Answer:
The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.
Explanation:
Concepts and reason
The concept required to solve this problem is Newton’s second law of motion.
Initially, write an expression for the force according to the Newton’s second law of motion. Later, rearrange the expression for the acceleration. Finally, substitute the value of the acceleration obtained to find the new force.
Fundamentals
According to the Newton’s second law of motion, the net force is equal to the product of the mass and the acceleration of an object. The expression for the Newton’s second law of motion is as follows:
F = maF=ma
Here, m is mass and a is the acceleration.
(a)
Rearrange the equation F = maF=ma for a.
a = \frac{F}{m}a=
m
F
Substitute 18,000 N for F and \left( {2300{\rm{ kg + 2400 kg}}} \right)(2300kg+2400kg) for m in the equation a = \frac{F}{m}a=
m
F
.
\begin{array}{c}\\a = \frac{{18,000{\rm{ N}}}}{{\left( {2300{\rm{ kg + 2400 kg}}} \right)}}\\\\ = \frac{{18,000{\rm{ N}}}}{{\left( {4700{\rm{ kg}}} \right)}}\\\\ = 3.83{\rm{ m/}}{{\rm{s}}^2}\\\end{array}
a=
(2300kg+2400kg)
18,000N
=
(4700kg)
18,000N
=3.83m/s
2
(b)
Substitute 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s
2
for a and 2400 kg for m in the equation F = maF=ma .
\begin{array}{c}\\F = \left( {2400{\rm{ kg}}} \right)\left( {3.83{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 9120{\rm{ N}}\\\end{array}
F=(2400kg)(3.83m/s
2
)
=9120N
Ans: Part a
The maximum possible acceleration the truck can give the SUV is 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s
2
.
Part b
The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.
The maximum possible acceleration the truck can give the SUV is equal to 4 m/s².
The force of the SUV's bumper on the truck's bumper is 10000N
What is acceleration?Acceleration of an object can be described as as the change in the velocity of an object w.r.t. time. The acceleration is a vector quantity, contains both magnitude and direction. Acceleration is the second derivative of position w.r.t. time and the first derivative of velocity w.r.t. time.
According to Newton's second law of motion, the force is equal to the product of acceleration and mass of an object.
F = ma
And, a = F/m
Given, the mass of the ruck , m = 2000 Kg
The mass of the SUV, M = 2500 Kg
The total mass of the both = 2000 + 2500 = 4500 Kg
The maximum force on the trick , F = 18000 N
The maximum acceleration of the truck can give the SUV:
[tex]a_{max} = \frac{F_{max}}{m+M}[/tex]
a = 18000/4500
a = 4 m/s²
The force of the SUV's bumper on the truck's bumper will be:
[tex]F_{max} -f= ma_{max}[/tex]
[tex]f= 18000-2000\times 4[/tex]
[tex]f =10000N[/tex]
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1.03 Transformation of energy flvs science question
Explanation:
the process of conversion of energy from one form to another is called transformation of energy.
A bar of steel has the minimum properties Se = 40 kpsi, Sy = 60 kpsi, and Sut = 80 ksi. The bar is subjected to a steady torsional stress of 15 kpsi and an alternating bending stress of 25 ksi. Find the factor of safety guarding against a static failure, and either the factor of safety guarding against a fatigue failure or the expected life of the part. For the fatigue analysis use Modified Goodman criterion.
Answer:
The correct solution is:
(a) 1.66
(b) 1.05
Explanation:
Given:
Bending stress,
[tex]\sigma_b = 25 \ kpsi[/tex]
Torsional stress,
[tex]\tau= 15 \ kpsi[/tex]
Yield stress of steel bar,
[tex]\delta_y = 60 \ kpsi[/tex]
As we know,
⇒ [tex]\sigma_{max}^' \ = \sqrt{\sigma_b^2 + 3 \gamma^2}[/tex]
[tex]= \sqrt{(25)^2+3(15)^2}[/tex]
[tex]=36.055 \ kpsi[/tex]
(a)
The factor of safety against static failure will be:
⇒ [tex]\eta_y = \frac{\delta_y}{\sigma_{max}^'}[/tex]
By putting the values, we get
[tex]=\frac{60}{36.055}[/tex]
[tex]=1.66[/tex]
(b)
According to the Goodman line failure,
[tex]\sigma_a = \sigma_b = 25 \ kpsi[/tex]
[tex]S_e = 40 \ kpsi[/tex]
[tex]\sigma_m = \sqrt{3} \tau[/tex]
[tex]=\sqrt{3}\times 15[/tex]
[tex]=26 \ kpsi[/tex]
[tex]Sut = 80 \ kpsi[/tex]
⇒ [tex]\frac{\sigma_a}{S_e} +\frac{\sigma_m}{Sut} =\frac{1}{\eta_y}[/tex]
[tex]\frac{25}{40}+\frac{26}{80}=\frac{1}{\eta_y}[/tex]
[tex]\eta_y = 1.05[/tex]
which planet composed entirely of hydrogen and helium?
Answer:
The composition of Jupiter is similar to that of the Sun—mostly hydrogen and helium. Deep in the atmosphere, pressure and temperature increase, compressing the hydrogen gas into a liquid. This gives Jupiter the largest ocean in the solar system—an ocean made of hydrogen instead of water.
what is threshold frequency?
Answer:
"the minimum frequency of radiation that will produce a photoelectric effect."
Explanation:
That answer was derived from gogle cuz my explanations was harder to explain but good luck
a stone is thrown vertically upwards with a velocity of 20 m per second determine the total time of flight of stone in air
Answer:
Explanation:
The best way to do this is to remember the rule about the halfway mark in a parabolic path. At a trajectory's half way point in its travels, it will be at its max height. To get the total time in the air, we take that time at half way and double it. Here's what we know that we are told:
initial velocity is 20 m/s
Here's what we know that we are NOT told:
a = -9.8 m/s/s and
final velocity is 0 at an object's max height in parabolic motion.
We will use the equation:
[tex]v=v_0+at[/tex] where v is final velocity and v0 is initial velocity. Filling in:
0 = 20 + (-9.8)t and
-20 = -9.8t so
t = 2 seconds. The stone reaches its max height 2 seconds after it is thrown; that means that after another 2 seconds it will be on the ground. Total air time is 4 seconds.
Tay quay OB quay đều quanh trục cố định đi qua O với vận tốc góc không đổi ω. Con lăn A chuyển động trong rãnh thẳng đứng. Tại vị trí trên hình vẽ thì thanh OB thẳng đứng, OA có phương nằm ngang. Hãy xác định vận tốc góc thanh AB, vận tốc của con lăn A; gia tốc góc của thanh AB, gia tốc của con lăn A. Cho ω = 1,5 rad/s, r = 1 m.
You have a 1 W light bulb in your lab. It puts out light of only 1 frequency. The wavelength of this light is 500nm. you set up a detector with a surface area of 1 square centimeter facing the light source at a distance of 100m.
Required:
a. Find the number of photons hitting the detector every second.
b. What is the maximum E field of the E M wave hitting the detector?
c. What is the maximum value of the B field of this E M wave?
d. How far away would you have to place the detector to only receive 1 photon per second from the light bulb?
Answer:
a) # _photon = 2.5 10¹⁸ photons / s, b) E = 10⁻² N / C, c) B = 3 10⁻¹¹ T
d) r= 2 10⁹ m
Explanation:
a) Let's solve this exercise in part, let's start by finding the energy of each photon using the Planck relation
E₀ = h f
c = λ f
E₀ = h c /λ
E₀ = 6.63 10⁻³³⁴ 3 10⁸/500 10⁻⁹
E₀ = 3.978 10⁻⁻¹⁹ J
Let's use a direct ratio rule to find the number of photons
#_foton = E / Eo
#_fototn = 1 / 3.978 10⁻¹⁹
# _photon = 2.5 10¹⁸ photons / s
b) The intensity received by the detector is related to the electric field
I = E²
Let's look for the intensity that the detector receives, suppose that the emission is shapeless throughout the space
I = P / A
P = I A
Let's use index 1 for the point on the bulb and index 2 for the point on the detector.
The area of a sphere is
A = 4π r²
P = I₁ A₁ = I₂ A₂
I₁ r₁² = I₂ r₂²
I₂ = I₁ r₁²/r₂²
I₂ = I₁ 1 / 100²
I₂ = I₁ 10⁻⁴
we must know the intensity at the output of the bulb suppose that I₁ = 1 J
I₂ = 10⁻⁴ J
let's look for the electric field
E =√I
E = √10⁻⁴
E = 10⁻² N / C
c) for the calculation of the magnetic field we use that the field is in phase
E / B = c
B = E / c
B = 10⁻² / 3 10⁸
B = 3 10⁻¹¹ T
d) Let's use a direct proportions rule if we fear 2.5 10¹⁸ photons in an area A = 4π R² where R = 100 m how many photons are there in the area of the detector r = 1 cm, A’= 10⁻⁴ m²
#_photons = 2.5 10¹⁸ A_detector / A_sphere
#_photons = 2.5 1018 10-4 / 4π 10⁴
#_photons = 2 10⁹ photons in the detector area
for the number of photons to decrease to 1, the radius of the sphere must be 2 10⁹ m
friction always opposes the _____
Answer:
Friction always opposes the motion
I HOPE ITS RIGHT IF NOT THEN SORRYHAVE A GREAT DAY :)
Which of the following is not an example of approximate simple harmonic motion
Answer:
where are the options
it's not full question
A crude approximation is that the Earth travels in a circular orbit about the Sun at constant speed, at a distance of 150,000,000 km from the Sun. Which of the following is the closest for the acceleration of the Earth in this orbit?
A. exactly 0 m/s2.
B. 0.006 m/s2.
C. 0.6 m/s2.
D. 6 m/s2.
E. 10 m/s2.
Answer:
The answer is "Option B".
Explanation:
[tex]r=15\times 10^{7}\ km\ = 15\times 10^{10}\ m\\\\w=\frac{2\pi}{1\ year}\\\\=\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec}\\\\a=w^2r\\\\=(\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec})^2 \times 15 \times 10^{10}\ \frac{m}{s^2}\\\\[/tex]
[tex]=5.940 \times 10^{-3} \ \frac{m}{s^2}\\\\=6 \times 10^{-3} \ \frac{m}{s^2}\\\\=0.006\ \frac{m}{s^2}\\\\[/tex]
The primary purpose of a switch in a circuit is to ___________.
A)either open or close a conductive path
B)change a circuit from parallel to series
C)change a circuit from series to parallel
D)store a charge for later use
Answer:
store a charge for later use
In the diagram, disk 1 has a moment of inertia of 3.4 kg · m2 and is rotating in the counterclockwise direction with an angular velocity of 6.1 rad/s about a frictionless rod passing through its center. A second disk rotating clockwise with an angular velocity of 9.3 rad/s falls from above onto disk 1. The two then rotate as one in the clockwise direction with an angular velocity of 1.8 rad/s. Determine the moment of inertia, in kg · m2, of disk 2.
Answer:
I = 3.6 kg•m²
Explanation:
Conservation of angular momentum
Let's assume CW is the positive direction
3.4(-6.1) + I(9.3) = 3.4(1.8) + I(1.8)
I(9.3 - 1.8) = 3.4(1.8 + 6.1)
I(7.5) = 3.4(7.9)
I = 3.4(7.9)/(7.5) = 3.5813333333...
The moment of inertia of the second disk will be [tex]I=3.58\ kg-m^2[/tex]
What is moment of inertia?The moment of inertia is defined as the product of mass of section and the square of the distance between the reference axis and the centroid of the section.
here it is given that
MOI of disk one [tex]I_1=3.4\ kg-m^2[/tex]
Angular velocity [tex]w_1=6.1\ \frac{rad}{s}[/tex]
Angular velocity of disk two [tex]w=1.8\ \frac{rad}{s}[/tex]
MOI of the disk two [tex]I=?[/tex]
The final angular velocity [tex]w_f= 1.8\ \frac{rad}{sec}[/tex]
Now from the conservation of the momentum the angular momentum before collision will be equal to the angular momentum after collision.
[tex]I_1w_1+I_2w_2=(I_1+I_2)w_f[/tex]
Now put the values in the formula
[tex](3.4\times 6.10)+(I_2\times 9.3)=(3.4+I_2)\times 1.8[/tex]
[tex]I_2=3.58\ kg-m^2[/tex]
Thus the moment of inertia of the second disk will be [tex]I=3.58\ kg-m^2[/tex]
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An auto mechanic needs to determine the emf and internal resistance of an old battery. He performs two measurements: in the first, he applies a voltmeter to the battery's terminals and reads 11.9 V;11.9 V; in the second, he applies an ammeter to the terminals and reads 16.1 A.16.1 A.
What are the battery's emf E and internal resistance r?
Answer:
Hence the battery's emf E is ε = 11.9 V.
The internal resistance is r = 0.739 ohms.
Explanation:
Now we know that
Voltage V = 11.9 V.
Current I = 16.1 A.
Hence this is an ideal voltmeter there are no current flows when the Voltmeter is applied.
ε = V + I r
∵ I = 0
ε = V
ε = 11.9 V
Then the ammeter is applied.
Let's take ( r ) to be the total resistance which is equal to internal resistance.
V = I r
r = [tex]\frac{V}{I}[/tex]
[tex]= \frac{11.9}{16.1}[/tex]
r = 0.739 ohms
The battery's emf (E) and internal resistance (r) are 11.9 Volts and 0.739 Ampere respectively.
Given the following data:
Voltage = 11.9 Volts.Current = 16.1 Amperes.To determine the battery's emf (E) and internal resistance (r):
How to calculate emf (E).For an ideal voltmeter, there isn't a flow of current and as such the current is equal to 0.
Mathematically, emf (E) is given by this formula:
[tex]E = V + IR[/tex]
Substituting the given parameters into the formula, we have;
[tex]E = 11.9 + 0R\\\\E = 11.9 + 0[/tex]
E = 11.9 Volts.
For the internal resistance (r):
Note: The total resistance is equal to internal resistance.
Applying Ohm's law, we have:
[tex]R = \frac{V}{I} \\\\R = \frac{11.9}{16.1}[/tex]
R = r = 0.739 Ampere.
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A car's bumper is designed to withstand a 6.12 km/h (1.7-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.210 m while bringing a 810 kg car to rest from an initial speed of 1.7 m/s.
Answer:
5572.8 N
Explanation:
Applying,
F = ma.............. Equation 1
Where F = Force, m = mass of the car, a = acceleration.
We can find a by applying,
v² = u²+2as............. Equation 2
Where v = final velocity, u = initial velocity, a = acceleration, = distance.
From the question,
Given: v = 0 m/s (come to rest), u = 1.7 m/s, s = 0.210 m
Substitute these value into equation 2
0² = 1.7²+2×0.21×a
a = -1.7²/(2×0.21)
a = -2.89/0.42
a = -6.88 m/s²
Also given: m = 810 kg
Substitute these value into equation 1
F = 810(-6.88)
F = -5572.8 N
Hence the force on the bumber is 5572.8 N
You and your friends find a rope that hangs down 11 m from a high tree branch right at the edge of a river. You find that you can run, grab the rope, swing out over the river, and drop into the water. You run at 2.0 m/s and grab the rope, launching yourself out over the river.
Required:
How long must you hang on if you want to drop into the water at the greatest possible distance from the edge?
Answer:
if you want to drop into the water at the greatest possible distance from the edge, you must hang for 1.662s.
Explanation:
The time period of the oscillation is,
[tex]T = 2\pi \sqrt{ \frac{I} {g }[/tex]
[tex]T = 2\pi \sqrt{\frac{11}{9.8} } \\\\T= 6.65 s[/tex]
This would be the time taken for the person to move from.
The duration of time he hangs over the river be one-fourth of the time period.
Here,
[tex]t= \frac{T}{4} \\\\t=\frac{6.65}{4}\\\\t = 1.662 s[/tex]
Hey guys,I hope u r gonna answer this question fast,SI system is extended from of MKS system.Why? I will be waiting for the answer. Good luck thank u
Answer:
Because SI system has fundamental units of MKS System
Answer:
Explanation: the unit of length ,mass , and time are same in both the system , thus, the SI system is the extended from of MKS system.
1. There is a famous intersection in Kuala Lumpur, Malaysia, where thousands of vehicles pass each hour. A 750 kg Tesla Model S traveling south crashes into a 1250 kg Ford F-150 traveling east. What are the initial speeds of each vehicle before collision if they stick together after crashing into each other and move at an angle of 320 and a common velocity of 18 m/s.
Solution :
Let the positive [tex]x-axis[/tex] is along the East and the positive [tex]y[/tex] direction is along the north.
Given :
Mass of the Tesla car, [tex]m_1[/tex] = [tex]750 \ kg[/tex]
Mass of the Ford car, [tex]m_2 = 1250 \ kg[/tex]
Now let the initial velocity of Tesla car in the south direction be = [tex]-v_1j[/tex]
The initial momentum of Tesla car, [tex]p_1 = -750 \ v_1[/tex]
Let the initial velocity of Ford car in the east direction be = [tex]v_2 \ i[/tex]
So the initial momentum of the Ford car is [tex]p_2=1250\ v_2 \ i[/tex]
Therefore, the initial velocity of both the cars is [tex]p_i = p_1+p_2[/tex]
[tex]=1250 \ v_2 \ i - 750\ v_1 \ j[/tex]
Now the final velocity of both the cars is [tex]v = 18 \ m/s[/tex]
So the vector form is :
[tex]v = 18\cos 32\ i-18 \sin 32 \ j[/tex]
[tex]= 15.26 \ i - 9.54 \ j[/tex]
Therefore the momentum after the accident is
[tex]p_f=(m_1+m_2) \times v[/tex]
[tex]=(750+1250) \times (15.26 \ i - 9.54 \ j)[/tex]
[tex]= 30520\ i -19080\ j[/tex]
According to the law of conservation of momentum, we know
[tex]p_i = p_f[/tex]
[tex]1250 \ v_2 \ i - 750\ v_1 \ j[/tex] [tex]= 30520\ i -19080\ j[/tex]
[tex]1250 \ v_2 = 30520[/tex]
[tex]v_2=24.4 \ m/s[/tex]
From, [tex]750\ v_1 = 19080[/tex]
We get, [tex]v_1=25.4 \ m/s[/tex]
Therefore the speed of Tesla car before collision = 25.4 m/s
The speed of ford car before collision = 24.4 m/s
F=(4i+3j)N acts on an object of mass m=2k.g and drags it by moving the object from origion to x=5m. Find the workdone on the object and the angle between the force and the displacement
Answer:
nnnjjdndbsnnshfhhgbfbdbdh
1:
Forces and Motion:Question 2
A car is travelling east, when suddenly a more massive car travelling
north hits it with a greater force. What is likely to happen to the car
that was originally travelling east?
Explanation:
the car will be brought back
Which circuit element is of special importance in AC circuits?
A. Resistor
B. Ammeter
C. Battery
D. Capacitor
Answer:
Explanation:capacitor
Answer:
Ammeter
Explanation:
pls mark me as a brainlist
The potential difference between the plates of a capacitor is 234 V. Midway between the plates, a proton and an electron are released. The electron is released from rest. The proton is projected perpendicularly toward the negative plate with an initial speed. The proton strikes the negative plate at the same instant the electron strikes the positive plate. Ignore the attraction between the two particles, and find the initial speed of the proton.
I have tried looking at the cramster.com solution manual and do not like the way it is explained. Simply put, I cannot follow what is going on and I am looking for someone who can explain it in plain man's terms and help me understand and get the correct answer. I am willing to give MAX karma points to anyone who can help me through this. Thank you kindly.
Answer:
The speed of proton is 2.1 x 10^5 m/s .
Explanation:
potential difference, V = 234 V
let the initial speed of the proton is v.
The kinetic energy of proton is
KE = q V
[tex]0.5 mv^2 = e V \\\\0.5\times 1.67\times 10^{-27} v^2 = 1.6\times 10^{-19} \times 234\\\\v=2.1\times 10^5 m/s[/tex]
A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force of 9.14 × 10-17 N. Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.
Answer:
a) The proton's speed is 5.75x10⁵ m/s.
b) The kinetic energy of the proton is 1723 eV.
Explanation:
a) The proton's speed can be calculated with the Lorentz force equation:
[tex] F = qv \times B = qvBsin(\theta) [/tex] (1)
Where:
F: is the force = 9.14x10⁻¹⁷ N
q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C
v: is the proton's speed =?
B: is the magnetic field = 3.28 mT
θ: is the angle between the proton's speed and the magnetic field = 17.6°
By solving equation (1) for v we have:
[tex]v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s[/tex]
Hence, the proton's speed is 5.75x10⁵ m/s.
b) Its kinetic energy (K) is given by:
[tex] K = \frac{1}{2}mv^{2} [/tex]
Where:
m: is the mass of the proton = 1.67x10⁻²⁷ kg
[tex] K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV [/tex]
Therefore, the kinetic energy of the proton is 1723 eV.
I hope it helps you!
How can I solve the following?
In (Figure 1), let V = 15.0 V and C1=C2=C3= 24.2 μF.
Part A: How much energy is stored in the capacitor network as shown in (Figure 1)?
Part B: How much energy would be stored in the capacitor network if the capacitors were all in series?
Part C: How much energy would be stored in the capacitor network if the capacitors were all in parallel?
Answer:
Part A - 4.084 mJ
Part B - 0.908 mJ
Part C - 8.168 mJ
Explanation:
Part A: How much energy is stored in the capacitor network as shown in (Figure 1)?
Since capacitors C₂ and C₃ are in series, their equivalent capacitance is C',
1/C' = 1/C₂ + 1/C₃ (Since C₁ = C₂ = C₃ = C)
1/C' = 1/C + 1/C
1/C' = 2/C
C' = C/2
Since C' is in parallel with C₁, the equivalent capacitance for the circuit is C" = C₁ + C' = C + C/2 = 3C/2
C" = 3C/2
The energy stored in the circuit, W = 1/2C"V² where C" = equivalent capacitance = 3C/2 and V = voltage = 15.0 V
W = 1/2C"V²
W = 1/2(3C/2)V²
W = 3CV²/4
since C = 24.2 μF = 24.2 × 10⁻⁶ F
W = 3CV²/4
W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/4
W = 3 × 24.2 × 10⁻⁶ F × 225 V²/4
W = 16335/4 × 10⁻⁶ FV²
W = 4083.75 × 10⁻⁶ J
W = 4.08375 × 10⁻³ J
W = 4.08375 mJ
W ≅ 4.084 mJ
Part B: How much energy would be stored in the capacitor network if the capacitors were all in series?
If the capacitors are connected in series, their equivalent resistance is C'
and 1/C' = 1/C₁ + 1/C₂ + 1/C₃
Since C₁ = C₂ = C₃ = C
1/C' = 1/C + 1/C + 1/C
1/C' = 3/C
C' = C/3
The energy stored in the circuit, W = 1/2C'V² where C' = equivalent capacitance = C/3 and V = voltage = 15.0 V
W = 1/2C'V²
W = 1/2(C/3)V²
W = CV²/6
since C = 24.2 μF = 24.2 × 10⁻⁶ F
W = CV²/6
W = 24.2 × 10⁻⁶ F (15.0 V)²/6
W = 24.2 × 10⁻⁶ F × 225 V²/6
W = 5445/6 × 10⁻⁶ FV²
W = 907.5 × 10⁻⁶ J
W = 0.9075 × 10⁻³ J
W = 0.9075 mJ
W ≅ 0.908 mJ
Part C: How much energy would be stored in the capacitor network if the capacitors were all in parallel?
If the capacitors are connected in parallel, their equivalent resistance is C'
and C' = C₁ + C₂ + C₃
Since C₁ = C₂ = C₃ = C
C' = C + C + C
C' = 3C
The energy stored in the capacitor network, W = 1/2C'V² where C' = equivalent capacitance = 3C and V = voltage = 15.0 V
W = 1/2C'V²
W = 1/2(3C)V²
W = 3CV²/2
since C = 24.2 μF = 24.2 × 10⁻⁶ F
W = 3CV²/2
W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/2
W = 3 × 24.2 × 10⁻⁶ F × 225 V²/2
W = 16335/2 × 10⁻⁶ FV²
W = 8167.5 × 10⁻⁶ J
W = 8.1675 × 10⁻³ J
W = 8.1675 mJ
W ≅ 8.168 mJ
Một bơm cánh gạt thủy lực có lưu lượng thực là 20lít/phút, tạo áp suất 230 bar, tốc độ bơm là 1400 vòng/phút. Biết công suất đầu vào là 10kW và hiệu suất cơ là 88%.
a) Tính hiệu suất thể tích của bơm
b) Tính thể tích riêng của bơm (cm/vòng). Câu 3 (2,5đ): Thiết kế hệ thống truyền động khí nén để điều khiển 02 xylanh tác động đơn, sử dụng 02 van đảo chiều 3/2 tác động bằng nút nhấn, 02 van tiết lưu - một chiều. Trình bày nguyên lý làm việc của của hệ thống.
Answer:
English please?
Explanation:
Explain in english?
You attach a 2.30 kg weight to a horizontal spring that is fixed at one end. You pull the weight until the spring is stretched by 0.500 m and release it from rest. Assume the weight slides on a horizontal surface with negligible friction. The weight reaches a speed of zero again 0.400 s after release (for the first time after release). What is the maximum speed of the weight (in m/s)
Answer: [tex]3.92\ m/s[/tex]
Explanation:
Given
Mass of the attached object is [tex]m=2.3\ kg[/tex]
Spring is stretched by [tex]A=0.5\ m[/tex]
Speed reaches zero after [tex]t=0.4\ s[/tex]
Speed is zero at the extremities of the S.H.M motion that is
[tex]\Rightarrow \dfrac{T}{2}=0.4\\\\\Rightarrow T=0.8\ s[/tex]
Time period of motion is [tex]0.8\ s[/tex] which can also be given by
[tex]\Rightarrow \omega T=2\pi\\\\\Rightarrow \omega=\dfrac{2\pi }{T}\\\\\Rightarrow \omega =\dfrac{2\pi }{0.8}\\\\\Rightarrow \omega=\dfrac{5\pi }{2}[/tex]
Maximum speed for S.H.M. is [tex]v_{max}=A\omega[/tex]
[tex]\Rightarrow v_{max}=0.5\times 2.5\pi\\\Rightarrow v_{max}=3.92\ m/s[/tex]
Consider a box with two gases separated by an impermeable membrane. The membrane can move back and forth, but the gases cannot penetrate the membrane. The left side is filled with gas A and the right side is filled with gas B. We will assume that equipartition applies to both gases, but gas A has an excluded volume due to large molecules so its entropy has a different formula.
SA=NAkln(VA+ bNA)+f(UA,NA)
SB=NBkln(VB)+f(UB,NB)
Required:
If NA= 1 moles, NB = 2 moles, the total volume of the box is 1 m3, and b= 4 × 10-4 m3/mole, then find the equilibrium value of VA by maximizing the total entropy.
Answer:
The answer is "[tex]0.3336\ m^3[/tex]"
Explanation:
Using the Promideal gas law:
[tex]P_A=P_B\\\\P_A(V_A-\eta_A b)= \eta_A RT......(1)\\\\P_B V_B=\eta_B \bar{R}T........(2)\\\\From (1) \zeta (2)\\\\[/tex]
[tex]\frac{\eta_A}{V_A-\eta_A b}=\frac{\eta B}{V B}\\\\ \frac{V A- \eta_A b}{V B}=\frac{\eta A}{\eta B }\\\\ \frac{V A-b}{V B}=\frac{1}{2}\\\\V A+V B=1\\\\V B =1- V A\\\\\frac{V A-b}{1-V A}=\frac{1}{2}\\\\2V A-2b=1-V A\\\\3 V A=1+2b\\\\V A=\frac{1+2b}{3}\\\\[/tex]
[tex]=\frac{1+2(4\times 10^{-4})}{3}\\\\=0.3336\ m^3[/tex]
The equilibrium value of Va is 0.3336 m³.
Ideal gas lawThe equilibrium value of Va is determine by applying ideal gas law as shown below;
Pressure of gas A = Pressure of gas B
Pa = Pb
Pa(Va - nab) = naRT----(1)
PbVb = nbRT -----(2)
Solve equation (1) and (2)
[tex]\frac{P_b}{RT} = \frac{n_b}{V_b} \\\\\frac{P_b}{P_a(V_a- n_ab)/n_a} = \frac{n_b}{V_b}\\\\\frac{n_a}{V_a - n_ab} = \frac{n_b}{V_b} \\\\\frac{V_a - n_ab}{V_b} = \frac{n_a}{n_b} \\\\\frac{V_a - b}{V_b} = \frac{1}{2}[/tex]
Va + Vb = 1
Vb = 1 - Va
[tex]\frac{V_a - b}{1 - V_a} = \frac{1}{2}[/tex]
2Va - 2b = 1 - Va
3Va = 1 + 2b
[tex]V_ a = \frac{1 + 2b}{3} \\\\V_a = \frac{1 + (2 \times 4\times 10^{-4})}{3} \\\\V_a = 0.3336 \ m^3[/tex]
Thus, the equilibrium value of Va is 0.3336 m³.
Learn more about equilibrium value here: https://brainly.com/question/22569960
The resistance of the light bulb changed as the voltage (and current) changed. Why does this resistance change occur?
Explain how newton's first law of motion follows from second law?
Answer:
Newton's First Law of Motion states that a body will stay at rest or continue its path with constant velocity unless an external force acts upon it. Newton's Second Law of Motion states that the net force that acts upon a body is equal to the mass of the body multiplied by the acceleration due to the net force.
I’m a photoelectric effect, which property of the incident light determines how much kinetic energy the ejected electrons have ?
A) brightness
B) frequency
C) size of the beam
D) none of the above
Answer:
b = frequency
A car can go from 0 to 60 m/s in 5 seconds. What is the acceleration?
A) 50 m/s^2
B) 6 m/s^2
C) 12 m/s^2
D) 300 m/s^2
I think it’s 12 because I did the difference divided by 5 but some places said there was no acceleration
Answer:
12
Explanation:
i agree the answer is 12 because the acceleration is given by the difference in the velocity divided by the time taken
a=v-u/t
60-0/5
=12m/s²
I hope this helps
