Answer:
The constant rate of acceleration required in order to accomplish this is 1.921 feet per square second.
Explanation:
Let suppose that car accelerates uniformly in a rectilinear motion. Given that initial and final speeds and travelled distances are known, then the acceleration needed by the vehicle ([tex]a[/tex]), measured in feet per square second, is determined by the following kinematic formula:
[tex]a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot \Delta x }[/tex] (1)
Where:
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds, measured in feet per second.
[tex]\Delta x[/tex] - Travelled distance, measured in feet.
If we know that [tex]v_{o} = 34.907\,\frac{ft}{s}[/tex], [tex]v_{f} = 67.027\,\frac{ft}{s}[/tex] and [tex]\Delta x = 852\,ft[/tex], then acceleration needed to accomplish the task is:
[tex]a = 1.921\,\frac{ft}{s^{2}}[/tex]
The constant rate of acceleration required in order to accomplish this is 1.921 feet per square second.
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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describe why people are better off not consuming an additional good or service if the marginal cost is greater than the marginal benefit.
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:
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.
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]
What 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.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].
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]
1. A particle is projected vertically upwards with a velocity of 30 ms from a point 0. Find (a) the maximum height reached(b) the time taken for it to return to 0 (c) the taken for it to be 35m below 0
Assuming the particle is in free fall once it is shot up, its vertical velocity v at time t is
v = 30 m/s - g t
where g = 9.8 m/s² is the magnitude of the acceleration due to gravity, and its height y is given by
y = (30 m/s) t - 1/2 g t ²
(a) At its maximum height, the particle has 0 velocity, which occurs for
0 = 30 m/s - g t
t = (30 m/s) / g ≈ 3.06 s
at which point the particle's maximum height would be
y = (30 m/s) (3.06 s) - 1/2 g (3.06 s)² ≈ 45.9184 m ≈ 46 m
(b) It takes twice the time found in part (a) to return to 0 height, t ≈ 6.1 s.
(c) The particle falls 35 m below its starting point when
-35 m = (30 m/s) t - 1/2 g t ²
Solve for t to get a time of about t ≈ 7.1 s
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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two identical balls are rolling down a hill ball 2 is rolling faster than ball 1 which ball has more kinetic energy
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
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
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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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
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.
An unstrained horizontal spring has a length of 0.29 m and a spring constant of 180 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.021 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.
Answer:
A) they both have the same algebraic sign
B)6.377×10^-6 C
Explanation:
From columb's law, the force acting on both charges can be expressed as
F=( kq1*q2)/r^2
Where F= electrostatic force
r= distance between the charges
q1 and q2= charges
The force acting on a spring can be expressed as
F= kx..................eqn(2)
Where
K= spring constant = 180 N/m.
x= stretch of the string= 0.021m
Substitute the values into eqn (2)
F= (180×0.021)
F= 3.78N
If we compare with spring force,
Hence, F( electrostatic) = 3.78N
From
F=( kq1*q2)/r^2 ..............eqn(1)
Where
r= (0.29 m + 0.021m)= 0.311m
K= the electrostatic constant= 8.99×10^9 kg⋅m3⋅s−2⋅C−2.
If we substitute the values we have
Since the charges are the same, then
kq1 and q2 equals "q"
3.78= (8.99×10^9 ×q^2)/(0.311)^2
Making q^2 subject of the formula
3.78× (0.311)^2 = 8.99×10^9 × q^2
q^2= [(0.311)^2 × 3.78]/8.99×10^9
q^2= 40.668×10^-12
q=√40.668×10^-12
q= 6.377×10^-6 C
(a) the possible algebraic signs
They have the same algebraic sign
(b) the magnitude of the charges.
6.377×10^-6 C
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
ts
An electric light bulb mixer is used for 19.2 seconds. In that time, it transfers 1
536 J of energy.
Calculate the power output of the cake mixer.
A
I DONT KNOW
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
Leticia leaves the grocery store And walks 150 M’s to parking lot then she turns 90° to the right and walks an additional 70 M’s to her car what is the magnitude of displacement of her car from the grocery store at exit
Answer:
Explanation:
Its 165.5m
What do you think about the attached scenario?
A piece of aluminum with a mass of 3.05 g initially at a temperature of 10.8 °C is heated to a temperature of 20.
Assume that the specific heat of aluminum is 0.901 J/(g°C).
How much heat was needed for this temperature change to take place?
Answer:
25.3J
Explanation:
Given parameters:
Mass of aluminum = 3.05g
Initial temperature = 10.8 °C
Final temperature = 20 °C
Specific heat = 0.9J/g °C
Unknown:
Amount of heat needed for the temperature to change = ?
Solution:
To solve this problem, we use the expression:
H = m C Ф
H is the amount of heat
m is the mass
C is the specific heat capacity
Ф is the change in temperature
H = 3.05 x 0.901 x (20 - 10.8) = 25.3J
a 250.0 g snowball of radius 4.00 cm starts from rest at the top of the peak of a roof and rolls down a section angled at 30.0 degrees
Answer:
The response to this question is as follows:
Explanation:
The whole question and answer can be identified in the file attached, please find it.
The force diagram of all the forces acting on the snowball include the normal force acting upwards, the weight of the snowball acting downwards and the frictional force acting horizontal.
The given parameters;
mass of the snow ball, m = 250 gradius of the snow ball, r = 4 cmangle of inclination of the plane, θ = 30 ⁰The force diagram of all the forces acting on the snowball is calculated as follows;
↑ N
⊕ → F
↓ W
Where;
N is the normal force on the snowballF is the frictional force on the snowballW is the weight of the ballThus, the force diagram of all the forces acting on the snowball include the normal force acting upwards, the weight of the snowball acting downwards and the frictional force acting horizontal.
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A 3 kg ball moving to the right at 4 m/sec collides with a 4 kg ball moving to the right at 2 m/sec. Find the final velocities of the balls in m/sec if the coefficient of restitution is 0.6.
A. 2.2, 3.4
B. 1, 2
C. 4, 5
D. 6, 8
Answer:
Option A
Explanation:
To solve this problem we need to apply the momentum conservation, and analyze the data.
For this problem, I will call the initial velocities as V₁ and V₂, while the final velocities will be V₃ and V₄.
According to the momentum principle, this states the following:
m₁V₁ + m₂V₂ = m₁V₃ + m₂V₄ (1)
From this equation we can write an expression in function of V₃ and V₄. We also know that coefficient of restitution is 0.6. Knowing this, we can write the expression that will help us to solve for the final velocities:
e = V₄ - V₃ / 2 (2)
With both expressions we can solve for the final velocities. Let's use (1) first and see what we can simplify first by replacing the given data:
(3*4) + (4*2) = 3V₃ + 4V₄
12 + 8 = 3V₃ + 4V₄
20 = 3V₃ + 4V₄ (3)
This is all we can do for now. Let's use (2) now:
0.6 = V₄ - V₃ / 2
1.2 = V₄ - V₃
V₄ = 1.2 + V₃ (4)
Now, we can replace (4) into (3), and then, solve for V₃:
20 = 3V₃ + 4(1.2 + V₃)
20 = 3V₃ + 4.8 + 4V₃
15.2 = 7V₃
V₃ = 15.2 / 7
V₃ = 2.17 m/sWe have the value of one final velocity, let's see the other one.
V₄ = 1.2 + V₃
V₄ = 1.2 + 2.17
V₄ = 3.37 m/sThe closest values to these results are in option A, so this will be the correct option.
Hope this helps
, 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
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]
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 = 625:18If 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:
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!!!!!!