Answer:
0.2812
Explanation:
Given that
mass of skater 1, m1 = 45 kg
mass of skater 2, m2 = 60 kg
speed of skater 1, v1 = 0.375 m/s
To attempt this question, we would be using the Law of conservation of momentum That says the momentum is constant, before and after the movement.
Thus, momentum p = mv
Law of conservation of momentum infers that,
m1v1 = m2v2
Now we proceed to substitute our values into the formula.
45 * 0.375 = 60 * v2
v2 = 16.875 / 60
v2 = 0.2812 m/s
Therefore the speed of the second skater has to be 0.2812 m/s
g Assume you are a farsighted person who has a near point distance of 40 (cm). If you use a converging contact lens with focal length of 10 (cm). What is nearest distance you can vision with you contacts now?
Answer:
object distance p = 13.33 cm
Explanation:
For this problem of finding the image of an object we must use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distances to the object and the image, respectively.
In this case they indicate the focal length f = 10 cm, since the person has hyperopia, the image must be formed q = 40 cm, let's find where the object is (p)
1 / p = 1 / f - 1 / q
1 / p = 1/10 - 1/40
1 / p = 0.075
p = 13.33 cm
At sea level, at a latitude where , a pendulum that takes 2.00 s for a complete swing back and forth has a length of 0.993 m. What is the value of g in m/s2 at a location where the length of such a pendulum is 0.970 m
Answer:
a) The value of g at such location is:
[tex]g=9.8005171\,\frac{m}{s^2}[/tex]
b) the period of the pendulum with the length is 0.970 m is:
[tex]T=1.9767 sec[/tex]
Explanation:
Recall the relationship between the period (T) of a pendulum and its length (L) when it swings under an acceleration of gravity g:
[tex]L=\frac{g}{4\,\pi^2} \,T^2[/tex]
a) Then, given that we know the period (2.0 seconds), and the pendulum's length (L=0.993 m), we can determine g at that location:
[tex]g=\frac{4\,\pi^2\,L}{T^2}\\g=\frac{4\,\pi^2\,0.993}{(2)^2}\\g=\pi^2\,(0.993)\,\frac{m}{s^2} \\g=9.8005171\,\frac{m}{s^2}[/tex]
b) for this value of g, when the pendulum is shortened to 0.970 m, the period becomes:
A plane electromagnetic wave travels northward. At one instant, its electric field has a magnitude of 9.6 V/m and points eastward. What are the magnitude (in T) and direction of the magnetic field at this instant?
Answer:
The values is [tex]B = 3.2 *10^{-8} \ T[/tex]
The direction is out of the plane
Explanation:
From the question we are told that
The magnitude of the electric field is [tex]E = 9.6 \ V/m[/tex]
The magnitude of the magnetic field is mathematically represented as
[tex]B = \frac{E}{c}[/tex]
where c is the speed of light with value
[tex]B = \frac{ 9.6}{3.0 *10^{8}}[/tex]
[tex]B = 3.2 *10^{-8} \ T[/tex]
Given that the direction off the electromagnetic wave( c ) is northward(y-plane ) and the electric field(E) is eastward(x-plane ) then the magnetic field will be acting in the out of the page (z-plane )
what path would an object have to take to have the distance and the displacement to be equal
Answer:
When an object move in a straight line without moving back.
Explanation:
Distance is covered by an object is the magnitude of length from one position to the another. It is a scalar quantity.
While displacement is the distance covered in a specific direction. Displacement is a vector quantity. It has both magnitude and direction.
If an object move in a straight path without going back, then, the magnitude of distance will be the same with the magnitude of displacement.
Both distance and displacement are measured in the same unit which is metres.
Therefore, an object have to take a straight path without going back to have the distance and the displacement equal.
Find the angle in degrees for the third-order maximum for 577 nm wavelength yellow light falling on a diffraction grating having 1,420 lines per centimeter.
Answer:
θ = 0.14°
Explanation:
Here we will use the grating equation. The grating equation is as follows:
mλ = d Sin θ
where,
θ = angle = ?
m = order number = 3
λ = wavelength of light = 577 nm = 5.77 x 10⁻⁷ m
d = spacing between slits = 1/(1420 lines/cm) = 7.042 x 10⁻⁴ m
Therefore, using these values, we get:
(3)(5.77 x 10⁻⁷ m) = (7.042 x 10⁻⁴ m)Sin θ
Sin θ = (3)(5.77 x 10⁻⁷ m)/(7.042 x 10⁻⁴ m)
Sin θ = 2.46 x 10⁻³
θ = Sin⁻¹(2.46 x 10⁻³)
θ = 0.14°
A) Hooke's law is described mathematically using the formula Fsp = -ku. Which statement is correct about the spring force, Fsp?
A.It is a vector quantity
B.It is the force doing the push or pull,
C.It is always a positive force.
D.It is larger than the applied force.
1. Which example best describes a restoring force?
B) the force applied to restore a spring to its original length
2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?
C) The spring exerts a restoring force to the left and returns to its equilibrium position.
3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?
D) 1 m
4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?
D)It is a vector quantity.
5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?
A) It decreases in magnatude.
Hope this Helps!! Sorry its late
A rod bent into the arc of a circle subtends an angle 2θ at the center P of the circle (see below). If the rod is charged uniformly with a total charge Q, what is the electric field at P? (Assume Q is positive. For the magnitude, use the following as necessary: ε0, Q, R, and θ.)
Answer:
Qsinθ/4πε₀R²θ
Explanation:
Let us have a small charge element dq which produces an electric field E. There is also a symmetric field at P due to a symmetric charge dq at P. Their vertical electric field components cancel out leaving the horizontal component dE' = dEcosθ = dqcosθ/4πε₀R² where r is the radius of the arc.
Now, let λ be the charge per unit length on the arc. then, the small charge element dq = λds where ds is the small arc length. Also ds = Rθ.
So dq = λRdθ.
Substituting dq into dE', we have
dE' = dqcosθ/4πε₀R²
= λRdθcosθ/4πε₀R²
= λdθcosθ/4πε₀R
E' = ∫dE' = ∫λRdθcosθ/4πε₀R² = (λ/4πε₀R)∫cosθdθ from -θ to θ
E' = (λ/4πε₀R)[sinθ] from -θ to θ
E' = (λ/4πε₀R)[sinθ]
= (λ/4πε₀R)[sinθ - sin(-θ)]
= (λ/4πε₀R)[sinθ + sinθ]
= 2(λ/4πε₀R)sinθ
= (λ/2πε₀R)sinθ
Now, the total charge Q = ∫dq = ∫λRdθ from -θ to +θ
Q = λR∫dθ = λR[θ - (-θ)] = λR[θ + θ] = 2λRθ
Q = 2λRθ
λ = Q/2Rθ
Substituting λ into E', we have
E' = (Q/2Rθ/2πε₀R)sinθ
E' = (Q/θ4πε₀R²)sinθ
E' = Qsinθ/4πε₀R²θ where θ is in radians
________ is a thermodynamic function that increases with the number of energetically equivalent ways to arrange components of a system to achieve a particular state.
Answer:
entropy
Explanation:
Two identical rooms in a house are connected by an open doorway. The temperatures in the two rooms are maintained at different values. Which room contains more air
Answer:
The room with the lower temperature
Explanation:
Using
PV=nRT
Since both the rooms same volume and are connected, so they will have same pressure
PV=nRT=constant
nT=Constant/R=constant
If T is more n has to be less
Thus, lower the temperature, more the number molecules.
Four charges each of magnitude 15 µC are arranged on the corners of a square of side 5 cm. What is the total potential energy of the system?
Answer:
-105J
Explanation:
See attached file
A fisherman in a stream 39 cm deep looks downward into the water and sees a rock on the stream bed. How deep does the stream appear to the fisherman
Answer:
30cm
Explanation:
assume that the eyes are substantially above the water so that sin(theta) is approximately theta.
( small angle approximation).
The point at which a ray leaving the fish hits the surface of the water is x to the side of the centreline and the depth of the water is d
x/d = sin( angle of incidence)
if the apparent depth of the water is h then
x/h = sin( angle of refraction)
and applying snells law
1 sin ( theta air) = 1.33 sin( theta water)
1 * x/h = 1.33 * x/d
d/h = 1.33
or h/d = 1/1.33
h/39 = 1.33
h = 39 /1.33 so that is the apparent depth of the stream assuming:-
1. Your eyes are almost directly overhead
and
2. your eyes are a significant distance above the surface of the water.
x/d = 1.33 x/h
h/d =39/1.3
= 30cm
Structures on a bird feather act like a diffraction grating having 8500 lines per centimeter. What is the angle of the first-order maximum for 577 nm light shone through a feather?
Answer:
29.5°
Explanation:
To find the distance d
d = 1E10^-2/8500lines
= 1.17x 10-6m
But wavelength in first order maximum is 577nm
and M = 1
So
dsin theta= m. Wavelength
Theta= sin^-1 (m wavelength/d)
= Sin^-1 ( 1* 577 x10^-8m)/1.17*10^-6
= 493*10^-3= sin^-1 0.493
Theta = 29.5°
The electric field 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge?
Answer:
2.1×10¹⁸ C
Explanation:
Using,
E = kq/r²...................... Equation 1
Where E = Electric field, q = charge, r = distance, k = coulombs constant.
make q the subject of the equation
q = Er²/k.................. Equation 2
Given: E = 180000 N/C, r = 2.8 cm = 0.028 m
Constant: k = 9×10⁹ Nm²/C².
Substitute these values into equation 2
q = 180000(9×10⁹)/0.028²
q = 2.1×10¹⁸ C
Hence the object charge is 2.1×10¹⁸ C
Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by a clock on board the spacecraft, and then return home at the same speed. Upon return, the people on earth will have advanced exactly 120 years into the future. According to special relativity, how fast must you travel
Answer:
I must travel with a speed of 2.97 x 10^8 m/s
Explanation:
Sine the spacecraft flies at the same speed in the to and fro distance of the journey, then the time taken will be 6 months plus 6 months
Time that elapses on the spacecraft = 1 year
On earth the people have advanced 120 yrs
According to relativity, the time contraction on the spacecraft is gotten from
[tex]t[/tex] = [tex]t_{0} /\sqrt{1 - \beta ^{2} }[/tex]
where
[tex]t[/tex] is the time that elapses on the spacecraft = 120 years
[tex]t_{0}[/tex] = time here on Earth = 1 year
[tex]\beta[/tex] is the ratio v/c
where
v is the speed of the spacecraft = ?
c is the speed of light = 3 x 10^8 m/s
substituting values, we have
120 = 1/[tex]\sqrt{1 - \beta ^{2} }[/tex]
squaring both sides of the equation, we have
14400 = 1/[tex](1 - \beta ^{2} )[/tex]
14400 - 14400[tex]\beta ^{2}[/tex] = 1
14400 - 1 = 14400[tex]\beta ^{2}[/tex]
14399 = 14400[tex]\beta ^{2}[/tex]
[tex]\beta ^{2}[/tex] = 14399/14400 = 0.99
[tex]\beta = \sqrt{0.99}[/tex] = 0.99
substitute β = v/c
v/c = 0.99
but c = 3 x 10^8 m/s
v = 0.99c = 0.99 x 3 x 10^8 = 2.97 x 10^8 m/s
Niobium metal becomes a superconductor when cooled below 9 K. Its superconductivity is destroyed when the surface magnetic field exceeds 0.100 T. In the absence of any external magnetic field, determine the maximum current a 5.68-mm-diameter niobium wire can carry and remain superconducting.
Answer:
The current is [tex]I = 1420 \ A[/tex]
Explanation:
From the question we are told that
The diameter of the wire is [tex]d = 5.68 \ mm = 0.00568 \ m[/tex]
The magnetic field is [tex]B = 0.100 \ T[/tex]
Generally the radius of the wire is mathematically evaluated as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{ 0.00568}{2}[/tex]
[tex]r = 0.00284 \ m[/tex]
Generally the magnetic field is mathematically represented as
[tex]B = \frac{\mu_o * I}{ 2 \pi r }[/tex]
=> [tex]I =\frac{ B * 2 \pi r }{\mu_o}[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4 \pi *10^{-7} N/A^2[/tex]
substituting values
=> [tex]I =\frac{ 0.100 * 2 * 3.142 * 0.00284 }{ 4 \pi * 10^{-7}}[/tex]
=> [tex]I = 1420 \ A[/tex]
Violet light of wavelength 400 nm ejects electrons with a maximum kinetic energy of 0.860 eV from sodium metal. What is the binding energy of electrons to sodium metal?
Answer:
Binding Energy = 2.24 eV
Explanation:
First, we need to find the energy of the photon of light:
E = hc/λ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light = 400 nm = 4 x 10⁻⁷ m
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(4 x 10⁻⁷ m)
E = (4.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)
E = 3.1 eV
Now, from Einstein's Photoelectric Equation:
E = Binding Energy + Kinetic Energy
Binding Energy = E - Kinetic Energy
Binding Energy = 3.1 eV - 0.86 eV
Binding Energy = 2.24 eV
A brass road is 2cm long at instance to what is the lense for a temperature rise of 100k, If the expansivity of brass is 18x10^-6/k^-1
The length of the brass at a temperature rise of 100 K is 2.0036 m
From the question given above, the following data were obtained:
Original length (L₁) = 2 m
Temperature rise (ΔT) = 100 K
Coefficient of linear expansion (α) = 18×10¯⁶ K¯¹
Final length (L₂) =?The final length of the brass can be obtained as follow:
α = L₂ – L₁ / L₁ΔT
18×10¯⁶ = L₂ – 2 / (2 × 100)
18×10¯⁶ = L₂ – 2 / 200
Cross multiply
L₂ – 2 = 18×10¯⁶ × 200
L₂ – 2 = 0.0036
Collect like terms
L₂ = 0.0036 + 2
L₂ = 2.0036 m
Thus, the length of the brass at a temperature rise of 100 K is 2.0036 m
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e. Your father bought you a pair of shoes. When you wore the shoes, you realized there was a problem. The shoes were too long Why might such a problem arise and how can it be mitigated?
The problem arose due to a difference in length. This was due to father not knowing the exact length of shoe used by the son. And this can be mitigated by the use of shoe fillers.
The length of an object implies how long the object is. And it is one of the fundamental unit of quantities measured in SI unit of meters.
Considering the given question, it can be observed that the father do not know the exact length of shoe that would fit the son appropriately. Thus the realized problem of the pair of shoes too long arose due to difference in length of the pair of shoes and the son's leg. This variation would not have occurred if the exact length of pair of shoes has been bought.
To mitigate this little problem, shoe fillers can be used.
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Using a conventional two-slit apparatus with light of wavelength 605 nm, 34 bright fringes per centimeter are observed on a screen 3.1 m away. What is the slit separation
Answer:
d = 6.38 x 10⁻³ m = 6.38 mm
Explanation:
Since, the no. of bright fringes is 34 in a centimeter, therefore, the fringe spacing must be equal to:
Fringe Spacing = Δx = 1 cm/34
Δx = 0.0294 cm = 2.94 x 10⁻⁴ m
But, the formula for fringe spacing in a double slit experiment is:
Δx = λL/d
where,
λ = wavelength of light = 605 nm = 6.05 x 10⁻⁷ m
L = Distance between screen and slits = 3.1 m
d = slit separation = ?
Therefore,
2.94 x 10⁻⁴ m = (6.05 x 10⁻⁷ m)(3.1 m)/d
d = (18.755 x 10⁻⁷ m²)/(2.94 x 10⁻⁴ m)
d = 6.38 x 10⁻³ m = 6.38 mm
name a device that converts mechanical energy into electrical energy.
Answer:
Electric generator is the device that converts mechanical energy into electrical energy
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm^2, separated by a distance of 1.70 mm. A 25.0-V potential difference is applied to these plates. Calculate: a. the electric field between the plates b. the surface charge density c. the capacitance d. the charge on each plate.
Answer:
(a) 1.47 x 10⁴ V/m
(b) 1.28 x 10⁻⁷C/m²
(c) 3.9 x 10⁻¹²F
(d) 9.75 x 10⁻¹¹C
Explanation:
(a) For a parallel plate capacitor, the electric field E between the plates is given by;
E = V / d -----------(i)
Where;
V = potential difference applied to the plates
d = distance between these plates
From the question;
V = 25.0V
d = 1.70mm = 0.0017m
Substitute these values into equation (i) as follows;
E = 25.0 / 0.0017
E = 1.47 x 10⁴ V/m
(c) The capacitance of the capacitor is given by
C = Aε₀ / d
Where
C = capacitance
A = Area of the plates = 7.60cm² = 0.00076m²
ε₀ = permittivity of free space = 8.85 x 10⁻¹²F/m
d = 1.70mm = 0.0017m
C = 0.00076 x 8.85 x 10⁻¹² / 0.0017
C = 3.9 x 10⁻¹²F
(d) The charge, Q, on each plate can be found as follows;
Q = C V
Q = 3.9 x 10⁻¹² x 25.0
Q = 9.75 x 10⁻¹¹C
Now since we have found other quantities, it is way easier to find the surface charge density.
(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e
σ = Q / A
σ = 9.75 x 10⁻¹¹ / 0.00076
σ = 1.28 x 10⁻⁷C/m²
An electric motor and a generator are similar in that both
A. Produce mechanical energy.
B. Use electricity.
C. Transform energy into a different form.
D. Create energy.
C. transform energy into a different from.
I hope this helps
By what angle should the second polarized sheet be rotated relative to the first to reduce the transmitted intensity to one-half the intensity that was transmitted through both polarizing sheets when aligned
Answer:
θ = 45º
Explanation:
The light that falls on the second polarized is polarized, therefore it is governed by the law of Maluz
I = I₀ cos² θ
in the problem they ask us
I = ½ I₀
let's look for the angles
½ I₀ = I₀ cos² θ
cos θ = √ ½ = 0.707
θ = cos 0.707
θ = 45º
A simple arrangement by means of which e.m.f,s. are compared is known
Answer:
A simple arrangement by means of which e.m.f,s. are compared is known as?
(a)Voltmeter
(b)Potentiometer
(c)Ammeter
(d)None of the above
Explanation:
For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the efficiency of the human body is 25%, and that he lifts the barbell at a constant speed. Show all work and include proper unit for your final answer.
a) In applying the energy equation (ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W) to the system consisting of the earth, the barbell, and the athlete,
1. Which terms (if any) are positive?
2. Which terms (if any) are negative?
3. Which terms (if any) are zero?
b) Determine the energy output by the athlete in SI unit.
c) Determine his metabolic power in SI unit.
d) Another day he performs the same task in 1.2 s.
1. Is the metabolic energy that he expends more, less, or the same?
2. Is his metabolic power more, less, or the same?
Answer:
Explanation:
(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)
ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .
ΔK = 0
ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .
ΔUg = positive
ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .
ΔUs = 0
ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence
ΔEth = negative .
b )
work done by athlete
= 400 x 2 = 800 J
energy output = 800 J
c )
It is 25% of metabolic energy output of his body
so metalic energy output of body
= 4x 800 J .
3200 J
power = energy output / time
= 3200 / 1.6
= 2000 W .
d )
1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .
2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .
A) Applying the energy equation
The positive terms is : ΔUg The negative terms is : ΔEth The zero term are : ΔK and ΔUsB) The energy output by the athlete is ; 800 Joules
C) The metabolic power is : 2000 w
D) When he performs the task in 1.2 s
The metabolic energy he expends is : the same His metabolic power is : moreGiven data :
Weight of barbell = 400 N
Height = 2.0 m
Time = 1.6 secs
efficiency of the human body = 25%
Speed = constant
A) From the energy equation the ΔK is zero because the athlete is lifting the barbell at a constant speed. ΔUg is positive because as the weight is lifted its potential energy increases. ΔEth ( change in energy of earth ) is negative because it exerts a force in opposite direction to displacement
B) Determine the energy output of the athlete
weight of barbell * Height = 400 * 2 = 800 J
C) Determine the metabolic power
Metabolic power = energy output / Time
where ; energy output = 4 * 800 = 3200
∴ Metabolic power = 3200 / 1.6
= 2000 w
D) When performs same task at 1.2 s
The metabolic energy he expends is the same and His metabolic power is more
Hence we can conclude that the answers to your questions are as listed above
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The tibia is a lower leg bone (shin bone) in a human. The maximum strain that the tibia can experience before fracturing corresponds to a 1 % change in length.
A. Young's modulus for bone is about Y = 1.4 x 10 N/m². The tibia (shin bone) of a human is 0.35 m long and has an average cross-sectional area of 2.9 cm. What is the effective spring constant of the tibia?
B. If a man weighs 750 N, how much is the tibia compressed if it supports half his weight?
C. What is the maximum force that can be applied to a tibia with a cross-sectional area, A = 2.90 cm?
Answer:
a
[tex]k = 11600000 N/m[/tex]
b
[tex]\Delta L = 3.2323 *10^{-5} \ m[/tex]
c
[tex]F = 3750.28 \ N[/tex]
Explanation:
From the question we are told that
The Young modulus is [tex]E = 1.4 *10^{10} \ N/m^2[/tex]
The length is [tex]L = 0.35 \ m[/tex]
The area is [tex]2.9 \ cm^2 = 2.9 *10^{-4} \ m ^2[/tex]
Generally the force acting on the tibia is mathematically represented as
[tex]F = \frac{E * A * \Delta L }{L}[/tex] derived from young modulus equation
Now this force can also be mathematically represented as
[tex]F = k * \Delta L[/tex]
So
[tex]k = \frac{E * A }{L}[/tex]
substituting values
[tex]k = \frac{1.4 *10^{10} * 2.9 *10^{-4} }{ 0.35}[/tex]
[tex]k = 11600000 N/m[/tex]
Since the tibia support half the weight then the force experienced by the tibia is
[tex]F_k = \frac{750 }{2} = 375 \ N[/tex]
From the above equation the extension (compression) is mathematically represented as
[tex]\Delta L = \frac{ F_k * L }{ A * E }[/tex]
substituting values
[tex]\Delta L = \frac{ 375 * 0.35 }{ (2.9 *10^{-4}) * 1.4*10^{10} }[/tex]
[tex]\Delta L = 3.2323 *10^{-5} \ m[/tex]
From the above equation the maximum force is
[tex]F = \frac{1.4*10^{10} * (2.9*10^{-4}) * 3.233*10^{-5} }{ 0.35}[/tex]
[tex]F = 3750.28 \ N[/tex]
A 26-g rifle bullet traveling 220 m/s embeds itself in a 3.8-kg pendulum hanging on a 2.7-m-long string, which makes the pendulum swing upward in an arc, Determine the vertical and horizontal component of the pendulum's maximum displacement
Answer:
displacements are 0.776m, 0.114m
Explanation:
We were given mass of 26-g rifle bullet , then we can convert to Kg since
Momentum is conserved here.
The initial momentum before impact = (Mi * Vi)
Where Mi= initial given mass
Vi=initial velocity given
= 0.026 * 220 = 5.72 kgm/s
The final momentum after impact is (Mf * Vf )
Mf= final mass
5.72=( 3.82* Vf )
= 5.72/ 3.82
= 1.497 m/s
the speed of the pendulum bob with bullet afterwards= 1.497 m/s
the total energy after the collision is the addition of the kinetic energy of the bob+bullet and the potential energy of the bob and bullet, potential energy can be taken as zero.
M = 3.82 kg the mass of the bob containing the bullet
E(total) = ¹/₂MV² = 1/2 * (3.82kg)*(1.497m/s)² = 4.280J
When the Bob got to highest point the kinetic energy is zero and the potential energy is due to the increase in height of the bob, and the addition of the potential and kinetic energies still equal the total energy from before
E(total) = Mgh + 0 = Mgh = 4.280J
solving for h and substituting,
h = 4.280 J/(9.8m/s^2*3.82kg) = 0.114 m
Since the height is found,we the angle of the pendulum at the top of the swing can also be determined
A = arccos[(2.7 - 0.114) / 2.7] or A = 16.71degrees
Since A is known, the displacement along the horizontal axis can be calculated as
x = 2.7* sin(A) = 0.776m
therefore, displacement is 0.776m, 0.114m
the vertical and horizontal component of the pendulum's maximum displacement are displacement is 0.776m, 0.114m
(b) State the unit of volume in S.I. system and define it. 4.
Answer:
the cubic meter (m3)
Explanation:
Volume is the measure of the 3-dimensional space occupied by matter, or enclosed by a surface, measured in cubic units. The SI unit of volume is the cubic meter (m3),
Answer:
please answer all component's
Explanation:
please answer all component's
A single force acts on a particle situated on the positive x axis. The torque about the origin is in the negative z direction. The force might be:_______.
A. in the positive y direction
B. in the negative y direction
C. in the positive x direction
D. in the negative x direction
A single force acts on a particle situated on the positive x axis. The torque about the origin is in the negative z direction. The force might be in the negative y direction. Thus, option B is correct.
To determine the force that could generate a torque in the negative z direction, we need to consider the right-hand rule for cross products. The torque vector, denoted by τ, is given by the cross product of the position vector, r, and the force vector, F:
[tex]τ = r × F[/tex]
In this case, the position vector, r, points along the positive x-axis. The negative z-direction torque indicates that the force vector must be perpendicular to both the position vector and the negative z-axis.
Using the right-hand rule, we can determine that the force vector must be in the negative y-direction, which is option B.
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¿Qué resistencia debe ser conectada en paralelo con una de 20 Ω para hacer una
resistencia combinada de 15 Ω?
Answer:
60 Ω
Explanation:
R(com) = 15 Ω
1/R(com) = 1/R1 + 1/R2 + 1/R3 ..... + 1/Rn
1/15 = 1/20 + 1/R2
1/R2 = 1/15 - 1/20
1/R2 = (4 - 3) / 60
1/R2 = 1/60
R2 = 60 Ω
así, la combinada de resistencia necesaria es 60 Ω