Explanation:
Given data:
mass of the car = 1500 kg
maximum friction force = 7000 N
initial velocity v_i = 20 m/s ( it is not given in the question just an assumption)
final velocity v_f = 0 m/s
[tex]\begin{array}{l}
\sum F_{y}=M g-F_{n}=0 \\
\sum F_{x}=-F_{s}=m a_{x} \\
-F_{s}=m a_{x}
\end{array}[/tex]
[tex]a_{x}=\frac{-F_{s}}{m}=\frac{-7000}{1500}[/tex]
[tex]a_{x}=-4.7 \mathrm{~m} / \mathrm{s}^{2}[/tex]
Now we can find the distance from this formula:
[tex]v_{f x}^{2}=v_{i x}^{2}+2 a_{x}(\Delta x)[/tex]
[tex]0=20^{2}+(2 \times-4.7 \times \Delta x)[/tex]
[tex]20^{2}=9.4 \Delta x[/tex]
[tex]\Delta x=\frac{20^{2}}{9.4}=42.55 \mathrm{~m}[/tex]
So, the shortest distance in which the car can stop safely without kidding
=42.55 m
A person pushes down on a lever with a force of 100 N. At the other end of the lever, a force of 200 N lifts a heavy object. What is the mechanical advantage of the lever?
A. 1/2, because the object will be lifted half the distance
B. -1, because the direction changes
C. 2, because the output force is twice the input force
D. 1, because the same amount of work is done
Answer:
Explanation:
C 200÷100=2
Output ÷ Input= MA
16+109=................
Answer:
=125 tysm for points
Explanation:
Answer: 125
Explanation: I good at addition
what is energy? list the three mechanical forms of energy and their associated equations.
Answer:
mm hope that helps
Explanation:
What are the 3 forms of energy?
Forms of energy
Potential energy.
Kinetic energy.
When the bowling ball has fallen halfway down the building (height = 20 m), it has a speed of 19.8 m/s.
How much potential energy does the bowling ball have?
How much kinetic energy does the bowling ball have?
How much total energy (potential + kinetic) does the bowling ball have?
Of the bowling ball’s total energy, is more in the form of potential or kinetic energy?
Answer:
I think the answer is 19.8 potential energy
Explanation:
NONE.
Which of the following is a concern about the use of wind energy?
A. Wind farms produce a large amount of air and water pollution.
B. Wind farms may pose a threat to birds and bats.
O
C. Wind farms produce radioactive waste.
D. Wind farms often change the local weather patterns.
SUBMIT
Use of wind energy may pose a threat to birds and bats.
What is wind energy?The power of wind is harnessed to generate renewable energy through the use of wind turbines. These turbines are designed to capture the kinetic energy of the wind and convert it into electricity by rotating their blades. This process involves the movement of turbines, which in turn drives a generator that produces electricity. Wind energy is an eco-friendly and sustainable form of energy that does not emit any harmful gases or pollutants into the atmosphere. As countries worldwide seek to reduce their dependence on fossil fuels and transition towards a low-carbon economy, wind energy is gaining popularity as a clean source of electricity.
Which is a concern about the use of wind energy?The implementation of wind energy has raised concerns about its impact on wildlife, specifically birds and bats. The use of wind turbines can lead to collisions and habitat disruptions for these animals. Despite this, studies have shown that wind turbines cause fewer fatalities for birds and bats in comparison to other human-related sources such as buildings, power lines, and vehicles. To mitigate the impact on wildlife, measures like radar technology to detect animal movement and shutting down turbines during migration periods are being employed..
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Suppose you drop a superball of massMand a marble of mass m(both treated as point masses) from a heighthwith the marble just on top of the superball. A superbowl has essentially elastic collisions with hard objects. Thesuperball hits the floor, rebounds and then collides with the marble. How highdoes the marble go if all the motion is vertical
Answer:
h '= [ ( \frac{ M-m}{M +m } )+ 2 (\frac{M}{M+m})]² h
Explanation:
Let's analyze this problem, first the two bodies travel together, second the superball bounces, third it collides with the marble and fourth the marble rises to a height h ’
let's start by finding the velocity of the two bodies just before the collision, let's use the concepts of energy
starting point. Starting point
Em₀ = U = m g h
final point. Just before the crash
Em_f = K = ½ m v²
as there is no friction the mechanical energy is conserved
Em₀ = Em_f
mg h = ½ m v²
v = √2gh
this speed is the same for the two bodies.
Second point. The superball collides with the ground, this process is very fast, so we will assume that the marble has not collided, let's use the concept of conservation of moment
initial instant. Just when the superball starts contacting the ground
p₀ = M v
this velocity is negative because it points down
final instant. Just as the superball comes up from the floor
p_f = M v '
the other body does not move
p₀ = p_f
- m v = M v '
v ’= -v
Therefore, the speed of the asuperbola is the same speed with which it arrived, but in the opposite direction, that is, upwards.
Let's use the subscript 1 for the marble and the subscript 2 for the superball
Third part. The superball and the marble collide
the system is formed by the two bodies, so that the forces during the collision are internal and the moment is conserved
initial instant. Moment of shock
p₀ = M [tex]v_{1'}[/tex]+ m v_2
final instant. When the marble shoots out.
P_f = Mv_{1f'}+ m v_{2f}
p₀ = p_
M v_{1'}+ m v_2 = M v_{1f'} + m v_{2f}
M (v_{1'} - v_{1f'}) = -m (v_2 - v_{2f})
in this expression we look for the exit velocity of the marble (v2f), as they indicate that the collision is elastic the kinetic nerve is also conserved
K₀ = K_f
½ M v_{1'}² + m v₂² = M v_{1f'}² + ½ m v_{2f}²
M (v_{1'}² - v_{1f'}²) = - m (v₂² - v_{2f}²)
Let's set the relation (a + b) (a-b) = a² - b²
M (v_{1'} + v_{1f'}) (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})
let's write our two equations
M ( v_{1'} - v_{1f'}) = -m (v₂ - v_{2f}) (1)
M (v_{1'} + v_{1f'}) (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})
if we divide these two expressions
(v_{1'}+ v_{1f'}) = (v₂ + v_{2f} )
we substitute this result in equation 1 and solve
v_{1f'}= (v₂ + v_{2f}) - v_{1'}
M (v_{1'} - [(v₂ + v_{2f}) - v_{1'}] = -m (v₂ - v_{2f})
-M v₂ - M v_{2f1'} + 2M v_{1'} = m v₂ - m v_{2f}
-M v_{2f} -m v_{2f} = m v₂ -M v₂ + 2M v_{1'}
v_{2f} (M + m) = - v₂ (M-m) + 2 M v_{1'}
v_{2f} = - [tex]( \frac{ M-m}{M +m } )[/tex]) v₂ + 2 [tex](\frac{M}{M+m})[/tex] v_{1'}
now we can substitute the velocity values found in the first two parts
[tex]v_{2f}[/tex] = - ( \frac{ M-m}{M +m } ) √2gh + 2(\frac{M}{M+m}) √2gh
we simplify
v_{2f} = [( \frac{ M-m}{M +m } ) + 2 (\frac{M}{M+m})] [tex]\sqrt{2gh}[/tex]
let's call the quantity in brackets that only depends on the masses
A = ( \frac{ M-m}{M +m } )+ 2 (\frac{M}{M+m})]
v_{2f}= A \sqrt{2gh}
in general, the marble is much lighter than the superball, so its speed is much higher than the speed of the superball
finally with the conservation of energy we find the height that the marble reaches
Starting point
Emo = K = ½ mv_{2f}²
Final point
Emf = U = m g h'
Em₀ = Em_f
½ m v_{2f}² = m g h ’
h ’= ½ v_{2f}² / g
h ’= ½ (A \sqrt{2gh})² / g
h ’= A² h
h '= [ ( \frac{ M-m}{M +m } )+ 2 (\frac{M}{M+m})]² h
The impulse given to a body of mass 1.5 kg, is 6.0 kg
• m•s-?
If the body was initially at rest, what
will its resulting kinetic energy be? Give your answer in J without units.
Show work
Answer:
12J
Explanation:
Given parameters:
Mass = 1.5kg
Impulse = 6kgm/s
let us start first by find the velocity with which this body moves;
Impulse = mass x velocity
Velocity = Impulse / mass = 6/ 1.5 = 4m/s
Initial velocity = 0m/s
Unknown:
Resulting kinetic energy = ?
Solution:
To solve this problem use the formula below:
K.E = [tex]\frac{1}{2}[/tex] m (v - u)²
m is the mass
v is the final velocity
u is the initial velocity
So;
K.E = [tex]\frac{1}{2}[/tex] x 1.5 x (4 - 0)²
K.E = 1.5 x 8 = 12J
Your __________ environment includes the people you spend time with.
Answer:
social
Explanation:
Your social environment includes the people you spend time with – your family, your friends, classmates and other people in your community. Your social environment is healthier when you choose friends who show concern for their own health and yours.
please help!!!!!! 26 points
A cars mass is 950kg and it travels at a speed of 35 m/s when it rounds a flat curve of radius 215 m.
a. Determine the value of the frictional force exerted on the car.
b. Determine the value of the coefficient of friction between the tires and the road.
(a) It's the force of (static) friction that keeps the car on the road and prevents it from skidding, and this friction is directed toward the center of the curve.
Recall that centripetal acceleration has a magnitude a of
a = v ² / R
where
v = tangential speed
R = radius of the curve
so that
a = (35 m/s)² / (215 m) ≈ 5.69767 m/s² ≈ 5.7 m/s²
Parallel to the road, the only force acting on the car is friction. So by Newton's second law, we have
∑ F = Fs = m a
where
Fs = magnitude of static friction
m = mass of the car
Then
Fs = (950 kg) (5.7 m/s²) ≈ 5412.79 N ≈ 5400 N
(b) Perpendicular to the road, the car is in equilbrium, so its weight and the normal force of the road on the car are equal in magnitude. By Newton's second law,
N - W = 0
where
N = magnitude of normal force
W = weight
so that
N = W = m g = (950 kg) (9.8 m/s²) = 9310 N
Friction is proportional to the normal force by a factor of µ, the coefficient of static friction:
Fs = µ N
Assuming 35 m/s is the maximum speed the car can travel without skidding, we find
µ = Fs / N = (5400 N) / (9310 N) ≈ 0.581395 ≈ 0.58
According to Newton's law of universal gravitation, which statements are true?
As we move to higher altitudes, the force of gravity on us decreases.
O As we move to higher altitudes, the force of gravity on us increases,
O As we gain mass, the force of gravity on us decreases.
O Aswe gain mass, the force of gravity on us increases.
DAs we move faster, the force of gravity on us increases.
Which has the most mass?
O The Moon
O A Pencil
O Your teacher.
O Earth
Answer:
Earth
lol... ....
Your friend, a world-class long jumper, is trapped on the roof of a burning building. His only escape route is to jump to the roof of the next building. Fortunately for him, he is in telephone contact with you, a Physics 161 student, for advice on how to proceed. He has two options. He can jump to the next building by using the long-jump technique where he jumps at 45o to the horizontal. Or, he can take his chances by staying where he is in the hopes that the fire department will rescue him. You learn from the building engineers that the next building is 10 m away horizontally and the roof is 3 m below the roof of the burning building. You also know that his best long-jump distance is 7.9 m. What do you advise him to do
Answer:
y = -2.69 m
the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.
Explanation:
his problem must be solved with the missile launch equations.
Let's start by looking for the jumper's initial velocity
R = v₀² sin 2θ / g
for the long jump the angle used is tea = 45º, in the exercise they indicate that the best record is R = 7.9m
v₀² = R g / sin 2te
v₀ = [tex]\sqrt{ \frac{7.9 \ 9.8}{1 }[/tex]
v₀ = 8.80 m / s
Now suppose you jump with this speed to get to the other building, let's use trigonometry for the components of the speed
sin 45 = [tex]v_{oy}[/tex] /v₀
cos 45 = v₀ₓ / v₀
v_{oy} = v₀ sin 45
v₀ₓ = v₀ cos 45
v_{oy} = 8.8 sin 45 = 6.22 m / s
v₀ₓ = 8.8 cos 45 = 6.22 m / s
now let's calculate the sato with these speeds
x = [tex]v_{ox}[/tex] t
the minimum jump is x = 10 m
t = x / v₀ₓ
t = 10 / 6.22
t = 1.61 s
let's find the vertical distance for this time
y = v_{oy} t - ½ g t²
where zero is placed on the jump building
y = 6.22 1.61 - ½ 9.8 1.61²
y = -2.69 m
Let's analyze this result, the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.
Two identical cylinders with a movable piston contain 0.7 mol of helium gas at a temperature of 300 K. The temperature of the gas in the first cylinder is increased to 564 K at constant volume by doing work W1 and transferring energy Q1 by heat. The temperature of the gas in the second cylinder is increased to 564 K at constant pressure by doing work W2 while transferring energy Q2 by heat.
A. Find ΔEint, 1, Q1, and W1 for the process at constant volume.
B. Find ΔEint, 2, Q2, and W2 for the process at constant pressure.
Answer:
Explanation:
A ) At constant volume :
ΔEint = n Cv x ΔT , n is no of moles , Cv is specific heat at constant volume , ΔT is increase in temperature .
For helium Cv = 3/2 R = 1.5 x 8.3 J = 12.45 J
ΔEint = .7 x 12.45 x ( 564 - 300 )
= 2300.76 J .
W₁ = 0 because volume is constant so work done by gas is zero .
Q₁ = ΔEint = 2300.76 J
B )
At constant pressure
Q₂ = n Cp Δ T , Cp is specific heat at constant pressure .
For monoatomic gas ,
Cp = 5/2 R = 2.5 x 8.3 J = 20.75 J
Q₂ = .7 x 20.75 x 264 J
= 3834.6 J
W₂ = work done by gas
= PΔV = nRΔT
= .8 x 8.3 x 264
= 1752.96 J
ΔEint = Q₂ - W₂
= 3834.6 - 1752.96
= 2081.64 J.
ΔEint, 1, Q1, and W1 for the process at constant volume. and ΔEint, 2, Q2, and W2 for the process at constant pressure is mathematically given as
a)
dE1= 2300.76 J .
W1=0 as Volume is constant
Q1= 2300.76 J as Q= dE1
b)
Q2= 3834.6 J
W2= 1752.96 J
dE2= 2081.64 J.
What is ΔEint, Q1 and W1 for the process at constant volume and pressure?a)
Generally, the equation for the Constant Volume is mathematically given as
dE = n Cv x dT
Where
Cv = 3/2
R = 1.5 x 8.3 J
R= 12.45 J
Therefore
dE = 0.7 x 12.45 x ( 564 - 300 )
dE1= 2300.76 J .
W1=0 as Volume is constant
Q1= 2300.76 J as Q= dE1
b)
Generally
Q2 = n Cp dT
Where
Cp = 5/2
R = 2.5 x 8.3 J
R= 20.75 J
Hemce
Q2 = 0.7 x 20.75 x 264 J
Q2= 3834.6 J
For Work done
W=PdV
W= nRdT
Therefore
W= 0.8 x 8.3 x 264
W2= 1752.96 J
Hence
dE = Q₂ - W₂
dE= 3834.6 - 1752.96
dE2= 2081.64 J.
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Which of these represent approaches to psychological science? (Choose every correct answer.)
Behavioral
Chemical
Investigative
Metaphysical
Sociocultural
Cognitive
Humanistic
Answer:
cognitive, humanistic, behavioral, sociocultural
Explanation:
Behavioral, sociocultural, cognitive, and humanistic are approaches to psychological science.
Psychology is a term to refer to the discipline that focuses on the study of various topics related to human thought such as:
The conductMental processes of individuals and human groups in different situations,Human experienceDue to the above, several subdisciplines have emerged that focus on the study of each of the topics. For example:
Behavioral psychology: focused on the study of human behavior.
Sociocultural psychology: focused on the study of human behavior and thought in different social situations.
Cognitive psychology: focused on mental processes related to learning.
Humanistic psychology: focused on the study of human thought from a comprehensive approach.
According to the above, options A, E, F, and G are correct because they mention different sub-disciplines of psychology while the other options mention terms that are not related to sub-disciplines or psychological sciences.
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When a rattlesnake strikes, its head accelerates from rest to a speed of 22 m/s in 0.48 seconds. Assume for simplicity that the only moving part of the snake is its head of mass 170 g. How much (average) power does the rattlesnake need to accelerate its head that fast? Answer in units of W.
Answer:
P = 85.72 W
Explanation:
Given that,
Initial speed, u = 0
Final speed, v = 22 m/s
Time, t = 0.48 s
Mass, m = 170 g = 0.17 kg
Let a be the acceleration of the rattlesnake.
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{22-0}{0.48}\\\\a=45.84\ m/s^2[/tex]
Let x is the displacement of a rattlesnake. It can be given by :
[tex]x=ut+\dfrac{1}{2}at^2\\\\x=0+\dfrac{1}{2}\times 45.84\times (0.48)^2\\\\x=5.28\ m[/tex]
The power of the rattlesnake is given by :
[tex]P=\dfrac{W}{t}\\\\P=\dfrac{m\times a\times x}{t}\\\\P=\dfrac{0.17\times 45.84\times 5.28}{0.48}\\\\P=85.72\ W[/tex]
So, the power of the rattlesnake is 85.72 W.
a car accelerates from 2 m/s to 28m/s at a constant rate of 3 m/s^2. How far does it travel while accelerating?
Answer:
Distance, S = 130m
Explanation:
Given the following data;
Initial velocity = 2m/s
Final velocity = 28m/s
Acceleration = 3m/s²
To find the distance, we would use the third equation of motion.
V² = U² + 2aS
Substituting into the equation, we have;
28² = 2² + 2*3*S
784 = 4 + 6S
6S = 784 - 4
6S = 780
S = 780/6
Distance, S = 130m
When an object with an electric charge of is from an object with an electric charge of , the force between them has a strength of . Calculate the strength of the force between the two objects if they are apart. Round your answer to significant digits.
The question is incomplete, the complete question is;
When an object with an electric charge of −7.0μC is 5.0cm from an object with an electric charge of 4.0μC, the force between them has a strength of 100.7N. Calculate the strength of the force between the two objects if they are 1.7cm apart. Round your answer to 2 significant digits
Answer:
865.1 N
Explanation:
F1 = Kq1q2/r1^2 ---------1
F2 = Kq1q2/r2^2 -------2
We have that;
r1 = 5cm
r2 =1.7 cm
F1 = 100.7 N
Comparing equations 1 and 2
F2 = F1r1^2/r2^2
F2 = 100.7N[(5cm)^2/(1.7cm)^2]
F2= 865.1 N
Energy from the Sun is transferred from the Earth’s surface to the atmosphere, resulting in
atmospheric convection currents that produce winds. How do physical properties of the air
contribute to convection currents?
a -The warmer air sinks because it is more dense than cooler air.
b -The warmer air rises because it is more dense than cooler air.
c- The warmer air sinks because it is less dense than cooler air.
d -The warmer air rises because it is less dense than cooler air.
A tank, in the shape of a cube with each side measuring 2 meters, is half-full of water. Above the water is air that is pressurized to 5 kPa. Calculate the resultant force on one side of the tank due to the air and water (ignore atmospheric pressure) and determine the height of this force above the bottom of the tank.
Answer:
Explanation:
The tank is half full so height of water level is at 1 m . Centre of gravity of water will be at height of .5 m . Pressure will act at this point .
Pressure = h d g where h is height of centre of mass of water column , d is density of water and g is acceleration due to gravity .
Pressure of water column = .5 x 10³ x 9.8 = 4.9 k Pa .
Air is pressurized to 5 kPa so
resultant pressure on one side of the tank due to the air and water
= 4.9 + 5 kPa = 9.9 kPa .
Total force on one face = pressure x area of one face under water
= 9.9 x 10³ x .5 x 2²
= 19.8 kN .
How can force-time and force-displacement graphs be used to find the impulse or work done?
A. Area under force-time graph & Area under force-displacement graph
B. Area under force-displacement graph & Area under force-time graph
C. Gradient of force-time graph & Gradient of force-displacement graph
D. Gradient of force-displacement graph & Gradient of force-time graph
Answer:
A. Area under force-time graph & Area under force-displacement graph
Explanation:
To find the impulse or work done the area under force-time graph and area under force-displacement graph will give us these respective values.
Impulse = Force x time
Work done = Force x displacement
When we plot a graph of force and time, the area under it is the impulse.
When a graph of force and displacement is plotted, the area under is the work done.
A student is driving through a mountainous region where the road is at some times flat, at some times inclined upward, and at some time inclined downward. The student maintains a speed of 20 m/s on the roadway, but is required to make an emergency stop on the three sepearte occasions. On levels roadway, it takes 25 m to stop. On a downward-sloping roadway, it takes 40 m to stop. On an upward-sloping roadway, it takes 18 m to stop. Explain why the stopping distances are different. (Focus answer using work and energy, other concepts may be used as well but be sure work and energy are included.)
Answer:
Explanation:
It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .
At levelled road , for stoppage
Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .
At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .
At upward inclined road , for stoppage
Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force = (friction force + gravitational force ) x displacement .
Hence displacement is less .
At downward slopping road , friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle also .
At downward inclined road , for stoppage
Kinetic energy of vehicle + work done by gravitational force = Work done by frictional force = friction force x displacement .
Hence displacement is more .
Hence displacement is more in the downward slopping.
What is Displacement?Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.
It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .
At levelled road , for stoppage
Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .
At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .
At upward inclined road , for stoppage
Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force = (friction force + gravitational force ) x displacement .
Hence displacement is less .
At downward slopping road , friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle also .
At downward inclined road , for stoppage
Kinetic energy of vehicle + work done by gravitational force = Work done by frictional force = friction force x displacement .
Hence displacement is more in the downward slopping.
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Please help. I'm stuck!
What is the mass of a catamaran moving at 7.65 m/s that has a momentum of 530145 kg x m/s?
As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock. As the waves recede, they carry the sediment away. In this scenario, which process represents weathering, and which process represents erosion?
Answer:
WEATHERING is represented by the scenario (As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock).
Erosion is represented by the scenario (As the waves recede, they carry the sediment away).
Explanation:
A wave is a disturbance which travels through a medium and transfers energy from one point to another. When wind blows over a water body like the ocean, ocean waves are formed. As the generated energy from the wind is transported through the water by the waves, the can hit against rocks on the shores leading to its break down with time. WEATHERING occurs when tiny bit of sediments from rocks are loosened due to the impact of ocean waves.
Erosion can be described as the wearing away of the earth's surface due to the impact of wind, rainfall ( water) or waves. There are different types of erosion which is classified according it's cause of formation.
Wave erosion occurs when sediments such as sand, shell and other materials are carried to the shoreline by ocean waves. This erodes the shore over time as the sediments act like sandpapers.
An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the center of the fluid surface is depressed. At what angular velocity
Answer:
Explanation:
To find the angular velocity of the tank at which the bottom of the tank is exposed
From the information given:
At rest, the initial volume of the tank is:
[tex]V_i = \pi R^2 h_i --- (1)[/tex]
where;
height h which is the height for the free surface in a rotating tank is expressed as:
[tex]h = \dfrac{\omega^2 r^2}{2g} + C[/tex]
at the bottom surface of the tank;
r = 0, h = 0
∴
[tex]h = \dfrac{\omega^2 r^2}{2g} + C[/tex]
0 = 0 + C
C = 0
Thus; the free surface height in a rotating tank is:
[tex]h=\dfrac{\omega^2 r^2}{2g} --- (2)[/tex]
Now; the volume of the water when the tank is rotating is:
dV = 2π × r × h × dr
Taking the integral on both sides;
[tex]\int \limits ^{V_f}_{0} \ dV = \int \limits ^R_0 \times 2 \pi \times r \times h \ dr[/tex]
replacing the value of h in equation (2); we have:
[tex]V_f} = \int \limits ^R_0 \times 2 \pi \times r \times ( \dfrac{\omega ^2 r^2}{2g} ) \ dr[/tex]
[tex]V_f = \dfrac{ \pi \omega ^2}{g} \int \limits ^R_0 \ r^3 \ dr[/tex]
[tex]V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{r^4}{4} \Big]^R_0[/tex]
[tex]V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{R^4}{4} \Big] --- (3)[/tex]
Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.
Then [tex]V_f = V_i[/tex]
Replacing equation (1) and (3)
[tex]\dfrac{\pi \omega^2}{g}( \dfrac{R^4}{4}) = \pi R^2 h_i[/tex]
[tex]\omega^2 = \dfrac{4g \times h_i }{R^2}[/tex]
[tex]\omega =\sqrt{ \dfrac{4g \times h_i }{R^2}}[/tex]
[tex]\omega = \sqrt{\dfrac{4 \times 9.81 \ m/s^2 \times 0.7 \ m}{(0.5)^2} }[/tex]
[tex]\omega = \sqrt{109.87 }[/tex]
[tex]\mathbf{\omega = 10.48 \ rad/s}[/tex]
Finally, the angular velocity of the tank at which the bottom of the tank is exposed = 10.48 rad/s
a 14n force is applied for 0.33 seconds, calculate the impulse
Answer:
4.62 N-s
Explanation:
recall that the formula for impulse is given by
Impulse = Force x change in time
in our case, we are given
Force = 14 N
change in time = 0.33s
Simply substituting the above into the equation for impulse, we get
Impulse = Force x change in time
Impulse = 14 x 0.33
= 4.62 N-s
[tex]\\ \sf\longmapsto Impulse=Force(Time)[/tex]
[tex]\\ \sf\longmapsto Impulse=14(0.33)[/tex]
[tex]\\ \sf\longmapsto Impulse=4.62Ns[/tex]
What is the purpose of a tractor?
To plow fields.
To harvest wheat.
To haul loads of manure, hay etc.
To power other machines.
to plow fields beacuse it's much easier to plow on tractor than on foot
One disadvantage to experimental research is that experimental conditions do not always reflect reality.
Please select the best answer from the choices provided
T
F
Answer:
It's true I took the test on Edge.
Explanation:
Answer:
True
Explanation:
Got it right on edg
What do I know about dance movements?
Answer:
Hi lol
Explanation:
follow me lol
sarivigakarthi this is my account...
An astronaut with a mass of 75 kg stands on Mars (gravity = 3.7 m/s2). How much does the astronaut weigh?
Answer:
277.5 N.
Explanation:
From the question given above, the following data were obtained:
Mass (m) of astronaut = 75 kg
Acceleration due to gravity on Mar (gₘ) = 3.7 m/s²
Weight of astronaut on Mar (Wₘ) =.?
Weight of an object is simply defined by the following equation:
Weight (W) = mass × acceleration due to gravity (g)
W = m × g
With the above formula, we can obtain the weight of the astronaut on Mar as follow:
Mass (m) of astronaut = 75 kg
Acceleration due to gravity on Mar (gₘ) = 3.7 m/s²
Weight of astronaut on Mar (Wₘ) =.?
Wₘ = m × gₘ
Wₘ = 75 × 3.7
Wₘ = 277.5 N
Thus, the weight of the astronaut on Mar is 277.5 N