Answer:
257 kN.
Explanation:
So, we are given the following data or parameters or information in the following questions;
=> "A jet transport with a landing speed
= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"
= > The distance = 425 m along the runway with constant deceleration."
=> "The total mass of the aircraft is 140 Mg with mass center at G. "
We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"
Step one: determine the acceleration;
=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.
=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.
Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).
= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).
= 257 kN.
The reaction N under the nose wheel B towards the end of the braking interval = 257 kN
Given data :
Landing speed of Jet = 200 km/h
Distance = 425 m
Total mass of aircraft = 140 Mg with mass center at G
Determine the reaction N under the nose of wheel B First step : calculate the value of the Jet accelerationJet acceleration = 1 / (2 *425) * (200² - 60² ) * 1 / (3.6)²
= 3.3 m/s²
Next step : determine the reaction N under the nose of WheelReaction N = Total mass of aircraft * jet acceleration* 1.2 = 15N - (9.8*2.4* 140). ----- ( 1 )
∴ Reaction N = 140 * 3.3 * 1.2 = 15 N - ( 9.8*2.4* 140 )
Hence Reaction N = 257 KN
We can conclude that the The reaction N under the nose wheel B towards the end of the braking interval = 257 kN
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A 11.7-Ω resistor is made from a coil of copper wire whose total mass is 13.5 g . The resistivity of copper is 1.68×10−8Ω⋅m, the density of copper is 8.9×103kg/m3.
a) What is the diameter of the wire?
b) What is the length of the wire?
Answer:
a) d = 7.62 10⁻⁶ m, b) l = 3.25 10⁴ m
Explanation:
Resistance is expressed by the formula
R = ρ l / A (1)
density is defined by
density = m / V
the volume of a wire is the cross section by the length
V = A l
we substitute
density = m / A l
A = m / density l
we substitute in 1
R = ρ l density l / m
R =ρ density l² / m
l = √ (R m /ρ density)
let's calculate the cable length
l = √(11.7 13.5 10⁻³ / (1.68 10⁻⁸ 8.9 10³))
l = √(10.56 10⁸)
l = 3.25 10⁴ m
now we can find the cable diameter with the density equation
A = m / density l
A = 13.5 10⁻³ / (8.9 10³ 3.25 10⁴)
A = 4,557 10⁻¹¹ m²
the area of the circle is
A = π r² = π d² / 4
d = √ (4A /π)
d = √ (4 4,557 10⁻¹¹/π)
d = 7.62 10⁻⁶ m
The diameter of the wire is 0.202m and the length of the wire is 4.47*10^-5m
Data;
Resistor = 11.7Ωmass = 13.5gresistivity of copper = 1.68 * 10 ^-8 Ω.mdensity of copper = 8.9*10^3 kg/m^3Resistivity of CopperThe resistivity of copper is calculated by
[tex]R = \frac{\rho L}{A}\\[/tex]
Let's calculated the volume of the wire first;
[tex]\rho = \frac{mass}{volume} \\volume = \frac{mass}{density} \\volume = \frac{13.5*10^-^3}{8.9*10^3} \\v = 1.52*10^-6m^3[/tex]
The diameter of the wire will be
[tex]R = \frac{\rho L}{A}\\ R = \frac{\rho LA}{A^2}\\ R = \frac{\rho V}{A^2} \\A^2 = \frac{\rho V}{R}\\ A^2 = \frac{8.9*10^3 * 1.516*10^-^6}{11.7} \\A^2 = 0.0011\\A = \sqrt{0.0011} \\A = 0.034m^2[/tex]
Taking the area
[tex]a = \pi r^2\\0.034 = 3.14 * r^2\\r^2 = \frac{0.034}{3.14} \\r^2 = 0.01083\\r = \sqrt{0.01083}\\ r = 0.104m\\d = 2r\\d = 2 * 0.104\\d = 0.208m[/tex]
Length of the wire can be calculated as
[tex]V = AL\\L = V/A\\L = \frac{1.52*10^-^6}{0.034}\\ L = 4.47*10^-^5m[/tex]
The diameter of the wire is 0.202m and the length of the wire is 4.47*10^-5m
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car travel at the speed of 20 km/hr for 2 hour and 60 km/hr for next 2 hour find average speed
The index of refraction of a certain material is 1.5. If I send red light (700 nm) through the material, what will the frequency of the light be in the material
Answer: [tex]4.29\times10^{14}\text{ Hz}[/tex]
Explanation:
Given: Speed of red light = 700 nm
= [tex]700\times10^{-9}[/tex] m
[tex]= 7\times10^{-7}[/tex] m
Frequency of red light = [tex]\dfrac{\text{Speed of light}}{\text{Speed of red light}}[/tex]
Speed of light = [tex]3\times10^8[/tex] m
Then, Frequency of red light = [tex]\dfrac{3\times10^8}{7\times10^{-7}}[/tex]
[tex]=0.429\times10^{8-(-7)}=0.429\times10^{15}\\\\=4.29\times10^{14}\ Hz[/tex]
Hence, Frequency of red light = [tex]4.29\times10^{14}\text{ Hz}[/tex]
The frequency of the light be in the material is [tex]4.29\times10^{14}\text{ Hz}[/tex].
Where do most metamorphic processes take place?
Answer:
Most metamorphic processes take place deep underground, inside the earth's crust.
Explanation:
During metamorphism, protolith chemistry is mildly changed by increased temperature (heat), a type of pressure called confining pressure, and/or chemically reactive fluids. hope this helps you :)
Science activity
Imagine that some settlers have left Earth and gone to the Moon, taking
their recipe books with them. The first cake they baked was a disaster. It had
far too little moisture and was about six times the size they had expected.
the cake recipe was:
1.25 N butter
1.50 N sugar
4 eggs
1.50 N flour
20 ml milk
ANALYTICAL THINKING
Q. Why was the cake so big? Why was it se
dry?
Answer:
Answer in explanation
Explanation:
The reason for the big size and less moisture of the cake is due to difference in weight of the ingredients on the surface of moon. So, the same has the lesser weight on the surface of the moon than it has on the surface of earth. Or in other words, The same weight of the ingredients will have greater mass and thus the greater quantity on the surface of earth than the surface of earth. For example, on earth 1.25 N butter will have a mass:
m = W/g = 1.25 N/(9.8 m/s²) = 0.13 kg
But, on moon:
m = W/g = 1.25 N/(1.625 m/s²) = 0.77 kg
Hence, it is clear that the mass of the same weight of the substance becomes 6 times greater on the surface of moon. This explains why the cake was so big.
Now, coming to the second part about the dryness of the cake. The main and only source of moisture in recipe is the eggs bu the eggs are taken in a quantity of numbers. So they are exactly the same on moon as well. While all the other ingredients are increased, the same amount of eggs are not sufficient to provide them with enough moisture. Hence the cake was dry.
When you replace helium in a balloon with less-dense hydrogen, does the buoyant force on the balloon change if the balloon remains the same size?
Answer:
No change
Explanation:
First, we hare to understand what we mean by buoyant force.
Archimedes Principle states that
"the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object."
Hot air balloons rise into the air as a result of the density of the air inside the balloon is less dense i.e warmer air than the air outside the balloon i.e cooler air. This is basically how the balloons work. Now applying this to our question
Replacing Helium with less dense Hydrogen will make no difference to the buoyant force because the volume of the balloon did not change. The buoyant force depends on the weight of the displaced air, and not on the force causing the displacement.
Equal but opposite charges Q are placed on the square plates of an air-filled parallel-plate capacitor. The plates are then pulled apart to twice their original separation, which is small compared to the dimensions of the plates. Which of the following statements about this capacitor are true?
A) The energy stored in the capacitor has doubled.
B) The energy density in the capacitor has increased.
C) The electric field between the plates has increased.
D) The potential difference across the plates has doubled.
E) The capacitance has doubled.
Answer:
A& D
Explanation:
See attached file
Three resistors, 6.0-W, 9.0-W, 15-W, are connected in parallel in a circuit. What is the equivalent resistance of this combination of resistors?
Answer:
2.9Ω
Explanation:
Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
Where;
Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.
Note that;
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
Therefore;
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω
The equivalent resistance of this combination of resistors is 2.9Ω.
Calculation of the equivalent resistance:The combined resistance in such arrangement of resistors is provided by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
here.
Req means the equivalent resistance and R1, R2, R3
.Rn means the resistance of individual resistors interlinked in parallel.
Also,
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
So,
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω
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Two factors that regulate (control) glandular secretion.
Answer:
The factors include age and puberty
Explanation:
Glandular secretion release chemicals such as hormones in response to the body’s metabolic needs.
As an individual ages , the metabolic rate of the body also reduces . This is due to the stress and ageing of the cells of the body. This explains why glandular secretion is optimal with young people and Lower in older people. It also explains why the immune system of a young person is mostly stronger than older people.
Puberty is another factor which affects glandular secretion as during puberty there is usually a high amount of hormonal changes due to high levels of secretions of some hormones. These hormones could however inhibit the other glandular secretions.
10 pts! :) If Kyla picks up a grocery bag, using 10 N of force to lift it 1.5 m off the floor, how much work did Kyla do on the bag?
Explanation:
work = force x Distance
w = 10 x 1.5 = 15Nm
The amount of work done by Kyla in lifting the bag is 15 J.
What is meant by work done ?Work done on an object is defined as the cross product of the force applied on the object and the vertical displacement of the object.
Here,
Force applied by Kyla to pick up the bag, F = 10 N
Vertical displacement of the bag, s = 1.5 m
The work done by Kyla in lifting the bag,
W = F x s
W = 10 x 1.5
W = 15 J
Hence,
The amount of work done by Kyla in lifting the bag is 15 J.
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Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person in second with apparent frequency of 3400 Hertz what was the speed of cars
Answer:
v ’= 21.44 m / s
Explanation:
This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s
f ’= f (v + v₀) / (v-[tex]v_{s}[/tex])
where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer
in this exercise both the source and the observer are moving, we will assume that both have the same speed,
v₀ = v_{s} = v ’
we substitute
f ’= f (v + v’) / (v - v ’)
f ’/ f (v-v’) = v + v ’
v (f ’/ f -1) = v’ (1 + f ’/ f)
v ’= (f’ / f-1) / (1 + f ’/ f) v
v ’= (f’-f) / (f + f’) v
let's calculate
v ’= (3400 -3000) / (3000 +3400) 343
v ’= 400/6400 343
v ’= 21.44 m / s
If your brain is 0.4 m higher than your heart when you are standing, how much lower is your blood pressure at your brain than it is at your heart? The density of blood plasma is about 1025 kg/m3 and a typical maximum (systolic) pressure of the blood at the heart is 120 mg of Hg (= 16 kP = 1.6 × 104 N/m2). Give your answer in mg of Hg.
Answer:
The correct answer is 88.84 mmHg.
Explanation:
The pressure differential between the brain and the heart while standing up will be 120 - rho × g (gravity) × h, here h is the distance from the brain to the heart. The h is 40 cm or 0.4 m.
rho×g×h = 1060 kg/m³×9.8 m/s²×0.4m
= 4155 Pa
Now converting Pa to mmHg we get:
4155 Pa × 760 mmHg / 1.01325 × 10⁵ Pa
= 31.16 mmHg
Thus, the pressure in the brain now is 120 - 31.16
= 88.84 mmHg (hypotension)
One hundred turns of insulated copper wire are wrapped around an iron core of cross-sectional area 0.100m2. As the magnetic field along the coil axis changes from 0.5 T to 1.00T in 4s, the voltage induced is:
Answer:
The voltage induced in the coil is 1.25 V.
Explanation:
Given;
number of turns, N = 100 turns
cross sectional area of the copper coil, A = 0.1 m²
initial magnetic field, B₁ = 0.5 T
final magnetic field, B₂ = 1.00 T
duration of change in magnetic field, dt = 4 s
The induced emf in the coil is calculated as;
[tex]emf = -N\frac{\delta \phi}{\delta t} \\\\emf = - N (\frac{\delta B}{\delta t}) A\\\\emf = -N (\frac{B_1 -B_2}{\delta t} )A\\\\emf = N(\frac{B_2-B_1}{\delta t} )A\\\\emf = 100(\frac{1-0.5}{4} )0.1\\\\emf = 1.25 \ Volts[/tex]
Therefore, the voltage induced in the coil is 1.25 V.
Two long parallel wires are separated by 11 cm. One of the wires carries a current of 54 A and the other carries a current of 45 A. Determine the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current.
Explanation:
It is given that,
The separation between two parallel wires, r = 11 cm = 0.11 m
Current in wire 1, [tex]q_1=54\ A[/tex]
Current in wire 2, [tex]q_2=45\ A[/tex]
Length of wires, l = 4.3 m
We need to find the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current. The magnetic force per unit length is given by :
[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 54\times 45\times 4.3}{2\pi \times 0.11}\\\\F=0.0189\ N[/tex]
So, the magnetic force on a 4.3 m length of the wire on both of currents is F=0.0189 N.
A 22g bullet traveling 210 m/s penetrates a 2.0kg block of wood and emerges going 150m/s. If the block were stationary on a frictionless plane before the collision, what is the velocity of the block after the bullet passes through
Answer:
The final velocity of the block after the bullet passes through is 0.66 meters per second.
Explanation:
The interaction between the bullet and the block of woods is a clear example of a perfectly inelastic collision, which can be modelled after the Principle of Momentum Conservation. There are no external forces exerted on the bullet-block system. The equation describing the collision is described below:
[tex]m_{B}\cdot v_{B,o} + m_{W}\cdot v_{W,o} = m_{B}\cdot v_{B,f} + m_{W}\cdot v_{W,f}[/tex]
Where:
[tex]m_{B}[/tex], [tex]m_{W}[/tex]- Masses of the bullet and the block of wood, measured in kilograms.
[tex]v_{B,o}[/tex], [tex]v_{W,o}[/tex] - Initial speeds of the bullet and the block of wood, measured in meters per second.
[tex]v_{B,f}[/tex], [tex]v_{W,f}[/tex]- Final speeds of the bullet and the block of wood, measured in meters per second.
The final speed of the block is cleared:
[tex]v_{W,f} = \frac{m_{B}\cdot (v_{B,o}-v_{B,f})+m_{W}\cdot v_{W,o}}{m_{W}}[/tex]
[tex]v_{W,f} = v_{W,o} + \frac{m_{B}}{m_{W}} \cdot (v_{B,o}-v_{B,f})[/tex]
If [tex]v_{W,o} = 0\,\frac{m}{s}[/tex], [tex]m_{B} = 0.022\,kg[/tex], [tex]m_{W} = 2\,kg[/tex], [tex]v_{B,o} = 210\,\frac{m}{s}[/tex] and [tex]v_{B,f} = 150\,\frac{m}{s}[/tex], then the final velocity of the block after the bullet passes through is:
[tex]v_{W,f} = 0\,\frac{m}{s}+\left(\frac{0.022\,kg}{2\,kg}\right)\cdot \left(210\,\frac{m}{s}-150\,\frac{m}{s} \right)[/tex]
[tex]v_{W,f} = 0.66\,\frac{m}{s}[/tex]
The final velocity of the block after the bullet passes through is 0.66 meters per second.
A solid 200-g block of lead and a solid 200-g block of copper are completely submerged in an aquarium filled with water. Each block is suspended just above the bottom of the aquarium by a thread. Which of the following is true?
A. More information is needed to choose the correct answer.
B. The buoyant force on the copper block is greater than the buoyant force on the lead block.
C. The buoyant force on the lead block is greater than the buoyant force on the copper block.
D. The buoyant force is the same on both blocks.
Answer:
B. The buoyant force on the copper block is greater than the buoyant force on the lead block.
Explanation:
Given;
mass of lead block, m₁ = 200 g = 0.2 kg
mass of copper block, m₂ = 200 g = 0.2 kg
density of water, ρ = 1 g/cm³
density of lead block, ρ₁ = 11.34 g/cm³
density of copper block, ρ₂ = 8.96 g/cm³
The buoyant force on each block is calculated as;
[tex]F_B = mg(\frac{density \ of \ fluid}{density \ of \ object} )[/tex]
The buoyant force of lead block;
[tex]F_{lead} = 0.2*9.8(\frac{1}{11.34} )\\\\F_{lead} = 0.173 \ N[/tex]
The buoyant force of copper block
[tex]F_{copper} = 0.2*9.8(\frac{1}{8.96})\\\\F_{copper} = 0.219 \ N[/tex]
Therefore, the buoyant force on the copper block is greater than the buoyant force on the lead block
A force acting on an object moving along the x axis is given by Fx = (14x - 3.0x2) N where x is in m. How much work is done by this force as the object moves from x = -1 m to x = +2 m?
Answer:
72J
Explanation:
distance moved is equal to 3m.then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
The answer is 72J.
Distance moved is equal to 3m.
Then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
Is there any definition of force?A force is a push or pulls upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.
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Consider an electromagnetic wave where the electric field of an electromagnetic wave is oscillating along the z-axis and the magnetic field is oscillating along the x-axis.
Required:
In what directions is it possible that the wave is traveling?
Answer:
The wave is traveling in the y axis direction
Explanation:
Because the wave will always travel in a direction 90° to the magnetic and electric components
A microwave oven uses 1,200 watts. It runs a total of 6 hours during the corse of a week. How much energy does the microwave oven use during the week?? a 4.8 kilowatt-hours. b 5.0 kilowatt-hours. c 7.2 kilowatt-hours. d 6.5 kilowatt-hours.
Answer:
7.2 kilowatts
Explanation:
Energy= power x time
where:
power = 1,200watts to kilowatts is 1.2kw
time = 6hours
therefore,
energy = 1.2kw x 6hrs
=7.2 kilowatts
You push a shopping cart full of groceries. The shopping cart has a mass of 32
kg.
a. What is the weight of your shopping cart?
b. How much force must you apply to give the cart an acceleration of 1.3 m/s2?
c. If you push with a force of 200 N, what is the acceleration of the cart?
d. You are driving home from the store. If your car has an acceleration of 4.1 m/s2 and its engine is applying a forward force of 7000 N, what is the mass of your car?
Answer:
a) 320N (if you take gravity as 10ms^-2) or 313.6N (if you take gravity as 9.8ms^-2
b) 41.6N
c) 6.25ms^-2
d) 1707.317 kg
Explanation:
a) W=mg
W= 32 x 10 or 9.8
W = 320N or 313.6N
b) F=ma
F= 32 x 1.3
F= 41.6N
c) F=ma
200 = 32 x a
a= 6.25ms^-2
d) F=ma
7000= m x 4.1
m= 1707.317 kg
I am not completely sure about the d) part because I dont whether you will be taking Friction and Normal Reaction too. As per my knowledge, I think no, as no angles nor the gradient of the floor/road is mentioned here.
a) The weight of your shopping cart is 313.6N .
b) The required force is 41.6N
c) The acceleration of the car is 6.25ms^-2
d) The mass of the car is 1707.317 kg.
What is acceleration?Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).
a) The weight of the shopping cart: W=mg
W= 32 x 9.8 N
W = 313.6N
b) The required force is to be applied: F=ma
F= 32 x 1.3 N.
F= 41.6N.
c) Let the acceleration of the cart is a.
Then, force: F=ma
200 = 32 x a
a= 6.25ms^-2
Hence, the acceleration of the cart is 6.25ms^-2.
d) Let the mass of the car is m.
Force applied on the car: F=ma
7000= m x 4.1
m= 1707.317 kg
The mass of the car is 1707.317 kg.
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Find an article online or application in your daily life involving rotating objects and physics.
Answer:
the planet Earth is a good example
Water enters a typical garden hose of diameter 0.016 m with a velocity of 3 m/s. Calculate the exit velocity of water from the garden hose when a nozzle of diameter 0.0050 m) is attached to the end of the hose in units of m/s.
Answer:
v₂ = 306.12 m/s
Explanation:
We know that the volume flow rate of the water or any in-compressible liquid remains constant throughout motion. Therefore, from continuity equation, we know that:
A₁v₁ = A₂v₂
where,
A₁ = Area of entrance pipe = πd₁²/4 = π(0.016 m)²/4 = 0.0002 m²
v₁ = entrance velocity = 3 m/s
A₂ = Area of nozzle = πd₂²/4 = π(0.005 m)²/4 = 0.0000196 m²
v₂ = exit velocity = ?
Therefore,
(0.0002 m²)(3 m/s) = (0.0000196 m²)v₂
v₂ = (0.006 m³/s)/(0.0000196 m²)
v₂ = 306.12 m/s
The temperature coefficient of resistivity for copper is 0.0068 (C°)-1. If a copper wire has a resistance of 104 Ω at 20°C, what is its resistance 80°C?
Answer:
R₈₀ = 146.43 Ω
Explanation:
The resistance of a resistor depends upon many factors. One of the main factors of the change in resistance of a resistor is the change in temperature. The formula for the resistance at a temperature other than 20°C is given as follows:
R₈₀ = R₀(1 + αΔT)
where,
R₈₀ = Resistance of wire at 80°C = ?
R₀ = Resistance of wire at 20° C = 104 Ω
α = Temperature coefficient of resistance for copper = 0.0068 °C⁻¹
ΔT = T₂ - T₁ = 80°C - 20°C = 60°C
Therefore,
R₈₀ = (104 Ω)[1 + (0.0068°C⁻¹)(60°C)]
R₈₀ = 146.43 Ω
A 1-kilogram mass is attached to a spring whose constant is 18 N/m, and the entire system is then submerged in a liquid that imparts a damping force numerically equal to 11 times the instantaneous velocity. Determine the equations of motion if the following is true?
a. the mass is initially released from rest from a point 1 meter below the equilibrium position
b. the mass is initially released from a point 1 meter below the equilibrium position with an upward velocity of 11 m/s
Answer:
Let [tex]x(t)[/tex] denote the position (in meters, with respect to the equilibrium position of the spring) of this mass at time [tex]t[/tex] (in seconds.) Note that this question did not specify the direction of this motion. Hence, assume that the gravity on this mass can be ignored.
a. [tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].
b. [tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].
Explanation:
Let [tex]x[/tex] denote the position of this mass (in meters, with respect to the equilibrium position of the spring) at time [tex]t[/tex] (in seconds.) Let [tex]x^\prime[/tex] and [tex]x^{\prime\prime}[/tex] denote the first and second derivatives of [tex]x[/tex], respectively (with respect to time [tex]t[/tex].)
[tex]x^\prime[/tex] would thus represent the velocity of this mass.[tex]x^{\prime\prime}[/tex] would represent the acceleration of this mass.Constructing the ODEConstruct an equation using [tex]x[/tex], [tex]x^\prime[/tex], and [tex]x^{\prime\prime}[/tex], with both sides equal the net force on this mass.
The first equation for the net force on this mass can be found with Newton's Second Law of motion. Let [tex]m[/tex] denote the size of this mass. By Newton's Second Law of motion, the net force on this mass would thus be equal to:
[tex]F(\text{net}) = m\, a = m\, x^{\prime\prime}[/tex].
The question described another equation for the net force on this mass. This equation is the sum of two parts:
The restoring force of the spring: [tex]F(\text{spring}) = -k\, x[/tex], where [tex]k[/tex] denotes the constant of this spring.The damping force: [tex]F(\text{damping}) = - 11\,x^\prime[/tex] according to the question. Note the negative sign in this expression- the damping force should always oppose the direction of motion.Assume that there's no other force on this mass. Combine the restoring force and the damping force obtain an expression for the net force on this mass:
[tex]F(\text{net}) = -k\, x - 11\, x^\prime[/tex].
Combine the two equations for the net force on this mass to obtain:
[tex]m\, x^{\prime\prime} = -k\, x - 11\, x^\prime[/tex].
From the question:
Size of this mass: [tex]m = 1\; \rm kg[/tex].Spring constant: [tex]k = 18\; \rm N \cdot m^{-1}[/tex].Hence, the equation will become:
[tex]x^{\prime\prime} = -18\, x - 11\, x^\prime[/tex].
Rearrange to obtain:
[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex].
Finding the general solution to this ODE[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex] fits the pattern of a second-order homogeneous ODE with constant coefficients. Its auxiliary equation is:
[tex]m^2 + 11\, m + 18 = 0[/tex].
The two roots are:
[tex]m_1 = -2[/tex], and[tex]m_2 = -9[/tex].Let [tex]c_1[/tex] and [tex]c_2[/tex] denote two arbitrary real constants. The general solution of a second-order homogeneous ODE with two distinct real roots [tex]m_1[/tex] and [tex]m_2[/tex] is:
[tex]x = c_1\, e^{m_1\cdot t} + c_2\, e^{m_2\cdot t}[/tex].
For this particular ODE, that general solution would be:
[tex]x = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex].
Finding the particular solutions to this ODENote, that if [tex]x(t) = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex] denotes the position of this mass at time [tex]t[/tex], then [tex]x^\prime(t) = -2\,c_1\, e^{-2 t} -9\, c_2\, e^{-9 t}[/tex] would denote the velocity of this mass at time
The position at time [tex]t = 0[/tex] would be [tex]x(0) = c_1 + c_2[/tex].The velocity at time [tex]t = 0[/tex] would be [tex]x^\prime(0) = -2\, c_1 - 9\, c_2[/tex].For section [tex]\rm a.[/tex]:
[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = -\frac{9}{7} \\ &c_2 = \frac{2}{7}\end{aligned}\right.[/tex].
Hence, the particular solution for section [tex]\rm a.[/tex] will be:
[tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].
Similarly, for section [tex]\rm b.[/tex]:
[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = \frac{2}{7} \\ &c_2 = -\frac{9}{7}\end{aligned}\right.[/tex].
Hence, the particular solution for section [tex]\rm b.[/tex] will be:
[tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].
Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero? =
Answer:
x = -3 cm
Explanation:
The electrical potential is the sum of the potentials of each charge
V = k ∑ [tex]q_{i} / r_{i}[/tex]
let's apply this to our case where the potential is V = 0 for x = 0
0 = k (q₁ / (x₁-0) + q₂ / (x₂-0) + q₃ / (x₃-0))
in our case
q₁ = + 2.0 10⁻⁶ C
q₂ = - 6.0 10⁻⁶ C
q₃ = + 3.0 10⁻⁶ C
x₁ = -1.0 cm = 1.0 10⁻² m
x₂ = +2.0 cm = 2.0 10⁻² m
we substitute in the equation
0 = k (2 10⁻⁶ / 1 10⁻² - 6 10⁻⁶ / 2 10⁻² + 3 10⁻⁶ / x)
3 10⁻⁶ / x = 2 10⁻⁴ - 3 10⁻⁴
3 10⁻⁶ / x = -1 10⁻⁴
x = - 3 10⁻² m
x = -3 cm
A 43.0-g toy car is released from rest on a frictionless track with a vertical loop of radius R. The initial height of the car is h = 4.05R.
Required:
a. What is the speed of the car at the top of the vertical loop?
b. What is the magnitude of the normal force acting on the car at the top of the vertical loop?
Answer:
A.) 909 cm/s
B.) 33075 N
Explanation:
A.) Given that the
Mass M = 43 g
Height h = 4.05 R
Radius r = R
At the top of the loop, the maximum potential energy P.E = mgh
Substitutes m and h into the formula where g = 9.8 m/s^2 = 9610.517 cm/s^2
P.E = 43 × 9610.517 × 4.05R
P.E = 1673671.536R J
According to conservative of energy
The maximum P.E = maximum K.E
But K.E = 1/2mv^2
1673671.536R = 1/2mv^2
Substitutes for mass m into the formula
1673671.536R = 1/2× 4.05R × v^2
The R will cancel out
Cross multiply
4.05 v^2 = 3347343.072
V^2 = 3347343.072 / 4.05
V^2 = 826504.4622
V = sqrt( 826504.4622)
V = 909 cm/s
B.) At the top of the loop, the centripetal force = the sum of the normal force N and the weight W of the car. That is,
MV^2/R = N + W
Make N the subject of formula
N = mv^2/ R - W
Where W = mg
Substitute all the parameters into the formula
N = (4.05R × 909^2) /R - 4.05R × 9610.517
N = 3346438.05 - 38922.59
N = 3307515 N
A 12 kg box is pulled across the floor with a 48 N horizontal force. If the force of friction is 12 N, what is the acceleration of the box?
Answer:
The acceleration of the box is 3 m/s²
Explanation:
Given;
mass of the box, m = 12 kg
horizontal force pulling the box forward, Fx = 48 N
frictional force acting against the box in opposite direction, Fk = 12 N
The net horizontal force on the box, F = 48 N - 12 N
The net horizontal force on the box, F = 36 N
Apply Newton's second law of motion to determine the acceleration of the box;
F = ma
where;
F is the net horizontal force on the box
a is the acceleration of the box
a = F / m
a = 36 / 12
a = 3 m/s²
Therefore, the acceleration of the box is 3 m/s²
A charged particle moves through a region of space at constant speed, without deflecting. From this, one can conclude that in this region: Select the correct answer
a) There must be no magnetic field in the region There could be electric and magnetic fields, oriented perpendicular to each other.
b) There could be electric and magnetic fields, oriented in opposite directions There must be no magnetic field and no electric field in the region.
c) There could be electric and magnetic fields, oriented in the same direction Your Ans
Answer:
There could be electric and magnetic fields, oriented in opposite directions
Explanation:
Lorentz force, is the force that may be exerted on a body of a specified magnitude of charge q, moving with a velocity v, in a magnetic field B and in an electric field of intensity E. This Lorentz force is given by; F= qE+qvBsin ϕ
However, if the motion of the particle is opposite to the magnetic field such a that ϕ = 0, then there is no net magnetic force on the charge and it moves freely, with a constant velocity and in a straight line. Hence, there is no magnetic field in the region.
The charge moves with constant speed due to same direction of magnetic and electric field.
There could be electric and magnetic fields that is oriented in the same direction or the other reason is that there is no magnetic field and electric field in that region where the charge moves. If the electric and magnetic field are present at the same direction then it means that it applies no force on the charge.
This is due to more distance from the charge as well as the charge travels away from the field occupies by the magnetic and electric field so we can conclude that the charge moves with constant speed due to same direction of magnetic and electric field.
Learn more: https://brainly.com/question/17132472
1. The frequency of a wave defines
O A. the minimum height of a wave.
O B. the maximum height of a wave.
O C. how fast the wave is moving in cycles per second.
D. the height of the wave at a given point.
Answer:
The answer is CExplanation:
Frequency, in physics, the number of waves that pass a fixed point in unit time; also, the number of cycles or vibrations undergone during one unit of time by a body in periodic motion. ... See also angular velocity; simple harmonic motion.
What would be the correct value of m in the interference equation if the growth process were started again and the diamond layer were grown to three times the thickness t
Answer:
To calculate the correct value of m if the thickness t is grown 3 times again we can deduce that:
2*(3u)*t = m* lambda
Making m the subject of the formula will give the formula:
m= 6*u*t/ Lambda
Given: Lambda= 633*10^9 while u and t are unknown
Therefore the value of m can be calculated given the formula below:
m= 6*u*t/ 633*10^9
Explanation:
To calculate the correct value of m if the thickness t is grown 3 times again we can deduce that:
2*(3u)*t = m* lambda
Making m the subject of the formula will give the formula:
m= 6*u*t/ Lambda
Given: Lambda= 633*10^9 while u and t are unknown
Therefore the value of m can be calculated given the formula below:
m= 6*u*t/ 633*10^9