tex]e^{ikx}[/tex]
.d φ dx = i k [tex]e^{ikx}[/tex]
We replace
p = h’ k [tex]e^{ikx}[/tex]
i i = -1
The exponential is a sine and cosine function, so its measured value is zero, so the average moment is zero
p = 0
b) φ = cos kx
p = h’ k sen kx
The average sine function is zero,
p = 0
c) φ = [tex]e^{-ax^{2} }[/tex]
d φ / dx = -a 2x [tex]e^{-ax^{2} }[/tex]
.p = i a g ’2x [tex]e^{-ax^{2} }[/tex]
The average moment is
p = (p₂ + p₁) / 2
p = i a h ’(-∞ + ∞)
p = ∞
A force F produces an acceleration a on an object of mass m. A force 3F is exerted on a second object, and an acceleration 8a results. What is the mass of the second object in terms of m?
a. 3m
b. 9m
c. 24m
d. (3/8)m
e. (8/3)m
To solve this problem we will proceed to use Newton's second law for which the mass (m) multiplied by the acceleration (a) is defined as the Force (F) applied on a body, mathematically that is,
[tex]F = ma[/tex]
According to the statement for the first object, the acceleration is,
[tex]a = \frac{F}{m} \rightarrow 1^{st} Object[/tex]
For the second object the acceleration is,
[tex]8a = \frac{3F}{m_2} \rightarrow 2^{nd} Object[/tex]
Solving for the mass of the second object,
[tex]m_2 = \frac{3F}{8a}[/tex]
[tex]m_2 = \frac{3F}{8(F/m)}[/tex]
[tex]m_2 = \frac{3}{8}m[/tex]
Therefore the correct answer is D.
A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 14.5 m/s, and the distance from the limb to the level of the saddle is 3.29 m.
(a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move?
(b) How long is he in the air?
Answer:
For Part A:
ΔX=11.8813 m
Part B:
[tex]t=0.8194 s[/tex]
Explanation:
Note: In order to find Part A we first have to find Part B i.e time
Data given:
[tex]V_i=0[/tex]
a=-9.8m/s^2 (-ve is because ranch is falling down)
Δy=-3.29m (-ve is because ranch is falling down)
Second equation of Motion:
Δy=[tex]V_i*t+\frac{1}{2}g*t^2[/tex]
V_i=0, Equation will become
Δy=[tex]\frac{1}{2}g*t^2[/tex]
[tex]t=\sqrt{2Y/g} \\t=\sqrt{2*-3.29/-9.8} \\t=0.8194 s[/tex]
For Part A:
Again:
Second equation of Motion:
ΔX=[tex]V_i*t+\frac{1}{2}a*t^2[/tex]
Since velocity is constant a=0
V_i=14.5m/s, t=0.8194 sec
ΔX=[tex]V_i*t[/tex]
ΔX=[tex]14.5*0.8194[/tex]
ΔX=11.8813 m
Part B: (Calculated above)
Data given:
[tex]V_i=0[/tex]
a=-9.8m/s^2 (-ve is because ranch is falling down)
Δy=-3.29m (-ve is because ranch is falling down)
Second equation of Motion:
Δy=[tex]V_i*t+\frac{1}{2}g*t^2[/tex]
V_i=0, Equation will become
Δy=[tex]\frac{1}{2}g*t^2[/tex]
[tex]t=\sqrt{2Y/g} \\t=\sqrt{2*-3.29/-9.8} \\t=0.8194 s[/tex]
In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. Pluto already was known to have a large satellite, Charon, orbiting at 19,600 km with an orbital period of 6.39 days.Part AAssuming that the satellites do not affect each other, find the orbital periods T1 and T2 of the two small satellites without using the mass of Pluto.Enter your answers numerically separated by a comma.
Answer:
24.48 days
37.7 days
Explanation:
r = Radius
s denotes satellite
C denotes Charon
Time period is given by
[tex]T=\dfrac{2\pi r^{1.5}}{\sqrt{2GM}}[/tex]
So,
[tex]T\propto r^{1.5}[/tex]
[tex]\dfrac{T_C}{T_{s1}}=\dfrac{r_c^{1.5}}{r_{s1}^{1.5}}\\\Rightarrow T_{s1}=\dfrac{T_Cr_{s1}^{1.5}}{r_C^{1.5}}\\\Rightarrow T_{s1}=\dfrac{6.39\times 86400\times {48000000}^{1.5}}{19600000^{1.5}}\\\Rightarrow T_{s1}=2115886.41242\ s\\\Rightarrow T_{s1}=24.48\ days[/tex]
The time period of the first satellite is 24.48 days
[tex]T_{s2}=\dfrac{T_Cr_{s2}^{1.5}}{r_C^{1.5}}\\\Rightarrow T_{s2}=\dfrac{6.39\times 86400\times {64000000}^{1.5}}{19600000^{1.5}}\\\Rightarrow T_{s2}=3257620.23942\ s\\\Rightarrow T_{s2}=37.7\ days[/tex]
The time period of the second satellite is 37.7 days
A 200 kg chunck of lead falls from a hight of 30m and smashes into a rigid floor. calculate the increse in internal engery
Answer:
[tex]\Delta E=58800\ J[/tex]
Explanation:
Given:
mass of lead chunk, [tex]m=200\ kg[/tex]
height of the fall, [tex]h= 30\ m[/tex]
Assuming the collision to be perfectly inelastic the whole kinetic energy of the chunk is lost after the collision.Here when the chunk is at the given height it has a potential energy which on falling transforms into kinetic energy of the chunk.So, increase in the internal energy of the system after the collision is :
[tex]\Delta E=m.g.h[/tex]
[tex]\Delta E=200\times 9.8\times 30[/tex]
[tex]\Delta E=58800\ J[/tex]
A negative charge of -0.510 μC exerts an upward 0.600-N force on an unknown charge that is located 0.300 m directly below the first charge. What are (a) the value of the unknown charge (magnitude and sign); (b) the magnitude and direction of the force that the unknown charge exerts on the −0.550μC charge?
By employing Coulomb's Law and additional physical principles like Newton's third law, it becomes possible to determine the characteristics of an unknown electrical charge from the force it experiences and exerts.
Explanation:The question concerns two charges, one of which is negative and known (-0.510 µC), and the other is unknown. You're asked to find the value of this unknown charge and the force it exerts on the known charge. To do this, we would use Coulomb's Law, which states that the electric force between two charges is directly proportional to the product of their charges, and inversely proportional to the square of the distance separating them.
So, using the formula F = k*q1*q2/r^2, where F is the force, k is Coulomb's constant (approximately 9 * 10^9 N.m^2/C^2), q1 and q2 are the charges, and r is the distance between them, we can solve for the unknown charge(q2) - which gives us q2 = Fr^2/(kq1). Substituting the provided values in the question, we should be able to solve it.
For part (b), the sign of the force that the unknown charge exerts on the known charge would be exactly opposite in direction to the force it experiences, as per Newton's third law (equal and opposite reactions). However, we'd need to know the magnitude of the unknown charge to calculate the actual force.
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The electrons in a particle beam each have a kinetic energy K. What is the magnitude of the electric field that will stop these electrons in a distance d?
Answer:
Electric field magnitude
E = K/qd
Where
K = kinetic energy of electron
d = electron distance
q = charge
Explanation:
Given the relationship between workdone and energy
Work-energy theorem:
Net workdone = Energy change
W = ∆E
In this case
W = ∆K.E
And,
∆K.E = K(final) - K(initial)
To stop the kinetic energy | K(final) = 0
K(initial) = K (given)
∆K.E = 0 - K = -K
Let the electric force on the electron has magnitude F.
And
W = -Fd = ∆K.E = -K
-Fd = -K
F = K/d .....1
The magnitude of the electric field E that can stop these electron in a distance d:
E = F/q ......2
Where q is the charge on electron.
substituting equation 1 to 2
E = (K/d)/q = K/qd
E = K/qd
A drag racer, starting from rest, travels 6.0 m in 1.0 s. Suppose the car continues this acceleration for an additional 4.0 s. How far from the starting line will the car be? We assume that the acceleration is constant, and the initial speed is zero, so the displacement will scale as the square of the time. After 1.0 s, the car has traveled 6.0 m; after another 4.0 s, a total of 5.0 s will have elapsed. The initial elapsed time was 1.0 s, so the elapsed time increases by a factor of 5. The displacement thus increases by a factor of 5^2, or 25. The total displacement is Delta x = 25(6.0 m) = 150 m This is a big distance in a short time, but drag racing is a fast sport, so our answer makes sense.
Answer:
x = 150.0 m
Explanation:
If the acceleration is constant, we can find the value of the acceleration, starting from rest, applying the following kinematic equation:
x = 1/2*a*t² = 6.0 m
Solving for a:
a = 2*x / t² = 2*6.0 m / 1.0s² = 12 m/s²
Now, during the following 4.0 s, the car continues moving with this acceleration, but its initial velocity is not zero anymore, but the speed at 1.0 s, which is just 12 m/s (as it accelerates 12 m/s each second), so we can write again the same kinematic equation, taking into account that initial velocity for the second part, as follows:
x = x₀ + v₀*t + 1/2*a*t² = 6.0 m + 12m/s*4.0s + 1/2*12 m/s²*(4.0)s² = 150.0 m
⇒ x = 150.0 m
Hold a small piece of paper (e.g., an index card) flat in front of you. The paper can be thought of as a part of a larger plane surface.
A. What single line could you use to specify the orientation of the plane of the paper (i.e., so that someone else could hold the paper in the same, or in a parallel, plane)?
B. The area of a flat surface can be represented by a single vector, called the area vector A. What does the direction of the vector represent? What would you expect the magnitude of the vector to represent?
C. Place a large piece of graph paper flat on the table. Describe the direction and magnitude of the area vector, A, for the entire sheet of paper. Describe the direction and magnitude of the area vector, dA, for each of the individual squares that make up the sheet.
D. Fold the graph paper twice so that it forms a hollow triangular tube. Can the entire sheet be represented by a single vector with the characteristics you defined above? If not, what is the minimum number of area vectors required?
E. Form the graph paper into a tube as shown. Can the orientation of each of the individual squares that make up the sheet of graph paper still be represented by dA vectors as inabove? Explain.
F. What must be true about a surface or a portion of a surface in order to be able to associate a single area vector A with that surface?
Answer:
Explanation:
(a). The line used to specify the orientation of the plane of paper is the line normal to the plane of sheet of paper
(b). The direction of the vector represents the normal to the lat surface while the Magnitude represents the area of flat surface.
(c). Say the area of each smaller square is 1 square unit, then the area of graph paper is 64 square units. Direction of this area vector is given by a unit vector perpendicular to the graph sheet. If X and Y axes are in the plane of paper, then unit vector normal to the sheet of paper is K. Hence the complete vector is 64 K sq. units.
Area vector of each individual square is 1 squ. unit. where all these individual squares are parallel as vectors.
(d). Absolutely.
the entire sheet can be represented by a single vector. Its area vector is the sum of area vectors of three flat sides of triangular tube.
(e) NO.
Orientation of the individual squares is not the same for all squares. They cannot be represented by the same vector when compared to part C above, because they are in different directions even tough their magnitude are same.
(f) To represent a surface with a single area vector, divide the surface in to as many as possible flat pieces (if necessary infinitely large number of infinitesimally small pieces). Find the area vectors of all pieces. Add all the area vectors to obtain the single area vector resenting the complete surface.
But since the process can be done for any surface, any surface can be represented by a single area vector.
i hope this helps, cheers
The normal vector is perpendicular to the flat surface while the area vector is the direction in which the plane is embedded in 3 dimensions.
Normal Vector:A vector that is perpendicular to the plane of the surface. So a normal vector will be used to specify the plane of the paper.The magnitude of the flat surface represents the area while the vector represents the normal. Area vector:An area vector is an area (magnitude) with direction.
Therefore, the normal vector is perpendicular to the flat surface while the area vector is the direction in which the plane is embedded in 3 dimensions.
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You place a point charge q = -4.00 nC a distance of 9.00 cm from an infinitely long, thin wire that has linear charge density 3.00×10−9C/m. What is the magnitude of the electric force that the wire exerts on the point charge?
The question is about calculating the electric force on a point charge near an infinitely long wire using the concepts of electric fields and Coulomb's law within the domain of Physics.
Explanation:The question involves calculating the magnitude of the electric force that an infinitely long, thin wire exerts on a point charge placed at a certain distance from it. This falls under the subject of Physics, specifically dealing with electrostatics and the concept of electric fields and forces.
First, we need to find the electric field due to the wire at the location of the point charge. The electric field E from an infinitely long wire is given by the formula E = (2kλ)/r, where k is Coulomb's constant (8.99 × 109 Nm2/C2), λ is the linear charge density of the wire, and r is the distance from the wire to the point where the electric field is being calculated.
After calculating E, we can find the force exerted on the charge q using F = qE. Since we're given q = -4.00 nC and λ = 3.00×10−9C/m, with r = 9.00 cm, we can substitute these values into the formulas to calculate the electric field and then the force acting on the point charge.
The magnitude of the electric force exerted by the wire on the point charge is 2.40 × 10⁻⁶ N.
To find the magnitude of the electric force that an infinitely long, thin wire with linear charge density λ = 3.00×10⁻⁹ C/m exerts on a point charge q = -4.00 nC placed a distance of 9.00 cm from the wire, we use the principle that an electric field E is created by a line of charge.
The electric field created by an infinitely long, thin wire is given by:
E = (2kλ)/r
where k = 8.99 × 10⁹ Nm²/C²
λ = 3.00×10⁻⁹ C/m , charge density
r = 9 × 10⁻² m
Substitute the given values:
E = (2 × 8.99 × 10⁹ Nm²/C² × 3.00×10⁻⁹ C/m ) / 9 × 10⁻² m
Calculating this, we get:
E ≈ 600 N/C
The force F on the point charge q due to the electric field E is given by:
F = q × E
Since q = -4.00 nC = -4.00 × 10⁻⁹ C:
F = (-4.00 × 10⁻⁹ C) × 600 N/C ≈ -2.40 × 10⁻⁶ N
Therefore, the magnitude of the electric force is 2.40 × 10⁻⁶N.
A uniform charge density of 500nC/m^3 is distributed throughout a spherical volume (radius 16cm). Consider a cubical (4cm along each edge) surface completely inside the sphere. Determine the electric flux through the surface of the cube.
Answer:
[tex]3.61581\ Nm^2/C[/tex]
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
V = Volume of cube = [tex]0.04^3[/tex]
[tex]\rho[/tex] = Charge density = [tex]500\ nC/m^3[/tex]
Electric flux is given by
[tex]\phi=\dfrac{Q}{\epsilon_0}\\\Rightarrow \phi=\dfrac{\rho V}{\epsilon_0}\\\Rightarrow \phi=\dfrac{500\times 10^{-9}\times 0.04^3}{8.85\times 10^{-12}}\\\Rightarrow \phi=3.61581\ Nm^2/C[/tex]
The electric flux through the surface of the cube is [tex]3.61581\ Nm^2/C[/tex]
The electric flux through the surface of a cubical object inside a sphere can be found by using Gauss's Law. It is equal to the total charge enclosed by the cube, calculated by multiplying the charge density by the volume of the cube and then divided by the permittivity of free space. In this case, the electric flux is about 36.109 N.m^2/C.
Explanation:The subject of this question is physics, specifically dealing with the concept of electric charges and electric fields. To find the electric flux through the cubical surface placed inside the sphere, we can use Gauss's Law.
According to Gauss's Law, the electric flux through a closed surface is equal to the total charge enclosed by the surface divided by permittivity constant (εo).
In this case, we first need to find the charge enclosed by the cube, which we can find by multiplying the charge density with the volume of the cube. Since the charge density is 500nC/m^3 and the volume is 4cm x 4cm x 4cm (converted to m^3), the enclosed charge is 32nC. Then, divide this by the permittivity of free space, resulting in about 36.109 N.m^2/C, which represents the electric flux through the surface of the cube.
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If you place a glass cylinder in Wessin Oil, the view of the cylinder nearly disappeared when the eyedropper was full of Wessin Oil because _____________.
A. the indices of refraction matched
B. the light was reflecte
C. the light was absorbed
Answer:
A. the indices of refraction matched
Explanation:
The index of refraction, or refractive
index, is a measure of how fast light
rays travel through a given medium.
Alternatively, it could be said that
the refractive index is the measure of
the bending of a light ray when
passing from one medium to
another. Mathematically, it can be
represented as a ratio between two
different velocities – the velocity of
light in vacuum and the velocity of
light in a given medium.
For example, try putting a pencil in a jar full of water. If you look at the pencil from above, it would look as though the pencil has bent in the water. That happens due to the refraction of light. It occurs because as light rays enter water, they slow down, as the speed of light in water is lower than the speed of light in air. The magnitude of how much a medium refracts a light ray is determined by the index
If you place a glass cylinder in Wessin Oil, the view of the cylinder nearly disappeared when the eyedropper was full of Wessin Oil because the indices of refraction matched here. Thus, the correct option is A.
What is the refraction of light?Refraction is the phenomena of bending of light, which also happens with the rays of sound, water and other waves as it passes from one transparent substance into another substance or medium. This phenomena of bending of light waves by the refraction makes it possible for the people to have lenses, magnifying glasses, prisms, and rainbows. Even, our eyes depend upon this bending of light for the visibility and other effects of light.
If we place a glass cylinder in Wessin Oil, the view of the cylinder nearly disappeared when the cylinder nearly disappeared when the eyedropper was full of Wessin Oil because the indices of refraction matched.
Therefore, the correct option is A.
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Photons of what minimum frequency are required to remove electrons from gold? The work function for gold is 4.8 eV. 1 eV = 1.6 x 10 -19 J. h = 6.626 x 10 -34 J s.a. 7.3 x10^14 Hz b. 6.5x10^15 Hz c. 1.2 x 10^15 Hz d. 4.6 x 10^14 Hz e. 3.8 x10^17 Hz
Answer:
c. 1.2×10¹⁵ Hz
Explanation:
Work Function: This is the minimum amount of energy a photon requires to liberate an electron from the surface of a metal.
Mathematically, it can be represented as
E' = hf' ................................... Equation 1
Where E' = work function of gold, f' = minimum frequency ( threshold frequency), h = Planck's constant
Making f' the subject of the equation,
f' = E'/h................................. Equation 2
Given: E' = 4.8 ev = 4.8×1.6×10⁻¹⁹ J = 7.68×1.6×10⁻¹⁹ J, h = 6.626×10⁻³⁴Js.
Substituting into equation 2
f' = 7.68×10⁻¹⁹/ 6.626×10⁻³⁴
f' = 1.16×10¹⁵ Hz.
f' ≈ 1.2×10¹⁵ Hz
Thus the minimum frequency = 1.2×10¹⁵ Hz
The right option is c. 1.2×10¹⁵ Hz
Using Planck's equation, the minimum frequency of photons required to remove an electron from gold, given a work function of 4.8 eV, is approximately 1.15 * 10^15 Hz or 1.15 PHz.
Explanation:
The question is asking for the minimum frequency of photons capable of removing electrons from gold. This is related to the photoelectric effect, where electrons can be ejected from a metal surface by incident light. In this process, energy of the incoming light is absorbed by the electron, which can then be ejected if the light's energy is greater than the binding energy of the electron, which is also known as the work function.
The work function for gold is given as 4.8 eV. To calculate the minimum frequency, we need to employ Planck's equation, E = hf, which implies frequency, f = E/h. Given that E is the energy of the photon which must be equal to the work function (4.8 eV or 4.8 * 1.6 x 10^-19 J) and h is Planck’s constant (6.626 x 10^-34 J s), we can solve for f. The calculation gives approximately 1.15 * 10^15 Hz or 1.15 PHz, which is not listed among the provided answer choices.
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Calculate the average power output (in watts) of a photodetector that collects 8.0 x 107 photons in 3.8 ms from monochromatic light of wavelength (a) 470 nm, the wavelength produced by some commercially available light-emitting diodes (LED), and (b) 780 nm, a wavelength produced by lasers that are commonly used in compact disc (CD) players. Hint: The total energy emitted by a source or collected by a detector in a given interval is its power multiplied by the time interval of interest (1 J = 1 W s).
Answer:
[tex]8.90392\times 10^{-9}\ W[/tex]
[tex]5.36518\times 10^{-9}\ W[/tex]
Explanation:
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
t = Time taken = 3.8 ms
[tex]\lambda[/tex] = Wavelength
n = Number of protons = [tex]8\times 10^7[/tex]
Power is given by
[tex]P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{nh\dfrac{c}{\lambda}}{t}\\\Rightarrow P=\dfrac{8\times 10^7\times 6.626\times 10^{-34}\times \dfrac{3\times 10^8}{470\times 10^{-9}}}{3.8\times 10^{-3}}\\\Rightarrow P=8.90392\times 10^{-9}\ W[/tex]
The power is [tex]8.90392\times 10^{-9}\ W[/tex]
[tex]P=\dfrac{nh\dfrac{c}{\lambda}}{t}\\\Rightarrow P=\dfrac{8\times 10^7\times 6.626\times 10^{-34}\times \dfrac{3\times 10^8}{780\times 10^{-9}}}{3.8\times 10^{-3}}\\\Rightarrow P=5.36518\times 10^{-9}\ W[/tex]
The power is [tex]5.36518\times 10^{-9}\ W[/tex]
6 A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 20 kPa in the condenser and a turbine inlet temperature of 700°C. The boiler is sized to provide a steam flow of 50 kg/s. Determine the power produced by the turbine and consumed by the pump.
To solve this problem we will start by differentiating the values in each of the states of matter. Subsequently through the thermodynamic tables we will look for the values related to the entropy, enthalpy and respective specific volumes. Through the relationship of Power defined as the product between mass and enthalpy and mass, specific volume and pressure, we will find the energetic values in the two states investigated. We will start defining the states
State 1
[tex]T_1 = 700\°C[/tex]
[tex]P_1 = 4 Mpa[/tex]
From steam table
[tex]h_1 =3906.41 KJ/Kg[/tex]
[tex]s_1 = 7.62 KJ/Kg.K[/tex]
Now
[tex]s_1 = s_2 = 7.62 KJ/Kg.K[/tex] As 1-2 is isentropic
State 2
[tex]P_2 = 20 Kpa[/tex]
[tex]s_2 = 7.62 KJ/Kg \cdot K[/tex]
From steam table
[tex]h_2 = 2513.33 KJ/Kg[/tex]
PART A) The power produced by turbine is the product between the mass and the enthalpy difference, then
[tex]Power = m \times (h_1-h_2)[/tex]
[tex]P = (50)(3906.41 - 2513.33)[/tex]
[tex]P = 69654kW[/tex]
b) Pump Work
State 3
[tex]P_3 = 20 Kpa[/tex]
[tex]\upsilon= 0.001 m^3/kg[/tex]
The Work done by the pump is
[tex]W= m\upsilon \Delta P[/tex]
[tex]W = (50)(0.001)(4000-20)[/tex]
[tex]W = 199kJ[/tex]
A small object A, electrically charged, creates an electric field. At a point P located 0.250 m directly north of A, the field has a value of 40.0 N/C directed to the south.
What is the charge of object A?
the magnitude of the charge of object A is[tex]\( 2.78 \times 10^{-10} \, \text{C} \).[/tex]
We can use the formula for electric field intensity ( E ) to calculate the charge of object A. The electric field intensity is given by:
[tex]\[ E = \frac{k \cdot |q|}{r^2} \][/tex]
Where:
- ( E ) is the electric field intensity (40.0 N/C),
- ( k) is Coulomb's constant [tex](\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),[/tex]
- [tex]\( |q| \)[/tex] is the magnitude of the charge on object A (what we're trying to find), and
- ( r ) is the distance from object A to point P (0.250 m).
First, rearrange the formula to solve for[tex]\( |q| \)[/tex]:
[tex]\[ |q| = \frac{E \cdot r^2}{k} \][/tex]
Now, substitute the given values into the formula:
[tex]\[ |q| = \frac{40.0 \, \text{N/C} \cdot (0.250 \, \text{m})^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2} \]\[ |q| = \frac{40.0 \times 0.0625}{8.99 \times 10^9} \, \text{C} \]\[ |q| = \frac{2.5}{8.99 \times 10^9} \, \text{C} \]\[ |q| = 2.78 \times 10^{-10} \, \text{C} \][/tex]
So, the magnitude of the charge of object A is[tex]\( 2.78 \times 10^{-10} \, \text{C} \).[/tex]
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656(red) and 486 (blue). Light from a hydrogen lamp illuminates a diffraction grating with 500 lines per mm, and the light is observed on a screen 1.50 behind the grating. What is the distance between the first-order red and blue fringes?
To find the distance between the first-order red and blue fringes formed by a diffraction grating, we calculate the diffraction angles using the grating equation and then use trigonometry to determine the positions of the fringes on the screen. The distance apart is the absolute difference between these positions.
Explanation:The student is asking about the distance between the first-order red and blue fringes that appear on a screen as a result of a diffraction pattern formed by light from a hydrogen discharge lamp passing through a diffraction grating. To answer this question, we apply the formula for diffraction maxima, d sin( heta) = m ext{ extlambda}, where d is the distance between the grating lines, ext{ extlambda} is the wavelength of light, ext{ exttheta} is the diffraction angle, and m is the order of the maximum.
Given that the grating has 500 lines per mm or 500,000 lines per meter, we can calculate d as 1/500,000 meters. Since we are looking for the first-order maxima (m=1), we can find the angles for each color using the formula sin( heta) = ext{ extlambda}/d. We can then use basic trigonometry to find the positions of the red and blue fringes on the screen, and their distance apart.
For the 656 nm red light:
ext{ exttheta}_{red} = sin^{-1}( ext{ extlambda}_{red}/d)
ext{ exttheta}_{red} = sin^{-1}(656 ext{ exttimes}10^{-9} m / (1/500,000 m))
For the 486 nm blue light:
ext{ exttheta}_{blue} = sin^{-1}( ext{ extlambda}_{blue}/d)
ext{ exttheta}_{blue} = sin^{-1}(486 ext{ exttimes}10^{-9} m / (1/500,000 m))
After calculating the angles, the positions on the screen for first-order maxima are found by L tan( heta), where L is the distance from the grating to the screen (1.50 m in this case).
The final distance between the red and blue fringes is the absolute difference between their positions on the screen.