greyhound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but
3 leaps of the hound are equal to 5 leaps of the hare. Compare the speed of the
hound and the hare,
need full solution:-
[tex]{\large{\bold{\rm{\underline{Given \; that}}}}}[/tex]
★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.
[tex]{\large{\bold{\rm{\underline{To\; find}}}}}[/tex]
★ The speed of the hound and the hare
[tex]{\large{\bold{\rm{\underline{Solution}}}}}[/tex]
★ The speed of the hound and the hare = 25:18
[tex]{\large{\bold{\rm{\underline{Full \; Solution}}}}}[/tex]
[tex]\dashrightarrow[/tex] As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.
So firstly let us assume a metres as the distance covered by the hare in one leap.
Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.
But 3 leaps of the hound are equal to 5 leaps of the hare.
Henceforth, (5/3)a meters is the distance that is covered by the hound.
Now according to the question,
Hound pursues a hare and takes 5 leaps for every 6 leaps of the hare..! (Same interval)
Now the distance travelled by the hound in it's 5 leaps..!
(5/3)a × 525/3a metresNow the distance travelled by the hare in it's 6 leaps..!
6a metresNow let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!
25/3a = 6a25/3 = 625:18What energy store is in the torch
BEFORE it gets switched on?
Answer:
Chemical energy
Explanation:
The energy in the torch is stored as chemical energy before the torch gets switch on.
The chemical energy energy in the battery of cell will power the cell and allows it to produce light.
Chemical energy is a form of potential energy. The electrolytes within the battery are capable of producing electric current. So the chemical energy is transformed into electrical energy which is used to produce the light of the torch.Which of the following is NOT a step used to perform a scientific inquiry
Answer:
b. Designing an uncontrolled experiment.
Explanation:
They always have it controlled.
Answer:
B. Designing an uncontrolled experiment.
Explanation:
Correct Answer!!!!!!
We should stress again that the Carnot engine does not exist in real life: It is a purely theoretical device, useful for understanding the limitations of heat engines. Real engines never operate on the Carnot cycle; their efficiency is hence lower than that of the Carnot engine. However, no attempts to build a Carnot engine are being made. Why is that
Answer:
The Carnot engine has zero power
Explanation:
Although theoretically the Carnot engine has more efficiency than the real engine. In practice however they tend to have zero power.
This is because all its processes are reversible (that is isothermic and adiabatic).
So the system equilibrates with its surroundings at every point in time. This makes work done very slow and the power generated is zero.
Carnot cycles requires attaining isothermal heat transfers which is quite difficult and take a long time. Also a pump that can handle liquid-vapour phase mixture will be required.
This is not practical.
A box with a mass of 2 kg is pushed by a 10 N force. The acceleration
is
_m/s^2?
Answer:
a = 5 m/s^2
Explanation:
First, we look at Newton's 2nd Law:
F = ma
We now plug in the values,
10 N = 2 kg * a
10 N/2 kg = a
5 m/s^2 = a
A mass m is gently placed on the end of a freely hanging spring. The mass then falls 33 cm before it stops and begins to rise. What is the frequency of the oscillation
Answer:
Explanation:
The mass falls by .33 m before it begins to rise . At that point loss of potential energy is equal to gain of elastic energy .
1/2 k x² = mgx
.5 x k x .33² = m x 9.8 x .33
k / m = 59.4
frequency of oscillation = [tex]\frac{1}{2\pi} \times\sqrt{\frac{k}{m} }[/tex]
= [tex]\frac{1}{2\pi} \times\sqrt{59.4}[/tex]
= 1.22 per second .
The diagram shows two balls before they collide.
2 balls with grey arrows pointing to them from the outside. The left ball has below it m subscript 1 = 0.6 kilograms v subscript 1 = 0.5 meters per second. The right ball has below it m subscript 2 = 0.5 kilograms v subscript 2 = negative 0.2 meters per second.
What is the momentum of the system after the collision?
1. 0.0 kg • m/s
2. 0.2 kg • m/s
3. 0.3 kg • m/s
4. 0.4 kg • m/s
Answer:
The Answer is B)0.2 kg • m/s
Explanation:
I made a 100 on my test. Sorry if I'm late but hope I helped.
Answer:
B. 0.2 kg x m/s
Explanation:
A solid sphere of radius R = 5 cm is made of non-conducting material and carries a total negative charge Q = -12 C. The charge is uniformly distributed throughout the interior of the sphere.
What is the magnitude of the electric potential V at a distance r = 30 cm from the center of the sphere, given that the potential is zero at r = [infinity] ?
Answer:
V= -3.6*10⁻¹¹ V
Explanation:
Since the charge is uniformly distributed, outside the sphere, the electric field is radial (due to symmetry), so applying Gauss' Law to a spherical surface at r= 30 cm, we can write the following expression:[tex]E* A = \frac{Q}{\epsilon_{0} } (1)[/tex]
At r= 0.3 m the spherical surface can be written as follows:[tex]A = 4*\pi *r^{2} = 4*\pi *(0.3m)^{2} (2)[/tex]
Replacing (2) in (1) and solving for E, we have:[tex]E = \frac{Q}{4*\pi *\epsilon_{0}*r^{2} } = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)^{2} y} (3)[/tex]
Since V is the work done on the charge by the field, per unit charge, in this case, V is simply:V = E. r (4)Replacing (3) in (4), we get:[tex]V =E*r = E*(0.3m) = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)} = -3.6e11 V (5)[/tex]
V = -3.6*10¹¹ Volts.The electrical potential module will be [tex]-3.6*10^-^1^1 V[/tex]
We can arrive at this answer as follows:
To answer this, we owe Gauss's law. This is because the charge is evenly distributed across the sphere. This will be done as follows:[tex]E*A=\frac{Q}{^E0} \\\\\\A=4*\pi*r^2[/tex]
Solving these equations will have:[tex]E=\frac{Q}{4*\pi*^E0*r^2} \\E= \frac{(9e9N*m2/c2)*(-12C)}{(0.3m)^2y}[/tex]
As we can see, the electric potential is carried out on the field charge. In this case, using the previous equations, we can calculate the value of V as follows:[tex]V=E*r\\V=E*0.3m= \frac{(9e9N*m^2/C2)*(-12C)}{0.3m} \\V= -3.6*10^-^1^1 V.[/tex]
More information about Gauss' law at the link:
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Electron cloud configuration for
Answer:
electrons are located around the nucleus of an atom.
Explanation:
Electron configurations describe where electrons are located around the nucleus of an atom. For example, the electron configuration of lithium, 1s²2s¹, tells us that lithium has two electrons in the 1s subshell and one electron in the 2s subshell.
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m
Complete Question
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?
Answer:
[tex]H_m=1.65m[/tex]
[tex]H_E=1.16307m[/tex]
Explanation:
From the question we are told that
Mass of ball [tex]M=2kg[/tex]
Length of string [tex]L= 2m[/tex]
Wind force [tex]F=13.2N[/tex]
Generally the equation for [tex]\angle \theta[/tex] is mathematically given as
[tex]tan\theta=\frac{F}{mg}[/tex]
[tex]\theta=tan^-^1\frac{F}{mg}[/tex]
[tex]\theta=tan^-^1\frac{13.2}{2*2}[/tex]
[tex]\theta=73.14\textdegree[/tex]
Max angle =[tex]2*\theta= 2*73.14=>146.28\textdegree[/tex]
Generally the equation for max Height [tex]H_m[/tex] is mathematically given as
[tex]H_m=L(1-cos146.28)[/tex]
[tex]H_m=0.9(1+0.8318)[/tex]
[tex]H_m=1.65m[/tex]
Generally the equation for Equilibrium Height [tex]H_E[/tex] is mathematically given as
[tex]H_E=L(1-cos73.14)[/tex]
[tex]H_E=0.9(1+0.2923)[/tex]
[tex]H_E=1.16307m[/tex]
If a person visits an exercise facility, buys a new piece of fitness/sporting equipment,
or just starts planning to be active, which of the five stages of change for physical
activity are they at?
Planning
Maintenance
Precontemplation
Contemplation
Answer:planning
Explanation:
The person is in the stage of planning due to its action of planning to be active.
What is planning stage?The person is in the planning stage among the five stages of change for physical activity because the person just started planning to be active not yet started the activity. If a person is in the state of looking thoughtfully at something for a very long time then it is said to be Contemplation.
While on the other hand, if a person is in a stage in which there is no intention to change behavior in the foreseeable future then it is called precontemplation so we can conclude that the person is in the stage of planning due to its action of planning to be active.
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If an ice cube with the mass of 5.0 grams melts in a closed system such as a closed glass jar what is the mass of the liquid water after the ice cube completely melts
According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity. Do you think this equation is valid in any system of units
This question is incomplete, the complete question is;
According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula; h= (0.04 to 0.09)(D/d)⁴V²/2g
where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity.
Do you think this equation is valid in any system of units
Answer:
YES, the equation is a general equation that is valid in any system of units
Explanation:
Given the data in the question;
h = (0.04 to 0.09)(D/d)⁴ × [tex]\frac{V^{2} }{2g}[/tex]
so
[ N.m/N ] = (0.04 to 0.09) ( m/m)² × (m²/s²)1/2 × (s²/m)
[ N.L/N ] = (0.04 to 0.09) ( L⁴/L⁴) × (L²/T²)1/2 × (T²/L)
∴ [ L ] = (0.04 to 0.09) [L]
So as each term in the equation must have the same dimensions, the constant term (0.04 to 0.09) must be without dimension.
Therefore, YES, the equation is a general equation that is valid in any system of units
two identical balls are rolling down a hill ball 2 is rolling faster than ball 1 which ball has more kinetic energy
Consider two points in an electric field. The potential at point 1, V1, is 24V. The potential at point 2, V2, is 154V. A proton is moved from point 1 to point 2.
(a) Write an equation for the change of electric potential energy AU of the proton, in terms of the symbols given and the charge of the proton e.
(b) Find the numerical value of the change of the electric potential energy in electron volts (eV).
(c) Express v2, the speed of the electron at point 2, in terms of AU, and the mass of the electron me.
(d) Find the numerical value of v2 in m/s
Answer:
[tex]\triangle U=-e (V_2-V_1)[/tex]
[tex]\triangle U=130eV[/tex]
[tex]V_2=\sqrt{ \frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)}[/tex]
Explanation:
From the question we are told that
The potential at point 1, [tex]V_1 = 24V[/tex]
The potential at point 2, [tex]V_2 = 154V[/tex]
a)Generally work done by proton is given as
[tex]w=-\triangle U[/tex]
[tex]e\triangle V=-\triangle U[/tex]
[tex]\triangle U=-e (V_2-V_1)[/tex]
Generally the Equation for the change of electric potential energy AU of the proton, in terms of the symbols given and the charge of the proton e is mathematically given as
[tex]\triangle U=-e (V_2-V_1)[/tex]
b)Generally the electric potential energy in electron volts (eV). is mathematically given as
[tex]\triangle U=-e (154-24)V[/tex]
[tex]|\triangle U| =|-e (130)V|[/tex]
[tex]\triangle U=130eV[/tex]
c) Generally according to the law of conservation of energy
[tex](K.E+P.E)_1=(K.E+P.E)_2[/tex]
[tex]\frac{1}{2}meV_1^2+eV_1 =\frac{1}{2}mev_2^2+eV_2[/tex]
[tex]V_2^2=\frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)[/tex]
[tex]V_2=\sqrt{ \frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)}[/tex]
6. What is the lowest temperature on the Kelvin scale? What happens to matter when it
reaches this temperature?
7. What is different about the degrees on the Fahrenheit and Kelvin scales and the Celsius
and Kelvin scales?
A primary coil has 360 turns, while secondary has 120 turns in a transformer. The output voltage is 220V. What is the input voltage, and what type of transformer is it
Answer:
550V
Step - Down transformer
Explanation:
Given parameters:
Number of turns in primary coil = 300 turns
Secondary turns = 120 turns
Output voltage = 220V
Unknown:
Input voltage = ?
Type of transformer = ?
Solution:
To solve this problem, we use the expression below:
[tex]\frac{V_{out} }{V_{in} }[/tex] = [tex]\frac{Ns}{Np}[/tex]
So insert the parameters and find Vin;
[tex]\frac{220}{Vin}[/tex] = [tex]\frac{120}{300}[/tex]
120Vin = 220 x 300
Vin = [tex]\frac{220 x 300}{120}[/tex] = 550V
Since the input voltage is greater than the output voltage, this is step - down transformer.
If a person weighs 140 lb'on Earth, their mass in kilograms is
Answer:
70 kg
Explanation:
divide it by 2
Hope this helped!
Answer:
63.502932 Kilograms
Explanation:
Sam heaves a 16lb shot straight upward, giving it a constant upward acceleration from rest of 35 m/s^2 for 64.0 cm. He releases it 2.20m above the ground. You may ignore air resistance.
(a) What is the speed of the shot when Sam releases it?
(b) How high above the ground does it go?
(c ) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?
Answer:
6.69 m/s
4.483 m
1.42s
Explanation:
Given that:
Initial Velocity, u = 0
Final velocity, v =?
Acceleration, a = 35m/s²
1.) using the relation :
v² = u² + 2as
v² = 0 + 2(35) * 64*10^-2m
v² = 70 * 0.64
v = sqrt(44.8)
v = 6.693
v = 6.69 m/s
B.) height from the ground, h0 = 2.2
How high ball went , h:
Using :
v² = u² + 2as
Upward motion, g = - ve
0 = 6.69² + 2(-9.8)*(h - 2.2)
0= 6.69² - 19.6(h - 2.2)
44.7561 + 43.12 - 19.6h = 0
19.6h = 44.7561 - 43.12
h = 87.8761 / 19.6
h = 4.483 m
C.)
vt - 0.5gt² = h - h0
6.69t - 0.5(9.8)t²
6.69t - 4.9t² = 1.83 - 2.2
-4.9t² + 6.69t + 0.37 = 0
Using the quadratic equation solver :
Taking the positive root:
1.4185 = 1.42s
A particle has a velocity that is 90.% of the speed of light. If the wavelength of the particle is 1.5 x 10^-15 m, calculate the mass of the particle
Answer:
[tex]m=1.63\times 10^{-27}\ kg[/tex]
Explanation:
The velocity of a particle is 90% of the speed of light.
The wavelength of the particle is [tex]1.5\times 10^{-15}\ m[/tex]
We need to find the mass of the particle.
The formula for the wavelength of a particle is given by :
[tex]\lambda=\dfrac{h}{mv}[/tex]
h is Planck's constant
v is 90% of speed of light
m is mass of the particle
[tex]m=\dfrac{h}{\lambda v}\\\\m=\dfrac{6.63\times 10^{-34}}{1.5\times 10^{-15}\times 0.9\times 3\times 10^8}\\\\m=1.63\times 10^{-27}\ kg[/tex]
So, the mass of the particle is [tex]1.63\times 10^{-27}\ kg[/tex].
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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, puck 1 of mass m1 ! 0.20 kg is sent sliding across a frictionless lab bench, to undergo a one-dimensional elastic collision with stationary puck 2. Puck 2 then slides off the bench and lands a distance d from the base of the bench. Puck 1 rebounds from the collision and slides off the opposite edge of the bench, landing a distance 2d from the base of the bench. What is the mass of puck 2
Answer:
1 kg
Explanation:
Assuming that,
Δx(2) = v(2)t, where Δx(2) = d and v(2) = 2m1 / (m1 + m2) v1i
On the other hand again, if we assume that
Δx(1) = v(1)t, where Δx(1) = -2d, and v(1)t = m1 - m2 / m1 + m2 v1i
From the above, we proceed to dividing Δx(2) by Δx(1), so that we have
d/-2d = [2m1 / (m1 + m2) v1i] / [m1 - m2 / m1 + m2 v1i], this is further simplified to
1/-2 = [2m1 / (m1 + m2)] / [m1 - m2 / m1 + m2]
1/-2 = 2m1 / (m1 + m2) * m1 + m2 / m1 - m2
1/-2 = 2m1 / m1 - m2, if we cross multiply, we have
m1 - m2 = -2 * 2m1
m1 - m2 = -4m1
m2 = 5m1
From the question, we're told that m1 = 0.2 kg, if we substitute for that, we have
m2 = 5 * 0.2
m2 = 1 kg
An atom undergoes nuclear decay, but its atomic number is not changed.
What type of nuclear decay did the atom undergo?
A. Gamma decay
B. Beta decay
C. Nuclear fission
D. Alpha decay
Answer:
A. Gamma decay
Explanation:
A form of nuclear decay in which the atomic number is unchanged is a gamma decay.
The atom has undergone a gamma decay.
In a gamma decay, no changes occur to the mass and atomic number of the substance.
Gamma rays have zero atomic and mass numbers. When they cause decay, they cause no change to the mass and atomic numbers. They simply produce gamma rays during such reactions and these rays are very energetic.Mary is trying to pull Julie on a sled across a flat snowy field. Mary pulls on the rope attached to the sled. Her pulling force is directed horizontally. Julie weighs 109 pounds. The sled weights 12 pounds. If the coefficient of static friction between the sled runners and the snow is 0.42, how much force must Mary pull with (in lbs) to start moving the sled
Answer: F = 498.04 lbs
Explanation: The forces acting on the sled and Julie are show in the figure below. In it, we notice that, for the sled and Julie to go accross the field, they only need force of friction, because, force of friction is a force that resists the relative motion of surfaces.
Force of friction is given by the formula
[tex]F_{f}=\mu.F_{N}[/tex]
where
μ is coefficient of friction
[tex]F_{N}[/tex] is normal force
Normal force is the force the surface exerts on the object. It is always perpendicular and a force of contact.
In the case of the sled, since it is on a horizontal plane, Normal Force has the same magnitude of Gravitational Force. So
[tex]F_{N}=m.g[/tex]
Coefficient of friction is how much friction exists between two surfaces.
Rearraging friction force is
[tex]F_{f}=\mu.m.g[/tex]
Mass for this system is the sum of Julie and the sled, therefore
m = 109 + 12
m = 121 lb
Calculating Friction Force:
[tex]F_{f}=0.42.121.9.8[/tex]
[tex]F_{f}=[/tex] 498.04 lbs
LBS is a unit of measurement referred as pound by weight.
In conclusion, force Mary needs to start moving the sled is 498.04 lbs
Match each vocabulary word with the correct definition. 1. measure of how quickly velocity is changing 2. speed in a given direction 3. force that resists moving one object against another 4. measure of the pull of gravity on an object 5. tendency of an object to resist a change in motion 6. size friction acceleration velocity magnitude inertia weight friction
1. measure of how quickly velocity is changing . . . acceleration
2. speed in a given direction . . . velocity
3. force that resists moving one object against another . . . friction
4. measure of the pull of gravity on an object . . . weight
5. tendency of an object to resist a change in motion . . . inertia
6. size . . . magnitude
1. measure of how quickly velocity is changing is acceleration.
2. speed in a given direction is velocity.
3. force that resists moving one object against another is friction.
4. measure of the pull of gravity on an object is weight.
5. tendency of an object to resist a change in motion is inertia.
What is acceleration?Acceleration has the term used in mechanics to describe the pace at which the velocity of an object varies over time. Acceleration is a vector quantity (in that they have magnitude and direction). The direction of an object's acceleration is determined by the direction of the net force acting on it.
It is a vector quantity with an SI unit is m/s² and the dimension formula is LT⁻². A massive body will accelerate or alter its velocity at a constant rate when a constant force is applied to it, according to Newton's second law. In the simplest case, when a force is applied to an object at rest, it accelerates in the force's direction.
Therefore, 1. measure of how quickly velocity is changing is acceleration.
2. speed in a given direction is velocity.
3. force that resists moving one object against another is friction.
4. measure of the pull of gravity on an object is weight.
5. tendency of an object to resist a change in motion is inertia.
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a wooden block is cut into two pieces, one with three times the mass of the other. a depression is made in both faces of the cut so that a fire cracker can be placed in it and the block is reassembled. the reassembled block is set on rough surface and the fuse is lit. when the fire cracker explodes, the two blocks separate. what is the ratio of distances traveled by blocks?
Answer:
1/9
Explanation:
Let A denote the bigger piece and let B denote the smaller piece.
We are told that one with three times the mass of the other.
Therefore, we have;
M_a = 3M_b
Firecracker is placed in the block and it explodes and thus, momentum is conserved.
Thus;
V_ai = V_bi = 0
Where V_ai is initial velocity of piece A and V_bi is initial velocity of piece B.
Since initial momentum equals final momentum, we have;
P_i = P_f
Thus;
0 = (M_a × V_af) + (M_b × V_bf)
Since M_a = 3M_b, we have;
(3M_b × V_af) + (M_b × Vbf) = 0
Making V_af the subject, we have;
V_af = -⅓V_bf
The kinetic energy gained by each block during the explosion will later be lost due to the negative work done by friction. Thus;
W_f = -½M_b•(v_bf)²
Now, let's express the work is in terms of the force and the distance.
Thus;
W_f = F_f × Δx × cos 180°
Frictional force is also expressed as μmg
Thus;
W_f = -μM_b × g × Δx
Earlier, we saw that;
W_f = -½M_b•(v_bf)²
Thus;
-½M_b•(v_bf)²= -μM_b × g × Δx
Δx = (v_bf)²/2μg
Let the distance travelled by block A be Δx_a and that travelled by B be Δx_b
Thus;
Δx_a/Δx_b = ((v_ba)²/2μg)/((v_bf)²/2μg)
Δx_a/Δx_b = ((v_af)²/((v_bf)²)
Δx_a/Δx_b = (-⅓V_bf)²/(V_bf)²
Δx_a/Δx_b = 1/9
Calculate the work done to raise a charge of 25 coulombs through an emf of 8 volts.
1) 3
2) 200
Corrected, it's 2) 200
QUESTION 4.
If
you have 2 randomly selected vectors like R and R;
Show that R. RX 5) = 0
(102)
Answer:
Follows are the solution to this question:
Explanation:
Please find the correct question in the attachment file.
Let:
[tex]\overrightarrow{R}= R_i\hat{i}+R_j\hat{j}+R_k\hat{k}\\\\\overrightarrow{S}= S_i\hat{i}+S_j\hat{j}+S_k\hat{k}\\\\[/tex]
Calculating the value of [tex]\overrightarrow{R} \times \overrightarrow{S}:[/tex]
[tex]\to \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{K}\\R_i&R_j&R_k\\S_i&S_j&S_k\end{array}\right | = \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j][/tex]
Calculating the value of [tex]\overrightarrow{R} \cdot (\overrightarrow{R} \times \overrightarrow{S}):[/tex]
[tex]\to (R_i\hat{i}+R_j\hat{j}+R_k\hat{k}) \cdot ( \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j])[/tex]
by solving this value it is equal to 0.
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?
A turbofan operates at 25,000 ft and moves at 815 ft/s. It ingests 1.2 times the amount of air into the fan than into the core, which all exits through the fan exhaust. The fuel-flow-to-core airflow ratio is 0.0255. The exit densities of the fan and core are 0.00154 and 0.000578 slugs/ft3, respe~tively. The exit pressures from the fan and core are 10.07 and 10.26 psia, respectively. The developed thrust is 10,580 !bf, and the exhaust velocities from the fan and core are 1147 and 1852 ft/s, respectively. (a) Find the ingested air mass flow rate for the core and TSFC. (b) What are the exit areas of the fan and core nozzles
Answer:
a)
Mass flow rate of core = [tex]m_{e}[/tex] = 60.94 Kg/s
Mass flow rate of fan = [tex]m_{s}[/tex] = 73.12 kg/s
TSFC = 3.301 x [tex]10^{-5}[/tex]
b)
Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]
Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]
Explanation:
Data Given:
Height = 25000 ft
Vehicle velocity = [tex]u_{a}[/tex] = 815 ft/s = 248.41 m/s
[tex]m_{s} = 1.2m_{e}[/tex]
[tex]m_{f}[/tex] = 0.0255[tex]m_{e}[/tex]
Where,
[tex]m_{s}[/tex] = Mass flow rate of fan
[tex]m_{e}[/tex] = Mass flow rate of core
F = Thrust
Density of core = [tex]D_{e}[/tex] = 0.000578 slugs/[tex]ft^{3}[/tex] = 0.2979 kg/[tex]m^{3}[/tex]
Density of fan = [tex]D_{s}[/tex] = 0.00154 slugs/[tex]ft^{2}[/tex] = 0.7937 kg/[tex]m^{3}[/tex]
Ambient Pressure of Fan = [tex]P_{s}[/tex] = 10.07 Psi = 69430.21 Pa
Ambient Pressure of core = [tex]P_{e}[/tex] = 10.26 Psi = 70740.2 Pa
Thrust = F = 10580 lbf = 47062.2 N
Velocity of fan = [tex]u_{s}[/tex] = 1147 ft/s = 349.6 m/s
Velocity of core = [tex]u_{e}[/tex] = 1852 ft/s = 564.5 m/s
At the height of 25000 ft, P = 37600 [tex]P_{a}[/tex]
Now,
we have:
[tex]m_{e}[/tex] = [tex]u_{e}[/tex] x [tex]D_{e}[/tex] x [tex]A_{e}[/tex]
Plugging in the values, we get:
[tex]m_{e}[/tex] = 168.16 [tex]A_{e}[/tex] Equation 1
And,
[tex]m_{s}[/tex] = [tex]D_{s}[/tex] x [tex]A_{s}[/tex] x [tex]u_{s}[/tex]
[tex]m_{s}[/tex] = 277.5 [tex]A_{s}[/tex] Equation 2
As, we know,
[tex]m_{s} = 1.2m_{e}[/tex]
[tex]m_{s}[/tex] = 277.5 [tex]A_{s}[/tex]
And now for Thrust, we have:
F = [tex]A_{e}[/tex] x ([tex]P_{e}[/tex] - [tex]P_{a}[/tex] ) + [tex]A_{s}[/tex] x ([tex]P_{s}[/tex] - [tex]P_{a}[/tex] ) + [tex]m_{e}[/tex]x ([tex]u_{e}[/tex] - [tex]u_{a}[/tex] ) + [tex]m_{s}[/tex] x ([tex]u_{s}[/tex] - [tex]u_{a}[/tex] ) Equation 3
Now, substitute equation 1 and 2 in equation 3, we get:
Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]
Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]
Mass flow rate of core = [tex]m_{e}[/tex] = 60.94 Kg/s
Mass flow rate of fan = [tex]m_{s}[/tex] = 73.12 kg/s
TSFC = Thrust Specific Fuel Consumption = fuel mass flow rate / Thrust
TSFC = [tex]m_{f}[/tex]/F
And,
[tex]m_{f}[/tex] = 0.0255[tex]m_{e}[/tex]
[tex]m_{e}[/tex] = 60.94
[tex]m_{f}[/tex] = 0.0255 x 60.94
[tex]m_{f}[/tex] = 1.55397
TSFC = [tex]m_{f}[/tex]/F
TSFC = 1.55397/47062.2
TSFC = 3.301 x [tex]10^{-5}[/tex]
Low TSFC = High efficiency
High TSFC = Low efficiency
a)
Mass flow rate of core = [tex]m_{e}[/tex] = 60.94 Kg/s
Mass flow rate of fan = [tex]m_{s}[/tex] = 73.12 kg/s
TSFC = 3.301 x [tex]10^{-5}[/tex]
b)
Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]
Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]