Answers
When a ball of mass m makes a head-on elastic collision with a second ball (at rest) and rebounds in the opposite direction with a speed equal to one-fourth its original speed, then mass of the second ball having v/3 is velocity after collision is 9m/4.
What is momentum ?Momentum is defined as mass times velocity of body. it is denoted by p and its SI unit is Kg.m/s. It has both magnitude and direction. it is a vector quantity. it tells about the moment of the body. it is denoted by p and expressed in kg.m/s. mathematically it is written as p = mv. A body having zero velocity or zero mass has zero momentum. its dimensions is [M¹ L¹ T⁻¹]. Momentum is conserved throughout the motion.
initial momentum = final momentum
Given,
mass of first body m₁ = m
initial velocity of first body = v₁' = v
final velocity of first body = v₁'' =v/4
mass of second body m₂ = ?
initial velocity of second body = v₂' = 0
final velocity of second body = v₂'' = v/3
According to conservation of momentum,
initial momentum = final momentum
m₁v₁' + m₂v₂' = m₁v₁'' + m₂v₂''
putting al above values
m₁v + 0 = m₁v/4 + m₂v/3
m₁v - m₁v/4 = m₂v/3
m (1 - 1/4)v = m₂v/3
3m/4 = m₂/3
m₂ = 9m/4
Hence mass of the second body is 9m/4.
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Related Questions
state four law of photoelectric effect
Answers
Answer:
LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.
---------------------------------------------
LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place. This frequency is called threshold frequency.
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LAW 3: For a frequency greater than the threshold frequency, the kinetic energy of photoelectrons is dependent upon frequency or wavelength but not on the intensity of light.
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LAW 4: Photoelectric emission is an instantaneous process. The time lag between incidence of radiations and emission of electron is 10^-9 seconds.
Explanation:
Answer:
LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation. ... LAW 4: Photoelectric emission is an instantaneous process.
All charged objects exert a force that can cause other charges to move. What is the force that
charged objects give off called? What else can it do?
Answers
Answer:
exerts force
Explanation:
The accumulation of excess electric charge on an object is called static electricity. ... An electric field surrounds every electric charge and exerts the force that causes other electric charges to attract or repel. Electric fields are represented by arrows showing the electric field would make a positive charge move.
At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power generation potential of a wind turbine with 80-m-diameter (D) blades at that location. Take the air density to be 1.25 kg/m3. The mechanical energy of air per unit mass is kJ/kg. The power generation potential of the wind turbine is kW.
Answers
Answer:
[tex]0.05\ \text{kJ/kg}[/tex]
[tex]3141.6\ \text{kW}[/tex]
Explanation:
v = Velocity of wind = 10 m/s
A = Swept area of blade = [tex]\dfrac{\pi}{4}d^2[/tex]
d = Diameter of turbine = 80 m
[tex]\rho[/tex] = Density of air = [tex]1.25\ \text{kg/m}^3[/tex]
Wind energy per unit mass of air is given by
[tex]E=\dfrac{v^2}{2}\\\Rightarrow E=\dfrac{10^2}{2}\\\Rightarrow E=50\ \text{J/kg}[/tex]
The mechanical energy of air per unit mass is [tex]0.05\ \text{kJ/kg}[/tex]
Power is given by
[tex]P=\rho AvE\\\Rightarrow P=1.25\times \dfrac{\pi}{4}\times 80^2\times 10\times 50\\\Rightarrow P=3141592.65=3141.6\ \text{kW}[/tex]
The power generation potential of the wind turbine is [tex]3141.6\ \text{kW}[/tex].
A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. A 77.0-kg water-skier has an initial speed of 6.3 m/s. Later, the speed increases to 10.9 m/s. Determine the work done by the net external force acting on the skier.
Answers
Answer:
the work done by the net external force acting on the skier is 3046.12 J.
Explanation:
Given;
initial speed of the water skier, u = 6.3 m/s
final speed of the water skier, v = 10.9 m/s
mass of the water skier, m = 77 kg
The work done by the net external force is calculated as;
W = ΔK.E
[tex]W = \frac{1}{2} m(v^2 - u^2)\\\\W = \frac{1}{2} \times \ 77.0(10.9^2 - 6.3^2)\\\\ W= 3046.12 \ J[/tex]
Therefore, the work done by the net external force acting on the skier is 3046.12 J.
What energy store is in the torch
BEFORE it gets switched on?
Answers
Answer:
Chemical energy
Explanation:
The energy in the torch is stored as chemical energy before the torch gets switch on.
The chemical energy energy in the battery of cell will power the cell and allows it to produce light.
Chemical energy is a form of potential energy. The electrolytes within the battery are capable of producing electric current. So the chemical energy is transformed into electrical energy which is used to produce the light of the torch.A mysterious crate has shown up at your place of work, Firecracker Company, and you are told to measure its inertia. It is too heavy to lift, but it rolls smoothly on casters. Getting an inspiration, you lightly tape a 0.60-kg iron block to the side of the crate, slide a firecracker between the crate and the block, and light the fuse. When the firecracker explodes, the block goes one way and the crate rolls the other way. You measure the crate's speed to be 0.058 m/s by timing how long it takes to cross floor tiles. You look up the specifications of the firecracker and find that it releases 7 J of energy. That's all you need, and you quickly calculate the inertia of the crate.
What is that inertia?
Answers
Answer:
the inertia of the crate is (49.67 kg)r²
Explanation:
Given the data in the question;
First; we will use the law of conservation of momentum to determine the mass of the crate;
m₁v₁ - m₂v₂ = 0
given that; m₁ = 0.60 kg and v₂ = 0.058 m/s
we substitute
0.60 × v₁ = m₂ × 0.058 = 0
m₂ = 0.60v₁ / 0.058 ----------- EQU 1
Next, we use the energy conservation relation to find the velocity
According to conservation of energy;
1/2m₁v₁² + 1/2m₂v₂² = 7 J
we substitute
1/2×0.60×v₁² + 1/2×m₂×(0.058)² = 7 J
0.3v₁² + 0.001682m₂ = 7 J ----- EQU 2
substitute value of m₂ form equ 1 into equ 2
0.3v₁² + 0.001682(0.60v₁ / 0.058) = 7 J
0.3v₁² + 0.0174v₁ = 7 J
0.3v₁² + 0.0174v₁ - 7 J = 0
we solve the quadratic equation;
{ x = [-b±√( b² - 4ac)] / 2a }
v₁ = [-0.0174 ±√( 0.0174² - 4×0.3×-7)] / 2×0.3
= [-0.0174 ±√(8.4003)] / 0.6
= [-0.0174 ± 2.8983 ] / 0.6
= -4.8595 or 4.8015 but{ v₁ ≠ - }
so v₁ = 4.8015 m/s ≈ 4.802 m/s
next we input value of v₁ into equation 1
m₂ = (0.60×4.8015) / 0.058
m₂ = 2.8809 / 0.058
m₂ = 49.67 kg
So, the moment of inertia of the crate will be;
I₂ = m₂r²
we substitute value of m₂
I₂ = (49.67 kg)r²
Therefore, the inertia of the crate is (49.67 kg)r²
2. Mrs. Stern is standing still on rollerblades on a frictionless floor in the middle of the A-gym while
carrying heavy textbooks. How can she use the textbooks to get herself moving?
Answers
Answer:
If she bends forward
Explanation:
because the equilibrium of gravity will not stay the same causing her to move forward
Bill is not only good at riding a bike with no hands, but he can also ride the bike with no hands and facing backwards while the bike goes forward. Bill is riding his bike in such a manner while playing catch with Betty who is stationary on the ground and facing Bill. Bill is on the bike (facing backwards toward Betty) and traveling away from Betty at a speed of 12.0 m/s when he throws a ball to Betty at a speed of 17.8 m/s relative to the bike. What is the velocity of the ball as measured by Betty
Answers
Answer:
5.8 m/s
Explanation:
Let v = velocity of bike relative to Betty = -12.0 m/s (since the bike is moving away from betty).
u = velocity of ball relative to bike = + 17.8 m/s
and V = velocity of ball relative to Betty.
So, by Galilean relativity,
V = v + u
V = -12.0 m/s + 17.8 m/s
V = 5.8 m/s
So, the velocity of the ball as measured by Betty is 5.8 m/s
When monochromatic light passes through the interface between two unknown materials at an angle θ where 0∘<θ<90∘, no changes in the direction of propagation of light are observed. What can be said about the two materials? Check all that apply. View Available Hint(s) Check all that apply. The two materials have matching indexes of refraction. The second material through which light propagates has a lower index of refraction. The second material through which light propagates has a higher index of refraction. The two materials are identical.
Answers
Answer:
the correct one is the first, the refractive index of the two materials must be the same
Explanation:
When a beam of light passes through two materials, it must comply with the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where n₁ and n₂ are the refractive indices of each medium.
In this case, it indicates that the light does not change direction, so the input and output angle of the interface must be the same,
θ₁ = θ₂ = θ
substituting
n₁ = n₂
therefore the refractive index of the two materials must be the same
When reviewing the answers, the correct one is the first
A three-phase line, which has an impedance of (2 + j4) ohm per phase, feeds two balanced three-phase loads that are connected in parallel. One of the loads is Y-connected with an impedance of (30 + j40) ohm per phase, and the other is connected with an impedance of (60 - j45) ohm per phase. The line is energized at the sending end from a 60-Hz, three-phase, balanced voltage source of 120 √3V (rms, line-to-line).
Determine:
a. the current, real power and reactive power delivered by the sending-end source
b. the line-to-line voltage at the load
c. the current per phase in each load
d. the total three-phase real and reactive powers absorbed by each load and by the
Answers
Answer:
hello your question has a missing information
The other is Δ-connected with an impedance of (60 - j45) ohm per phase.
answer : A) 5A ∠0° ,
p( real power ) = 1800 and Q ( reactive power ) = 0 VAR
B) 193.64 v
C) current at load 1 = 2.236 A , current at load 2 = 4.472 A
D) Load 1 : 450 watts(real power ) , 600 VAR ( reactive power )
Load 2 : 1200 watts ( real power ), -900 VAR ( reactive power )
Explanation:
First convert the Δ-connection to Y- connection attached below is the conversion and pre-solution
A) determine the current, real power and reactive power delivered by the sending-end source
current power delivered (Is) = 5A ∠0°
complex power delivered ( s ) = 3vs Is
= 3 * 120∠0° * 5∠0° = 1800 + j0 ---- ( 1 )
also s = p + jQ ------ ( 2 )
comparing equation 1 and 2
p( real power ) = 1800 and Q ( reactive power ) = 0 VAR
B) determine Line-to-line voltage at the load
Vload = √3 * 111.8
= 193.64 v
c) Determine current per phase in each load
[tex]I_{l1} = Vl1 / Zl1[/tex]
= [tex]\frac{111.8<-10.3}{50<53.13}[/tex] = 2.236∠ 63.43° A hence current at load 1 = 2.236 A
[tex]I_{l2} = V_{l2}/Z_{l2}[/tex]
= [tex]\frac{111.8<-10.3}{25<-36.87}[/tex] = 4.472 ∠ 26.57° A hence current at load 2 = 4.472 A
D) Determine the Total three-phase real and reactive powers absorbed by each load
For load 1
3-phase real power = [tex]3I_{l1} ^{2} R_{l1}[/tex] = 3 * 2.236^2 * 30 = 450 watts
3-phase reactive power = [tex]3I_{l1} ^{2} X_{l1}[/tex] = 3 * 2.236^2 * 40 = 600 VAR
for load 2
3-phase real power = [tex]3I_{l1} ^{2} R_{l2}[/tex] = 1200 watts
3-phase reactive power = [tex]3I_{l1} ^{2} X_{l2}[/tex] = -900 VAR
The sum of load powers and line losses, 1800 W+ j0 VAR and The line voltage magnitude at the load terminal is 193.64 V.
(a) The impedance per phase of the equivalent Y,
[tex]\bar{Z}_{2}=\frac{60-j 45}{3}=(20-j 15) \Omega[/tex]
The phase voltage,
[tex]\bold { V_{1}=\frac{120 \sqrt{3}}{\sqrt{3}}=120 VV }[/tex]
Total impedance from the input terminals,
[tex]\bold {\begin{aligned}&\bar{Z}=2+j 4+\frac{(30+j 40)(20-j 15)}{(30+j 40)+(20-j 15)}=2+j 4+22-j 4=24 \Omega \\&\bar{I}=\frac{\bar{V}_{1}}{\bar{Z}}=\frac{120 \angle 0^{\circ}}{24}=5 \angle 0^{\circ} A\end{aligned} }[/tex]
The three-phase complex power supplied [tex]\bold {=\bar{S}=3 \bar{V}_{1} \bar{I}^{*}=1800 W}[/tex]
P =1800 W and Q = 0 VAR delivered by the sending-end source.
(b) Phase voltage at load terminals will be,
[tex]\bold {\begin{aligned}\bar{V}_{2} &=120 \angle 0^{\circ}-(2+j 4)\left(5 \angle 0^{\circ}\right) \\&=110-j 20=111.8 \angle-10.3^{\circ} V\end{aligned} }[/tex]
The line voltage magnitude at the load terminal,
[tex]\bold{\left(V_{ LOAD }\right)_{L-L}=\sqrt{3} 111.8=193.64 V(V }[/tex]
(c) The current per phase in the Y-connected load,
[tex]\bold {\begin{aligned}&\bar{I}_{1}=\frac{\bar{V}_{2}}{\bar{Z}_{1}}=1-j 2=2.236 \angle-63.4^{\circ} A \\&\bar{I}_{2}=\frac{\bar{V}_{2}}{\bar{Z}_{2}}=4+j 2=4.472 \angle 26.56^{\circ} A\end{aligned} }[/tex]
The phase current magnitude,
[tex]\bold {\left(I_{p h}\right)_{\Delta}=\frac{I_{2}}{\sqrt{3}}=\frac{4.472}{\sqrt{3}}=2.582 }[/tex]
(d) The three-phase complex power absorbed by each load,
[tex]\bold {\begin{aligned}&\bar{S}_{1}=3 \bar{V}_{2} \bar{I}_{1}^{*}=430 W +j 600 VAR \\&\bar{S}_{2}=3 \bar{V}_{2} \bar{I}_{2}^{*}=1200 W -j 900 VAR\end{aligned}}[/tex]
The three-phase complex power absorbed by the line is
[tex]\bold{\bar{S}_{L}=3\left(R_{L}+j X_{L}\right) I^{2}=3(2+j 4)(5)^{2}=150 W +j 300 VAR }[/tex]
Since, the sum of load powers and line losses,
[tex]\bold {\begin{aligned}\bar{S}_{1}+\bar{S}_{2}+\bar{S}_{L} &=(450+j 600)+(1200-j 900)+(150+j 300) \\&=1800 W +j 0 VAR\end{aligned} }[/tex]
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during a typical afternoon thunderstorm in the summer, an area of 66.0 km2 receives 9.57 108 gal of rain in 18 min. how many inches of rain fell during this 18 min period
Answers
Answer:
2.16 inch
Explanation:
area under water = 66 km²
= 66 x ( 3280.84 x 12 )² inch²
= 1.023 x 10¹¹ sq inch
volume of rain = 9.57 x 10⁸ gallon = 9.57 x 10⁸ x 231 inch³
= 2.21 x 10¹¹ inch³
If depth of rainfall be t
volume of rain = surface area x depth
= 1.023 x 10¹¹ x t
So ,
1.023 x 10¹¹ x t = 2.21 x 10¹¹
t = 2.16 inch
Passage of an electric current through a long conducting rod of radiusriand thermalconductivitykrresults in uniform volumetric heating at a rate ofq. The conducting rodis wrapped in an electrically nonconducting cladding material of outer radiusroandthermal conductivitykc, and convection cooling is provided by an adjoining fluid. Forsteady-state conditions, write appropriate forms of the heat equations for the rod andcladding. Express appropriate boundary conditions for the solution of these equations.
Answers
Answer:
a) For radial heat transfer to be zero along the perfectly insulated adiabatic surface; [tex]\frac{dT_{y} }{dr}[/tex][tex]|_{r-0}[/tex] = 0
b) For constant temperature; [tex]T_{y}[/tex]([tex]r_{i}[/tex]) = [tex]T_{C}[/tex]([tex]r_{i}[/tex])
c) The heat transfer in the conducting rod and the cladding material is the same, i.e; [tex]k_{r}[/tex][tex]\frac{dT_{y} }{dr}[/tex] [tex]|_{ri}[/tex] = [tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{ri}[/tex]
d) The convection surface conduction by cooling fluid will be;
[tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{r0}[/tex] = h( [tex]T_{c}[/tex]( [tex]r_{0}[/tex] ) - [tex]T_{\infty}[/tex] )
Explanation:
Given the data in question;
we write the general form of the heat conduction equation equation in cylindrical coordinates with internal heat generation.
1/r[tex]\frac{d}{dr}[/tex]( kr[tex]\frac{dT}{dr}[/tex] ) + 1/r² [tex]\frac{d}{d\beta }[/tex]( ( k[tex]\frac{dT}{dr}[/tex] ) + [tex]\frac{d}{dz}[/tex]( k[tex]\frac{dT}{dr}[/tex]) + q = 0
where radius of cylinder is r, thermal conductivity of the cylinder is k, and q is heat generated in cylinder.
Now, Assume one dimensional heat conduction
lets substitute the condition for conducting rod with steady state condition.
[tex]k_{y}[/tex]/r [tex]\frac{d}{dr}[/tex]( r[tex]\frac{dT_{y} }{dr}[/tex] ) + q = 0
Apply the conditions for cladding by substituting 0 for q
[tex]\frac{d}{dr}[/tex]( r[tex]\frac{dT_{r} }{dr}[/tex] ) = 0
Apply the following boundary conditions;
a) For radial heat transfer to be zero along the perfectly insulated adiabatic surface;
[tex]\frac{dT_{y} }{dr}[/tex][tex]|_{r-0}[/tex] = 0
b) For constant temperature
[tex]T_{y}[/tex]([tex]r_{i}[/tex]) = [tex]T_{C}[/tex]([tex]r_{i}[/tex])
c) The heat transfer in the conducting rod and the cladding material is the same, i.e
[tex]k_{r}[/tex][tex]\frac{dT_{y} }{dr}[/tex] [tex]|_{ri}[/tex] = [tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{ri}[/tex]
d) The convection surface conduction by cooling fluid will be;
[tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{r0}[/tex] = h( [tex]T_{c}[/tex]( [tex]r_{0}[/tex] ) - [tex]T_{\infty}[/tex] )
The diagram shows two balls before they collide.
2 balls with grey arrows pointing to them from the outside. The left ball has below it m subscript 1 = 0.6 kilograms v subscript 1 = 0.5 meters per second. The right ball has below it m subscript 2 = 0.5 kilograms v subscript 2 = negative 0.2 meters per second.
What is the momentum of the system after the collision?
1. 0.0 kg • m/s
2. 0.2 kg • m/s
3. 0.3 kg • m/s
4. 0.4 kg • m/s
Answers
Answer:
The Answer is B)0.2 kg • m/s
Explanation:
I made a 100 on my test. Sorry if I'm late but hope I helped.
Answer:
B. 0.2 kg x m/s
Explanation:
what measurement do geologists use to find absolute age
Answers
Answer:
see below :)
Explanation:
Radiometric dating.
1. A particle is projected vertically upwards with a velocity of 30 ms from a point 0. Find (a) the maximum height reached(b) the time taken for it to return to 0 (c) the taken for it to be 35m below 0
Answers
Assuming the particle is in free fall once it is shot up, its vertical velocity v at time t is
v = 30 m/s - g t
where g = 9.8 m/s² is the magnitude of the acceleration due to gravity, and its height y is given by
y = (30 m/s) t - 1/2 g t ²
(a) At its maximum height, the particle has 0 velocity, which occurs for
0 = 30 m/s - g t
t = (30 m/s) / g ≈ 3.06 s
at which point the particle's maximum height would be
y = (30 m/s) (3.06 s) - 1/2 g (3.06 s)² ≈ 45.9184 m ≈ 46 m
(b) It takes twice the time found in part (a) to return to 0 height, t ≈ 6.1 s.
(c) The particle falls 35 m below its starting point when
-35 m = (30 m/s) t - 1/2 g t ²
Solve for t to get a time of about t ≈ 7.1 s
While investigating Kirchhoff's Laws, you begin observing a blackbody, such as a star, from Earth using advanced technology that can analyze spectra. While pointing it at the star with nothing between you and the star, you observe a full spectrum. You come back and repeat this same experiment a year later using the same star, except this time you observe an absorption spectrum. What is the most likely explanation for this
Answers
Answer:
the second time there is a gas between you and the star,
Explanation:
When you observe the star for the first time you do not have a given between you and the star, therefore you observe the emission spectrum of the same that is formed by lines of different intensity and position that indicate the type and percentage of the atoms that make up the star.
When you observe the same phenomenon for the second time there is a gas between you and the star, this gas absorbs the wavelengths of the star that has the same energies and the atomisms and molecular gas, therefore these lines are not observed by seeing a series of dark bands,
The information obtained from the two spectra is the same, the type of atoms that make up the star
A cheerleader of mass 55 kg stand on the shoulders of a football player of mass 86 kg. The football player is standing in a soft, thin layer of mud that does not permit air under his shoes. If each of his shoes has an area of 264 cm2, calculate the absolute pressure exerted on the surface underneath one of the shoes. Answer in Pascal, assuming g = 9.80 m/s2 and atmospheric pressure is 101,000 Pa.
Answers
A flat circular mirror of radius 0.100 m is lying on the floor. Centered directly above the mirror, at a height of 0.920 m, is a small light source. Calculate the diameter of the bright circular spot formed on the 2.70 m high ceiling by the light reflected from the mirror.
Answers
Answer:
the diameter of the bright circular spot formed is 0.787 m
Explanation:
Given that;
Radius of the flat circular mirror = 0.100 m
height of small ight source = 0.920 m
high ceiling = 2.70 m
now;
Diameter(mirror) = 2×r = 2 × 0.100 = 0.2 m
D(spot) = [Diameter(mirror) × ( 2.70m + 0.920 m)] / 0.920 m
so
D(spot) = 0.2m × 3.62m / 0.920 m
D(spot) = 0.724 m / 0.920 m
D(spot) = 0.787 m
Therefore, the diameter of the bright circular spot formed is 0.787 m
2. Why are numbers better than words in a science experiment?
Answers
Why words are more important than numbers: ... Words on the other hand are harder to manipulate, they also tell you why someone voted a particular way and to improve your delivery and thus your customer satisfaction you need to understand the why's...
#pglubestiehere
A compact car has a mass of 1310 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that the mass is equally distributed over the four springs.
Required:
a. What is the spring constant of each spring if the empty car bounces up and down 2.0 times each second?
b. What will be the car’s oscillation frequency while carrying four 70 kg passengers?
Answers
Answer:
a) k= 3232.30 N / m, b) f = 4,410 Hz
Explanation:
In this exercise, the car + spring system is oscillating in the form of a simple harmonic motion, as the four springs are in parallel, the force is the sum of the 4 Hocke forces.
The expression for the angular velocity is
w = √k/m
the angular velocity is related to the period
w = 2π / T
we substitute
T = 2[tex]\pi[/tex] √m/ k
a) empty car
k = 4π² m / T²
k = 4 π² 1310/2 2
k = 12929.18 N / m
This is the equivalent constant of the short springs
F1 + F2 + F3 + F4 = k_eq x
k x + kx + kx + kx = k_eq x
k_eq = 4 k
k = k_eq / 4
k = 12 929.18 / 4
k= 3232.30 N / m
b) the frequency of oscillation when carrying four passengers.
In this case the plus is the mass of the vehicle plus the masses of the passengers
m_total = 1360 + 4 70
m_total = 1640 kg
angular velocity and frequency are related
w = 2pi f
we substitute
2 pi f = Ra K / m
in this case the spring constant changes us
k_eq = 12929.18 N / m
f = 1 / 2π √ 12929.18 / 1640
f = π / 2 2.80778
f = 4,410 Hz
Compare and Contrast the Following:
1. Pitch and Loudness
2. Infrasonic and Ultrasonic
3. Luminous and Nonluminous
It's okay if you answer only one of these.
Thanks in Advance
Answers
Answer:
pitch and loudnesss
thanks for your question
QUESTION 4.
If
you have 2 randomly selected vectors like R and R;
Show that R. RX 5) = 0
(102)
Answers
Answer:
Follows are the solution to this question:
Explanation:
Please find the correct question in the attachment file.
Let:
[tex]\overrightarrow{R}= R_i\hat{i}+R_j\hat{j}+R_k\hat{k}\\\\\overrightarrow{S}= S_i\hat{i}+S_j\hat{j}+S_k\hat{k}\\\\[/tex]
Calculating the value of [tex]\overrightarrow{R} \times \overrightarrow{S}:[/tex]
[tex]\to \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{K}\\R_i&R_j&R_k\\S_i&S_j&S_k\end{array}\right | = \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j][/tex]
Calculating the value of [tex]\overrightarrow{R} \cdot (\overrightarrow{R} \times \overrightarrow{S}):[/tex]
[tex]\to (R_i\hat{i}+R_j\hat{j}+R_k\hat{k}) \cdot ( \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j])[/tex]
by solving this value it is equal to 0.
20
A person walks 2.0 m east, then turns and goes 4.0 m west, then turns
and goes back 6.0 m east. What is that person's total displacement?
(Remember to include the correct units) *
Your answer
Answers
Which of the following is a vector quantity?
speed
distance
acceleration
Answers
For example, if person A is the observer, then he/she can predict the distance or speed or time by simply looking at it. But at the same time, the observer can't predict the acceleration of a moving object as he/she can't determine the direction of the object. So, without a direction, the acceleration can't be predicted. For example, displacement, force, velocity, momentum, etc. are vector quantities.
◙ But, Distance, speed and time are only specified with their magnitude. For example, work, volume, density, mass, etc. don't need a direction for their representation.
So, (c)Acceleration is the answer.
What day of the year is solar time the same as sidereal time?
Answers
Answer:
I think the answers March 21
Answer:
Once a year, mean solar time and sidereal time will be the same.
the luminous flux of a torch of intensity 50 cd is?
Answers
Answer:
i dont know i am right but here Luminous intensity is defined as dI=dΨλ / dΩ, where dΨλ is the luminous flux (light energy flux in watts per m2) emitted within a solid angle dΩ. The light energy flux may be expressed in terms of the incident x-ray energy flux and the x-ray absorption and conversion properties of the scintillator(7,8,9).
Explanation:
The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero. (a) Determine the forces and and the couple (b) Determine the sum of the moments about the right end of the beam. (c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the
Answers
This question is incomplete, the complete question is;
The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.
(a) Determine the forces and and the couple
(b) Determine the sum of the moments about the right end of the beam.
(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?
Answer:
a)
the x-component of the force at A is [tex]A_{x}[/tex] = 0
the y-component of the force at A is [tex]A_{y}[/tex] = 400 N
the couple acting at A is; [tex]M_{A}[/tex] = 146 N-m
b)
the sum of the momentum about the right end of the beam is; ∑[tex]M_{R}[/tex] = 0
c)
the equivalent force acting at the left end is; F = -400J ( N)
the couple acting at the left end is; M = - 146 N-m
Explanation:
Given that;
The sum of the forces acting on the beam is zero ∑f = 0
Sum of the moments about the left end of the beam is also zero ∑[tex]M_{L}[/tex] = 0
Vector force acting at A, [tex]F_{A}[/tex] = [tex]A_{x}i[/tex] + [tex]A_{y}j[/tex]
Now, From the image, we have;
a)
∑f = 0
[tex]F_{A}[/tex] - 600j + 200j = 0i + 0j
[tex]A_{x}i[/tex] + [tex]A_{y}j[/tex] - 600j + 200j = 0i + 0j
[tex]A_{x}i[/tex] + ([tex]A_{y}[/tex] - 400)j = 0i + 0j
now by equating i- coefficients'
[tex]A_{x}[/tex] = 0
so, the x-component of the force at A is [tex]A_{x}[/tex] = 0
also by equating j-coefficient
[tex]A_{y}[/tex] - 400 = 0
[tex]A_{y}[/tex] = 400 N
hence, the y-component of the force at A is [tex]A_{y}[/tex] = 400 N
we also have;
∑[tex]M_{L}[/tex] = 0
[tex]M_{A}[/tex] - ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0
[tex]M_{A}[/tex] - 30 N-m - 228 N-m + 112 Nm = 0
[tex]M_{A}[/tex] - 146 N-m = 0
[tex]M_{A}[/tex] = 146 N-m
Therefore, the couple acting at A is; [tex]M_{A}[/tex] = 146 N-m
b)
The sum of the moments about right end of the beam is;
∑[tex]M_{R}[/tex] = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)([tex]A_{y}[/tex] ) + [tex]M_{A}[/tex]
∑[tex]M_{R}[/tex] = (108 N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m
∑[tex]M_{R}[/tex] = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m
∑[tex]M_{R}[/tex] = 0
Therefore, the sum of the momentum about the right end of the beam is; ∑[tex]M_{R}[/tex] = 0
c)
The 600-N force, the 200-N force and the 30 N-m couple by a force F which is acting at the left end of the beam and a couple M.
The equivalent force at the left end will be;
F = -600j + 200j (N)
F = -400J ( N)
Therefore, the equivalent force acting at the left end is; F = -400J ( N)
Also couple acting at the left end
M = -(30 N-m) + (0.560 m)( 200N) - ( 0.380 m)( 600 N)
M = -(30 N-m) + (112 N-m) - ( 228 N-m))
M = 112 N-m - 258 N-m
M = - 146 N-m
Therefore, the couple acting at the left end is; M = - 146 N-m
PLZ FAST!!
Compare and contrast microscopic and macroscopic energy transfer. Give at least three comparisons for each. THX
Answers
Answer:
Macroscopic energy is energy at a level of system while microscopic energy is energy at the level of atoms and molecules
Explanation:
1. Macroscopic energy is possessed by a system as whole while microscopic energy is possessed by its constituents’ atoms or molecules.
2. The common form of macroscopic energy is Kinetic and potential energy while the microscopic form of energy are atomic forces due its random, disordered motion and due to intermolecular forces
3. At microscopic level we consider behaviour of every molecule and in macroscopic approach we consider gross or average effects of various molecular infractions
Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 554 N/C. If the particles are free to move, what are their speeds (in m/s) after 51.6 ns
Answers
Answer:
the speed of electron is 5.021 x 10⁶ m/s
the speed of proton is 2733.91 m/s
Explanation:
Given;
magnitude of electric field, E = 554 N/C
charge of the particles, Q = 1.6 x 10⁻¹⁹ C
time of motion, t = 51.6 ns = 51.6 x 10⁻⁹ s
The force experienced by each particle is calculated as;
F = EQ
F = (554)(1.6 x 10⁻¹⁹)
F = 8.864 x 10⁻¹⁷ N
The speed of the particles are calculated as;
[tex]F = ma\\\\F = \frac{mv}{t} \\\\v = \frac{Ft}{m} \\\\v_e = \frac{Ft}{m_e}\\\\v_e = \frac{(8.864 \times 10^{-17})(51.6\times 10^{-9})}{9.11 \times \ 10^{-31}}\\\\v_e = 5.021 \times 10^{6} \ m/s[/tex]
[tex]v_p = \frac{Ft}{m_p}\\\\v_p = \frac{(8.864 \times 10^{-17})(51.6\times 10^{-9})}{1.673 \times \ 10^{-27}}\\\\v_p = 2733.91 \ m/s[/tex]
At an airport, two business partners both walk at 1.5 m/sm/s from the gate to the main terminal, one on a moving sidewalk and the other on the floor next to it. The partner on the moving sidewalk gets to the end in 60 ss, and the partner on the floor reaches the end of the sidewalk in 90s.
Required:
What is the speed of the sidewalk in the Earth reference frame?
Answers
Answer:
[tex]v=0.8m/s[/tex]
Explanation:
From the question we are told that
Distance [tex]d=1.5m/sm/s[/tex]
Time [tex]t_1=60s[/tex]
Time [tex]t_2=90s[/tex]
Generally the the equation for the distance traveled is mathematically given as
[tex]d=vt[/tex]
[tex]d=1.5*90[/tex]
[tex]d=138m[/tex]
Generally equation for speed of side walk is mathematically given as
[tex]d=(v+u)t[/tex]
[tex]v=\frac{d}{t}-u[/tex]
[tex]v=\frac{138}{60}-1.5[/tex]
[tex]v=0.8m/s[/tex]