Answer:
619.2 Hz and 582.5Hz
Explanation:
Given that
Frequency of the generator, F(s) = 600 Hz
Length if the rope, r = 1 m
Speed of whirling, w = 100 rpm
For starters, we find the speed of the generator, v(g)
v(g) = r.w
It should be noted, however, that we are to convert the speed of whirling from rpm to rad/s, thus
100 rpm = 100 * 2π/60 rad/s
100 rpm = 200π/60
100 rpm = 620.4 / 60
100 rpm = 10.47 rad/s
Now, we use this to find the speed of the generator
v(g) = r.w
v(g) = 1 * 10.47
v(g) = 10.47 m/s
When approaching the generator, the frequency is calculated as
F(a) = F(s) / [1 - (v(g)/v)]
F(a) = 600 / [1 - (10.47/343)]
F(a) = 600 / 0.969
F(a) = 619.2 Hz
On the other hand, the receding generator frequency is
F(r) = F(s) / [1 + (v(g)/v)]
F(r) = 600 / [1 + (10.47/343)]
F(r) = 600 / 1.030
F(r) = 582.5 Hz
Therefore, we can conclude that the highest and lowest frequency is 619.2 Hz and 582.5 Hz respectively
An electron starts from rest in a vacuum, in a region of strong electric field. The electron moves through a potential difference of 36 volts. What is the kinetic energy of the electron in electron volts (eV)
Answer:
The kinetic energy is [tex]KE = 5.67*10^{-18} \ J[/tex]
Explanation:
From the question we are told that
The potential difference is [tex]\Delta V = 36 \ volts[/tex]
The potential energy of the end is mathematically represented as
[tex]PEs = - q * \Delta V[/tex]
q is the charge on an electron with a constant value of [tex]q = 1.60 *10^{-19} \ C[/tex]
substituting values
[tex]PE = - 1.60*10^{-19} * 36[/tex]
[tex]PE = - 5.67*10^{- 18} \ J[/tex]
Now from the law of energy conservation
The [tex]PE_e = KEe[/tex]
Where [tex]KE _e[/tex] is the potential energy at the end
So
[tex]KE = 5.67*10^{-18} \ J[/tex]
The negative sign is not includes because kinetic energy can not be negative
Classify the bending of light as exhibited by the ray diagrams. According to your data, is light refracted away from or toward the normal as it passes at an angle into a medium with a higher index of refraction?
Answer:
the ray of light should approach normal
Explanation:
When light passes through two means of different refractive index, it fulfills the equation
n₁ sin θ₁ = n₂ sin θ₂
where index 1 and 2 refer to each medium
In this problem, they tell us that light passes to a medium with a higher index, which is why
n₁ <n₂
let's look for the angle in the second half
sinθ₂ = n₁ /n₂ sin θ₁
θ₂ = sin⁻¹ (n₁ /n₂ sin θ₁)
let's examine the angle argument the quantity n₁ /n₂ <1 therefore the argument decreases, therefore the sine and the angle decreases
Consequently the ray of light should approach normal
A fluid is forced through a pipe of changing cross section as shown. In which section would the pressure of the fluid be a minimum?
Answer:
straight
Explanation:
because it is good in that direction
An ultrasound machine uses 1.64 × 105 watts of power. If it draws 12.0 amps of current, what is the resistance?
Answer:
R = 1138.9 Ω
Explanation:
Hello,
In this case, for the given power (P) and current (I), we can compute the resistance (R) via:
R = P / I²
Thus, we obtain:
R = 1.64x10⁵ W / (12.0 A)²
R = 1138.9 Ω
Best regards.
An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the speed remained the same, by what factor would its centripetal acceleration change?
Answer:
The centripetal acceleration changed by a factor of 0.5
Explanation:
Given;
first radius of the horizontal circle, r₁ = 500 m
speed of the airplane, v = 150 m/s
second radius of the airplane, r₂ = 1000 m
Centripetal acceleration is given as;
[tex]a = \frac{v^2}{r}[/tex]
At constant speed, we will have;
[tex]v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1[/tex]
a₂ = 0.5a₁
Therefore, the centripetal acceleration changed by a factor of 0.5
Calculate the current through a 15.0-m long 20 gauge (having radius 0.405 mm) nichrome wire if it is connected to a 12.0-V battery. The resistivity of nichrome is 100 × 10-8 Ω ∙ m.
Given Information:
Radius of wire = r = 0.405 mm = 0.405×10⁻³ m
Length of wire = L = 15 m
Voltage = V = 12 V
Resistivity = ρ = 100×10⁻⁸ Ωm
Required Information:
Current = I = ?
Answer:
Current = I = 0.412 A
Explanation:
The current flowing through the wire can be found using Ohm's law that is
V = IR
I = V/R
Where V is the voltage across the wire and R is the resistance of the wire.
The resistance of the wire is given by
R = ρL/A
Where ρ is the resistivity of the wire, L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(0.405×10⁻³)²
A = 0.515×10⁻⁶ m²
So the resistivity of the wire is
R = ρL/A
R = (100×10⁻⁸×15)/0.515×10⁻⁶
R = 29.126 Ω
Finally, the current flowing through the wire is
I = V/R
I = 12/29.126
I = 0.412 A
Therefore, the current through a 15.0-m long 20 gauge nichrome wire is 0.412 A.
⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a) If the coefficient of static friction is 0.5, what minimum force magnitude is required from the rope to start the crate moving? (b) If µk= 0.35, what is the magnitude of the initial acceleration of the crate?
Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2
a person lifts 4.5kg block up a vertical distance of 1.2meters and then carries it horizontally for 7.3meters. Calculate the Total Work done by the person and the block.
Answer:
The total work done by the person is given as = m g h
= 4.5kg x 9.8m/s²x1.2m
= 52.92J
This is the work done in moving the block in a vertical distance
However there is no work done when the block is moved in a horizontal direction since ko work is done against gravity.
Explanation:
A uniform solid disk and a uniform ring are place side by side at the top of a rough incline of height h.
a) If they are released from rest and roll without slipping, determine the velocity vring of the ring when it reaches the bottom.
b) Verify your answer by calculating their speeds when they reach the bottom in terms of h.
Explanation:
velocity of disc [tex]=\sqrt((gh)/0.75)[/tex]
lets call (h) 1 m to make it simple.
= 3.614 m/s
[tex]\sqrt((4/3) x 1 x 9.8) = 3.614[/tex] m/s pointing towards this:
[tex]4×V_d=\sqrt(4/3hg)[/tex]
[tex]V_h=\sqrt(hg)[/tex]
velocity of hoop=[tex]\sqrt(gh)[/tex]
lets call (h) 1m to make it simple again.
[tex]\sqrt(9.8 x 1) = 3.13[/tex] m/s
[tex]\sqrt(gh) = sqrt(hg)
so [tex]4×V_d= \sqrt(4/3hg)V_h=\sqrt(hg)[/tex]
The disc is the fastest.
While i'm on this subject i'll show you this:
Solid ball [tex]=0.7v^2= gh[/tex]
solid disc [tex]= 0.75v^2 = gh[/tex]
hoop [tex]=v^2=gh[/tex]
The above is simplified from linear KE + rotational KE, the radius or mass makes no difference to the above formula.
The solid ball will be the faster of the 3, like above i'll show you.
solid ball: velocity [tex]=\sqrt((gh)/0.7)[/tex]
let (h) be 1m again to compare.
[tex]\sqrt((9.8 x 1)/0.7) = 3.741[/tex] m/s
solid disk speed [tex]=\sqrt((gh)/0.75)[/tex]
uniform hoop speed [tex]=\sqrt(gh)[/tex]
solid sphere speed [tex]=\sqrt((gh)/0.7)[/tex]
Which statements must be true about the surface of a charged conductor in which no charge is moving?
1. The electric potential is constant over the surface.
2. The electric field is constant at the surface.
3. The electric field is zero at the surface.
4. The electric field is perpendicular to the surface.
5. The electric potential of the surface is zero.
Answer:
true 1 and 4
Explanation:
En un conductor que la carga no se mueve.
1. True. Debe ser una superficie equipotencial, para no haya fuerzas sobre la carga
2. False. Si existe un campo eléctrico, debe existir una fuerza, por lo tanto con pueden estar quitas las cargas
3. False. No puede ser cero ya que existe una carga finita
4. True. Con la ley de Gauss podemos considerar la carga concentrador un punto en el centro de la orbita; por lo tanto existe un campo eléctrico perpendicular a la superficie
5. False si el potencial es ero no hay carga
What happens when you increase the number of slits per millimeter (decrease the spacing between slits)?
Answer:
Increasing the number of slits not only makes the diffraction maximum sharper, but also much more intense. If a 1 mm diameter laser beam strikes a 600 line/mm grating, then it covers 600 slits and the resulting line intensity is 90,000 x that of a double slit. Such a multiple-slit is called a diffraction grating.
key points that can be found in the realist philosophical position
Answer:
Key points that can be found in the realist philosophical position are as follows:
The view that we observe or identify is real, truly out there.The objects which are identified are independent of someone's perceptions, linguistic practices, conceptual scheme, and beliefs.Quantum mechanics is an example of philosophical realism that claims world is mind-independent.How much electromagnetic energy is contained in each cubic meter near the Earth's surface if the intensity of sunlight under clear skies is 1000 W/m2
Answer:
344.8 x10^-8J/m³
Explanation:
Using=> energy intensity/ speed oflight
= 1000/2.9x10^8
= 344.8 x10^-8J/m³
The electromagnetic energy is 344.8 x10⁻⁸J/m³
We have to use the formula which says
Electromagnetic energy = energy intensity/ speed of light
We are given intensity as 1000 W/m²
Electromagnetic energy = 1000/2.9 x 10⁸
= 344.8 x 10⁻⁸J/m³
Therefore the electromagnetic energy is contained in each cubic meter will be 344.8 x 10⁻⁸J/m³
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6a. A special lamp can produce UV radiation. Which two statements
describe the electromagnetic waves emitted by a UV lamp? *
They have a higher frequency than X-rays.
They have the same wave speed as visible light
They have a longer wavelength than microwaves.
They have a lower frequency than gamma rays.
They have a greater wave speed than radio waves.
Answer:
The correct options are:
B) They have the same wave speed as visible light
D) They have a lower frequency than gamma rays.
Explanation:
B) Ultraviolet rays, commonly known as UV rays, are a type of electromagnetic ways. As electromagnetic waves, in the layman's term, are all kinds of life that can be identified, all electromagnetic waves (UV rays, visible light, infrared, radio etc) all travel with the same velocity, that is the speed of light, given as v = 3 × 10⁸ m/s
D) The frequency of all electromagnetic rays can be found by electromagnetic spectrum (picture attached below).
We can clearly see in the picture that the frequencies of UV rays lie at about 10¹⁵ - 10¹⁶ Hz which is lower than the frequency of Gamma ray, which lie at about 10²⁰ Hz.
At what minimum speed must a roller coaster be traveling when upside down at the top of a 7.4 m radius loop-the-loop circle so the passengers will not fall out?
Answer:
v = 8.5 m/s
Explanation:
In order for the passengers not to fall out of the loop circle, the centripetal force must be equal to the weight of the passenger. Therefore,
Weight = Centripetal Force
but,
Weight = mg
Centripetal Force = mv²/r
Therefore,
mg = mv²/r
g = v²/r
v² = gr
v = √gr
where,
v = minimum speed required = ?
g = 9.8 m/s²
r = radius = 7.4 m
Therefore,
v = √(9.8 m/s²)(7.4 m)
v = 8.5 m/s
Minimum speed for a roller coaster while travelling upside down so that the person will not fall out = 8.5 m/s
For a roller coaster be traveling when upside down the Force balance equation can be written for a person of mass m.
In the given condition the weight of the person must be balanced by the centrifugal force.
and for the person not to fall out centrifugal force must be greater than or equal to the weight of the person
According to the Newton's Second Law of motion we can write force balance
[tex]\rm mv^2/r -mg =0 \\\\mg = mv^2 /r (Same\; mass) \\\\\\g = v^2/r\\\\v = \sqrt {gr}......(1)[/tex]
Given Radius of loop = r = 7.4 m
Putting the value of r = 7.4 m in equation (1) we get
[tex]\sqrt{9.8\times 7.4 } = \sqrt{72.594} = 8.5\; m/s[/tex]
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Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P P and volume V V satisfy the equation P V = C PV=C, where C C is a constant. Suppose that at a certain instant the volume is 1000 cm 3 1000 cm3, the pressure is 80 kPa 80 kPa, and the pressure is increasing at a rate of 40 kPa/min 40 kPa/min. At what rate is the volume decreasing at this instant?
Answer:
the volume decreases at the rate of 500cm³ in 1 min
Explanation:
given
v = 1000cm³, p = 80kPa, Δp/t= 40kPa/min
PV=C
vΔp + pΔv = 0
differentiate with respect to time
v(Δp/t) + p(Δv/t) = 0
(1000cm³)(40kPa/min) + 80kPa(Δv/t) = 0
40000 + 80kPa(Δv/t) = 0
Δv/t = -40000/80
= -500cm³/min
the volume decreases at the rate of 500cm³ in 1 min
Given that the voltage from a wall outlet is AC if you were to correctly use your digital multimeter to measure the wall outlet voltage to be 120 V, what is the maximum voltage level provided by the wall outlet
Answer:
V_{max} = 169.7 V
Explanation:
In an alternating voltage system, the value measured by the measurement equipment is the effective voltage, that is, the average voltage over a period
[tex]V_{rms}[/tex] = [tex]V_{max}[/tex] / √2
in our voltmeter measure a V_{rms} = 120 V
V_{max} = √2 V_{rms}
V_{max} = 120 √ 2
V_{max} = 169.7 V
this is the peak or maximum voltage of the wave
A small merry-go round is spinning about its center in a clockwise direction at 1.6 r a d s and with rotational kinetic energy of 580 J. Calculate the moment of inertia of the merry-go round.
Answer:
The moment of inertia of the merry-go round is 453.125 kg.m²
Explanation:
Given;'
angular velocity of the merry-go round, ω = 1.6 rad/s
rotational kinetic energy, K = 580 J
Rotational kinetic energy is given as;
K = ¹/₂Iω²
Where;
I is the moment of inertia of the merry-go round
[tex]I = \frac{2K}{\omega^2} \\\\I = \frac{2*580}{1.6^2} \\\\I = 453.125 \ kg.m^2[/tex]
Therefore, the moment of inertia of the merry-go round is 453.125 kg.m²
Since the small merry-go round is spinning about its center in a clockwise direction, its moment of inertia is equal to 453.13 [tex]Kgm^2[/tex]
Given the following data:
Angular velocity = 1.6 rad/sRotational kinetic energy = 580 Joules.To calculate the moment of inertia of the small merry-go round:
Mathematically, the rotational kinetic energy of an object is giving by the formula:
[tex]E_{rotational} = \frac{1}{2} Iw^2[/tex]
Where:
I is the moment of inertia.w is the angular speed.Making moment of inertia (I) the subject of formula, we have:
[tex]I = \frac{2E_{rotational}}{w^2}[/tex]
Substituting the given parameters into the formula, we have;
[tex]I = \frac{2(580)}{1.6^2}\\\\I = \frac{1160}{2.56}[/tex]
Moment of inertia (I) = 453.13 [tex]Kgm^2[/tex]
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A cylinder of mass 8.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of 13.0 m/s.
(a) Determine the translational kinetic energy of its center of mass.
J
(b) Determine the rotational kinetic energy about its center of mass.
J
(c) Determine its total energy.
J
Answer:
b
Explanation:
determine the rotational kinetic energy about it's center of mass
A slender rod of length L has a varying mass-per-unit-length from the left end (x=0) according to dm/dx=Cx where C has units kg/m2. Find the total mass in terms of C and L, and then calculate the moment of inertia of the rod for an axis at the left end note: you need the total mass in order to get the answer in terms of ML^2
Answer:
ML²/6
Explanation:
Pls see attached file
The total mass is M = CL²/2, and the moment of inertia is I = ML²/2,
Moment of inertia:The length of the rod is L. It has a non-uniform distribution of mass given by:
dm/dx = Cx
where C has units kg/m²
dm = Cxdx
the total mass M of the rod can be calculated by integrating the above relation over the length:
[tex]M =\int\limits^L_0 {} \, dm\\\\M=\int\limits^L_0 {Cx} \, dx\\\\M=C[x^2/2]^L_0\\\\M=C[L^2/2]\\\\[/tex]
Thus,
C = 2M/L²
Now, the moment of inertia of the small element dx of the rod is given by:
dI = dm.x²
dI = Cx.x²dx
[tex]dI = \frac{2M}{L^2}x^3dx\\\\I= \frac{2M}{L^2}\int\limits^L_0 {x^3} \, dx \\\\I= \frac{2M}{L^2}[\frac{L^4}{4}][/tex]
I = ML²/2
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According to string theory, six space-time dimensions cannot be measured except as quantum numbers of internal particle properties because they are curled up in size of the order of
Answer:
Order of 10^(-35) m.
Explanation:
The string theory is a theoretical concept whereby the very small particles of particle physics are replaced by one dimensional objects which are called strings. This theory is also applicable to black hole physics, nuclear physics, cosmology, etc.
Now, according to string theory, six space-time dimensions cannot be measured except as quantum numbers of internal particle properties because they are curled up in size of the order of 10^(-35) m.
This is because the length of the scale is assumed to be on the order of the Planck length, or 10^(−35) meters which is the scale at which the effects of quantum gravity are usually believed to become very significant.
A certain superconducting magnet in the form of a solenoid of length 0.550 m can generate a magnetic field of 9.20 T in its core when its coils carry a current of 140 A. Find the number of turns in the solenoid.
Answer:
28,750 turns
Explanation:
According to the given situation the solution of number of turns in the solenoid is shown below:-
Number of turns in the solenoid is
[tex]B = \frac{\mu_0NI}{I}[/tex]
[tex]N = \frac{BI}{\mu_oI}[/tex]
[tex]N = \frac{(9.20)(0.550)}{4\pi\times10{-7}(140)}[/tex]
[tex]N = \frac{5.06}{0.000176}[/tex]
= 28,750 turns
Therefore for computing the number of turns in the solenoid we simply applied the above formula.
So, the right answer of number of turns in the solenoid is 28,750 turns.
A total electric charge of 2.00 nC is distributed uniformly over the surface of a metal sphere with a radius of 26.0 cm . The potential is zero at a point at infinity.
a) Find the value of the potential at 45.0 cm from the center of the sphere.
b) Find the value of the potential at 26.0 cm from the center of the sphere.
c) Find the value of the potential at 16.0 cm from the center of the sphere.
Answer:
a) 40 V
b) 69.23 V
c) 69.23 V
Explanation:
See attachment for solution
The sound intensity level of a horn is 80dB. What is the sound intensity? The threshold of sound is 1x10-12W/m2.
Answer:
1.0x10^-4 W/m^2 as sound intensity
Explanation:
Using
dB= 10log( I/Io)
Where Io= 10^-12W/m²
So dB=80dB
80= 10log(I/10^-12)
So
80/10= log (I/10^-12)
8= log (I/10^-12)
Taking the definition of log
10^8 = I/10^-12
I= 10^8 x10^ -12
I= 10^-4W/m² as sound intensity
Answer:
The sound intensity is 1 x 10⁻⁴ W/m²
Explanation:
Given;
sound intensity level, β = 80 dB
The threshold of sound or threshold intensity of hearing, I₀ = 1 x 10⁻¹² W/m²
Sound intensity level is given as;
[tex]\beta = 10 Log(\frac{I}{I_0} )[/tex]
where;
β is the intensity level (dB)
I₀ is threshold intensity of hearing (W/m²)
I is the sound intensity (W/m²)
[tex]\beta = 10Log(\frac{I}{I_0)} \\\\\frac{\beta}{10} = Log(\frac{I}{I_0})\\\\\frac{80}{10} = Log(\frac{I}{I_0})\\\\8 = Log(\frac{I}{I_0})\\\\10^{8} = \frac{I}{I_0}\\\\I = 10^{8} * I_0\\\\I = 10^{8} * 10^{-12} \ (W/m^2)\\\\I = 1*10^{-4} \ (W/m^2)[/tex]
Therefore, the sound intensity is 1 x 10⁻⁴ W/m²
an object of 30kg is in free fall in a vacuum where there is no air resistance. Determine the acceleration of the object.
We're missing one essential piece of information that we need in order to answer this question. You have not specified what planet the object is falling on. The answer depends on the gravitational acceleration on that planet, and they're all different.
Without that information, we'll just go ahead and assume that the object is falling to the surface of the Earth. Wherever on Earth this tense drama is unfolding, the acceleration of gravity is going to be around 9.8 m/s² everywhere.
So THAT's the object's acceleration if there is no air resistance. The object's MASS makes no difference. It doesn't matter whether the object is a sparrow feather or a school bus. Heavier objects DO NOT fall faster than light objects.
If there is no air resistance, then ALL objects fall with the same acceleration. It's called the "acceleration of gravity" on that planet or moon, and you can easily look it up. It's 9.8 m/s² on Earth, 1.62 m/s² on the Moon, 3.71 m/s² on Mars, 8.87 m/s² on Venus, and 24.8 m/s² on Jupiter.
A spherical shell rolls without sliding along the floor. The ratio of its rotational kinetic energy (about an axis through its center of mass) to its translational kinetic energy is:
Answer:
The ratio is [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]
Explanation:
Generally the Moment of inertia of a spherical object (shell) is mathematically represented as
[tex]I = \frac{2}{3} * m r^2[/tex]
Where m is the mass of the spherical object
and r is the radius
Now the the rotational kinetic energy can be mathematically represented as
[tex]RE = \frac{1}{2}* I * w^2[/tex]
Where [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{v}{r}[/tex]
=> [tex]w^2 = [\frac{v}{r}] ^2[/tex]
So
[tex]RE = \frac{1}{2}* [\frac{2}{3} *mr^2] * [\frac{v}{r} ]^2[/tex]
[tex]RE = \frac{1}{3} * mv^2[/tex]
Generally the transnational kinetic energy of this motion is mathematically represented as
[tex]TE = \frac{1}{2} mv^2[/tex]
So
[tex]\frac{RE}{TE} = \frac{\frac{1}{3} * mv^2}{\frac{1}{2} * m*v^2}[/tex]
[tex]\frac{RE}{TE} = \frac{2}{3}[/tex]
An ideal gas in a cubical box having sides of length L exerts a pressure p on the walls of the box. If all of this gas is put into a box having sides of length 0.5L without changing its temperature, the pressure it exerts on the walls of the larger box will be...
p.
2p.
4p.
8p.
12p.
Answer:
2P
Explanation:
See attached file
A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ball is also thrown at the same speed horizontally at the wall, but sticks to it upon hitting. [THE CLAY BALL / THE RUBBER BALL /NEITHER] (circle one) exerts a greater magnitude of impulse on the wall. Briefly explain (either in words or calculations)
Answer:
THE RUBBER BALL
Explanation:
From the question we are told that
The mass of the rubber ball is [tex]m_r = 2 \ kg[/tex]
The initial speed of the rubber ball is [tex]u = 3 \ m/s[/tex]
The final speed at which it bounces bank [tex]v - 3 \ m/s[/tex]
The mass of the clay ball is [tex]m_c = 2 \ kg[/tex]
The initial speed of the clay ball is [tex]u = 3 \ m/s[/tex]
The final speed of the clay ball is [tex]v = 0 \ m/s[/tex]
Generally Impulse is mathematically represented as
[tex]I = \Delta p[/tex]
where [tex]\Delta p[/tex] is the change in the linear momentum so
[tex]I = m(v-u)[/tex]
For the rubber is
[tex]I_r = 2(-3 -3)[/tex]
[tex]I_r = -12\ kg \cdot m/s[/tex]
=> [tex]|I_r| = 12\ kg \cdot m/s[/tex]
For the clay ball
[tex]I_c = 2(0-3)[/tex]
[tex]I_c = -6 \ kg\cdot \ m/s[/tex]
=> [tex]| I_c| = 6 \ kg\cdot \ m/s[/tex]
So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression, x = 8.00 cos 5t + π 8 where x is in centimeters and t is in seconds. (a) At t = 0, find the position of the piston. cm (b) At t = 0, find velocity of the piston. cm/s (c) At t = 0, find acceleration of the piston. cm/s2 (d) Find the period and amplitude of the motion. period s amplitude cm
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression, x = 8.00 cos (5t + π / 8) where x is in centimeters and t is in seconds. (a) At t = 0, find the position of the piston. cm (b) At t = 0, find velocity of the piston. cm/s (c) At t = 0, find acceleration of the piston. cm/s2 (d) Find the period and amplitude of the motion. period s amplitude cm
Answer:(a) 7.392cm
(b) -15.32 cm/s
(c) -184cm/s²
(d) 0.4πs and 8.00cm
Explanation:The general equation of a simple harmonic motion (SHM) is given by;
x(t) = A cos (wt + Φ) --------------(i)
Where;
x(t) = position of the body at a given time t
A = amplitude or maximum displacement during oscillation
w = angular velocity
t = time
Φ = phase constant.
Given from question:
x(t) = 8.00 cos (5t + π / 8) ---------------(ii)
(a) At time t = 0;
The position, x(t), of the body (piston) is given by substituting the value of t = 0 into equation (ii) as follows;
x(0) = 8.00 cos (5(0) + π / 8)
x(0) = 8.00 cos (π /8)
x(0) = 8.00 x 0.924
x(0) = 7.392 cm
Therefore, the position of the piston at time t = 0 is 7.392cm
(b) To get the velocity, v(t), of the piston at t = 0, first differentiate equation (ii) with respect to t as follows;
v(t) = [tex]\frac{dx(t)}{dt}[/tex]
v(t) = [tex]\frac{d(8.00cos(5t + \pi / 8 ))}{dt}[/tex]
v(t) = 8 (-5 sin (5t + π / 8))
v(t) = -40sin(5t + π / 8) --------------------(iii)
Now, substitute t=0 into the equation as follows;
v(0) = -40 sin(5(0) + π / 8)
v(0) = -40 sin(π / 8)
v(0) = -40 x 0.383
v(0) = -15.32 cm/s
Therefore, the velocity of the piston at time t = 0 is -15.32 cm/s
(c) To find the acceleration a(t) of the piston at t = 0, first differentiate equation (iii), which is the velocity equation, with respect to t as follows;
a(t) = [tex]\frac{dv(t)}{dt}[/tex]
a(t) = [tex]\frac{d(-40sin (5t + \pi /8))}{dt}[/tex]
a(t) = -200 cos (5t + π / 8)
Now, substitute t = 0 into the equation as follows;
a(0) = -200 cos (5(0) + π / 8)
a(0) = -200 cos (π / 8)
a(0) = -200 x 0.924
a(0) = -184.8 cm/s²
Therefore, the acceleration of the piston at time t = 0 is -184cm/s²
(d) To find the period, T, first, let's compare equations (i) and (ii) as follows;
x(t) = A cos (wt + Φ) --------------(i)
x(t) = 8.00 cos (5t + π / 8) ---------------(ii)
From these equations it can be deduced that;
Amplitude, A = 8.00cm
Angular velocity, w = 5 rads/s
But;
w = [tex]\frac{2\pi }{T}[/tex] [Where T = period of oscillation]
=> T = [tex]\frac{2\pi }{w}[/tex]
=> T = [tex]\frac{2\pi }{5}[/tex]
=> T = 0.4π s
Therefore, the period and amplitude of the piston's motion are respectively 0.4πs and 8.00cm
A tennis ball is thrown with a downward velocity of 2.00m/s off a cliff. It takes 6.10 seconds for it to hit the ground. How tall is the cliff?
Answer:
6.09m tall
Explanation:
use this;
U²=2as