Answer:
Vf = 1.41 m/s
Explanation:
The velocity of the disc just before the friend catches it, can be found out by using 3rd equation of motion, as follows:
2gh = Vf² + Vi²
where,
g = - 9.8 m/s² (negative sign for upward motion)
h = height = 5 m
Vi = Initial Velocity = 10 m/s
Vf = Final Velocity = ?
Therefore,
2(-9.8 m/s²)(5 m) = Vf² - (10 m/s)²
Vf² = 100 m²/s² - 98 m²/s²
Vf = √(2 m²/s²)
Vf = 1.41 m/s
Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of 10.0 m/s. At the start of a play, Matthews runs downfield at 45° with respect to the 50-yard line and covers 8.0 m in 1 s. He then runs straight down the field at 90° with respect to the 50-yard line for 12 m, with an elapsed time of 1.2 s. (a) What is Matthews' final displacement from the start of the play? (b) What is his average velocity?
Answer:
a) D_ total = 18.54 m, b) v = 6.55 m / s
Explanation:
In this exercise we must find the displacement of the player.
a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components
sin 45 = y₁ / d
cos 45 = x₁ / d
y₁ = d sin 45
x₁ = d sin 45
y₁ = 8 sin 45 = 5,657 m
x₁ = 8 cos 45 = 5,657 m
The second offset is d₂ = 12m at 90 of the 50 yard
y₂ = 12 m
x₂ = 0
total displacement
y_total = y₁ + y₂
y_total = 5,657 + 12
y_total = 17,657 m
x_total = x₁ + x₂
x_total = 5,657 + 0
x_total = 5,657 m
D_total = 17.657 i^+ 5.657 j^ m
D_total = Ra (17.657 2 + 5.657 2)
D_ total = 18.54 m
b) the average speed is requested, which is the offset carried out in the time used
v = Δx /Δt
the distance traveled using the pythagorean theorem is
r = √ (d1² + d2²)
r = √ (8² + 12²)
r = 14.42 m
The time used for this shredding is
t = t1 + t2
t = 1 + 1.2
t = 2.2 s
let's calculate the average speed
v = 14.42 / 2.2
v = 6.55 m / s
How does sodium (Na) becomes an ion?
6. A plane due to fly from Montreal to Edmonton required refueling. Because the fuel gauge on the aircraft was not working, a mechanic used a dipstick to determine that 7682 L of fuel were left on the plane. The plane required 22,300 kg of fuel to make the trip. In order to determine the volume of fuel required during the refueling, the pilot asked for the density of the fuel so he could convert a volume of fuel to a mass of fuel. The mechanic provided a factor of 1.77. Assuming that this factor was in metric units (kg/L), the pilot calculated the volume to be added as 4916 L. This volume of fuel was added and the plane subsequently ran out of fuel, but landed safely by gliding into Gimli Airport near Winnipeg. The error arose because the factor 1.77 was in units of pounds per liter (lbs/L). How many liters of fuel should have been added
Answer:
The amount of liters of fuel should have been added is 20093 L
Explanation:
Given that;
a mechanic used a dipstick to determine that 7682 L of fuel were left on the plane
i.e the volume of fuel left in the plane = 7682 L
Required fuel to make a trip = 22,300 kg of fuel
Also from the question; we are being told that in order for the pilot to determine the volume ; he asked for the density of the fuel and the mechanic said 1.77.
This volume of fuel was added and the plane subsequently ran out of fuel, but landed safely by gliding into Gimli Airport near Winnipeg. The error arose because the factor 1.77 was in units of pounds per liter (lbs/L).
Now; we can understand that the density of the fuel was 1.77 pound /litre.
SO , let convert 1.77 pound /litre to kg/Litre;
we all know that
1 pound = 0.4536 kg
1.77 pound/litre = x kg
If we cross multiply ; we will have:
1.77 pound/litre × 0.4536 kg = 1 pound × x kg
x kg = (1.77 pound/litre × 0.4536 kg) /1 pound
x = 0.802872 kg/litre
[tex]\mathbf{Density = \dfrac{mass}{volume}}[/tex]
where ;
mass = 22,300 kg of fuel
volume = unknown ???
density = 0.802872 kg/litre
making volume the subject of the formula from above; we have:
[tex]\mathbf{volume = \dfrac{mass}{Density}}[/tex]
[tex]\mathbf{volume = 22300 \ kg \ of \ fuel *\dfrac{1 \ litre }{0.802872 \ kg \ of \ fuel}}[/tex]
volume = 27775.28672 litre
volume [tex]\approx[/tex] 27775 L
Let not forget that we are being told as well that the volume of fuel left in the plane = 7682 L
Now;
The amount of liters of fuel should have been added is: = 27775 L - 7682 L
The amount of liters of fuel should have been added is 20093 L
Professional baseball player Nolan Ryan could pitch a baseball at approximately 160.0 km/h. At that average velocity, how long (in s) did it take a ball thrown by Ryan to reach home plate, which is 18.4 m from the pitcher's mound
Answer:
t = 0.414s
Explanation:
In order to calculate the time that the ball takes to reach home plate, you assume that the speed of the ball is constant, and you use the following formula:
[tex]t=\frac{d}{v}[/tex] (1)
d: distance to the plate = 18.4m
v: speed of the ball = 160.0km/h
You first convert the units of the sped of the ball to appropriate units (m/s)
[tex]160.0\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=44.44\frac{m}{s}[/tex]
Then, you replace the values of the speed v and distance s in the equation (1):
[tex]t=\frac{18.4m}{44.44m/s}=0.414s[/tex]
THe ball takes 0.414s to reach the home plate
The Pauli exclusion principle states that Question 1 options: the wavelength of a photon of light times its frequency is equal to the speed of light. no two electrons in the same atom can have the same set of four quantum numbers. both the position of an electron and its momentum cannot be known simultaneously very accurately. the wavelength and mass of a subatomic particle are related by . an electron can have either particle character or wave character.
Answer:
no two electrons in the same atom can have the same set of four quantum numbers
Explanation:
Pauli 's Theory of Exclusion specifies that for all four of its quantum numbers, neither two electrons in the same atom can have similar value.
In a different way, we can say that no more than two electrons can take up the identical orbital, and two electrons must have adversely spin in the identical orbital
Therefore the second option is correct
A double slit illuminated with light of wavelength 588 nm forms a diffraction pattern on a screen 11.0 cm away. The slit separation is 2464 nm. What is the distance between the third and fourth bright fringes away from the central fringe
Answer:
[tex]y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}[/tex]
Explanation:
Given that
Wavelength [tex]\lambda=588 \mathrm{nm}[/tex]
slit separation [tex]\mathrm{d}=2464 \mathrm{nm}[/tex]
slit screen distance [tex]\mathrm{D}=11 \mathrm{cm}[/tex]
We know that for double slit the maxima condition is that
[tex]\operatorname{dsin} \theta=m \lambda[/tex]
[tex]\sin \theta=\frac{m \lambda}{d}[/tex]
[tex]\theta=\sin ^{-1}\left(\frac{\mathrm{m} \lambda}{\mathrm{d}}\right)[/tex]
For small angle approximation, [tex]\sin \theta \approx \tan \theta \approx \theta[/tex]
[tex]\tan \theta=\frac{y_{m}}{D}[/tex]
[tex]y_{m}=D \times \tan \left[\sin ^{-1}\left(\frac{m \lambda}{d}\right)\right][/tex]
Now [tex]y_{4}[/tex] [tex]y_{4}=D \times \tan \left[\sin ^{-1}\left(\frac{4 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{4 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=35.22 \mathrm{cm}[/tex]
Again [tex]y_{3}=D \times \tan \left[\sin ^{-1}\left(\frac{3 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{3 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=11.27 \mathrm{cm}[/tex]
Hence [tex]y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}[/tex]
A 2.0-kg object moving at 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost by the system as a result of this collision.
Answer:
20 J
Explanation:
From the question, since there is a lost in kinetic energy, Then the collision is an inelastic collision.
m'u'+mu = V(m+m')........... Equation 1
Where m' = mass of the moving object, m = mass of the stick, u' = initial velocity of the moving object, initial velocity of the stick, V = common velocity after collision.
make V the subject of the equation above
V = (m'u'+mu)/(m+m')............. Equation 2
Given: m' = 2 kg, m = 8 kg, u' = 5 m/s, u = 0 m/s (at rest).
Substitute into equation 2
V = [(2×5)+(8×0)]/(2+8)
V = 10/10
V = 1 m/s.
Lost in kinetic energy = Total kinetic energy before collision- total kinetic energy after collision
Total kinetic energy before collision = 1/2(2)(5²) = 25 J
Total kinetic energy after collision = 1/2(2)(1²) +1/2(8)(1²) = 1+4 = 5 J
Lost in kinetic energy = 25-5 = 20 J
The collision is inelastic collision. As a result of collision the kinetic energy lost by the given system is 20 J.
Since there is a lost in kinetic energy, the collision is inelastic collision.
m'u'+mu = V(m+m')
[tex]\bold {V =\dfrac { (m'u'+mu)}{(m+m')} }[/tex]
Where
m' = mass of the moving object = 2 kg
m = mass of the stick = 8 kg,
u' = initial velocity of the moving object = 5 m/s
V = common velocity after collision= ?
u = 0 m/s (at rest).
put the values in the formula,
[tex]\bold {V = \dfrac {(2\times 5)+(8\times 0)}{(2+8)}}\\\\\bold {V = \dfrac {10}{10}}\\\\\bold {V = 1\\ m/s.}[/tex]
kinetic energy before collision
[tex]\bold { = \dfrac 1{2} (2)(5^2) = 25 J}[/tex]
kinetic energy after collision
[tex]\bold { = \dfrac 12(2)(1^2) + \dfrac 12(8)(1^2) = 5\ J}[/tex]
Lost in kinetic energy = 25-5 = 20 J
Therefore, As a result of collision the kinetic energy lost by the given system is 20 J.
To know more about Kinetic energy,
https://brainly.com/question/12669551
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm .What is the difference in maximum stored energy between the sprinters and the nonathlethes?
Complete Question:
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, [tex]31 {\rm {N}/{mm}}[/tex]. What is the difference in maximum stored energy between the sprinters and the nonathlethes?
Answer:
[tex]\triangle E = 12.79 J[/tex]
Explanation:
Sprinters' tendons stretch, [tex]x_s = 43 mm = 0.043 m[/tex]
Non athletes' stretch, [tex]x_n = 32 mm = 0.032 m[/tex]
Spring constant for the two groups, k = 31 N/mm = 3100 N/m
Maximum Energy stored in the sprinter, [tex]E_s = 0.5kx_s^2[/tex]
Maximum energy stored in the non athletes, [tex]E_m = 0.5kx_n^2[/tex]
Difference in maximum stored energy between the sprinters and the non-athlethes:
[tex]\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J[/tex]
If vector C is added to vector D, the result is a third vector that is perpendicular to D and has a magnitude equal to 3D. What is the ratio of the magnitude of C to that of D?
a) 1.3
b) 1.6
c) 1.8
d) 2.2
e) 3.2
Answer:
(e) 3.2
Explanation:
We are given that vector C and D.
Let R be the magnitude of C+D.
According to question
R=3D
We have to find the ratio of the magnitude of C to that of D.
By using right triangle property
[tex]C^2=R^2+D^2[/tex]
[tex]C^2=(3D)^2+D^2[/tex]
[tex]C^2=9D^2+D^2[/tex]
[tex]C^2=10D^2[/tex]
[tex]C=\sqrt{10D^2}=3.2D[/tex]
[tex]\frac{C}{D}=3.2[/tex]
Hence, the ratio of the magnitude of C to that of D=3.2
(e) 3.2
according to newtons second law of motion, what is equal to the acceleration of an object
Answer: According to Newtons second Law of motion ;
F = ma (Force equals mass multiplied by acceleration.)
The acceleration is directly proportional to the net force; the net force equals mass times acceleration; the acceleration in the same direction as the net force; an acceleration is produced by a net force
Explanation:
The Thomson model of a hydrogen atom is a sphere of positive charge with an electron (a point charge) at its center. The total positive charge equals the electronic charge e. Prove that when the electron is at a distance r from the center of the sphere of positive charge, it is attracted with a force F=\frac{e^2r}{4\pi\varepsilon_oR^3} where R is the radius of the sphere.
Answer:
E = (1 / 4π ε₀ ) q r / R³
Explanation:
Thomson's stable model that the negative charge is mobile within the atom and the positive charge is uniformly distributed, to calculate the force we can use Coulomb's law
F = K q₁ q₂ / r²
we used law Gauss
Ф = ∫ E .dA = q_{int} /ε₀
E 4π r² = q_{int} /ε₀
E = q_{int} / 4π ε₀ r²
we replace the charge inside
E = (1 / 4π ε₀ r²) ρ 4/3 π r³
E = ρ r / 3 ε₀
the density for the entire atom is
ρ = Q / V
V = 4/3 π R³
we substitute
E = (r / 3ε₀ ) Q 3/4π R³
E = (1 / 4π ε₀ ) q r / R³
Differences between regular and irregular
Answer:
Differences between regular and irregular objects are:
Regular object Those substances which have fixed geometrical shape are called regular objects.For example: Books,pencils etc.Irregular objectsThose substances which do not have fixed geometrical shape are called irregular object.For example: A piece of stone, pieces of broken glass etc.Hope this helps...
Good luck on your assignment..
The smallest shift you can reliably measure on the screen is about 0.2 grid units. This shift corresponds to the precision of positions measured with the best Earth-based optical telescopes. If you cannot measure an angle smaller than this, what is the maximum distance at which a star can be located and still have a measurable parallax
Answer:
The distance is [tex]d = 1.5 *10^{15} \ km[/tex]
Explanation:
From the question we are told that
The smallest shift is [tex]d = 0.2 \ grid \ units[/tex]
Generally a grid unit is [tex]\frac{1}{10}[/tex] of an arcsec
This implies that 0.2 grid unit is [tex]k = \frac{0.2}{10} = 0.02 \ arc sec[/tex]
The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as
[tex]d = \frac{1}{k}[/tex]
substituting values
[tex]d = \frac{1}{0.02}[/tex]
[tex]d = 50 \ parsec[/tex]
Note [tex]1 \ parsec \ \to 3.26 \ light \ year \ \to 3.086*10^{13} \ km[/tex]
So [tex]d = 50 * 3.08 *10^{13}[/tex]
[tex]d = 1.5 *10^{15} \ km[/tex]
When you skid to a stop on your bike, you can significantly heat the small patch of tire that rubs against the road surface. Suppose a person skids to a stop by hitting the brake on his back tire, which supports half the 90 kg combined mass of the bike and rider, leaving a skid mark that is 48 cm long. Assume a coefficient of kinetic friction of 0.80. How much thermal energy is deposited in the tire and the road surface?
Answer:
E = 169.34 J
Explanation:
First, we need to find the frictional force between the back tire and the road. For that purpose, we use the following formula:
f = μR = μW
f = μmg
where,
f = frictional force = ?
μ = coefficient of friction between tire and road = 0.8
g = 9.8 m/s²
m = mass supported by back tire = (0.5)(90 kg) = 45 kg
Therefore,
f = (0.8)(45 kg)(9.8 m/s²)
f = 352.8 N
Now, for the heat energy we use the formula of work. Because, thermal energy will be equal to work done by frictional force:
E = W = fd
where,
E = Thermal Energy = ?
f = frictional force = 352.8 N
d = displacement = 48 cm = 0.48 m
Therefore,
E = (352.8 N)(0.48 m)
E = 169.34 J
Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are charged with equal amount of opposite charges, ±17 µC. The charges on the plates face each other. Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates when the normal to the circle makes an angle of 4° with a line perpendicular to the plates. Note that this angle can also be given as 180° + 4°. N · m2/C
Answer:
Φ = 361872 N.m^2 / C
Explanation:
Given:-
- The area of the two plates, [tex]A_p = 180 cm^2[/tex]
- The charge on each plate, [tex]q = 17 * 10^-^6 C[/tex]
- Permittivity of free space, [tex]e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}[/tex]
- The radius for the flux region, [tex]r = 3.3 cm[/tex]
- The angle between normal to region and perpendicular to plates, θ = 4°
Find:-
Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.
Solution:-
- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:
[tex]A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2[/tex]
- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):
σ = [tex]\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\[/tex]
σ = 0.00094 C / m^2
- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.
[tex]E+ = E- = \frac{sigma}{2*e_o} \\\\[/tex]
- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:
[tex]E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\[/tex]
- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:
Φ = E_net * Ar * cos ( θ )
Φ = (106214689.26553) * (0.00342) * cos ( 5 )
Φ = 361872 N.m^2 / C
Is the friction of the pendulum (catch mechanism, support axis, etc.) a random or systematic error? Will this source of error cause your calculated velocity to be less than or greater than the actual velocity?
Answer:
l these errors believe that the speed of the system is less than that calculated
Explanation:
When we carry out any measurement in addition to the magnitude, the sources of uncertainty must also be analyzed.
We can have random uncertainties, correspondin
g to momentary errors, for example early warps during medicine, parallax errors, errors in the starting and ending points of the movement; I mean every possible random error. This error is the one that is analyzed and calculated in the statistical equations
There is another source of error, the systematic ones, these are much more complicated, they can be an error in the pendulum length, friction in the pendulum movement mechanism, deformities in the support systems, this errors are not analyzed by the statistic, in general They discover by looking at the results and comparing with the tabulated or real ones.
tith the explanation we see that the errors described are systematic.
In general these errors believe that the speed of the system is less than that calculated
A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the initial force that must be applied to the
Answer:
F₂ = 925.92 N
Explanation:
In a hydraulic lift the normal stress applied to one arm must be equally transmitted to the other arm. Therefore,
σ₁ = σ₂
F₁/A₁ = F₂/A₂
F₂ = F₁ A₂/A₁
where,
F₂ = Initial force that must be applied to narrow arm = ?
F₁ = Load on Wider Arm to be raised = 12000 N
A₁ = Area of wider arm = πr₁² = π(18 cm)² = 324π cm²
A₂ = Area of narrow arm = πr₂² = π(5 cm)² = 25π cm²
Therefore,
F₂ = (12000 N)(25π cm²)/(324π cm²)
F₂ = 925.92 N
In a hydraulic lift, if the pressure exerted on the liquid by one piston is increased by 100 N/m2 , how much additional weight can the other piston slowly lift if its cross sectional area is 25 m2
Answer:
The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.
Explanation:
By means of the Pascal's Principle, the hydraulic lift can be modelled by the following two equations:
Hydraulic Lift - Before change
[tex]P = \frac{F}{A}[/tex]
Hydraulic Lift - After change
[tex]P + \Delta P = \frac{F + \Delta F}{A}[/tex]
Where:
[tex]P[/tex] - Hydrostatic pressure, measured in pascals.
[tex]\Delta P[/tex] - Change in hydrostatic pressure, measured in pascals.
[tex]A[/tex] - Cross sectional area of the hydraulic lift, measured in square meters.
[tex]F[/tex] - Hydrostatic force, measured in newtons.
[tex]\Delta F[/tex] - Change in hydrostatic force, measured in newtons.
The additional weight is obtained after some algebraic handling and the replacing of all inputs:
[tex]\frac{F}{A} + \Delta P = \frac{F}{A} + \frac{\Delta F}{A}[/tex]
[tex]\Delta P = \frac{\Delta F}{A}[/tex]
[tex]\Delta F = A\cdot \Delta P[/tex]
Given that [tex]\Delta P = 100\,Pa[/tex] and [tex]A = 25\,m^{2}[/tex], the additional weight is:
[tex]\Delta F = (25\,m^{2})\cdot (100\,Pa)[/tex]
[tex]\Delta F = 2500\,N[/tex]
The additional mass needed for the additional weight is:
[tex]\Delta m = \frac{\Delta F}{g}[/tex]
Where:
[tex]\Delta F[/tex] - Additional weight, measured in newtons.
[tex]\Delta m[/tex] - Additional mass, measured in kilograms.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
If [tex]\Delta F = 2500\,N[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then:
[tex]\Delta m = \frac{2500\,N}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]\Delta m = 254.92\,kg[/tex]
The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.
A rocket car on a horizontal rail has an initial mass of 2500 kg and an additional fuel mass of 1000 kg. At time t0 the rocket motor is ignited and the rocket burns fuel at a rate of 95 kg/s. The exit speed of the exhaust gas relative to the rocket is 2900 m/s. Neglecting drag and friction forces, determine the acceleration and the velocity of the car at time t = 10 s. Plot the acceleration and velocity from time t0 to t = 10 s.
Answer: Acceleration of the car at time = 10 sec is 108 [tex]m/s^{2}[/tex] and velocity of the car at time t = 10 sec is 918.34 m/s.
Explanation:
The expression used will be as follows.
[tex]M\frac{dv}{dt} = u\frac{dM}{dt}[/tex]
[tex]\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt[/tex]
= [tex]u\int_{M_{o}}^{M_{f}} \frac{dM}{M}[/tex]
[tex]v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}[/tex]
[tex]v_{o} = 0[/tex]
As, [tex]v_{f} = u ln (\frac{M_{f}}{M_{o}})[/tex]
u = -2900 m/s
[tex]M_{f} = M_{o} - m \times t_{f}[/tex]
= [tex]2500 kg + 1000 kg - 95 kg \times t_{f}s[/tex]
= [tex](3500 - 95t_{f})s[/tex]
[tex]v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s[/tex]
Also, we know that
a = [tex]\frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}[/tex]
= [tex]\frac{u}{3500 - 95 t} \times (-95) m/s^{2}[/tex]
= [tex]\frac{95 \times 2900}{3500 - 95t} m/s^{2}[/tex]
At t = 10 sec,
[tex]v_{f}[/tex] = 918.34 m/s
and, a = 108 [tex]m/s^{2}[/tex]
7. A ball of mass m makes a head-on elastic collision with a second ball (at rest) and rebounds with a speed equal to 0.450 its original speed. What is the mass of the second ball
Answer:
mass of the second ball is 0.379m
Explanation:
Given;
mass of first ball = m
let initial velocity of first ball = u₁
let final velocity of first ball = v₁ = 0.45u₁
let the mass of the second ball = m₂
initial velocity of the second ball, u₂ = 0
let the final velocity of the second ball = v₂
Apply the principle of conservation of linear momentum;
mu₁ + m₂u₂ = mv₁ + m₂v₂
mu₁ + 0 = 0.45u₁m + m₂v₂
mu₁ = 0.45u₁m + m₂v₂ -------- equation (i)
Velocity for elastic collision in one dimension;
u₁ + v₁ = u₂ + v₂
u₁ + 0.45u₁ = 0 + v₂
1.45u₁ = v₂ (final velocity of the second ball)
Substitute in v₂ into equation (i)
mu₁ = 0.45u₁m + m₂(1.45u₁)
mu₁ = 0.45u₁m + 1.45m₂u₁
mu₁ - 0.45u₁m = 1.45m₂u₁
0.55mu₁ = 1.45m₂u₁
divide both sides by u₁
0.55m = 1.45m₂
m₂ = 0.55m / 1.45
m₂ = 0.379m
Therefore, mass of the second ball is 0.379m (where m is mass of the first ball)
The mass of the second ball is 0.379m and this can be determined by conserving the momentum.
Given :
A ball of mass 'm' makes a head-on elastic collision with a second ball (at rest) and rebounds with a speed equal to 0.450 its original speed.
In order to determine the mass of the second ball, apply conservation of linear momentum.
[tex]\rm m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
where [tex]m_1[/tex] is the mass of the first ball, [tex]m_2[/tex] is the mass of the second ball, [tex]u_1[/tex] is the initial velocity of the first ball, [tex]u_2[/tex] is the initial velocity of the second ball, [tex]\rm v_1[/tex] is the final velocity of the first ball, and [tex]\rm v_2[/tex] is the final velocity of the second ball.
Now, substitute the known terms in the above formula.
[tex]\rm mu_1+0=0.45u_1m+m_2v_2[/tex]
[tex]\rm mu_1=0.45u_1m+m_2v_2[/tex] ---- (1)
For elastic collision, the velocity is given by:
[tex]\rm v_1+u_1=v_2+u_2[/tex]
[tex]\rm 0.45u_1+u_1=0+v_2[/tex]
[tex]\rm v_2 = 1.45u_1[/tex]
Now, substitute the value of [tex]\rm v_2[/tex] in the equation (1).
[tex]\rm mu_1=0.45u_1m+1.45u_1m_2[/tex]
[tex]\rm 0.55mu_1=1.45m_2u_1[/tex]
[tex]\rm m_2=0.379m[/tex]
So, the mass of the second ball is 0.379m.
For more information, refer to the link given below:
https://brainly.com/question/19689434
A 26 kg child is coasting at 2.0 m/s over flat ground in a 5.0 kg wagon. The child drops a 1.5 kg ball from the side of the wagon. What is the final speed (in m/s) of the child and wagon
Answer:
Final speed = 2.067 m/s
Explanation:
We are told that the child weighs 26 kg.
Also, that the wagon weighs 5kg.
Thus,initial mass of the child and wagon with ball is;
m_i = 26 + 5 = 31 kg.
Also, we are told that the child now dropped 1.5 kg ball from the wagon. So,
Final mass is;
m_f = 26 + 5 - 1 = 30 kg
Now, from conservation of linear momentum, we know that;
Initial momentum = final momentum
Thus;
m_i * v_i = m_f * v_f
Where v_i is initial velocity and v_f is final velocity.
Making v_f the subject, we have;
v_f = (m_i * v_i)/m_f
We are given that initial velocity v_i = 2 m/s
Plugging in the relevant values, we have;
v_f = (31 * 2)/30
v_f = 2.067 m/s
A person is standing on an elevator initially at rest at the first floor of a high building. The elevator then begins to ascend to the sixth floor, which is a known distance h above the starting point. The elevator undergoes an unknown constant acceleration of magnitude a for a given time interval T. Then the elevator moves at a constant velocity for a time interval 4T. Finally the elevator brakes with an acceleration of magnitude a, (the same magnitude as the initial acceleration), for a time interval T until stopping at the sixth floor.
Answer:
The found acceleration in terms of h and t is:
[tex]a=\frac{h}{5(t_1)^2}[/tex]
Explanation:
(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)
We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.
Stage 1Constant acceleration, starts from rest.
Distance = [tex]y = \frac{1}{2}a(t_1)^2[/tex]
Velocity = [tex]v_1=at_1[/tex]
Stage 2Constant velocity where
Velocity = [tex]v_o=v_1=at_1[/tex]
Distance =
[tex]y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\[/tex]Stage 3Constant deceleration where
Velocity = [tex]v_0=v_1=at_1[/tex]
Distance =
[tex]y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2[/tex]
Total HeightTotal height = y₁ + y₂ + y₃
Total height = [tex]\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2[/tex]
AccelerationFind acceleration by rearranging the found equation of total height.
Total Height = h
h = 5a(t₁)²
[tex]a=\frac{h}{5(t_1)^2}[/tex]
4. How much force is required to stop a 60 kg person traveling at 30 m/s during a time of a)
5.0 seconds
b) 0.50 seconds
c) 0.05 seconds
Explanation:
F = ma, and a = Δv / Δt.
F = m Δv / Δt
Given: m = 60 kg and Δv = -30 m/s.
a) Δt = 5.0 s
F = (60 kg) (-30 m/s) / (5.0 s)
F = -360 N
b) Δt = 0.50 s
F = (60 kg) (-30 m/s) / (0.50 s)
F = -3600 N
c) Δt = 0.05 s
F = (60 kg) (-30 m/s) / (0.05 s)
F = -36000 N
360N, 3600N and 36000N forces are required to stop a 60 kg person traveling at 30 m/s during a time of a)5.0 seconds, b) 0.50 seconds, c)0.05 seconds respectively.
To find the force, we need to know about the mathematical formulation of force.
What is force?According to Newton's second law of motion, force is defined as mass times acceleration. Its SI unit is Newton (N).What is the mathematical formulation of force?Mathematically, it is written as
F= m×a= m×(∆V/∆t)
What is the force needed to stop 60 kg person traveling at 30 m/s during a time of a)5.0 seconds, b) 0.50 seconds, c)0.05 seconds?Here, initially the velocity of the person is 30m/s. But after applying the force, he came to rest. So his final velocity is 0 m/s. ∆V= 30m/s
When ∆t=5 seconds, F= 60×(30/5)=360N
When ∆t=0.5 seconds, F= 60×(30/0.5)=3600N
When ∆t=0.05 seconds, F= 60×(30/0.05)=36000N
Thus, we can conclude that 360N, 3600N and 36000N forces are required to stop a 60 kg person traveling at 30 m/s during a time of a)5.0 seconds, b) 0.50 seconds, c)0.05 seconds respectively.
Learn more about force here:
brainly.com/question/12785175
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A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.0 s later. You may ignore air resistance. If the height of the building is 20 m, what must the initial speed be of the first ball if both balls are to hit the ground at the same time
Answer:
4.9 m/s
Hope this helps! ;)
Explanation:
The probability of nuclear fusion is greatly enhanced when the appropriate nuclei are brought close together, but their mutual coulomb repulsion must be overcome. This can be done using the kinetic energy of high temperature gas ions or by accelerating the nuclei toward one another.
Required:
a. Calculate the potential energy of two singly charged nuclei separated by 1.00×10^−12m
b. At what temperature will atoms of a gas have an average kinetic energy equal to this needed electrical potential energy?
Answer:
a
[tex]PE = 2.3 *10^{-16} \ J[/tex]
b
[tex]T = 1.1 *10^{7} \ K[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 1.00 *10^{-12} \ m[/tex]
Generally the electric potential energy can be mathematically represented as
[tex]PE = \frac{k * q_1 q_2 }{d}[/tex]
Given that in a nuclei the only charged particle is the proton who charge is
[tex]p = 1.60 *10^{-19} \ C[/tex]
Hence
[tex]q_1 = q_2 = 1.60 *10 ^{-19} \ C[/tex]
And k is the coulomb constant with values [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.N/A2[/tex]
So we have that
[tex]PE = \frac{9*10^9 * (1.60 *10^{-19})^2}{ 1.00*10^{-12}}[/tex]
[tex]PE = 2.3 *10^{-16} \ J[/tex]
The relationship between the electrical potential energy and the temperature is mathematically represented as
[tex]PE = \frac{3}{2} kT[/tex]
Here k is the Boltzmann's constant with value [tex]k = 1.38*10^{-23} JK^{-1}[/tex]
making T the subject
[tex]T = \frac{2}{3} * \frac{PE}{k}[/tex]
substituting values
[tex]T = \frac{2}{3} * \frac{2.30 *10^{-16}}{ 1.38 *10^{-23}}[/tex]
[tex]T = 1.1 *10^{7} \ K[/tex]
The largest graduated cylinder in my lab holds 2 L and has an inner diamter (the part that holds the water) of 8 cm. When it is full of 2 L of water, what is the pressure at the bottom of the cylinder in Pa?
Answer:
3924 Pa
Explanation:
Volume of cylinder = 2 L = 0.002 m^3 (1000 L = 1 m^3)
diameter of the inner cylinder = 8 cm = 0.08 m (100 cm = 1 m)
radius of the inner cylinder = diameter/2 = 0.08/2 = 0.04 m
area of the inner cylinder = [tex]\pi r^{2}[/tex]
where [tex]\pi[/tex] = 3.142,
and r = radius = 0.04 m
area of inner cylinder = 3.142 x [tex]0.04^{2}[/tex] = 0.005 m^2
height h of the water in this cylinder = volume/area
h = 0.002/0.005 = 0.4 m
pressure at the bottom of the cylinder due to the height of water = pgh
where
p = density of water = 1000 kg/m^3
g = acceleration due to gravity = 9.81 m/s^2
h = height of water within this cylinder = 0.4 m
pressure = 1000 x 9.81 x 0.4 = 3924 Pa
In the slap shot video, and the professor said that the puck experienced "an average force of 100 pounds" which we now know is 445 Newtons of average force. The duration of this collision is .02 seconds. Calculate the Impulse in this puck/stick collision. The mass of the puck is .175 kg.
Answer:
I = 89 N s
Explanation:
Momentum is a concept that tells us how much the amount of movement of a system changes, it is described by the expression
I = ∫ F dt = F_average t
where F is the force and t is the time
let's calculate
I = 445 0.2
I = 89 N s
A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 43.0 vibrations in 33.0 s. Also, a given maximum travels 424 cm along the rope in 15.0 s. What is the wavelength
Answer:
0.218
Explanation:
Given that
Total vibrations completed by the wave is 43 vibrations
Time taken to complete the vibrations is 33 seconds
Length of the wave is 424 cm = 4.24 m
to solve this problem, we first find the frequency.
Frequency, F = 43 / 33 hz
Frequency, F = 1.3 hz
Also, we find the wave velocity. Which is gotten using the relation,
Wave velocity = 4.24 / 15
Wave velocity = 0.283 m/s
Now, to get our answer, we use the formula.
Frequency * Wavelength = Wave Velocity
Wavelength = Wave Velocity / Frequency
Wavelength = 0.283 / 1.3
Wavelength = 0.218
A mercury barometer reads 745.0 mm on the roof of a building and 760.0 mm on the ground. Assuming a constant value of 1.29 kg/m3 for the density of air, determine the height of the building
Answer:
The height of the building is 158.140 meters.
Explanation:
A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:
[tex]\Delta P \propto \rho \cdot \Delta h[/tex]
[tex]\Delta P = k \cdot \rho \cdot \Delta h[/tex]
Where:
[tex]\Delta P[/tex] - Manometric pressure difference, measured in kilopascals.
[tex]\rho[/tex] - Fluid density, measured in kilograms per cubic meter.
[tex]\Delta h[/tex] - Height difference, measured in meters.
Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:
[tex]\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}[/tex]
[tex]\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}[/tex]
Where:
[tex]\Delta h_{air}[/tex] - Height difference of the air column, measured in meters.
[tex]\Delta h_{Hg}[/tex] - Height difference of the mercury column, measured in meters.
[tex]\rho_{air}[/tex] - Density of air, measured in kilograms per cubic meter.
[tex]\rho_{Hg}[/tex] - Density of mercury, measured in kilograms per cubic meter.
If [tex]\Delta h_{Hg} = 0.015\,m[/tex], [tex]\rho_{air} = 1.29\,\frac{kg}{m^{3}}[/tex] and [tex]\rho_{Hg} = 13600\,\frac{kg}{m^{3}}[/tex], the height difference of the air column is:
[tex]\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)[/tex]
[tex]\Delta h_{air} = 158.140\,m[/tex]
The height of the building is 158.140 meters.
158.13m
Explanation:
Force exerted over a unit area is called Pressure. Also, in a given column of air, the pressure(P) is given as the product of the density(ρ) of the air, the height(h) of the column of air and the acceleration due to gravity(g). i.e
P = ρhg
Let;
Pressure measured at the roof top = ([tex]P_{R}[/tex])
Pressure measured at the ground level = ([tex]P_{G}[/tex])
Pressure at the ground level = Pressure at the roof + Pressure at the column height of air.
[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P ---------------(i)
(a) P = ρhg -----------(***)
But;
ρ = density of air = 1.29kg/m³
h = height of column of air = height of building
g = acceleration due to gravity = 10m/s²
Substitute these values into equation (***)
P = 1.29 x h x 10
P = 12.9h Pa
(b) [tex]P_{G}[/tex] = ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{ground} }[/tex] x g ------------(*)
But;
ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³
h[tex]_{(mercury)_{ground} }[/tex] = height of mercury on the ground = 760.0mm = 0.76m
g = acceleration due to gravity = 10m/s²
Substitute these values into equation (*)
[tex]P_{G}[/tex] = 13600 x 0.76 x 10
[tex]P_{G}[/tex] = 103360 Pa
(c) [tex]P_{R}[/tex] = ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{roof} }[/tex] x g --------------(**)
But;
ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³
h[tex]_{(mercury)_{roof} }[/tex] = height of mercury on the roof = 745.0mm = 0.745m
g = acceleration due to gravity = 10m/s²
Substitute these values into equation (**)
[tex]P_{R}[/tex] = 13600 x 0.745 x 10
[tex]P_{R}[/tex] = 101320 Pa
(d) Now that we know the values of P, [tex]P_{G}[/tex] and [tex]P_{R}[/tex] , let's substitute them into equation (i) as follows;
[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P
103360 = 101320 + 12.9h
Solve for h;
12.9h = 103360 - 101320
12.9h = 2040
h = [tex]\frac{2040}{12.9}[/tex]
h = 158.13m
Therefore, the height of the building is 158.13m
A uniformly charged sphere has a potential on its surface of 450 V. At a radial distance of 8.1 m from this surface, the potential is 150 V. What is the radius of the sphere
Answer:
The radius of the sphere is 4.05 m
Explanation:
Given;
potential at surface, [tex]V_s[/tex] = 450 V
potential at radial distance, [tex]V_r[/tex] = 150
radial distance, l = 8.1 m
Apply Coulomb's law of electrostatic force;
[tex]V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}[/tex]
[tex]450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r} \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m[/tex]
Therefore, the radius of the sphere is 4.05 m