Answer:
A) [tex]k=34867.3384\ N.m^{-1}[/tex]
B) [tex]\omega'\approx84\ Hz[/tex]
Explanation:
Given:
mass of car, [tex]m=1380\ kg[/tex]
A)
frequency of spring oscillation, [tex]f=1.6\ Hz[/tex]
We knkow the formula for spring oscillation frequency:
[tex]\omega=2\pi.f[/tex]
[tex]\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f[/tex]
[tex]\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6[/tex]
[tex]k_{eq}=139469.3537\ N.m^{-1}[/tex]
Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.
So, the stiffness of each spring is (as they are identical):
[tex]k=\frac{k_{eq}}{4}[/tex]
[tex]k=\frac{139469.3537}{4}[/tex]
[tex]k=34867.3384\ N.m^{-1}[/tex]
B)
given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:
[tex]\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }[/tex]
[tex]\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }[/tex]
[tex]\omega'\approx84\ Hz[/tex]
the position of a crate sliding down a ramp is given by x=0.25t^3 m, z = 6-0.75t^5/2 m, where t is in seconds. determine the magnitude of the crates velocity and acceleration when t = 2 seconds
Answer:
v = 30.15 m/s
a = 60.07 m/s2
Explanation:
Velocity is derivative of position with respect to time
[tex]v_x = x' = 3*0.25t^2 = 0.75t^2[/tex]
[tex]v_z = z' = 5*0.75t^4/2 = 1.875t^4[/tex]
Acceleration is derivative of velocity with respect to time
[tex]a_x = v_x' = 2*0.75 t = 1.5t[/tex]
[tex]a_z = v_z' = 4*1.875t^3 = 7.5t^3[/tex]
At t = 2 seconds
[tex]v_x = 0.75*2^2 = 3m/s[/tex]
[tex]v_z = 1.875*2^4 = 30m/s[/tex]
[tex]v = \sqrt{v_x^2 + v_z^2} = \sqrt{3^2 + 30^2} = \sqrt{909} = 30.15 m/s[/tex]
[tex]a_x = 1.5*2 = 3 m/s^2[/tex]
[tex]a_z = 7.5*2^3 = 60 m/s^2[/tex]
[tex]a = \sqrt{a_x^2 + a_z^2} = \sqrt{3^2 + 60^2} = \sqrt{3609} = 60.07 m/s[/tex]
An isolated, parallel‑plate capacitor carries a charge Q . If the separation between the plates is doubled, the electrical energy stored in the capacitor will be:a. Unchanged.b. Halved.c. Doubled.d. Quartered.e. Quadrupled.
Answer:
option C
Explanation:
Given,
Change on the capacitor = Q
Separation is doubled
Energy stored in the capacitor,E = ?
we know,
[tex]E = \dfrac{Q^2}{2C}[/tex]
and [tex]C = \dfrac{\epsilon_0A}{d}[/tex]
now,
[tex]E = \dfrac{Q^2}{2\epsilon A}\ d[/tex].......(1)
where d is the separation between the two plates.
now, when the separation is doubled
[tex]E' = \dfrac{Q^2}{2\epsilon A}\ (2d)[/tex]
[tex]E' = 2(\dfrac{Q^2}{2\epsilon_0 A}\ d)[/tex]
From equation (1)
E' = 2 E
Hence, the energy stored in the capacitor is doubled if the separation is increased.
The correct answer is option C.
A tire has a tread pattern with a crevice every 2.00 cm. Each crevice makes a single vibration as the tire moves. What is the frequency of these vibrations if the car moves at 30.0 m/s?
Answer:
Frequency of the vibration due to crevice is 1500 Hz.
Explanation:
Frequency is defined as rate of vibration per second. Its S.I. unit is s⁻¹ or Hz.
According to the problem, we have to find number of crevice car makes in 1 second.
Speed of car, u = 30 m/s
Distance covered by car in 1 second, d = 30 m
For every 0.02 m, one crevice occurred by the tire of car.
Number of crevice occurred in 1 second by the car = [tex]\frac{30}{0.02}[/tex]
= 1500
Since, each crevice makes a single vibration. Thus, the frequency of these vibrations is 1500 Hz.
The frequency of the vibrations generated by the tire treads in this case, given that each crevice is 2.00 cm apart and the car moves at 30.0 m/s, is 1500 Hz.
Explanation:In this scenario, it is important to understand that the frequency of the vibrations is determined by the speed of the tire and the tread pattern. Here, the crevice, which causes the vibration, occurs every 2.00 cm (or 0.02 m). This means for every meter the car moves, 0.02 m into 1 m gives a total of 50 vibrations. Therefore, at a speed of 30.0 m/s, we will have 50 vibrations/m multiplied by 30 m/s, giving a frequency of 1500 Hz or vibrations per second.
Frequency in this question refers to the number of times the vibration from the crevices occur in a unit of time (in this case, per second). This concept is crucial in understanding wave motions and vibration patterns, and is often applied in physics-related disciplines.
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"A boat that can travel at 4.0 km/h in still water crosses a river with a current of 2.0 km/h. At what angle must the boat be pointed upstream (that is, relative to its actual path) to go straight across the river?
Answer:
30 degrees
Explanation:
The boat to go straight across the river uptream will have to make angle of 30 degree with the resultant velocity vector of boat.
What is vector law of addition ?
Vector addition is governed by the triangle law which states that when two vectors are represented by two triangle sides with their order of magnitude and direction, the resultant vector's magnitude and direction will be represented by the third triangle side.
It is given that speed of boat in still water v₁ = 4 km/h
speed of current v₂ = 2 km/h
Let the relative speed of boat with respect to current be = v km/h
and the angle made by v₁ with v be = θ
No find the angle θ with which the boat to go straight across the river uptream by triangle law of vector addition as shown below :
[tex]\begin{aligned}\theta &=\text{sin}^{-1}\left ( \frac{2}{4} \right )\\&= 30^{0}\end{aligned}[/tex]
Therefore, the boat to go straight across the river uptream will have to make angle of 30 degree with the resultant velocity vector of boat.
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The sound intensity from a jack hammer breaking concrete is 2.0W/m2 at a distance of 2.0 m from the point of impact. This is sufficiently loud to cause permanent hearing damage if the operator doesn't wear ear protection. What are (a) the sound intensity and (b) the sound intensity level for a person watching from 50 m away?
Answer:
(a) [tex]I_{1}=3.2*10^{-3}W/m^{2}[/tex]
(b) [tex]\beta =95dB[/tex]
Explanation:
Given data
Distance r₁=50 m
Distance r₂=2 m
Intensity I₂=2.0 W/m²
To find
(a) The Sound Intensity I₁
(b) The Sound Intensity level β
Solution
For (a) the Sound Intensity I₁
[tex]\frac{I_{1} }{I_{2}}=\frac{(r_{2})^{2} }{(r_{1})^{2} }\\I_{1} =I_{2}(\frac{(r_{2})^{2} }{(r_{1})^{2} })\\I_{1}=(2.0W/m^{2} )(\frac{(2m)^{2} }{(50m)^{2} })\\I_{1}=3.2*10^{-3}W/m^{2}[/tex]
For (b) the Sound Intensity level β
The Sound Intensity level β is calculated as follow
[tex]\beta =(10dB)log_{10}(\frac{I}{I_{o} } )\\\beta =(10dB)log_{10}(\frac{3.2*10^{-3}W/m^{2} }{1.0*10^{-12} W/m^{2} } )\\\beta =95dB[/tex]
Calculate the energy of the quantum involved in the excitation of (i) an electronic oscillation of period 1.0 fs, (ii) a molecular vibration of period 10 fs, (iii) a pendulum of period 1.0 s. Express the results in joules and kilojoules per mole.
Answer:
a) E = 3.99 × 10⁵J/mol = 3.99 × 10² KJ/mol = 399KJ = 400KJ/mol
b) E = 3.99 × 10⁴ J/mol = 3.99 × 10¹ KJ/mol = 39.9KJ/mol = 40 KJ/mol
c) E = (3.99 × 10^-10) J/mol = 3.99 × 10⁻⁷ KJ/mol
Explanation:
E = hf where E = energy, H = Planck's constant = 6.62607004 × 10⁻³⁴ J.s and f = 1/time period
a) Period = 1 fs = 1 × 10⁻¹⁵ s
f = 1/(10⁻¹⁵ ) = 10¹⁵ Hz
E = hf = 6.62607004 × 10⁻³⁴ × 10¹⁵ = 6.63 × ⁻10¹⁹ J = 6.63 × 10⁻¹⁶ KJ
In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol
E = 3.99 × 10⁵J/mol = 3.99 × 10² KJ/mol = 399KJ = 400KJ/mol
b) Period = 10 fs = 10 × 10⁻¹⁵ = 10⁻¹⁴ s
f = 1/period = 1/10⁻¹⁴ = 10¹⁴ Hz
E = hf = 6.62607004 × 10⁻³⁴ × 10¹⁴ = 6.63 × 10^-20 J = 6.63 × 10⁻¹⁷ KJ
In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol
E = 3.99 × 10⁴ J/mol = 3.99 × 10¹ KJ/mol = 39.9KJ/mol = 40 KJ/mol
c) Period = 1s
f = 1/period = 1.0 Hz
E = 6.62607004 × 10⁻³⁴ × 1 = 6.63 × 10⁻³⁴ J = 6.63 × 10⁻³¹ KJ.
In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol
E = (3.99 × 10^-10) J/mol = 3.99 × 10⁻⁷ KJ/mol
Answer:
(i) E = 6.626 X 10⁻¹⁹ J = 400 KJ/mol
(ii) E = 6.626 X 10⁻²⁰ J = 40 KJ/mol
(iii) E = 6.626 X 10⁻³⁴ J = 4 X 10⁻¹³ KJ/mol
Explanation:
Energy associated with excitation of a quantum is given as;
E = hf
where;
E is the energy of excitation
h is Planck's constant = 6.626 X 10⁻³⁴Js⁻¹
f is the threshold frequency in s⁻¹
Thus, E = h/t
Part (i)
E = (6.626 X 10⁻³⁴)/(1 X 10⁻¹⁵)
E = 6.626 X 10⁻¹⁹ J
In (KJ/mol) = 6.626 X 10⁻²² KJ X 6.022 X10²³ = 400 KJ/mol
Part (ii)
E = (6.626 X 10⁻³⁴)/(1 X 10⁻¹⁴)
E = 6.626 X 10⁻²⁰ J
In (KJ/mol) = 6.626 X 10⁻²³ KJ X 6.022 X10²³ = 40 KJ/mol
Part (iii)
E = (6.626 X 10⁻³⁴)/(1)
E = 6.626 X 10⁻³⁴ J
In (KJ/mol) = 6.626 X 10⁻³⁷ KJ X 6.022 X10²³ = 4 X 10⁻¹³ KJ/mol
A box is sitting on a 2 m long board at one end. A worker picks up the board at the end with the box so it makes an angle with the ground of 35o. The coefficient of static friction between the box and the board is 0.5 and the coefficient of kinetic friction is 0.3 . Will the box slide down the ramp when it is at 35o?
Answer:
Yes it will slide down the ramp
Explanation:
Let m be the mass of the box and gravitational acceleration g = 9.81m/s2, we can calculate the gravity that affects the box
W = mg
When the box is at 35 degree incline, this gravity is split into 2 components, 1 parallel to the incline (Wsin35) and the other one perpendicular with the incline (Wcos35).
The one perpendicular with the incline has an equal and opposite normal force of Wcos35
This normal force would dictate the static friction force where coefficient = 0.5. So the static friction is 0.5mgcos35
The box would slide when the parallel component of gravity wins over static friction force
mgsin35 > 0.5mgcos35
Since mg is positive we can cancel them out on both sides
sin35 > 0.5cos35
0.57 > 0.5*0.82
0.57 > 0.41
This is true so we can conclude that the box slides down the ramp
(a) If a flea can jump straight up to a height of 0.440 m, what is its initial speed as it leaves the ground? (b) How long is it in the air?
Answer
given,
height of the jump = 0.44 m
acceleration due to gravity, g = 9.8 m/s²
velocity at the height point = 0 m/s
initial speed = ?
Using equation of motion for speed calculation
v² = u² + 2 g h
0 = u² - 2 x 9.8 x 0.44
u = √8.624
u = 2.94 m/s
time taken to reach the highest point
v = u + g t
0 = 2.94 - 9.8 x t
t = 0.3 s
total time of flight will be equal to double of the time taken to reach the maximum height.
Total time = 2 x 0.3 = 0.6 s
A flea jumps straight up to a height of 0.440 m with an initial speed of 2.94 m/s and is in the air for 0.600 s.
A flea jumps straight up to a height (h) of 0.440 m. At the maximum height, the final speed (v) is zero. Given that the gravity (g) is 9.81 m/s², we can calculate the initial speed (u) using the following kinematic expression (We will assume y+ as the positive direction)
[tex]v^{2} = u^{2} + 2 \times g \times h \\\\(0m/s)^{2} = u^{2} + 2 \times (-9.81m/s^{2} ) \times 0.440m\\\\u = 2.94 m/s[/tex]
We can calculate the time (t) required to reach the maximum height using the following kinematic expression.
[tex]v = u + g \times t\\\\0m/s = 2.94m/s - 9.81m/s^{2} \times t\\\\t = 0.300 s[/tex]
The time in the air will be double the time to reach the maximum height.
[tex]t_{total} = 2 \times 0.300 s = 0.600 s[/tex]
A flea jumps straight up to a height of 0.440 m with an initial speed of 2.94 m/s and is in the air for 0.600 s.
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A good-quality measuring tape can be off by 0.42 cm over a distance of 28 m. What is its percent uncertainty? (Express your answer to the correct number of significant figures and proper units.)
Answer:
0.015%
Explanation:
Data provided in the question:
Length by which the measuring tape can be off, δL = 0.42 cm
Total measured length for which Error of δL is observed, L = 28 m
Now,
we know,
1 m = 100 cm
Thus,
28 m = 28 × 100 = 2800 cm
Percent uncertainty = [δL ÷ L] × 100%
= [0.42 ÷ 2800] × 100%
= 0.015%
A person is standing in an elevator that is moving downward and slowing down. Is the magnitude of the normal force on the person greater than, less than, or equal to the magnitude of the weight force on the person?
the magnitude of the normal force is greater than the magnitude of the weight of the person
Explanation:
As the elevator moves down, there are two forces acting on the person:
- The weight of the person, [tex]W=mg[/tex], where m is the mass of the person and g the acceleration due to gravity, acting downward
- The normal reaction exerted by the floor of the lift on the person, N, acting upward
This means that Newton's second law can be written as
[tex]\sum F = N-W= ma[/tex] (1)
where we chose upward as positive direction, and a is the acceleration of the elevator.
Here we know that the elevator is moving downward and it is slowing down: this means that the velocity is negative (upward), and the acceleration is in the opposite direction (upward), so the acceleration is positive.
Eq.(1) can be rewritten as
[tex]N=W+ma[/tex]
and as we said that [tex]a>0[/tex], this means that
[tex]N>W[/tex]
So the magnitude of the normal force is greater than the magnitude of the weight of the person.
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The magnitude of the normal force on a person in a downward-slowing elevator is greater than the magnitude of the weight force.
Explanation:When a person is standing in an elevator that is moving downward and slowing down, the magnitude of the normal force on the person is greater than the magnitude of the weight force on the person.
According to Newton's third law, the normal force and the weight force are equal in magnitude and opposite in direction. In this case, since the elevator is moving downward and slowing down, the acceleration of the person is in the upward direction. Therefore, the normal force exerted by the floor of the elevator on the person must be greater than the weight force in order to cause the upward acceleration.
It is important to note that if the elevator is in free-fall and accelerating downward at the acceleration due to gravity, the normal force becomes zero and the person appears to be weightless.
The total distance treaveled by a car moving in a straight line is as follows: After the first 7.0 minutes it has gone a total of 2.0 miles. After 14.0 minutes it has traveled a total of 4.5 miles. Finally at 21.0 minutes it has traveled a total of 6.0 miles. Find the average speed at: Time
Answer:
Average speed of the car will be 27.582 km/hr
Explanation:
We have given that in first 5 minutes distance traveled by car is 2 miles
After 14 minutes it has travel 4.5 miles
And finally after 21 minutes distance traveled by car is 6 miles
So total time of traveling t = 21 minutes
As we know that 1 hour = 60 minutes
So 21 minutes [tex]=\frac{21}{60}=0.35hour[/tex]
Total distance traveled = 6 miles
As 1 miles = 1.609 km
So 6 miles [tex]=6\times 1.609=9.654km[/tex]
Average speed is equal to the ratio of total distance and total time
So average speed [tex]v=\frac{9.654}{0.35}=27.582km/hr[/tex]
A closed system consists of 0.5 kmol of ammonia occupying a volume of 6 m3. Determine (a) the weight of the system, in N, and (b) the specific volume, in m3/kmol and m3/kg. Let g 5 9.81 m/s2.
Explanation:
It is known that the molecular weight of ammonia ([tex]NH_{3}[/tex]) is as follows.
Molecular weight ([tex]NH_{3}[/tex]) = [tex]14 + 3 \times 1[/tex] = 17
(a) Therefore, we will calculate the mass as follows.
[tex]0.5 kmol \times (\frac{1000 mol}{1 kmol}) \times (\frac{17 g}{1 mol})[/tex]
= 8500 g
Now, formula to calculate weight of the system in N is as follows.
Weight = mass × g
= [tex]8500 g \times (\frac{1 kg}{1000 g}) \times (9.8 m/s^{2})[/tex]
= 83.3 N (1 [tex]kg m/s^{2}[/tex] = 1 N)
Hence, the weight of the system is 83.3 N.
(b) Relation between specific volume and number of moles is as follows.
[tex]v (m^{3}/kmol) = \frac{V}{n}[/tex]
Therefore, calculate the specific volume as follows.
[tex]V_m = \frac{6 m^{3}}{0.5 k mol}[/tex]
= 12 [tex]m^{3}/k mol[/tex]
Also,
[tex]v (m^{3}/kmol) = \frac{V}{m}[/tex]
v = [tex]\frac{6 m^{3}}{8.5 kg}[/tex]
= 0.705882 [tex]m^{3}/kg[/tex]
Therefore, we can conclude that the value of specific volume is 12 [tex]m^{3}/k mol[/tex] and 0.705882 [tex]m^{3}/kg[/tex].
Answer:
a) [tex]w=83.385\ N[/tex]
b) [tex]\bar V=12\ m^3.kmol^{-1}[/tex]
[tex]\b V=0.7059\ m^3.kg^{-1}[/tex]
Explanation:
Given:
no. of moles of ammonia in a closed system, [tex]n=0.5\ kmol=500\ mol[/tex]
volume of ammonia, [tex]V=6\ m^3[/tex]
We know the molecular formula of ammonia: [tex]NH_3[/tex]
The molecular mass of ammonia:
[tex]M=14+3\times 1=18\ g.mol^{-1}[/tex]
Now the mass of given ammonia:
[tex]m=n.M[/tex]
[tex]m=500\times 17[/tex]
[tex]m=8500\ g=8.5\ kg[/tex]
a)
Now weight:
[tex]w=m.g[/tex]
[tex]w=8.5\times 9.81[/tex]
[tex]w=83.385\ N[/tex]
b)
Specific volume:
[tex]\bar V=\frac{6}{0.5}[/tex]
[tex]\bar V=12\ m^3.kmol^{-1}[/tex]
also
[tex]\b V=\frac{V}{m}[/tex]
[tex]\b V=\frac{6}{8.5}[/tex]
[tex]\b V=0.7059\ m^3.kg^{-1}[/tex]
Assume that a clay model of a lion has a mass of 0.225 kg and travels on the ice at a speed of 0.85 m/s. It hits another clay model, which is initially motionless and has a mass of 0.37 kg. Both being soft clay, they naturally stick together.
What is their final velocity?
Answer:
Final velocity will be equal to 0.321 m/sec
Explanation:
We have given mass of clay model of lion [tex]m_1=0.225kg[/tex]
Its speed is 0.85 m/sec, so [tex]v_1=0.85m/sec[/tex]
Mass of another clay model [tex]m_2=0.37kg[/tex]
It is given that second clay is motionless
So its velocity [tex]v_2=0m/sec[/tex]
Now according to conservation of momentum
Momentum before collision will be equal to momentum after collision
So [tex]m_1v_!+m_2v_2=(m_1+m_2)v[/tex], here v is velocity after collision
So [tex]0.225\times 0.85+0.37\times 0+(0.225+0.37)v[/tex]
[tex]0.595v=0.191[/tex]
v = 0.321 m/sec
So final velocity will be equal to 0.321 m/sec
A particle position in meters is given by the function x(t) = ct^4 + dt^2 + f. where c = 6m/s^4, d=8m/s^2, and f=-6m and t is in seconds.
Answer:
See explanation below.
Explanation:
For this case we have the following function:
[tex] x(t)= ct^4 +dt^2 +f[/tex]
Where [tex] c= 6 m/s^4 , d = 8m/s^2 , f=-6m [/tex]
If we replace those values we got:
[tex] x(t) = 6t^4 + 8t^2 -6[/tex]
If we want to find the position after t = 2.358 seconds for example we ust need to replace in the position function t = 2.358 and we got:
[tex] x(t=2.358) = 6(2.358)^4 + 8(2.358)^2 -6 \approx 224 m[/tex]
If we want to find the velocity we need to take the derivate of the position function and we got:
[tex] \frac{dx}{dt}=v(t) = 4ct^3 + 2dt[/tex]
[tex] v(t)= 24 t^3 + 16 t[/tex]
If we want to find the instantaneous velocity we just need to replace on v(t) a value for t
And the accelaration would be given by the second derivate of the position:
[tex] \frac{dv}{dt}= 72 t^2[/tex]
If we want to find the instantaneous acceleration we just need to replace on v(t) a value for t
Final answer:
Explaining the position function of a particle in meters at a given time using a specific function with provided values for constants.
Explanation:
Position Function: The position of a particle in meters at time 't' is described by the function x(t) = ct² + dt² + f, with given values for c, d, and f.
Explanation: To find the position of the particle using this function, substitute the values of c, d, f, and the specific time 't'. This will give you the position of the particle at that time.
Example: If c = 6m/s², d = 8m/s², f = -6m, and t = 2s, then you can plug in these values to find the position of the particle at t = 2s.
The net electric charge of an amber rod which has been rubbed with fur is called negative Group of answer choices because amber is an insulator by arbitrary convention so that the proton charge will be positive because like charges repel None of the above
Answer:
The right option is option E. None of the answer choices given are totally correct.
Explanation:
All insulators normally have an equal amount of positive and negative charges distributed on their surface.
The amber rod (an insulator) is called negative because after the coming together with fur (another insulator), the amber rod rubs off electrons from the fur onto itself and has an overall more negatively charged particles than positively charged particles on its surface.
The fur in turn becomes positive because it has more positive charges than negative on its surface.
So, the convention allows the now rubbed off amber rod to be called negative.
So, it is evident that none of the answer choices are totally correct, the right answer is more of a mix of some of the answer choices and more!
Hope this helps!!
You have a neutral balloon. What is its charge after 12000 electrons have been removed from it? The elemental charge is 1.6 × 10−19 C. Answer in units of µC.
Final answer:
Upon removing 12000 electrons from a neutral balloon, it will acquire a positive charge of 1.92 microcoulombs (1.92 µC). This is calculated by multiplying the number of electrons by the elemental charge and converting to the appropriate unit.
Explanation:
When 12000 electrons are removed from a neutral balloon, it obtains a positive charge because electrons carry a negative charge. To determine the charge the balloon now carries, we multiply the number of electrons removed by the elementary charge of an electron.
The charge (Q) on the balloon can be calculated using the formula:
Q = n × e
Where:
n is the number of removed electronse is the elementary charge per electron (1.6 × 10⁻¹⁹ C)Let's plug in the values into the formula:
Q = 12000 × 1.6 × 10⁻¹⁹ C
Q = 1.92 × 10⁻¹⁵ C
Converting coulombs (C) to microcoulombs (µC) by multiplying by 10¶ gives:
Q = 1.92 × 10⁻¹⁵ C × 10¶ µC/C
Q = 1.92 µC
The balloon will carry a charge of 1.92 µC after 12000 electrons have been removed.
charge densities such that are equal in magnitude but opposite in sign. The difference in potential between the plates is 490 V.
(a) Is the positive or the negative plate at the higher potential?
select
the positive plate
the negative plate
(b) What is the magnitude of the electric field between the plates?
____ kV/m
Answer:
a) The positive plate is at a higher potential than the negative plate.
b) Electric field between the plates = 5.76 KV/m
Explanation:
a) The positive plate is at a higher potential than the negative plate because of convention. We could define the negative plate to be "high potential." However, that convention has to be consistent. The positive plate is at the high potential, and that potential causes positive charges to flow from high potential to low potential. If we had defined the convention in the other direction, then the negative terminal would be the high potential, and that potential would cause negative charges to flow from high potential to low potential.
b) Electric field, E = potential difference across plate/distance or separation between the plates
E = V/d
V = 490V, d = 8.5cm = 0.085m
E = 490/0.085 = 5,764.706 V/m = 5.76 KV/m
Hope this helps!!!
A new, previously unknown, planet Vulcan was discovered in our solar system. We measure an orbital period of 103 Earth days for the new planet. We send some astronauts to travel to the planet and they measure the planet's gravitational acceleration to be 8.2 m/s2 on its surface. They also determine that the planet's radius is half the radius of Earth.
Q1. How far away from the sun is the new planet?
(A) 8.3 x 100 m
(B) 6.4 x 100 m
(C) 1.7 x 100 m
(D) 9.5 x 1010 m
(E) 2.3 x 1010 m
Q2. What is the mass of planet Vulcan?
(A) 1.3 x 1024 kg
(B) 5.4 x 1024 kg
(C) 9.8 x 1024 kg
(D) 0.2 x 102 kg
(E) 4.8 x 1024 kg
Q3. The astronauts also discover a moon that orbits planet Vulcan with a period of 63 days. How far away is the moon from the planet Vulcan?
(A) 1 x 108 m
(B) 20 x 108 m
(C) 460 x 108 m
(D) 9 x108 m
(E) 4 x 10 m
Q4. The astronauts visit the newly discovered moon to study it. They measure the gravitational acceleration on the surface of the moon to be 2.7 m/s. When their mission is finished, they try to escape from the moon. They measure that the minimum velocity to escape the moon is 3,000 m/s. What is the mass of the moon?
(A) 1.1 x 1023 kg
(B) 5.0 x 1023 kg
(C) 7.8 x 1023 kg
(D) 29 x 104 kg
(E) 370 x 1023 kg
Answer:
Q1. corect is B, Q2. it is A, Q3. E and Q4. A
Explanation:
Q1 For this exercise we can use Newton's second law where acceleration is centripetal.
F = m a
a = v² / r.
G m M / r² = m v² / r
G M / r = v²
The velocity has a constant magnitude whereby we can divide the length of the circular orbit (2π r) between the period
G M / r = (2π r / T)²
r³ = G M T2 / 4π²
Let's calculate
T = 103 day (24 h / 1 day) (3600 s / 1h) = 8,899 10⁶ s
r³ = 6.67 10⁻¹¹ 1.99 10³⁰ (8,899 10⁶) 2 / 4π²
r = ∛ (266.25 10³⁰)
r = 6.4 10¹⁰ m
The distance matches the value in part B
Q2 Astronauts have measured the acceleration of gravity, so we can use the second law with a body on the planet's surface
F = m g
G m M_p / R_p² = m g
G M_p / R_p² = g
M_p = g R_p² / G
They indicate that the radius of the planet is half the radius of the Earth
R_p = ½ R_earth
R_p = ½ 6.37 10⁶
R_p = 3.185 10⁶ m
Let's calculate
M_p = 8.2 (3,185 10⁶)² / 6.67 10⁻¹¹
M_p = 1.25 10²⁴ kg
The correct answer is A
Q3 We use Newton's second law again, with part Q1, where M is the mass of the planet and m is the mass of the moon
r³ = G M T² / 4π²
T = 63 days (24h / 1day) (3600s / 1h) = 5.443 10⁶ s
r³ = 6.67 10⁻¹¹ 1.25 10²⁴ (5.443 10⁶)² / 4π²
r = ∛ (62.56807 10²⁴)
r = 3.97 10⁸ m
The correct answer is E
Q4 To calculate this part let's use the conservation of mechanical energy,
Starting point The surface of the moon
Em₀ = K + U = ½ m v2 - G m M / r
Final point. Infinity with zero speed
[tex]Em_{f}[/tex] = 0
Em₀ = Em_{f}
½ m v² - G m M / R = 0
v² = 2 G M / r
M = v2 r / 2G
r = 2 G M / v²
Since we don't know the radius of the moon, we will also use the equation in part 2
M = g r² / G
r = √ GM / g
Let's replace
2G M / v² = √ G M / g
4 G M / v⁴ = 1 / g
M = v⁴ / (g 4G)
M = 3000⁴ / (2.7 4 6.67 10-11)
M = 1.12 10²³ kg
corract is A
A 67 Vrms source is placed across a load that consists of a 12 ohm resistor in series with an capacitor whose reactance is 5 ohms. Compute the following: a) The average power of the load b) The reactive power of the load c) The apparent power of the load d) The power factor of the load
a) The average true power is 318.3 W
b) The reactive power is 132.6 W
c) The apparent power is 344.8 W
d) The power factor is 0.92
Explanation:
a)
For a circuit made of a resistor and a capacitor, the average (true) power is given by the resistive part of the circuit only.
Therefore, the average true power is given by:
[tex]P=I^2R[/tex]
where
I is the current
R is the resistance
In this problem, we have
V = 67 V (rms voltage)
[tex]R=12 \Omega[/tex] is the resistance of the load
[tex]X=5\Omega[/tex] is the reactance of the circuit
First we have to find the impedance of the circuit:
[tex]Z=\sqrt{R^2+X^2}=\sqrt{12^2+5^2}=13 \Omega[/tex]
Then we can find the current in the circuit by using Ohm's law:
[tex]I=\frac{V}{Z}=\frac{67}{13}=5.15 A[/tex]
Therefore, the average true power is
[tex]P=I^2R=(5.15)^2(12)=318.3 W[/tex]
b)
The reactive power of a circuit consisting of a resistor and a capacitor is the power given by the capacitive part of the circuit.
Therefore, it is given by
[tex]Q=I^2X[/tex]
where
I is the current
X is the reactance of the circuit
In this circuit, we have
[tex]I=5.15 A[/tex] (current)
[tex]X=5 \Omega[/tex] (reactance)
Therefore, the reactive power is
[tex]Q=(5.15)^2(5)=132.6W[/tex]
c)
In a circuit with a resistor and a capacitor, the apparent power is given by both the resistive and capacitive part of the circuit.
Therefore, it is given by
[tex]S=I^2Z[/tex]
where
I is the current
Z is the impedance of the circuit
Here we have
I = 5.15 A (current)
[tex]Z=13 \Omega[/tex] (impedance)
Therefore, the apparent power is
[tex]S=I^2 Z=(5.15)^2(13)=344.8 W[/tex]
d)
For a circuit with a resistor and a capacitor, the power factor is the ratio between the true power and the apparent power. Mathematically:
[tex]PF=\frac{P}{S}[/tex]
where
P is the true power
S is the apparent power
For this circuit, we have
P = 318.3 W (true power)
S = 344.8 W (apparent power)
So, the power factor is
[tex]PF=\frac{318.3}{344.8}=0.92[/tex]
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How do astronomers set about looking for extrasolar planets?
Explanation:
Implicit techniques for the discovery of extra-solar planets are used by astronomers. Evidence from the radial velocity of a change in the rotation of the star indicates the presence of a planet. If, with reference to Earth, the inclination of the planet's orbit is viewed as an edge-on, the detection of light originating from the star throughout its planetary transit would then be a verification.
A hot air balloon uses the principle of buoyancy to create lift. By making the air inside the balloon less dense then the surrounding air, the balloon is able to lift objects many times its own weight. A large hot air balloon has a maximum balloon volume of 2280 m3 and the required lift is 2500 N (rough estimate of the weight of the equipment and passengers). Calculate the temperature of the air inside the balloon which will produce the required lift. Assume that the outside air temperature is (exactly) 0◦ C and that air is an ideal gas under these conditions. What factor limit the maximum altitude attainable by this method for a given load?
Using the principle of buoyancy and the ideal gas law equation, we can calculate the temperature needed inside a hot air balloon to provide the required lift. The maximum altitude of the balloon is limited by air temperature, pressure, and other environmental factors.
Explanation:The principle of buoyancy used by hot air balloons to create lift is based on Archimedes' Principle, which states that the buoyant or lifting force exerted on a body immersed in a fluid equals the weight of the fluid the body displaces. In order for a hot air balloon to gain lift, the air inside the balloon must be heated to a temperature that makes it less dense than the surrounding cooler air. This difference in air density is what actually provides the uplift.
To calculate the required temperature inside the balloon, we can use the ideal gas law equation P₁V₁ / P₂V₂. Given the values in the question and the atmospheric pressure, we can derive the temperature inside the balloon required to provide the needed lift.
The maximum altitude attainable by a hot air balloon is limited by the ambient air temperature and pressure, which decrease as altitude increases. If the balloon's lift decreases (because it can no longer heat the air inside it sufficiently to provide lift), the balloon will stop ascending. Wind currents and weather conditions can also affect the maximum altitude.
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A friend is writing a science fiction screenplay about an asteroid on a collision course with Earth. She asks you to calculate some numbers so her scenario will be correct. Astronauts will attach a rocket engine to the asteroid in an attempt to divert it. The asteroid is moving at 21 km/s. The rocket will provide an acceleration of 0.035 km/s2 at a right angle to the original motion. The rocket only has enough fuel to provide this acceleration for 40 seconds. Will this change the direction of the asteroid’s motion by at least 22°, enough to miss Earth and save civilization? (15 pts, according to Grading for problem solving, see reverse side. Your group should submit one analysis, with all group member’s names, either on the Problem solving framework or the plain white paper provided.)
Answer:
No, the deviation in the path of asteroid is not by 22°
Explanation:
Given:
velocity of asteroid, [tex]v_a=21\ km.s^{-1}[/tex]
acceleration of the rocket, [tex]a=0.035\ km.s^{-2}[/tex]
time of acceleration, [tex]t=40\ s[/tex]
Now, the final velocity of the asteroid:
using the equation of motion,
[tex]v=u+a.t[/tex]
where:
[tex]v=[/tex] final velocity
[tex]u=[/tex] initial velocity in the direction
[tex]v=0+0.035\times 40[/tex]
[tex]v=1.4\ km.s^{-1}[/tex]
Now direction of the resultant velocity:
[tex]\tan\beta=\frac{v_a}{v}[/tex]
[tex]\tan\beta=\frac{21}{1.4}[/tex]
[tex]\beta=86.186^{\circ}[/tex]
So, the deviation in the asteroid:
[tex]\theta=90-\beta[/tex]
[tex]\theta=90-86.186[/tex]
[tex]\theta=3.814^{\circ}[/tex]
The first dancer in the line is 10 m from the speaker playing the music; the last dancer in the line is 120 m from the speaker.
Approximately how much time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer? Select the best answer from the choices provided.
a) 1/6 seconds
b) 1/4 seconds
c) 1/2 seconds
d)1/3 seconds
e) 1/5 seconds
Answer:
1/3
Explanation:
Take the displacement and divide it by the speed of light. Meters will cancel leaving you with seconds. :)
110m / 343m/s = 0.32069971 seconds or 1/3 seconds
The time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer 1/3 second ( approx.). So, option (d) is correct.
What is velocity?The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.
Given The first dancer in the line is 10 m from the speaker playing the music; the last dancer in the line is 120 m from the speaker.
So, distance between the first dancer in the line and the last dancer in the line be = 120 m -10 m = 110 m
Velocity of sound in air be = 343 m/s.
Hence, The time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer = distance/velocity = 110/343 second = 1/3 second ( approx.).
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An advertisement claims that a particular automobile can "stop on a dime". What net force would actually be necessary to stop an automobile of mass 970 kgkg traveling initially at a speed of 52.0 km/hkm/h in a distance equal to the diameter of a dime, which is 1.8 cmcm ?
Answer:
F=5618278.8 N
Explanation:
Given that
m = 970 kg
Initial speed ,u = 52 km/h
u = 14.44 m/s ( 1 km/h = 0.277 m/s)
Distance d= 1.8 cm = 0.018 m
The final speed ,v = 0 m/s
We know that
v² = u² + 2 a d
a=Acceleration
0² = 14.44² + 2 x a x 0.018
[tex]a=-\dfrac{14.44^2}{2\times 0.018}\ m/s^2[/tex]
a=5792.04 m/s²
We know that
Force = Mass x acceleration
F= m a
F= - 5792.04 x 970 N
F= - 5618278.8 N
Therefore the magnitude of the force will be 5618278.8 N.
F=5618278.8 N
10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when it crosses over onto a floor that has a coefficient of kinetic friction = 0.35. How far will the puck travel across the floor before it stops?
Answer:
[tex] d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m [/tex]
Explanation:
For this case we can use the second law of Newton given by:
[tex] \sum F = ma[/tex]
The friction force on this case is defined as :
[tex] F_f = \mu_k N = \mu_k mg [/tex]
Where N represent the normal force, [tex]\mu_k [/tex] the kinetic friction coeffient and a the acceleration.
For this case we can assume that the only force is the friction force and we have:
[tex] F_f = ma[/tex]
Replacing the friction force we got:
[tex] \mu_k mg = ma[/tex]
We can cancel the mass and we have:
[tex] a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}[/tex]
And now we can use the following kinematic formula in order to find the distance travelled:
[tex] v^2_f = v^2_i - 2ad[/tex]
Assuming the final velocity is 0 we can find the distance like this:
[tex] d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m [/tex]
The hockey puck will travel approximately 58.5 meters across the floor before it stops.
Explanation:To solve this problem, we will use the concepts of kinetic energy and the work-energy theorem.
Initially, the hockey puck has a kinetic energy of KE1 = 0.5*m*v^2 = 0.5*0.3 kg*(20 m/s)^2 =60 J
As it travels across the floor, the friction does work on it until it stops. The work done by the friction force is W = μ*m*g*d, where: μ is the coefficient of kinetic friction (0.35), m is the mass (0.3 kg), g is the gravitational constant (9.8 m/s^2), and d is the distance traveled.
The work done by the friction is equal to the initial kinetic energy (work-energy theorem), so we can solve for d: 60 J = 0.35 * 0.3 kg * 9.8 m/s^2 * d. Simplifying this equation gives d = 60 J / (0.35 * 0.3 kg * 9.8 m/s^2) = 58.5 m
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A mass m = 3.4 kg is at the end of a horizontal spring of spring constant k = 105 N/m on a frictionless horizontal surface. The block is pulled, stretching the spring a distance A = 6.5 cm from equilibrium, and released from rest.
(a) Write an equation for the angular frequency of the oscillation
(b) Calculate the angular frequency o of the oscillation in rad/seconds
Answer:
[tex]w=\sqrt{\frac{k}{m} }[/tex]
b. [tex]5.6rad/s[/tex]
Explanation:
a. from the spring-mass system which is explicitly describe by hooks law
from
F=-kx
which is in comparison to newtons general law of motion
F=ma
where the displacement x is expressed as
[tex]x=Asin(wt)\\[/tex]
and the acceleration is the second derivative of the displacement
[tex]a=-Aw^{2}sin(wt)\\[/tex]
hence final expression after substituting for the acceleration and the displacement is expressed as
[tex]w=\sqrt{\frac{k}{m} }[/tex]
b. for k=105N/m and m=3.4kg
we have the angular frequency to be
[tex]w=\sqrt{\frac{105}{3.4}}\\\\w=5.6rad/s[/tex]
Final answer:
The angular frequency (ω) can be found using the formula ω = sqrt(k/m). By plugging in the given values of mass (3.4 kg) and spring constant (105 N/m), the angular frequency is calculated to be approximately 5.56 rad/s.
Explanation:
The subject question pertains to simple harmonic motion (SHM) and specifically relates to the oscillatory motion of a mass attached to a horizontal spring. The question involves determining the angular frequency and calculating it in radians per second.
Part (a)
To write an equation for the angular frequency (ω) of the oscillation, we use the equation:
ω = sqrt(k/m)
where ω is the angular frequency, k is the spring constant, and m is the mass.
Part (b)
Plugging in the given values:
m = 3.4 kg
k = 105 N/m
We find ω using the formula:
ω = sqrt(105 N/m / 3.4 kg) = sqrt(30.8824 s-2) ≈ 5.56 rad/s
Therefore, the angular frequency of the oscillation is approximately 5.56 rad/seconds.
A 2 kg block rests on a 34o incline. If the coefficient of static friction is 0.2, how much additional force, F, must be applied to keep the block from sliding down the incline?
Answer:
Explanation:
Reaction force of inclined surface = mg cosθ
= Friction force acting in upward direction = μ x mg cosθ
If F be force required in upward direction to keep the block at rest on the plane
F + μ x mg cosθ = mg sinθ
F = mg sinθ - μ x mg cosθ
F = mg( sinθ - μ cosθ)
= 2 x 9.8 ( sin34 - 0.2 cos34 )
= 19.6 ( .559 - .1658)
= 7.7 N
This is the minimum force required .
The additional force that needs to be applied to keep a 2 kg block from sliding down a 34 degree incline, given a coefficient of static friction of 0.2, is 7.89 N.
Explanation:To find out how much additional force, F, needs to be applied to keep the 2 kg block from sliding down the 34° incline, we first need to calculate the gravitational force pulling the block down the incline. This force is given by F_gravity = m * g * sin(theta), where m = 2 kg, g = 9.8 m/s² (roughly the acceleration due to gravity on Earth), and theta = 34 degrees. Thus, F_gravity = 2 kg * 9.8 m/s² * sin(34°) = 11.10 N.
Next, we calculate the static frictional force that is preventing the block from sliding down the incline. This force is given by F_friction = m * g * cos(theta) * mu_s, where mu_s = 0.2 is the coefficient of static friction. Thus, F_friction = 2 kg * 9.8 m/s² * cos(34°) * 0.2 = 3.21 N.
The additional force, F, that must be applied to keep the block from sliding down the incline is the difference between the gravitational force and the frictional force. That's F = F_gravity - F_friction = 11.10 N - 3.21 N = 7.89 N.
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Consider a concentric tube heat exchanger. Assuming there is no fin attached to any of the surfaces and considering negligible fouling, what would be the appropriate equation for calculating the overall heat transfer coefficient?
Answer:
[tex]Q = T1 - T2 (KA)/L[/tex]
Explanation:
Where Q is the co-efficient of heat transfer.
T-1 is initial temperature of exchanger
T-2 is final temperature of sink where has to to be dissipated
k is the co-efficient of thermal conductivity
A is the total area of exchanger surface...
and L is the total length of exchanger...
An amount of work W is done on an object of mass m initially at rest, and as result it winds up moving at speed v. Suppose instead it were already moving at speed v and the same amount of work W was done on it. What would be its final speed
Answer:
Explanation:
Given
W amount of work is done on the system such that it acquires v velocity after operation(initial velocity)
According to work energy theorem work done by all the forces is equal to change in kinetic energy of object
[tex]W=\frac{1}{2}mv^2---1[/tex]
where m=mass of object
v=velocity of object
When the object is already have velocity v then the final speed is given by work energy theorem
[tex]W=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2-----2[/tex]
From 1 and 2 we get
[tex]\frac{1}{2}mv^2=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2[/tex]
[tex]2\times \frac{1}{2}mv^2=\frac{1}{2}mv_f^2[/tex]
[tex]v_f^2=2v^2[/tex]
[tex]v_f=\sqrt{2}v[/tex]
A solid sphere and a hollow sphere have the same mass and radius. The two spheres are spun with matching angular velocities. Which statement is true? I. The hollow has the greater angular momentum II. The solid sphere has the greater angular momentum. III. The angular momentum is the same for both spheres. IV. The moment of inertia is the same for both spheres.
Answer:
I. The hollow sphere has the greater angular momentum.
Explanation:
Given that the mass and radius of hollow sphere and solid sphere are same. Let the mass and radius of two spheres be m and r respectively. The two spheres are rotating having same angular velocity ω .
Moment of inertia of solid sphere, I₁ = [tex]\frac{2}{5}\times{m}r^{2}[/tex]
Moment of inertia of hollow sphere, I₂ = [tex]\frac{2}{3}\times{m}r^{2}[/tex]
Since, I₁ and I₂ are not equal. Therefore, the statement iv is wrong.
The relation between angular momentum, moment of inertia and angular velocity is :
L = Iω
Let L₁ and L₂ be the angular momentum of solid sphere and hollow sphere respectively.
L₁ = I₁ω and L₂ = I₂ω
As ω is same for both spheres but I₂ is greater than I₁, hence L₂ is greater than L₁.
Therefore, statement I is correct that the hollow sphere has the greater angular momentum.
The angular momentum of the hollow sphere is greater than that of solid sphere.
The moment of inertia of solid sphere is given as follows;
[tex]I_{ss} = \frac{2}{5} mr^2 = 0.4mr^2[/tex]
The moment of inertia of hollow sphere is given as follows;
[tex]I_{hs} = \frac{2}{3} mr^2 = 0.67 mr^2[/tex]
The angular momentum of each sphere is calculated as follows;
[tex]L = I\omega \\\\L_{ss} = 0.4mr^2 \omega \\\\L_{hs} = 0.67 mr^2 \omega[/tex]
Thus, we can conclude that the angular momentum of the hollow sphere is greater than that of solid sphere.
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