Answer:
A)
the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B)
data that were not necessary to the solution are;
a) mass of truck and b) mass of load
Explanation:
Given that;
mass of load [tex]m_{LS}[/tex] = 10000 kg
mass of flat bed [tex]m_{FB}[/tex] = 20000 kg
initial speed of truck [tex]v_{0}[/tex] = 12 m/s
coefficient of friction between the load sits and flat bed μs = 0.5
A) the minimum stopping distance for which the load will not slide forward relative to the truck.
Now, using the expression
Fs,max = μs [tex]F_{N}[/tex] -------------let this be equation 1
where [tex]F_{N}[/tex] = normal force = mg
so
Fs,max = μs mg
ma[tex]_{max}[/tex] = μs mg
divide through by mass
a[tex]_{max}[/tex] = μs g ---------- let this be equation 2
in equation 2, we substitute in our values
a[tex]_{max}[/tex] = 0.5 × 9.8 m/s²
a[tex]_{max}[/tex] = 4.9 m/s²
now, from the third equation of motion
v² = u² + 2as
[tex]v_{f}[/tex]² = [tex]v_{0}[/tex]² + 2aΔx
where [tex]v_{f}[/tex] is final velocity ( 0 m/s )
a is acceleration( - 4.9 m/s² )
so we substitute
(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx
0 = 144 m²/s² - 9.8 m/s²Δx
9.8 m/s²Δx = 144 m²/s²
Δx = 144 m²/s² / 9.8 m/s²
Δx = 14 m
Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B) data that were not necessary to the solution are;
a) mass of truck and b) mass of load
Which of the following is an example of an object with kinetic energy?
a. a plane lifting off of the runway
b. a bobsled perched at the top of a run
c. a snowball tumbling down a hill
d. both a and c
Answer:
A and C
Explanation:
Both have mass and are in motion
Which of these statements is true about the effect of a force exerted upon an object?
A. A large force always produces a large change in the object’s momentum.
B. A small force always produces a large change in the object’s momentum.
C. A small force applied over a long time interval can produce a large change in the object’s momentum.
D. A large force produces a large change in the object’s momentum only if the force is applied over a very short time interval.
Answer:
D. A large force produces a large change in the object’s momentum only if the force is applied over a very short time interval.
Explanation:
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
[tex] Momentum = mass * velocity [/tex]
Also, the impulse of an object is given by the formula;
[tex] Impulse = force * time [/tex]
In accordance with the impulse-momentum theorem, the statement which is true about the effect of a force exerted upon an object is that a large force produces a large change in the object’s momentum only if the force is applied over a very short time interval.
A piece of gold aluminum alloy weighs 49N. When suspended from a balance and submerged in water it weighs 39.2. What is the weight of gold in the alloy of relative density of gold if 19.3 and aluminum is 2.5
Answer:20.8
Explanation:
A 0.242 g sample of potassium is heated in oxygen. The result is 0.292 g of a crystalline compound. What is the formula of this compound?
A.
KO3
B.
KO2
C.
KO
D.
K2O
Answer:
Hello there Dude answer is B :D hope it helped mark me brainliest.
The formula of the compound formed has been [tex]\rm \bold {K_2O}[/tex]. Thus, option D is correct.
The sample of potassium has mass of 0.242 g. Since, the substance has been heated in the presence of oxygen, the gain in the weight has been corresponds with the mass of oxygen.
The given sample has:
Mass of potassium, [tex]m_K=0.242\;\text g[/tex]
Mass of heated sample, [tex]m_S=0.292\;\text g[/tex]
The mass of oxygen ([tex]m_O[/tex]) in the sample has been given as:
[tex]m_O=m_S-m_K[/tex]
Substituting the values:
[tex]m_O=0.292\;-\;0.242\;\text g\\m_O=0.05\;\text g[/tex]
The mass of oxygen in the sample has been 0.05 g.
The moles (M) of compounds in the sample has been given as:
[tex]M=\dfrac{m}{mwt}[/tex]
Where, m has been the mass of the compound, and
mwt has been the molecular weight of the compound.
The moles of potassium ([tex]M_K[/tex]) has been given as:
[tex]M_K=\dfrac{0.242}{39.098}\\M_K=0.006\;\text mol[/tex]
The moles of oxygen ([tex]M_O[/tex]) has been given as:
[tex]M_O=\dfrac{0.05}{16}\\M_O=0.003\;\text mol[/tex]
The molecular compound has been formed with Potassium and oxygen in the ratio of their moles as:
[tex]\rm \dfrac{K}{O}=\dfrac{0.006}{0.003}\\ \dfrac{K}{O}= \dfrac{2}1}[/tex]
Thus, the molecular formula of the compound has been [tex]\rm \bold {K_2O}[/tex]. Thus, option D is correct.
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What did Thomas Edison create that made him most famous today? I know it is something about light, but I do not believe he was first. May someone tell me his most famous accomplishment in easy words to understand? Thanks
Explanation:
Thomas Edison invented the incandescent light bulb which is kind of like the light bulbs we use today. but they have been improved throughout the years.
hope this is simple and understandable
Answer:
One of the most famous and prolific inventors of all time, Thomas Alva Edison exerted a tremendous influence on modern life, contributing inventions such as the incandescent light bulb, the phonograph, and the motion picture camera, as well as improving the telegraph and telephone.
Explanation:
Your well-wisher
The potential energy of a 40-kg cannonball is 14000 J. How high was the cannon ball to have this much potential energy?
Answer:
35.71 m
Explanation:
Potential energy is calculated using this formula:
PE = mghwhere m = mass (kg). g = gravitational acceleration on Earth (9.8 m/s²), h = height (m)We are given 3 out of the 4 variables in this problem. We want to solve for h, the height of the cannon ball.
List out the known variables:
PE = 14,000 J m = 40 kg g = 9.8 m/s² h = ? mSubstitute these values into the potential energy formula.
14,000 = (40)(9.8)h 14,000 = -392h h = 35.7142857143The cannonball was 35.71 m high to have a potential energy of 14,000 J.
The amplitude of a wave
determines the volume of a
sound.
True
O False
How did Einstein’s and Newton’s theories differ in terms of explaining the cause of gravity?
thank you
Answer:
Newton's theory identified mass as the factor that causes gravity. On the other hand, Einstein's theory identified the curvature of space-time as the factor that causes gravity.
Answer:
Hey mate...
Explanation:
This is ur answer....
In the 17th century Newton concluded that objects fall because they are pulled by Earth's gravity. Einstein's interpretation was that these objects do not fall. According to Einstein, these objects and Earth just freely move in a curved spacetime and this curvature is induced by mass and energy of these objects.
Hope it helps you,
mark me as the brainliest.....
Pls answer my 2 recent questions by going on my account...
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Name the principle which states that energy
cannot be created or destroyed, merely
transferred from one form to another:
Answer:
the Laws of Thermodynamics
Explanation:
these laws states that no form of energy can be created by anyone or anything, without a previous and equal input of energy being put in, that energy can only be transferred from object to object and through different forms
Example: the way a heater runs to warm up an area, that is the process of electrical energy, or energy stored in propane being converted into heat energy
this law is universally implied and has been proven on multiple accounts to be true, in no way can you create energy out of thin air, all you can do is transform and transfer it
hope this is what you was going for, very good point in science
this is one of two correct answers, the other answer to this question is also correct
A woman pushes a 35.0 kg object at a constant speed for 10.8 m along a level floor, doing 280 J of work by applying a constant horizontal force of magnitude F on the object. (a) Determine the value of F (in N). (Enter the magnitude.) N (b) If the worker now applies a force greater than F, describe the subsequent motion of the object. The object's speed would increase with time. The object's speed would remain constant over time. The object would slow and come to rest. (c) Describe what would happen to the object if the applied force is less than F. The object's speed would increase with time. The object's speed would remain constant over time. The object would slow and come to rest.
Answer:
a) F= 25.9 N
b) The object's speed would increase with time.
c) The object would slow and come to rest.
Explanation:
a)
By definition of work, this is the process through which a force applied on an object, produces a displacement of the object.If the force is constant, and the displacement is parallel to the direction of the force, this work is just the product of the applied force times the distance, as follows:[tex]W = F_{app} * d (1)[/tex]
We can solve for Fapp, replacing W and d by their values:[tex]F_{app} =\frac{W}{d} = \frac{280J}{10.8m} = 25.9 N (2)[/tex]
b)
If the object is moving at constant speed when it is applied a force F, this means that this force must be compensated by an equal opposite force, in this case, the kinetic friction force.Since this force is constant while the object is moving, if we increase the force F making it larger, there will be a net force in the direction of the displacement, which will cause an acceleration that will increase the speed with time.c)
We have already said in b) that if the object is moving at constant speed, there must be an equal and opposite force to the applied force F, the kinetic friction force, which is constant, acting on the object.If we apply a force less than F, there will be a net force in the direction opposite to the displacement, that will cause a acceleration opposite to the displacement, which will make the object to slow down and eventually come to rest.NEED HELP ASAP TEST IS DUE TMR
What is the angular displacement of a wheel with a 3mm radius and an angular speed of 6 rad/s over a time period of 2.5 seconds?
Answer:
The answer should be uranus
Explanation:
is electricity matter
Answer:
Yes it is
Explanation:
Electricity is the positive and negative matter that's found in protons and electrons.
Answer:
yes
Explanation:
because electricity is a positive and negative proton
PLEASE CLICK ON THIS IMAGE I NEED HELP
Answer:
Second option
Explanation:
"Uniform" pretty much means the same thing happens.
During a blood transfusion, gravity is used to provide the pressure to overcome the blood pressure and force the flow through a small needle into a vein. Consider a case where the needle is 3.0 cm long and has an internal diameter of 0.75 mm. If the required rate of flow is 0.03 cm3/s and the blood pressure in the vein is 11.00 kPa higher than atmospheric pressure (Pa), how high should the bottle be placed above the needle
Density of blood = 1.06x10³
Viscosity = 4 mpas
Answer:
It should be 1.10 higher
Explanation:
L = 0.93
D = 0.75
R = 0.75/2 = 0.375
Q = 0.03x10^-3
Blood pressure = 11x10³
Pn = 4x10^-3
n = 4 x 10^-3
Density of blood = 1.06x10³
Pn - Pv = 8*Q*n*L/pi*r⁴
Pn - pv = 463.57pa
Pn - pv = 463.57pa
Make pn subject
Pn = Pv + 463.57pa ----1
Vn = Q/An
= 0.0679m/s
To get height above needle
Pn + 1/2pv²n = Pa + pgh ----2
We equate 1 and 2 together
We get
Pgh = 11466
To get h we divide through by pg
h = 11466/pg
h = 11466/(1.06x10³)x9.81
h = 1.1026
Approximately
Height = 1.1 meters
So it should be 1.10 meters higher
the full calculations are in the attachment.
thank you!
You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a constant speed of 1.6 m/s. The rope makes an angle of 35 degrees with the horizontal. What is the net force on the block
Answer:
Fnet = 0
Explanation:
Since the block slides across the floor at constant speed, this means that it's not accelerated.According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:[tex]F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N (1)[/tex]
In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:[tex]F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N (2)[/tex]
⇒ 169 N + Fn = Fg = 216 N (3)
This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:Fn = 216 N - 169 N = 47 N (4)when a constant force is applied to an object, the acceleration of the object varies inversely with its mass. When a certain constant force acts upon an object with a mass 12 kg, the acceleration of the object is 6 m/s. If the same force acts on another object whose mass is 9kg, what is the objects acceleration
Answer:
8 m/s²
Explanation:
From the question,
Since the same force act on both object,
F = ma = m'a'.............................. Equation 1
Where F = force action on the obeject, m = mass of the first object, a = acceleration of the first object, m' = mass of the second object, a' = acceleration of the second object.
make a' the subject of the equation
a' = ma/m'................... Equation 2
Given: m = 12 kg, a = 6 m/s², m' = 9 kg.
Substitute these values into equation 2
a' = 12(6)/9
a' = 8 m/s².
Hence the acceleration of the second object is 8 m/s²
the motion of a body with respect to another body is?
Answer:
Motion that changes the orientation of a body is called rotation. ... In both cases all points in the body have the same velocity (directed speed) and the same acceleration (time rate of change of velocity).
Answer:
its called relative speed
An amusement park ride called the Rotor debuted in 1955 in Germany. Passengers stand in the cylindrical drum of the Rotor as it rotates around its axis. Once the Rotor reaches its operating speed, the floor drops but the riders remain pinned against the wall of the cylinder. Suppose the cylinder makes 26.0 rev/min and has a radius of 3.70 m. 1) What is the coefficient of static friction between the wall of the cylinder and the backs of the riders
Answer:
μs = 0.36
Explanation:
While the drum is rotating, the riders, in order to keep in a circular movement, are accelerated towards the center of the drum.This acceleration is produced by the centripetal force.Now, this force is not a different type of force, is the net force acting on the riders in this direction.Since the riders have their backs against the wall, and the normal force between the riders and the wall is perpendicular to the wall and aiming out of it, it is easily seen that this normal force is the same centripetal force.In the vertical direction, we have two forces acting on the riders: the force of gravity (which we call weight) downward, and the friction force, that will oppose to the relative movement between the riders and the wall, going upward.When this force be equal to the weight, it will have the maximum possible value, which can be written as follows:[tex]F_{frmax} = \mu_{s}* F_{n} = m * g (1)[/tex]
where μs= coefficient of static friction (our unknown)As we have already said Fn = Fc.The value of the centripetal force, is related with the angular velocity ω and the radius of the drum r, as follows:[tex]F_{n} = m* \omega^{2} * r (2)[/tex]
Replacing (2) in (1), simplifying and rearranging terms, we can solve for μs, as follows:[tex]\mu_{s} = \frac{g}{\omega^{2} r} (3)[/tex]
Prior to replace ω for its value, is convenient to convert it from rev/min to rad/sec, as follows:[tex]\omega = 26.0 \frac{rev}{min} * \frac{1min}{60 sec} *\frac{2*\pi rad }{1 rev} = 2.72 rad/sec (4)[/tex]
Replacing g, ω and r in (3):[tex]\mu_{s} = \frac{g}{\omega^{2} r} = \frac{9.8m/s2}{(2.72rad/sec)^{2} *3.7 m} = 0.36 (5)[/tex]a sentence that describes how energy is related to work
Answer:
Energy should be transferred to an object in order to move it. ... This amount of energy transferred by the force to move an object is called work or work done. Thus, the relation between Work and Energy is direct. That is, the difference in the Kinetic energy of an object is work done by an object
At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 55 kg, 90 kg, and 42.5 kg respectively. Car A is moving to the right with a velocity vA = 2 m/s and car C has a velocity vC = 1.5 m/s to the left, but car B is initially at rest. The coefficient of restitution between each car is 0.8. Determine the final velocity of each car, after all impacts, assuming car A hits car B before car C does. Assume positive sign denoting forward motion and negative sign denoting backward motion.
Answer:
Vb = 0.334 m/s
Va = -1.265 m/s
Vc = 1.424 m/s
Explanation:
Favorite Answer
Initial momentum = 255(2) – 242.5(1.5) = 146.25
Final momentum = 255Va + 290Vb + 242.5 Vc = 146.25
Vb - Va = 0.8(2) = 1.6
Vc - Vb = 0.8(1.5) = 1.2
Va = Vb -1.6
Vc = Vb + 1.2
255(Vb -1.6) + 290Vb + 242.5(Vb + 1.2) = 146.25
255 Vb – 408 + 290 Vb + 242.5 Vb + 291 = 146.25
787.5 Vb = 263.25
Vb = 0.334 m/s
Va = Vb -1.6 = 0.334 – 1.6 = -1.265 m/s
Vc = Vb + 1.2 = 0.224 + 1.2 = 1.424 m/s
A plastic ball in a liquid is acted upon by its weight and by a buoyant force. The weight of the ball is 4 N. The buoyant force has a magnitude of 5 N and acts vertically upward. When the ball is released from rest, what is it's acceleration and direction? [2 pts] for a Free Body Diagram correctly labeled.
Answer:
The acceleration is 2.448 meters per square second and is vertically upward.
Explanation:
The Free Body Diagram of the plastic ball in the liquid is presented in the image attached below. By Second Newton's Law, we know that forces acting on the plastic ball is:
[tex]\Sigma F = F - m\cdot g = m\cdot a[/tex] (1)
Where:
[tex]F[/tex] - Buoyant force, measured in newtons.
[tex]m[/tex] - Mass of the plastic ball, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]a[/tex] - Net acceleration, measured in meters per square second.
If we know that [tex]F = 5\,N[/tex], [tex]m = 0.408\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the net acceleration of the plastic ball is:
[tex]a = \frac{F}{m} - g[/tex]
[tex]a= 2.448\,\frac{m}{s^{2}}[/tex]
The acceleration is 2.448 meters per square second and is vertically upward.
素 Example three
After an airplane takes off, it travels 10.4 km west, 8.7 km north, and 2.1 km up How far is it
from the take off point?
Solution
Answer:
Let R be the total distance
R²=10.4²+(8.7+2.1)²
R²=10.4²+10.8²
R²=224.8
You square root both sides
R=14.99km
If an airplane takes off, it travels 10.4 km west, 8.7 km north, and 2.1 km up, then using the Pythagorean theorem it is 14.99km from its take-off point.
What is Pythagoras' theorem?Pythagoras' theorem is a fundamental theorem in mathematics that relates to the sides of a right-angled triangle. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
In equation form, the theorem can be written as:
c² = a² + b²
Where
c =the length of the hypotenuse,
a and b = the lengths of the other two sides.
The theorem is named after the ancient Greek mathematician Pythagoras, who is credited with his discovery and proof. Pythagoras' theorem is widely used in mathematics and has many practical applications in fields such as architecture, engineering, and physics.
Here in the question
We can use the Pythagorean theorem to find the distance from the takeoff point:
Distance = √(West)² + (North+Up)²
Plugging in the given values, we get:
Distance = √(10.4)² +( 8.7 + 2.1)²
Distance ≈ 14.99km
Therefore, the airplane is approximately 14.99km away from the takeoff point.
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You wish to make a simple amusement park ride in which a steel-wheeled roller-coaster car travels down one long slope, where rolling friction is negligible, and later slows to a stop through kinetic friction between the roller coaster's locked wheels sliding along a horizontal plastic (polystyrene) track. Assume the roller-coaster car (filled with passengers) has a mass of 743.0 kg and starts 83.4 m above the ground. (a) Calculate how fast the car is going when it reaches the bottom of the hill. m/s (b) How much does the thermal energy of the system change during the stopping motion of the car
Answer:
(a) The car is going approximately 40.43 m/s at the bottom of the hill
(b) The thermal energy will increase by 607,268.76 J
Explanation:
In the question, we have;
The height of the roller coaster above ground = 83.4 m
The mass of the roller coaster, m = 743.0 kg
(a) By the conservation of energy principle, we have;
The potential energy at the top of the hill, P.E., is equal to the kinetic energy at the bottom of the hill, K.E.
∴ P.E. = K.E.
P.E. = m·g·h
Where;
m = The mass of the roller coaster = 743.0 kg
g = The acceleration due to gravity = 9.8 m/s²
h = The height of the roller coaster = 83.4 m
Therefore, we have;
P.E. = 743.0 kg × 9.8 m/s² × 83.4 m = 607,268.76 J
P.E. = 607,268.76 J
K.E. = 1/2·m·v²
∴ K.E. = 1/2 × 743.0 kg × v²
P.E. = K.E.
∴ P.E. = K.E. = 607,268.76 J
1/2 × 743.0 kg × v² = 607,268.76 J
v² = 607,268.76 J/(1/2 × 743.0 kg) = 1,634.64 m²/s²
v = √(1,634.64 m²/s²) ≈ 40.43 m/s
(b) Given that the material wheel moves along polystyrene track, the sound released will be minimal and almost all the kinetic energy will be converted to heat energy when the train stops, therefore, the thermal energy will increase by K.E. = 607,268.76 J
The thermal energy change of the system is 624,492 J.
We know that in the roller coaster, there is an energy transformation from gravitational potential energy to kinetic energy. As such we can write;
mgh = 1/2mv^2
Where we cancel out the mass from both sides, we are left with;
gh=0.5v^2
v= √gh/0.5
v = √10 × 83.4 m/0.5
v = 41 ms-1
Now the kinetic energy is converted also into heat energy hence;
Thermal energy change of the system = 1/2 mv^2 = 0.5 × 743.0 kg × ( 41 ms-1)^2 = 624,492 J
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in a class where the number of girls is 36% of the total number,there are 48 boys.how many students are there in the class?
Answer:
There are 75 people in the class. The number of boys is 48 and the number of girls is 27. The percentage of girls is 36% of 75.
Explanation:
Two girls are estimating each other's power. One runs up some step
ng each other's power. One runs up some steps, and the other times her. Here are their
results:
height of one step = 20 cm
number of steps = 36
mass of runner = 45 kg
time taken = 4.2 s
a .Calculate the runner's weight. (Acceleration due to gravity g=10m
b .Calculate the increase in the girl's gravitational potential energy as she runs up the steps.
c. Calculate her power. Give your answer in kilowatts (kW).
Answer:
A. 450 N
B. 3240 J
C. 0.77 KW
Explanation:
From the question given above, the following data were obtained:
Height of one step = 20 cm
Number of steps = 36
Mass of runner = 45 kg
Time taken = 4.2 s
Next, we shall convert 20 cm to metre (m). This can be obtained as follow:
100 cm = 1 m
Therefore,
20 cm = 20 cm × 1 m /100 cm
20 cm = 0.2 m
Next, we shall determine the total height. This can be obtained as follow:
Height of one step = 0.2 m
Number of steps = 36
Total height =?
Total height = 36 × 0.2
Total height = 7.2 m
A. Determination of the runner's weight.
Mass of runner (m) = 45 kg
Acceleration due to gravity (g) = 10 m/s²
Weight (W) =?
W = m × g
W = 45 × 10
W = 450 N
B. Determination of the increase in the potential energy.
At the ground level, the potential energy (PE₁) is 0 J.
Next, we shall determine the potential energy at 7.2 m. This can be obtained as follow:
Mass of runner (m) = 45 kg
Acceleration due to gravity (g) = 10 m/s²
Total height (h) = 7.2 m
Potential energy at height 7.2 m (PE₂) = ?
PE₂ = mgh
PE₂ = 45 × 10 × 7.2
PE₂ = 3240 J
Final, we shall determine the increase in potential energy. This can be obtained as follow:
Potential energy at ground (PE₁) = 0 J
Potential energy at height 7.2 m (PE₂) = 3240 J
Increase in potential energy =?
Increase in potential energy = PE₂ – PE₁
Increase in potential energy = 3240 – 0
Increase in potential energy = 3240 J
C. Determination of the power.
Energy (E) = 3240 J
Time (t) = 4.2 s
Power (P) =?
P = E/t
P = 3240 / 4.2
P = 771.43 W
Finally, we shall convert 771.43 W to kilowatt (KW). This can be obtained as follow:
1000 W = 1 KW
Therefore,
771.43 W = 771.43 W × 1 KW / 1000 W
771.43 W = 0.77 KW
Therefore, her power is 0.77 KW
Which image best illustrates diffraction?
Answer: A
Explanation:
A student at a polymer lab conducted a stress-relaxation experiment on a polymer, whose material response can be modeled by the Maxwell model. In the experiment, a 10-cm-long bar was stretched instantly until the stress reached 200 MPa. After 2 min, the measured stress in the bar was 160 MPa. a) Calculate the relaxation time constant of the polymer. b) If the initial stress was reached by stretching the bar by 16 mm, what is the viscosity of the dashpot in the Maxwell model
Answer: add 10 cm plus 200 mpa divided by 2 min then caculate that into 160 bars.
Explanation:
a) The relaxation time constant of the polymer is approximately 8.9 min.
b) The viscosity of the dashpot in the Maxwell model is approximately 4.55 GPa s.
What is the stress-relaxation model?The stress-relaxation response of the Maxwell model can be modeled by the following equation:
σ(t) = σ₀ exp(-t/τ)
where σ₀ is the initial stress, τ is the relaxation time constant, and t is the time.
a) The relaxation time constant of the polymer:
σ(2 min) = 160 MPa
σ₀ = 200 MPa
160 MPa = 200 MPa exp(-2 min/τ)
Taking the natural logarithm of both sides, we get:
ln(160/200) = -2 min/τ
Solving for τ, we get:
τ = -2 min / ln(160/200) ≈ 8.9 min
Therefore, the relaxation time constant of the polymer is approximately 8.9 min.
b) The displacement of the spring in the Maxwell model can be modeled by the following equation:
x(t) = (σ₀ / G) (1 - exp(-t/τ))
where G is the shear modulus of the polymer. We can use this equation and the given displacement of 16 mm to find the viscosity of the dashpot:
x(t) = 16 mm = 0.016 m
σ₀ = 200 MPa
τ = 6.09 min = 365.4 s
Assume that the shear modulus of the polymer is constant and use the given stress to find it:
G = σ₀ / γ
where γ is the strain induced by the stretching.
Since the bar was stretched by 16 mm and its original length was 10 cm (i.e., 100 mm), the strain is:
γ = 16 mm / 100 mm = 0.16
Therefore,
G = 200 MPa / 0.16 = 1250 MPa
Now we can solve for the viscosity of the dashpot:
η = σ₀τ / x(t)
= (200 MPa) (365.4 s) / (0.016 m)
≈ 4.55 GPa s
Therefore, the viscosity of the dashpot in the Maxwell model is approximately 4.55 GPa s.
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g Design an experiment you can use to determine the mass of the metal cylinder. When you explain your experiment, be sure to mention: What is the underlying model (equation) that you can use to determine the mass from your measurements
Answer:
m = [tex]\frac{k}{g}[/tex] x,
graph of x vs m
Explanation:
For this exercise, the simplest way to determine the mass of the cylinder is to take a spring and hang the mass, measure how much the spring has stretched and calculate the mass, using the translational equilibrium equation
F_e -W = 0
k x = m g
m = [tex]\frac{k}{g}[/tex] x
We are assuming that you know the constant k of the spring, if it is not known you must carry out a previous step, calibrate the spring, for this a series of known masses are taken and hung by measuring the elongation (x) from the equilibrium position, with these data a graph of x vs m is made to serve as a spring calibration.
In the latter case, the elongation measured with the cylinder is found on the graph and the corresponding ordinate is the mass
Each vertical line on the graph is 1 millisecond (0.001 s) of time. What is the period and
frequency of the sound waves?
Explanation:
Given that,
Each vertical line on the graph is 1 millisecond (0.001 s) of time.
We need to find the period and the frequency of the sound wave. The period of a wave is equal to the each vertical line on graph i.e. 0.001 s.
Let f be the frequency of the sound wave. So,
f = 1/T
i.e.
[tex]f=\dfrac{1}{0.001 }\\\\f=1000\ Hz[/tex]
So, the period and the frequency of the sound waves is 1 milliseond and 1000 Hz respectively.
an object is found to have weight of 16.7N on moon
what is it's
weight on earth where g= 10N/kg
Answer:
167 kg
Explanation:
On Earth it is 167 kg. We have to multiply the weight in the Moon by 10 because 1 kg on Earth, is nearly 10 N. 1 kg is 9,8 N in fact, but we get it as 10 N in general.
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