Answer:
The coefficient of friction between the cars and the road is 0.859.
Explanation:
The two cars collide each other inelastically, then we can determine the resulting velocity by the Principle of Momentum Conservation:
[tex]m_{A}\cdot v_{A} + m_{B}\cdot v_{B} = (m_{A} + m_{B})\cdot v[/tex] (1)
Where:
[tex]m_{A}[/tex], [tex]m_{B}[/tex] - Masses of the cars, in kilograms.
[tex]v_{A}[/tex], [tex]v_{B}[/tex] - Initial velocities of the cars, in meters per second.
[tex]v[/tex] - Velocity of the resulting system, in meters per second.
If we know that [tex]m_{A} = 2120\,kg[/tex], [tex]v_{A} = 13.4\,\frac{m}{s }[/tex], [tex]m_{B} = 2810\,kg[/tex] and [tex]v_{B} = 0\,\frac{m}{s}[/tex], then the velocity of the resulting system:
[tex]v = \frac{m_{A}\cdot v_{A}+m_{B}\cdot v_{B}}{m_{A}+m_{B}}[/tex]
[tex]v = \frac{(2120\,kg)\cdot \left(13.4\,\frac{m}{s} \right)+(2810\,kg)\cdot \left(0\,\frac{m}{s} \right)}{2120\,kg + 2810\,kg}[/tex]
[tex]v = 5.762\,\frac{m}{s}[/tex]
By Principle of Energy Conservation and Work-Energy Theorem, we understand that the initial translational kinetic energy ([tex]K[/tex]), in joules, is dissipated due to work done by friction ([tex]W_{f}[/tex]), in joules, that is to say:
[tex]K = W_{f}[/tex] (2)
[tex]\frac{1}{2}\cdot (m_{A}+m_{B})\cdot v^{2} = \mu\cdot (m_{A}+m_{B})\cdot g \cdot s[/tex]
[tex]\frac{1}{2}\cdot v^{2} = \mu \cdot g\cdot s[/tex] (2b)
Where:
[tex]\mu[/tex] - Coefficient of friction, no unit.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]s[/tex]- Travelled distance, in meters.
If we know that [tex]v = 5.762\,\frac{m}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]s = 1.97\,m[/tex], then the coefficient of friction is:
[tex]\mu = \frac{v^{2}}{2\cdot g\cdot s}[/tex]
[tex]\mu = \frac{\left(5.762\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.97\,m)}[/tex]
[tex]\mu = 0.859[/tex]
The coefficient of friction between the cars and the road is 0.859.
A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff?
Answer:
144 meters
Explanation:
it takes 6 seconds to hit the ground right and the ball lays off 24 m per second .
so by the time the ball hits the ground 6 seconds passed. so that means the cliff is 6.0×24=144
You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of 1.8 m/s as they reach the gate. For safety, the skiers should have a speed of no more than 28.0 m/s when they reach the bottom of the ramp. You determine that for a 75kg skier with good form, friction and air resistance will do total work of magnitude 3500 J on him during his run down the slope. What is the maximum height (h) for which the maximum safe speed will not be exceeded?
Answer:
44.6 m
Explanation:
From the law of conservation of energy, the total energy at the top of the ramp, E equals the total energy at the bottom of the ramp.
E = E'
U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at top of ramp = mgh where = height of ramp, K₁ = kinetic energy at top of ramp = 1/2mv₁² where v₁ = speed at top of ramp = 1.8 m/s, W₁ = work done by friction and air resistance at top of ramp = 0 J, U₂ = potential energy at bottom of ramp = 0 J(since the skier is at ground level h = 0), K₂ = kinetic energy at bottom of ramp = 1/2mv₂² where v₂ = speed at bottom of ramp = 28.0 m/s, W₁ = work done by friction and air resistance at bottom of ramp = 3500 J
Substituting the values of the variables into the equation, we have
U₁ + K₁ + W₁ = U₂ + K₂ + W₂
mgh + 1/2mv₁² + W₁ = U₂ + 1/2mv₂² + W₂
mgh + 1/2m(1.8 m/s)² + 0 J = 0 J + 1/2m(28 m/s)² + 3500 J
9.8 m/s² × 75 kg h + 1/2 × 75 kg (3.24 m²/s²) + 0 J = 0 J + 1/2 × 75 kg (784 m²/s²) + 3500 J
(735 kgm/s²)h + 75 kg(1.62 m²/s²) = 75 kg(392m²/s²) + 3500 J
(735 kgm/s²)h + 121.5 kgm²/s² = 29400 kgm²/s² + 3500 J
(735 kgm/s²)h + 121.5 J = 29400 J + 3500 J
(735 kgm/s²)h + 121.5 J = 32900 J
(735 kgm/s²)h = 32900 J - 121.5 J
(735 kgm/s²)h = 32778.5 J
h = 32778.5 J/735 kgm/s²
h = 44.6 m
So, the maximum height of the ramp for which the maximum safe speed will not be exceeded is 44.6 m.
you are given a set of facts regarding a lens : object heigh, and dostance to objects. Given this jnformation, how can you tell if you're dealing with a concave or convex lens
Answer:
concave curves inward like an hourglass and convex is an outward curve like a football
Explanation:
hope this helps
What is the relationship between organ systems and organs? organs are made from one type of organ system organ systems are made from one type of organ organs are made from different types of organ systems organ systems are made from different types of organs
A constant force moves an object along the line segment from to . Find the work done if the distance is measured in meters and the force in newtons.
This question is incomplete, the complete question is;
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A constant force F = 6i+8j-6k moves an object along a straight line from point (6, 0, -10) to point (-6, 7, 2).
Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.
Answer:
the work done is -88 J
Explanation:
Given the data in the question;
we know that;
Work done = F × S
where constant force F = ( 6i + 8j - 6k )
S = ( -6i + 7j + 2k ) - ( 6i + 0j - 10k )
S = ( (-6i - 6i) + (7j - 0j) + ( 2k - ( -10k) ) )
S = ( -12I + 7j + 12k )
so
Work force = ( 6i + 8j - 6k ) × ( -12I + 7j + 12k )
Work force = ( 6 × -12 ) + ( 8 × 7 ) + ( -6 × 12 )
Work force = -72 + 56 - 72
Work force = -88 J
Therefore, the work done is -88 J
g Calculate the final speed of a solid cylinder that rolls down a 5.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm.
Answer:
[tex]V=8.08m/s[/tex]
Explanation:
From the question we are told that:
Height[tex]h=5.00m[/tex]
Mass [tex]m=0.750kg[/tex]
Radius [tex]r=4.00cm=>0.04m[/tex]
Generally the equation for Total energy is mathematically given by
[tex]mgh=\frac{1}{2}mv^2+\frac{1}{2}Iw^2[/tex]
Therefore
[tex]V=\sqrt{\frac{4gh}{3}}[/tex]
[tex]V=\sqrt{\frac{4*9.8*5}{3}}[/tex]
[tex]V=8.08m/s[/tex]
what is the frequency of a wave related to
Answer:
Frequency is the number of complete oscillations or cycles or revolutions made in one second.
assuming a filament in a 120W light bulb acts like a prefect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of 6.4×10^_5m^2. The stefan- boltzmann constant is 5.67×10^-8W/(m2.k2) A. 12OOk B. 2400K C. 2100K
Answer:
T = 2398 K
Explanation:
To calculate the emission of the light bulb we use the law is Stefan
P = σ A e T⁴
as they indicate that the filament is a black body, the emissivity is equal to 1 (e = 1)
T = [tex]\sqrt[4]{\frac{P}{ \sigma A} }[/tex]
let's calculate
T =[tex]\sqrt[4]{\frac{120}{5.67 \ 10^{-8} \ 6.4 \ 10^{-5}} }[/tex]
T = [tex]\sqrt[4]{33.06878 \ 10^{12} }[/tex]
T = 2,398 10³ K
T = 2398 K
Help!!
A table is pushed across the floor for a distance of 32 m with a force of 320 N in 150 seconds. How much power was used?
A.70.2W
B.68.3W
C.56.7W
D.49.8W
Compute the work done on the table:
W = Fd = (320 N) (32 m) = 10,240 J
Divide this by the given time duration to get the power output:
P = W/∆t = (10,240 J) / (150 s) ≈ 63.3 W
Which of the following elements has the largest atomic radius?
Silicon
Aluminum
Sulfur
Phosphorous
Answer:
francium
Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.
A disk of charge is placed in the x-y plane, centered at the origin. The electric field along the axis of a positive disk of charge... points towards the disk along the z-axis. points away from the disk along the z-axis. always points in the positive z-direction. none of these choices
Answer:
Points away from the disk along the z-axis.
Explanation:
Along the axis of the disk, which is the z - axis, the total vertical electric field components of the charged disk sum up while the horizontal components cancel out. Thus, leaving only vertical components of electric field along the axis of the disk.
Since the disk is positively charged and electric field lines point away from a positive charge, the electric field along the axis of a positive disk of charge points away from the disk along the z-axis.
What is the friction force on a box that has a mass of 15kg as it slides across the floor. The coefficient of friction of the not very clean floor is 0.25
please explain everything including formula used
Answer:
36.75 N
Explanation:
Applying
F = mgμ................. Equation 1
Where F = Friction force on the box, m = mass of the box, g = acceleration due to gravity of the box, μ = coefficient of static friction
From the question,
Given: m = 15 kg, μ = 0.25
Constant: g = 9.8 m/s²
Substitute these values into equation 1
F = 15(9.8)(0.25)
F = 36.75 N
Hence the friction force on the box is 36.75 N
A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 9.41 m/s in 4.24 s. What is the magnitude of the linear impulse experienced by a 67.0 kg passenger in the car during this time
Answer:
the impulse experienced by the passenger is 630.47 kg
Explanation:
Given;
initial velocity of the car, u = 0
final velocity of the car, v = 9.41 m/s
time of motion of the car, t = 4.24 s
mass of the passenger in the car, m = 67 kg
The impulse experienced by the passenger is calculated as;
J = ΔP = mv - mu = m(v - u)
= 67(9.41 - 0)
= 67 x 9.41
= 630.47 kg
Therefore, the impulse experienced by the passenger is 630.47 kg
Problem
A charged particle is moving in the presence of uniform magnetic field. The mass of the particle
is m = 10−6 kg its charge is Q = 10−5 C and the magnetic field vector is B~ = (1T, 0, 0). At the
beginning the velocity vector of the particle is ~v0 = (12 m/s, 0, 5 m/s).
a.) How large will the x component of the velocity of the particle be in t = 2 s?
b.) Where will the particle be in t = 3.14 s?
c.) How large will the magnitude of the velocity be in t = 2.5 s?
Answer:
Answer is a I checked the work
A locomotive pulls 11 identical freight cars. The force between the locomotive and the first car is 150.0 kN, and the acceleration of the train is 2 m/s2. There is no friction to consider. 1) Find the force between the tenth and eleventh cars. (Express your answer to two significant figures.)
Answer:
The force between the 10 th car and the 11 th car is 13636.4 N.
Explanation:
Force, F = 150 kN
acceleration, a = 2 m/s^2
Let the mass of each car is m. \Total numbers of cars = 11
F = n m a
150000 = 11 x m x 2
m = 6818.18 kg
The force between the 10 th and 11 th car is
T = ma = 6818.18 x 2 = 13636.4 N
if p=2i+4j+3k and q=I+5j-2k,find P×q.
Answer:
[tex]p\times q=-23i+7j+6k[/tex]
Explanation:
We are given that
p=2i+4j+3k
q=i+5j-2k
We have to find pxq
We know that
[tex]p\times q=\begin{vmatrix} i&j &k\\ 2&4 & 3\\ 1& 5 & -2\end{vmatrix}[/tex]
[tex]p\times q=i(-8-15)-j(-4-3)+k(10-4)[/tex]
[tex]p\times q=-23i+7j+6k[/tex]
Hence,[tex]p\times q=-23i+7j+6k[/tex]
Your job is to lift 30 kgkg crates a vertical distance of 0.90 mm from the ground onto the bed of a truck. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Force and power. Part A How many crates would you have to load onto the truck in one minute for the average power output you use to lift the crates to equal 0.50 hphp
Answer:
The number of crates is 84580.
Explanation:
mass, m = 30 kg
height, h = 0.9 mm
Power, P = 0.5 hp = 0.5 x 746 W = 373 W
time, t = 1 minute = 60 s
Let the number of crates is n.
Power is given by the rate of doing work.
[tex]P = \frac{n m gh}{t}\\\373 =\frac{n\times 30\times9.8\times 0.9\times 10^{-3}}{60}\\\\n =84580[/tex]
A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a constant force of 2850 N and the coefficient of kinetic friction between the ground and the log is 0.45, determine the time for the log to reach a speed of 0.5 m/s
Answer:
0.1 s
Explanation:
The net force on the log is F - f = ma where F = force due to winch = 2850 N, f = kinetic frictional force = μmg where μ = coefficient of kinetic friction between log and ground = 0.45, m = mass of log = 300 kg and g = acceleration due to gravity = 9.8 m/s² and a = acceleration of log
So F - f = ma
F - μmg = ma
F/m - μg = a
So, substituting the values of the variables into the equation, we have
a = F/m - μg
a = 2850 N/300 kg - 0.45 × 9.8 m/s²
a = 9.5 m/s² - 4.41 m/s²
a = 5.09 m/s²
Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.
So, making t subject of the formula, we have
t = (v - u)/a
substituting the values of the variables into the equation, we have
t = (v - u)/a
t = (0.5 m/s - 0 m/s)/5.09 m/s²
t = 0.5 m/s ÷ 5.09 m/s²
t = 0.098 s
t ≅ 0.1 s
An empty 12,954 kg railroad car, traveling at a speed of 28 m/s strikes a partially filled 17,616 kg railroad car moving in the same direction at a speed of 5 m/s. What is the total momentum of the two railroad cars AFTER the collision?
Answer:
450792 kgm/s
Explanation:
by conservation of momentum,
total momentum AFTER collision = total momentum BEFORE collision
=mv+m'v'
=12954×28+17616×5
=450792 kgm/s
A 8.37*10^-5 F capacitor has 2.15*10^-4 C of charge on its plates. How much energy is stored on the capacitor
Answer:
2.76*10^-4
Explanation:
Trust me :)
Two thin conducting plates, each 56.0 cm on a side, are situated parallel to one another and 7.0 mm apart. If 10^-10 electrons are moved from one plate to the other, what is the electric field between the plates?
Answer:
[tex]E=576.5V/m[/tex]
Explanation:
From the question we are told that:
Length [tex]l=56.0cm=0.56m[/tex]
Distance apart [tex]d=7.0mm=0.007m[/tex]
Electron Transferred [tex]n=10^{-10}[/tex]
Therefore
Total Charge
Since Charge on each electron is
[tex]e=1.602*10^{-19}[/tex]
Therefore
[tex]T=1.602*10^{-19} *10^{10}[/tex]
[tex]T=1.602*10^{-9}[/tex]
Generally the equation for Charge density is mathematically given by
[tex]\rho=T/A[/tex]
Where
Area
[tex]A=0.56*0.56[/tex]
[tex]A=0.3136[/tex]
Therefore
[tex]\rho=1.602*10^{-9}/0.3136[/tex]
[tex]\rho=5.10*10^{-9}[/tex]
Generally the equation for Electric Field in the capacitor is mathematically given by
[tex]E=\frac{\rho}{e_0}[/tex]
[tex]E=\frac{5.10*10^{-9}}{8.85x10{-12}}[/tex]
[tex]E=576.5V/m[/tex]
A spring scale hung from the ceiling stretches by 6.1cm when a 2.0kg mass is hung from it. The 2.0kg mass is removed and replaced with a 2.8kg mass.What is the stretch of the spring?
differentiate between step up and step down transformer
Answer:
The main difference between the step-up and step-down transformer is that the step-up transformer increases the output voltage, while the step-down transformer reduces the output voltage.
If the loading is 0.4, the coinsurance rate is 0.2, the number of units of medical care is 100, and the number of units of medical care is 1. What is the premium of this insurance?
Answer:
72 is the premimum of the insurance.
Explanation:
Below is the given values:
The loading = 0.4
Coinsurance rate = 0.2
Number of units = 100
Total number of units = 100 * 0.4 = 40
Remaining units = 60 * 0.2 = 12
Add the 60 and 12 values = 60 + 12 = 72
Thus, 72 is the premimum of the insurance.
A swimmer heading directly across a river that is 200 m wide reaches the opposite bank in 6 min 40 s. During this swim, she is swept downstream 480 m. How fast can she swim in still water
Answer:
The speed of the swimmer in stil water is 0.5 m/s
Explanation:
Given;
total time taken to swim across = 6 mins 40 s = (6 x 60s) + 40 s = 400 s
width of the river, = 200 m
Please find the image attached for explanation.
You swing a bat and hit a heavy box with a force of 1273 N. The force the box exerts on the bat is Group of answer choices less than 1273 N if the box moves. exactly 1273 N whether or not the box moves. None of the above choices are correct. exactly 1273 N only if the box does not move. greater than 1273 N if the bat bounces back. greater than 1273 N if the box moves.
Answer:
exactly 1273 N whether or not the box moves.
Explanation:
In the case when the bat is swing and it is hitted to a heavy box having a force of 1273 N so here the force of the box that exert on the box should be accurately 1273 N even if the box is moved or not. As the third law of the newton should be equivalent & the opposite reaction
Therefore as per the given situation, the above represent the answer
Given that o.2i+bj+o.4k is a unit vector,what is the value of b?
Answer:
b = 0.89
Explanation:
The given vector is, [tex]A=0.2i+bj+0.4k[/tex]
A is a unit vector
We need to find the value of b.
For a unit vector, |A| = 1
So,
[tex]0.2^2+b^2+0.4^2=1\\\\0.04+b^2+0.16=1\\\\0.2+b^2=1\\\\b^2=1-0.2\\\\b=0.89[/tex]
So, th value of b is 0.89.
The cycle is a process that returns to its beginning, but it does not repeat
itself.
True
False
1. Convert the following length into meters
a. 123.50mm
b. 560cm
c. 100dm
d. 125.89km
A hoop rolls with constant velocity and without sliding along level ground. Its rotational kinetic energy is:______a- half its translational kinetic energyb- the same as its translational kinetic energyc- twice its translational kinetic energyd- four times its translational kinetic energy
Answer:
The same as its translational KE.
The easy way to do this is to make up numbers and use them.
So, I'll say m=2 and r=3. I will also say v=3 .
Rot. Inertia of a hoop is mr^2. So the rot KE is: 1/2 (mr^2)(w^2)
note: (1/2*I*w^2)
Translational kinetic energy is basically normal KE, so 1/2(m)(v^2)
Now, lets plug our made up values in:
Rot Ke : 1/2 (9*2)(3/3) *note w = v/r
Tran Ke: 1/2(2)(9)
Rot Ke: 9
Tran Ke: 9
9=9, same.
