(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.
(b) The force the ground exerts on the front set of wheels is 0.239 MN.
Center mass of the airplaneThe concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.
Some assumptionsThe wheels under the wind do not pass through the center line.The position of the front wheel is constant and it is zero mark (origin).The rear wheels are at 21.7 m markPosition of the center mass of the plane is calculated as follows;
Let the position of the center mass, Xcm = y
the center mass is 3 m in front of rear wheels, that is
21.7 - y = 3
y = 21.7 - 3
y = 18.7 m
Xcm = 18.7 m
Mass of the plane at the position of the rear wheelsLet the mass of the plane at front wheels = M1
Let the mass of the plane at rear wheels = M2
[tex]X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}[/tex]
[tex]18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg[/tex]
Force exerted by the ground on each rear wheelThere are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;
[tex]W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN[/tex]
Mass of the plane at the position of the front wheelM1 + M2 = 177,000
M1 = 177,000 - M2
M1 = 177,000 - 152,529.95
M1 = 24,470.05 kg
Force exerted by the ground on the front wheelW = mg
W = 24,470.05 x 9.8
W = 239,806.5 N = 0.239 MN
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A ballplayer catches the ball 4 seconds after throwing it straight up in the air. What speed did he throw it and how high did it go?
Answer:
Speed of the ball when it was thrown = 16.67 m/s
Height reached by the ball = 14.16 s
Explanation:
Given time in the ball comes to the same height from where the ball was thrown is 3.4 s.
Let us assume that the ball was thrown with a speed of u in the upward direction. Since only gravitational force acts on the particle in the downward direction, it has a constant acceleration in the downward direction.
We also know that a particle has the same magnitude of velocity when it is under a free fall at a fixed height but direction is opposite i.e., initially in the upward direction and finally in the downward direction.
https://tex.z-dn.net/?f=%5Ctherefore%20%5Cdfrac%7B-u-u%7D%7B3.4%7D%3D%20-9.8%5C%5C%5CRightarrow%20%5Cdfrac%7B2u%7D%7B3.4%7D%3D%209.8%5C%5C%5CRightarrow%202u%20%3D%209.8%5Ctimes%203.4%5C%5C%5CRightarrow%20u%20%3D%20%5Cdfrac%7B9.8%5Ctimes%203.4%7D%7B2%7D%5C%5C%5CRightarrow%20u%20%3D%2016.67%5C%20m%2Fs
Hence, the particle was thrown with a speed of 16.67 m/s.
Now, we also note that the particle has zero velocity at its maximum height.
So, from the initial position to the maximum height reached by the particle, we have
https://tex.z-dn.net/?f=0%5E2%20%3D%20u%5E2%2B2as%5C%5C%5CRightarrow%20s%20%3D%20%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2a%7D%5C%5C%5CRightarrow%20s%20%3D%20%5Cdfrac%7B0%5E2-16.67%5E2%7D%7B2(-9.8)%7D%5C%5C%5CRightarrow%20s%20%3D14.16%5C%20m
Hence, the maximum height reached by the ball is 14.16 m.
7. A 2.0 kg block, starting from rest, is pushed by a
constant force along a frictionless track. The
position of the block as a function of time is
recorded in the data above. The final momentum
of the block is
(A) 0.8 kgm/s
(B) 1.2 kgm/s
(C) 1.6 kgm/s
(D) 3.2 kgm/s
Answer:
(A) 0.8 kgm/s
Explanation:
because of the even ground it would only slow down
I WILL MARK BRAINLIEST IF YOU ANSWER THIS PHYSICS QUESTION!!!
A 21 kg child is riding a 5.9 kg bike with a velocity of 4.5 m/s to the northwest.
1. What is the momentum of the child and the bike together?
2. What is the momentum of the child?
3. What is the momentum of the bike?
Answer:
1. 121.05 kg m/s NW
2. 94.5 kg m/s NW
3. 26.55 kg m/s NW
Explanation:
As a conservation biologist for the Chesapeake Bay, you and your
colleagues have been conducting a research study that tracks the decrease
in the bald eagle population over the past few years.
What evidence can you find for the decrease in the bald eagle population?
As a conservation biologist for the Chesapeake Bay, you and your
colleagues have been conducting a research study that tracks the decrease
in the bald eagle population over the past few years.
What evidence can you find for the decrease in the bald eagle population?
A car of mass 1000 kg moves 3 km east in a straight line and then 4 km north. What is the total distance and displacement of the car from the initial position?
The net (resultant) force on the car is
Select one:
a) distance = 7 km and displacement = 5 km
b) distance = 5 km and displacement =7 km.
c) distance = 25 km and displacement =7 km.
d) distance = 7 km and displacement = 25 km
Answer:
a
Explanation:
Distance is simply the distance travelled which in this case would be 4km + 3km = 7km
To work out displacement, try to imagine the situation.
Draw a straight line to the east (label it 3) and then draw another line from the end of the first line upwards (label this one 4). Thus, you've created a right angles triangle. Now use pythagorean theorem to work out the displacement
4^2 + 3^2 = 25
sqrt 25 = 5 = displacement
Differences between fundamental and derived quantities
Answer:
Fundamental quantities are quantities that were created to measure a object/substance which are the basis on which derived quantities are formed, where as derived quantities are created by extracting variables from the fundamental quantities.
If an object accelerates from rest, with a constant acceleration of
10 m/s2, what will its velocity be after 2 s?
Answer: 20 m/s
Explanation:
Acceleration = change in velocity / time → 10 m/s^2 = final velocity - 0 (original velocity) / 2s → final velocity = 20 m/s
Explain the differences between horticulture and agriculture.
[tex] \tt{Difference \: Between \: \underline\red {Horticulture} \: And \: \underline\green{Agriculture}}[/tex]
[tex] \text{\underline\red {Horticulture}}[/tex]
The Cultivation of Fruits, Vegetables and Flowers for domestic and international markets are called Horticulture.
[tex] \text{\underline\green{Agriculture}}[/tex]
It is science, art and occupation of cultivating the soil, producing crops and livestock. It is systematic and controlled use of living organisms and the environment to improve the human condition.
Hope This HelpsWhich of the following particles is similar to a He nucleus?
alpha
beta
gamma
neutrino
Alpha
I hope this helps you
:)
Suppose a grower sprays (8.2x10^1) kg of water at 0 °C onto a fruit tree of mass 180 kg. How much heat is released by the water when it freezes?
Since there is no temperature change which drives heat flow, thus no heat will be released by the water.
Heat released by the water when it freezesThe heat released by the water when it freezes is calculated as follows;
Q = mcΔФ
where;
m is mass of waterc is specific heat capacity of waterΔФ is change in temperature = Фf - Фiwhen water freezes, the temperature, Фf = 0 °C
Q = 82 x 4200 x (0 - 0)
Q = 0
Since there is no temperature change which drives heat flow, thus no heat will be released by the water.
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What is an analogy of two different roads or rivers to compare a series and parallel circuit?
Answer:
In a series circuit, the same amount of current flows through all the components placed in it. On the other hand, in parallel circuits, the components are placed in parallel with each other due to which the circuit splits the current flow.
The terminal velocity of a 3 × 10^-5 raindrop is about 9 m/s. Assuming a drag force Fd = −bv, determine (a) the value of the constant b and (b) the time required for such a drop, starting from rest, to reach 63% of terminal velocity.
Based on the data provided, the value of the constant b is 3.27 × 10^-5 kg/s and the time required to reach 63% of terminal velocity is 0.58 s.
What is terminal velocity?The terminal velocity of a body is the velocity at which the body falls at constant velocity through a fluid.
For the falling raindrop, let positive direction be downwards and negative direction upwards,
mass of the raindrop, m = 3×10-5 kg velocity at time t, is v(t)terminal velocity, v0 = 9 m/sgravitational acceleration, g = 9.81 m/s²The raindrop experiences a downward gravitational force mg, and an upward drag force -bv.
The total force at a time t is given as
F(t) = mg - bv(t)a)
Terminal velocity is achieved then the total force is 0,
0 = mg - bv0
Therefore
b = mg/v0
Substitutingthe values:
b = (3 × 10^-5 × 9.8)/9
b = 3.27 × 10^-5 kg/s
b) Applying Newton's Second Law
F = ma
where
a = v/tF = mgTherefore,
mg = mv/t t = v/g
however, t is at 63% velocity
thus:
t = 0.63v/g
t = 0.63 × 9 /9.8
t = 0.58 s
Therefore, the value of the constant b is 3.27 × 10^-5 kg/s and the time required to reach 63% of terminal velocity is 0.58 s
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How fast must a proton move so that its kinetic energy is 70% of its total
energy?
I thought it would be 0.7c but that is wrong. I really don't know how to do this type of problem and my text book isn't any help.
Try this solution, all the details are in the attachment. note, the answer is marked with orange colour. If it is possible, check the provided solution in other sources.
Answer: ≈0.81c.
Define the following terms.
1. plate boundaries
2. seismic
3. seismic waves
4. body waves
5. surface waves
6. tectonic quakes
7. volcanic quakes
8. focus of earthquake
9. epicenter
10. fault
Answer:
I can only define a few of them, hope you don't mind.
Seismic- It means connected with or caused by earthquakes
Epicenter- It means the point on the earth's surface where the effects of an earthquake are most felt strongly.
Fault- A place where there is a break that is longer than usual in the layers of rock in the earth's crust.
Tectonic waves- Waves connected with the structure of the earth's surface.
Plate boundaries- The boundaries that differentiates the large sheets of rock (called PLATES) that form the earth's surface. Plate tectonics is the movement of the large sheets of rocks that form the earth's surface
[tex] \: \: \: \: \: \: \: \: \: [/tex]
How do the displacement, velocity, and acceleration of a runner change as he races from the starting
line toward the finish line along a straight path?
Answer:
maybe line toward the finish
Explanation:that question is confusing
Velocity is the pace at which your displacement is shifting; it describes how quickly and in which direction you are travelling. And acceleration, which is measured in meters per second, is the rate at which your velocity is changing (or meters per second squared).
What factor that affect the acceleration and velocity?Net force and mass are the two primary factors that impact how quickly an item accelerates. For instance, mass has an inverse relationship with acceleration while net force has a direct relationship.
The acceleration is also influenced by other elements like friction, air or fluid resistance, and pressure.
According to the second law, the mass of the item and the net force acting on it both impact how quickly an object accelerates.
Therefore, an object's acceleration is directly proportional to the net force applied on it and inversely proportional to its mass.
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As a block falls through the air by 40 meter it does work equal to -1800 joule. Determine the mass of a block.
Answer:
m = 4.5 kg
Explanation:
w = - 1800 j
Fd = - 1800 j
mgd = - 1800 j
m = - 1800 ÷(gd)
m = - 1800 ÷( 10×-40)
m = 4.5 kg
Two bodies of specific heats S1 and S2 having the same heat capacities are combined to form a single composite body. What is the specific heat of the composite body?
[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]
Heat capacity of body 1 :
[tex]\qquad \sf \dashrightarrow \:m_1s_1[/tex]
Heat capacity of body 2 :
[tex]\qquad \sf \dashrightarrow \:m_2s_2[/tex]
it's given that, the the head capacities of both the objects are equal. I.e
[tex]\qquad \sf \dashrightarrow \:m_1s_1 = m_2s_2[/tex]
[tex]\qquad \sf \dashrightarrow \:m_1 = \dfrac{m_2s_2}{s_1} [/tex]
Now, consider specific heat of composite body be s'
According to given relation :
[tex]\qquad \sf \dashrightarrow \:(m_1 + m_2) s' = m_1s_1 + m_2s_2[/tex]
[tex]\qquad \sf \dashrightarrow \:s' = \dfrac{ m_1s_1 + m_2s_2}{m_1 + m_2}[/tex]
[tex]\qquad \sf \dashrightarrow \:s' = \dfrac{ m_2s_2+ m_2s_2}{ \frac{m_2s_2}{s_1} + m_2 }[/tex]
[ since, [tex] m_2s_2 = m_1s_1 [/tex] ]
[tex]\qquad \sf \dashrightarrow \:s' = \dfrac{ 2m_2s_2}{ m_2(\frac{s_2}{s_1} + 1)}[/tex]
[tex]\qquad \sf \dashrightarrow \:s' = \dfrac{ 2 \cancel{m_2}s_2}{ \cancel{m_2}(\frac{s_2}{s_1} + 1)}[/tex]
[tex]\qquad \sf \dashrightarrow \:s' = \dfrac{ 2 s_2}{ (\frac{s_2 + s_1}{s_1} )}[/tex]
[tex]\qquad \sf \dashrightarrow \: s' = \dfrac{2s_1s_2}{s_1 + s_2} [/tex]
➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖
a 1. You found that the MCB was tied with a thread and the thread was fixed with a nail on the wall in your friend's house. i. Is it good idea to do this? ii. What could be the possible hazard of this? iii. What should have done to keep the circuit safe?
The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.
What is a miniature circuit breaker?A miniature circuit breaker is a circuit breaker that is used in homes as a means of guarding against damage to appliances due to a very high current.
This miniature circuit breaker is also harzardous in the sense that it could lead to an electrical fault related fire outbreak especially when it is being blown freely by wind as you tie it with a thread. Doing this a very bad idea because of the risk of a fire hazard.
The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.
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An unbalanced force acting on an object will cause it
A. increase in mass
B. decrease in mass
C. accelerate
D. remain at rest
Please help solve. Image down below
Answer:
thats is so hard
Explanation:
i cant answer that
9. aDefine refraction of light. Draw the ray diagram when an object is placed at C in a concave mirror Write any two natures of that image
Answer:
the process of bending of light when travling throw different mediume is called refraction of light.
If interstellar gas has a density of 1 atom/cm3, how big a volume of material must be used to make a star with the mass of the Sun
The volume of the material that must be used to make a star with the mass of the sun is 1.2×10⁵¹ m³.
What is volume?Volume is the amount of space occupied by an object or a plane figure.
To calculate the volume of the material that must be used to make a star with the mass of the sun, we use the formula below.
Formula:
D = m/V............ Equation 1Where:
D = Density of the interstellar gasm = mass of the sunV = Volume of the materialMake V the subject of the equation
V = m/D........... Equation 2From the question,
Given:
m = mass of the sun = 1.9891×10³⁰ kgD = 1 atom/cm³ = 1.66×10⁻²¹ kg/m³Substitute these values into equation 2
V = ( 1.9891×10³⁰)/(1.66×10⁻²¹)V = 1.2×10⁵¹ m³Hence, The volume of the material that must be used to make a star with the mass of the sun is 1.2×10⁵¹ m³.
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a
Which of these is a chemical
change?
A. water boiling
B. salt disolving
C. paper burning
Answer:
Burning coal and boiling water are both chemical changes. Burning coal is a chemical change, and boiling water is a physical change. Burning coal is a physical change, and boiling water is a chemical change.
Explanation:
A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume calorimeter. Consequently,
the temperature of the water rose by 4.20 ºC. If the heat capacity of the bomb plus water was 10.4 kJ/ºC,
calculate the molar heat of combustion of methanol.
Can someone please help me with this assignment, this is due today
Answer:
did you get it done if not lmk I will help you out tomorrow when I get up
Marla pushes on a bag of mulch with a force of 160 N. Dale helps her, pushing
with a force of 230 N in the same direction as Marla. The frictional force between
the bag and the floor is 90 N. What is the total force the bag of mulch feels? Draw
a diagram to scale
Answer:
300N
Explanation:
in your drawing make sure to put 230n and 160n on the same side with arrows facing the same direction on the opposite side of the box add an arrow with 0n and an arrow to represent friction
Someone please help me !!
Answer:25
Explanation: because higher means less kinetic energ
Consider a parallel-plate capacitor with plates of area A and with separation d.
Part A
Find F(V), the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor.
Express your answer in terms of given quantities and ϵ0.
View Available Hint(s)for Part A
F(V)
The magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor is determined as [tex]\frac{V^2 A \varepsilon _o }{2d^2}[/tex].
Magnitude of the force
The magnitude of the force each plate experiences due to the other plate is determined as follows;
F = U/d
where;
U is potential energy stored in the capacitor[tex]F = \frac{1}{2} \frac{Q^2}{C} \times \frac{1}{d} \\\\[/tex]
Q = CV
[tex]F = \frac{1}{2} \frac{C^2V^2}{C d} = \frac{CV^2}{2d}[/tex]
where;
C is the capacitanceThe capacitance is given as;
[tex]C = \frac{\varepsilon _o A }{d}[/tex]
[tex]F = \frac{\varepsilon _o A }{d} \times \frac{V^2}{2d} \\\\F = \frac{V^2 A \varepsilon _o }{2d^2}[/tex]
Thus, the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor is determined as [tex]\frac{V^2 A \varepsilon _o }{2d^2}[/tex].
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enumerate two ways that you practice to control manage noise pollution
1.
2.
please answer it correctly i really need it
nonsense-report
a small block with mass 0.04 kg is moving in the xy-plane. the net force on the block is describe by the potential energy function U(x,y) = (5.80 j/m2)x2 - (3.60 j/m3)y3. What are the magnitude of the acceleration of the block when it is at the point (x=0.30m , y=0.60m ) ?
Answer:
Explanation:
Use that:
[tex]F=-\vec{\nabla}U(x,y)=-\left(\hat{i} \frac{\partial U}{\partial x}+\hat{j}\frac{\partial U}{\partial y}\right)=-11.6x\hat{i}+10.8y^{2}\hat{j}[/tex]
Then use the 2nd Newton's Law of Motion:
[tex]\vec{a}=\frac{\vec{F}}{m}=\frac{-11.6x\hat{i}+10.8y^{2}\hat{j}}{0.04}=-290x\hat{i}+270y^{2}\hat{j}[/tex]
At x = 0.3 and y = 0.6, we can find the acceleration as:
[tex]\vec{a}=-87\hat{i}+97.2\hat{j}[/tex] (in SI unit)
Then the magnitude of the acceleration on that point is:
[tex]a=\sqrt{(-87)^{2}+(97.2)^{2}}\approx 130.44[/tex] (SI Unit)