Answer:
40m/s
Explanation:
v²=u²+2as
v²=0²+2(16)(50)
v²=160v=40m/s
On a low-friction track, a 0.66-kg cart initially going at 1.85 m/s to the right collides with a cart of unknown inertia initially going at 2.17 m/s to the left. After the collision, the 0.66-kg cart is going at 1.32 m/s to the left, and the cart of unknown inertia is going at 3.22 m/s to the right. The collision takes 0.010 s.
What is the unknown inertia?
What is the average acceleration of the heavier cart?
What is the average acceleration of the lighter cart?
Answer:
(a) the unknown inertia is 0.388 kg
(b) the average acceleration of the heavier cart is 317 m/s²
(c) the average acceleration of the lighter cart is 539 m/s²
Explanation:
Given;
mass of the first cart, m₁ = 0.66 kg
initial speed of the first cart, u₁ = 1.85 m/s
let the mass of the cart with unknown inertia be m₂
initial velocity of the second cart, u₂ = 2.17 m/s to the left
velocity of the first cart after collision, v₁ = 1.32 m/s to the left
velocity of the second cart after collision, v₂ = 3.22 m/s
time of collision, t = 0.010 s
(a) What is the unknown inertia?
Apply the principle of conservation of linear momentum, to determine the unknown inertia.
let leftward direction be negative direction
let rightward direction be positive direction
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.66(1.85) + m₂(-2.17) = 0.66(-1.32) + m₂(3.22)
1.221 - 2.17m₂ = -0.8712 + 3.22m₂
1.221 + 0.8712 = 3.22m₂ + 2.17m₂
2.0922 = 5.39m₂
m₂ = 2.0922 / 5.39
m₂ = 0.388 kg
The unknown inertia is 0.388 kg
(b) the average acceleration of the heavier cart
the heavier cart has a mass of 0.66 kg
[tex]a = \frac{v_1 - u_1}{t} \\\\a = \frac{-1.32 - 1.85}{0.01} \\\\a = -317 \ m/s^2\\\\|a| = 317 \ m/s^2[/tex]
(c) the average acceleration of the lighter cart;
the lighter cart has a mass of 0.388 kg
[tex]a = \frac{v_2 - u_2}{t} \\\\a = \frac{3.22 - (-2.17)}{0.01} \\\\a =\frac{3.22 \ +\ 2.17}{0.01} \\\\a= 539\ m/s^2[/tex]
take a picture of an object in your house, describe the
energy stores and transfers that happen with it. You can be as imaginative as you wish
with the object (choose something unusual), but the stores you identify and transfers
that happen must be real.
pls give me ideas of what to take a photo of for this I'm really stuck :(
A 2028 kg Oldsmobile traveling south on Abbott Road at 14.5 m/s is unable to stop on the ice covered intersection for a red light at Saginaw Street. The car collides with a 4146 kg truck hauling animal feed east on Saginaw at 9.7 m/s. The two vehicles remain locked together after the impact. Calculate the velocity of the wreckage immediately after the impact. Give the speed for your first answer and the compass heading for your second answer. (remember, the CAPA abbreviation for degrees is deg) -1.75
Answer:
v = 8.1 m/s
θ = -36.4º (36.4º South of East).
Explanation:
Assuming no external forces acting during the collision (due to the infinitesimal collision time) total momentum must be conserved.Since momentum is a vector, if we project it along two axes perpendicular each other, like the N-S axis (y-axis, positive aiming to the north) and W-E axis (x-axis, positive aiming to the east), momentum must be conserved for these components also.Since the collision is inelastic, we can write these two equations for the momentum conservation, for the x- and the y-axes:We can go with the x-axis first:[tex]p_{ox} = p_{fx} (1)[/tex]
⇒ [tex]m_{tr} * v_{tr}= (m_{olds} + m_{tr}) * v_{fx} (2)[/tex]
Replacing by the givens, we can find vfx as follows:[tex]v_{fx} = \frac{m_{tr}*v_{tr} }{(m_{tr} + m_{olds)} } = \frac{4146kg*9.7m/s}{2028kg+4146 kg} = 6.5 m/s (3)[/tex]
We can repeat the process for the y-axis:[tex]p_{oy} = p_{fy} (4)[/tex]
⇒[tex]m_{olds} * v_{olds}= (m_{olds} + m_{tr}) * v_{fy} (5)[/tex]
Replacing by the givens, we can find vfy as follows:[tex]v_{fy} = \frac{m_{olds}*v_{olds} }{(m_{tr} + m_{olds)} } = \frac{2028kg*(-14.5)m/s}{2028kg+4146 kg} = -4.8 m/s (6)[/tex]
The magnitude of the velocity vector of the wreckage immediately after the impact, can be found applying the Pythagorean Theorem to vfx and vfy, as follows:[tex]v_{f} = \sqrt{v_{fx} ^{2} +v_{fy} ^{2} }} = \sqrt{(6.5m/s)^{2} +(-4.8m/s)^{2}} = 8.1 m/s (7)[/tex]
In order to get the compass heading, we can apply the definition of tangent, as follows:[tex]\frac{v_{fy} }{v_{fx} } = tg \theta (8)[/tex]
⇒ tg θ = vfy/vfx = (-4.8m/s) / (6.5m/s) = -0.738 (9)
⇒ θ = tg⁻¹ (-0.738) = -36.4º
Since it's negative, it's counted clockwise from the positive x-axis, so this means that it's 36.4º South of East.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.
Learn more about physical activity here: https://brainly.com/question/1561572
what type of waves can only travel through a medium?
Answer:
Mechanical waves
Explanation:
Mechanical waves are the waves that can travel only through a medium. Mechanical waves are disturbance of matter and require medium to transfer the energy. There are three types of mechanical waves that include transverse wave, longitudinal wave and surface wave.
Some of the examples of mechanical waves are sound waves and seismic waves etcetera.
Hence, the correct answer is "Mechanical waves".
An empty elevator has a mass of 722 kg. It moves between floors at a maximum speed of 6.00 m/s. The elevator is stopped on the 20th floor of the building when someone pushes the call button in the lobby (the first floor). Assuming that it takes 15.25 meters for the elevator to reach its maximum speed, and assuming constant acceleration, calculate the tension in the cable as the elevator car begins to descend (during the period of constant acceleration). Take the acceleration due to gravity to be 9.81 m/s2. Show your work.
Answer:
Explanation:
The lift is going down with acceleration
Initial speed u = 0
Final speed v = 6 m/s
distance s = 15.25 m
acceleration a = ?
v² = u² + 2 a s
6² = 0 + 2 x a x 15.25
a = 1.18 m /s²
Elevator is going down with acceleration .
mg - T = ma where T is tension in the cable .
722 x 9.8 - T = 722 x 1.18
7075.6 - T = 851.96
T = 6223.64 N .
a car accidently rolls off a cliff. as it leaves the cliff it has a horizontal velocity of 13m/s it hits the ground 60m from the shoreline. calculate the height of the cliff
Answer:
104.59 m
Explanation:
From the question given above, the following data were obtained:
Horizontal velocity (u) = 13 m/s
Horizontal distance (s) = 60 m
Height of cliff =?
Next, we shall determine the time taken for the car to hit the ground. This can be obtained as follow:
Horizontal velocity (u) = 13 m/s
Horizontal distance (s) = 60 m
Time (t) =?
s = ut
60 = 13 × t
Divide both side by 13
t = 60 /13
t = 4.62 s
Finally, we shall determine the height of the cliff. This can be obtained as follow:
Time (t) = 4.62 s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) =?
h = ½gt²
h = ½ × 9.8 × 4.62²
h = 4.9 × 21.3444
h = 104.59 m
The, the height of the cliff is 104.59 m
A 5.7 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4.5 + 13.7 x − 1.5 x 2 , where Fx is in Newtons and x is in meters. Find the work done by this force on the particle as the particle moves from x = 0 m to x = 1.9 m. Answer in units of J.
Answer:
The work done by the force on the particle is 29.85 J.
Explanation:
The work is given by:
[tex] W = ^{x_{2}}_{x_{1}}\int F_{x} dx [/tex]
Where:
x₁: is the lower limit = 0 m
x₂: is the upper limit = 1.9 m
Fₓ: is the force in the horizontal direction = (4.5 + 13.7x - 1.5x²)N
[tex]W = ^{1.9}_{0}\int (4.5 + 13.7x - 1.5x^{2}) dx[/tex]
[tex] W = 4.5x|^{1.9}_{0} + \frac{13.7}{2}x^{2}|^{1.9}_{0} - \frac{1.5}{3}x^{3}|^{1.9}_{0} [/tex]
[tex] W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3} [/tex]
[tex] W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3} [/tex]
[tex]W = 29.85 J[/tex]
Therefore, the work done by the force on the particle is 29.85 J.
I hope it helps you!
Why does it rain more in West Ferris than in East Ferris? Explain your answer.
Answer:
This idea helps students explain why more rain forms over West Ferris than East Ferris. ... Therefore, when students explain that water vapor condenses higher in the atmosphere, they are actually explaining that water vapor condenses high in the troposphere, which is relatively low in the atmosphere.
Explanation:
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a particle with a charge of 5.5 x 10^-8 c is 3.5 cm from a particle with a charge of -2.3 x10^-8 c. the potential energy of this two particle system relative to the potential energy at infinite separation is:
Answer:
-32.5 * 10^-5 J
Explanation:
The potential energy of this system of charges is;
Ue = kq1q2/r
Where;
k is the Coulumb's constant
q1 and q2 are the magnitudes of the charges
r is the distance of separation between the charges
Substituting values;
Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)
Ue= -32.5 * 10^-5 J
The potential energy of this two particle system relative to the potential energy at infinite separation is [tex]\bold {-32.5x 10^-^5\ J}[/tex].
The potential energy of this system of charges,
[tex]\bold {Ue = k\dfrac{q1q2}{r}}[/tex]
Where;
k - Coulumb's constant
q1 and q2 - magnitudes of the charges
r - distance between the charges
Put the values in the equation,
[tex]\bold {Ue = 9.0x10^9\times \dfrac {5.5 x 10^{-8} C \times -2.3 x10^{-8} C}{3.5 \times 10^{-2}}}\\\\\bold {Ue= -32.5 x 10^-^5\ J}[/tex]
Therefore, the the potential energy of this two particle system relative to the potential energy at infinite separation is [tex]\bold {-32.5x 10^-^5\ J}[/tex].
To know more about charges,
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A stretched string has a mass per unit length of 5.00 g/cm and a tension of 10.0 N. A sinusoidal wave on this string has an amplitude of 0.12 mm and a frequency of 100 Hz and is traveling in the negative direction of an x axis. If the wave equation is of the form y(x, t) = ym sin(kx ± ωt), what are (a) ym, (b) k, (c) ω, and (d) the correct choice of sign in front of ω?
Answer:
0.12 mm ; 140.50 rad/m ; 628.32 rad/sec ; +
Explanation:
Given the wave equation of the form :
y(x, t) = ym sin(kx ± ωt)
Mas per unit length (u) = 5 g/cm = (5÷1000)kg / 0.01m) = 0.005kg/0.01m = 0.5kg/m
Tension, T = 10 N
Amplitude, A = 0.12 mm
Frequency, F = 100 Hz
Comparing with the general wave equation :
y = Asin(kx ± ωt)
A = amplitude = ym = 0.12 mm
2.) k = 2π / λ
Recall :
v = fλ
v = sqrt(T/u) = sqrt(10/0.5) = sqrt(20) = 4.472
λ = v/ f = 4.472 / 100 = 0.04472
Hence,
k = (2 * π) / 0.04472
k = 140.50 rad/m
3.) Angular frequency, ω
ω = 2πf = 2 * 3.14 * 100 = 628.32 rad/sec
4.) sign is +ve
Direction of wave propagation as given is in the negative x axis
A 75 kg window cleaner uses a 10 kg ladder that is 5.0 m long. He places one end on the ground 2.5 m from a wall, rests the upper end against a cracked Window, and climbs the ladder. He is 3.0 m up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder does not slip. When the window is on the verge of breaking, what are (a) the magnitude of the force on the window from the ladder, (b) the magnitude of the force on the ladder from the ground, and (c) the angle (relative to the horizontal) of that force on the ladder
Solution :
Given : Mass of ladder = 10 kg
Length of ladder = 5 m
Weight of window cleaner = 75 kg
a). Now equate the torque about the lowermost point of the ladder is given by :
[tex]$=10 \times 9.8 \times \frac{2.5}{2} + 75 \times 9.8 \times \frac{3}{5} \times 2.5 = N \times \sqrt{5^2 - 2.5^2}$[/tex]
Here, N = normal force that the glass exerts on the ladder
Therefore, [tex]$N = 282.9 \ N$[/tex]
= 280 N (in 2 significant figures)
b). Equate the forces along horizontal direction,
The horizontal component of the friction, [tex]$F_x = N = 282.9 \ N $[/tex]
The vertical component of the friction, [tex]$F_y = (10+75) \times 9.8$[/tex]
= 833 N
Therefore, the net frictional force, [tex]$F = \sqrt{F_x^2+F_y^2}$[/tex]
[tex]$F = \sqrt{(282.9)^2+(833)^2}$[/tex]
= 879.7 N
= 880 N (in 2 significant figures)
c). The angle the forces makes [tex]$= \tan \frac{833}{282.9} $[/tex]
[tex]$= 71.2 ^\circ $[/tex]
Therefore in 2 significant figures = [tex]$71 ^\circ$[/tex]
A child uses one hand to charge a balloon by rubbing it against her shirt. She then holds a rod in her other hand and finds that the balloon and rod, when brought close to one another, repel. Which one of the following is true?
a. the rod must be a conductor
b. the rod must be an insulator
c. it could be either
Answer:
B. The rod must be an insulator
Explanation:
We have that this rod must be an insulator. After this child rubbed the balloon, the balloon acquired static charge. So holding a rid against it is going to cause it to repel, this is to say it is repelling because the rod also is carrying some static charges. If this rid was a conductor, there would be no charge in its surface. The charge would have passed through her hand as it comes in contact.
One of the disadvantages of experimental research is that __________.
A.
it isn’t easily replicated
B.
it doesn’t often reflect reality
C.
the results aren’t generalizable
D.
conditions are not controllable
Please select the best answer from the choices provided
Answer:
B
Explanation:
A hi-lo lifts an 25 N skid to top of a pallet rack. The pallet rack is 3.6 meters tall. The hi-lo takes 12 seconds to get the skid on top. Calculate the power output of the hi-lo.
Answer:
7.5Watts
Explanation:
Given parameters:
Force of lift = 25N
Height = 3.6m
Time = 12s
Unknown:
Power output = ?
Solution:
Power is the rate at which work is done ;
Power = [tex]\frac{force x height }{time}[/tex]
Power = [tex]\frac{25 x 3.6}{12}[/tex] = 7.5Watts
I NEED HELP please answer it
12. A bag weighing 20 N CARRIED horizontally a distance of 35 m, How much
work is done on the bag in Joules? (Do not put units with your answer.) W=Fd *
Your answer
13. A child performs 10J of work in lifting a box 1 m in 2 seconds. How much
power did the child apply to the box? (Do not include units with your answer.)
P=W/t *
Your answer
Answer:
Explanation:
Well they told you the exact formula to use. Work is the force multiplied by the distance through which its applied.
W = (20N)(35m)
= 700 Joules
13.) Power is the amount of work done over the time through which the work is being done.
P = W/t
= 10J/2s
= 5J/s
Billiard ball A (0.35 kg) is struck such that it moves at 10 m/s toward a
second identical ball (Ball B). After the collision Ball A continues to move
in the same direction at 2 m/s. What is the magnitude of the velocity for
Ball B after the collision?
Before Collision:
10 m/s
A
After Collision:
2 m/s
O
Answer:
6m/s
Explanation:
Using the law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the momentum after collision.
Using the expression
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the final velocity after collision
Substitute the given values in the formula
0.35(10)+0.35(2) = (0.35+0.35)v
3.5+0.7 = 0.7v
4.2 = 0.7v
v = 4.2/0.7
v = 6m/s
Hence the magnitude of the velocity for Ball B after the collision is 6m/s
A wheel starts from rest and has an angular acceleration that is given by α (t) = (6.0 rad/s4)t2. After it has turned through 10 rev its angular velocity is:
Answer:
75 rad/s
Explanation:
The angular acceleration is the time rate of change of angular velocity. It is given by the formula:
α(t) = d/dt[ω(t)]
Hence: ω(t) = ∫a(t) dt
Also, angular velocity is the time rate of change of displacement. It is given by:
ω(t) = d/dt[θ(t)]
θ(t) = ∫w(t) dt
θ(t) = ∫∫α(t) dtdt
Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:
θ(t) = ∫∫α(t) dtdt
θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt
θ(t) = ∫[2t³]dt = t⁴/2 rad
θ(t) = t⁴/2 rad
At θ(t) = 10 rev = (10 * 2π) rad = 20π rad, we can find t:
20π = t⁴/2
40π = t⁴
t = ⁴√40π
t = 3.348 s
ω(t) = ∫α(t) dt = ∫6t² dt = 2t³
ω(t) = 2t³
ω(3.348) = 2(3.348)³ = 75 rad/s
The required value of angular velocity of wheel is 75 rad/s.
Given data:
The angular acceleration of the wheel is, [tex]\alpha (t) = 6.0 t^{2} \;\rm rad/s^{4}[/tex].
The turning rate of wheel is, n = 10 rev.
The angular acceleration is the time rate of change of angular velocity. It is given by the formula:
α(t) = d/dt[ω(t)
Clearly, the angular velocity is the single integral of angular acceleration. Then,
ω(t) = ∫a(t) dt
Also, angular velocity is the time rate of change of displacement. It is given by:
ω(t) = d/dt[θ(t)]
θ(t) = ∫w(t) dt
θ(t) = ∫∫α(t) dt dt
Since,
α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴.
Then solve by substituting the values as,
θ(t) = ∫∫α(t) dt dt
θ(t) = ∫∫6t² dt dt
θ(t) =∫[∫6t² dt]dt
θ(t) = ∫[2t³]dt = t⁴/2 rad
θ(t) = t⁴/2 rad
At θ(t) = 10 rev = (10 * 2π) rad = 20π rad, we can find t:
20π = t⁴/2
40π = t⁴
t = ⁴√40π
t = 3.348 s
Also,
ω(t) = ∫α(t) dt = ∫6t² dt = 2t³
ω(t) = 2t³ ......................................................(1)
Substitute the value of time in equation (1) as,
ω( t = 3.348) = 2(3.348)³ = 75 rad/s
Thus, we can conclude that the required value of angular velocity of wheel is 75 rad/s.
Learn more about the angular velocity here:
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A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case?
Answer:
cos 0 = 1.
Fs = 7×8 = 56 J
Explanation:
CAN YOU GUYS PLEASE ANSWER THIS QUICKLY THIS IS DUE IN AN HOUR AND IM GETTING WORRIED
I'm pretty sure I the third option C.
Explanation:
sorry if I'm wrong
kid shoots a BB gun directly upward at a ock of birds. The initial velocity of the BB is21 m/s. If the BB hits a bird at less than 1 m/s then it will not harm the bird. Calculate the minimumheight above the gun at which the birds may safely y. Use a magnitude of acceleration of 9.8m
Answer:
22.5 m
Explanation:
Using v² = u² - 2gy where u = initial velocity of BB = 21 m/s, v = final velocity of BB = 1 m/s (since this is the required speed of BB in which it will not harm the birds), g = acceleration due to gravity = 9.8 m/s² and y = minimum height of BB above the gun at which the birds may safely fly.
Substituting the values of the variables into the equation, we have
v² = u² - 2gy
(1 m/s)² = (21 m/s)² - 2(9.8 m/s²)y
collecting like terms, we have
(1 m/s)² - (21 m/s)² = - 2(9.8 m/s²)y
1 m²/s² - 441 m²/s² = -(19.6 m/s²)y
simplifying, we have
- 440 m²/s² = -(19.6 m/s²)y
dividing through by -19.6 m/s², we have
y = - 440 m²/s² ÷ -19.6 m/s²
y = 22.45 m
y ≅ 22.5 m
A sinusoidal wave is traveling on a string with speed 19.3 cm/s. The displacement of the particles of the string at x = 6.0 cm is found to vary with time according to the equation y = (2.6 cm) sin[1.8 - (5.8 s-1)t]. The linear density of the string is 5.0 g/cm. What are (a) the frequency and (b) the wavelength of the wave? If the wave equation is of the form
Answer:
Explanation:
equation of wave is given by the following equation
y = (2.6 cm) sin[1.8 - (5.8 s-1)t].
Comparing it with standard form of wave
y = A sin ( ωt - kx )
we get
ω = 5.8
2πn = 5.8
n = .92 per second
kx = 1.8
k x 6 = 1.8
k = 0.3
[tex]\frac{2\pi}{\lambda}[/tex] = 0.3
λ = 20.9 cm
What industry uses extremely large equipment?
O mining
O harvesting
O food packaging
O welding
Answer:
mining
Explanation:
because they use the most heavy duty machines
The _______ changes light energy into nerve signals using receptors called rods and cones. A. retina B. lens C. iris D. pupil
Answer:
A. Retina
Explanation:
An RC car is carrying a tiny slingshot with a spring constant of 85 N/m at 0.2 m off the ground at 5.6 m/s. The sling shot is pulled back 3.5 cm from a relaxed state and shoots a 25 g steel pellet in the same direction the car is moving. What is the velocity of the steel pellet relative to the ground as it leaves the sling shot
Answer:
The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.
Explanation:
Let suppose that RC car-slingshot-steelpellet is a conservative system, that is, that non-conservative forces (i.e. friction, air viscosity) can be neglected. The velocity of the steel pellet can be found by means of the Principle of Energy Conservation and under the consideration that change in gravitational potential energy is negligible and that the RC car travels at constant velocity:
[tex]\frac{1}{2}\cdot (m_{C}+m_{P})\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{C}\cdot v_{o}^{2} + \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]
[tex]\frac{1}{2}\cdot m_{P}\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]
[tex]m_{P}\cdot v_{o}^{2} + k\cdot x^{2} = m_{P}\cdot v^{2}[/tex]
[tex]v^{2} = v_{o}^{2} + \frac{k}{m_{P}}\cdot x^{2}[/tex]
[tex]v = \sqrt{v_{o}^{2}+\frac{k}{m_{P}}\cdot x^{2} }[/tex] (1)
Where:
[tex]v_{o}[/tex] - Initial velocity of the steel pellet, measured in meters per second.
[tex]v[/tex] - Final velocity of the steel pellet, measured in meters per second.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]m_{P}[/tex] - Mass of the steel pellet, measured in kilograms.
[tex]m_{C}[/tex] - Mass of the RC car, measured in kilograms.
[tex]x[/tex] - Initial deformation of the spring, measured in meters.
If we know that [tex]v_{o} = 5.6\,\frac{m}{s}[/tex], [tex]k = 85\,\frac{N}{m}[/tex], [tex]m_{P} = 0.025\,kg[/tex] and [tex]x = 0.035\,m[/tex], then the velocity of the steel pellet relative to the ground when it leaves the sling shot is:
[tex]v = \sqrt{\left(5.6\,\frac{m}{s} \right)^{2}+\frac{\left(85\,\frac{N}{m} \right)\cdot (0.035\,m)^{2}}{0.025\,kg} }[/tex]
[tex]v \approx 5.960\,\frac{m}{s}[/tex]
The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.
An air-track glider attached to a spring oscillates between the 10 cm mark and the 60 cm mark on the track. The glider completes 10 oscillations in 33 s. What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider
Answer:
a) Time period is 3.3 seconds
b) The frequency is 0.3030 Hz
c) amplitude is 0.25 m
d) maximum speed is 0.476 m/s
Explanation:
Given the data in the question;
a) Period
Time Period T = Time taken for one oscillation
T = 33s / 10 = 3.3 seconds
Therefore, Time period is 3.3 seconds
b) Frequency
we know that frequency is the inverse of time period
so;
Frequency f = 1/T = 1 / 3.3 s
Frequency f = 0.3030 Hz
Therefore, The frequency is 0.3030 Hz
c) amplitude
amplitude A = [tex]\frac{1}{2}[/tex]( 60 cm - 10 cm )
A = [tex]\frac{1}{2}[/tex] × 50 cm
A = 25 cm
A = 0.25 m
Therefore, amplitude is 0.25 m
d) maximum speed of the glider
maximum speed [tex]V_{max}[/tex] = ωA
and ω = 2π/T
so maximum speed [tex]V_{max}[/tex] = [tex]\frac{2\pi }{T}[/tex]A
so we substitute
so maximum speed [tex]V_{max}[/tex] = [tex]\frac{2\pi }{3.3}[/tex] × 0.25 m
so maximum speed [tex]V_{max}[/tex] = 0.476 m/s
Therefore, maximum speed is 0.476 m/s
The time period is 3.3 s, frequency is 0.303 Hz, amplitude is 0.25 m, and maximum speed is 0.47575 m/s.
Based on the given information,
• The air-track glider connected with a spring oscillates between the 10 cm mark and the 60 cm mark.
• In 33 seconds, the glider completes 10 oscillations.
There is a need to find, the period, frequency, amplitude, and maximum speed of the glider.
a) Time period (T) for one oscillation is,
[tex]\frac{33s}{10} = 3.3 s[/tex]
b) The frequency (f) is the reciprocal of the time period,
[tex]f = \frac{1}{T} =\frac{1}{3.3S} = 0.303 Hz[/tex]
c) The amplitude (A) is,
[tex]A = \frac{1}{2} (60 cm-10cm) = 25cm[/tex]
[tex]A = 0.25 m[/tex]
d) Maximum speed of the glider is,
[tex]Vmax = \frac{2\pi }{T} (0.25 m)[/tex]
[tex]Vmax = 0.47575 m/s[/tex]
Thus, time period is 3.3 s, frequency is 0.303 Hz, amplitude is 0.25 m, and maximum speed is 0.47575 m/s.
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Answer:
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Explanation:
Which of the following is a mixture?
a air
biron
Chydrogen
d nickel
Answer:
it will option option A hope it helps
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]