5 9 . 4
- 3 7 . 2
2 2 . 2
Explanation:
Use the algorithm method.
5 9 . 4
- 3 7 . 2
2 2 . 2
2 Therefore, 59.4-37.2=22.259.4−37.2=22.2.
22.2
22.2
A satellite is launched to orbit the Earth at an altitude of 2.90 x10^7 m for use in the Global Positioning System (GPS). Take the mass of the Earth to be 5.97 x 10^24 kg and its radius 6.38 x10^6 m.
Required:
What is the orbital period of this GPS satellite?
Answer:
[tex]T=66262.4s[/tex]
Explanation:
From the question we are told that:
Altitude [tex]A=2.90 *10^7[/tex]
Mass [tex]m=5.97 * 10^{24} kg[/tex]
Radius [tex]r=6.38 *10^6 m.[/tex]
Generally the equation for Satellite Speed is mathematically given by
[tex]V=(\frac{GM}{d} )^{0.5}[/tex]
[tex]V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}[/tex]
[tex]V=3354.83m/s[/tex]
Therefore
Period T is Given as
[tex]T=\frac{2 \pi *a}{V}[/tex]
[tex]T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}[/tex]
[tex]T=66262.4s[/tex]
need help pleaseee,question is in the pic
Explanation:
For engine 1,
Energy removed = 239 J
Energy added = 567 J
[tex]\eta_1=\dfrac{239}{567}\cdot100=42.15\%[/tex]
For engine 2,
Energy removed = 457 J
Energy added = 789 J
[tex]\eta_2=\dfrac{457}{789}\cdot100=57.92\%[/tex]
For engine 3,
Energy removed = 422 J
Energy added = 1038 J
[tex]\eta_3=\dfrac{422}{1038}\cdot100=40.65\%[/tex]
So, the engine 2 has the highest thermal efficiency.
Harmonics a.are components of a complex waveform. b.have frequencies that are integer multiples of the frequency of the complex waveform. c.are pure tones. d.have sinusoidal waveforms. e.all of the above
Answer:
b.have frequencies that are integer multiples of the frequency of the complex waveform
Explanation:
Please correct me if I am wrong
Which graph would be created by a pendulum with the greatest amplitude?
Answer:
Graph (c) would be created by a pendulum with the greatest amplitude.
Explanation:
The amplitude of a wave is the greatest displacement covered by an object. It refers to the maximum amount of displacement of a particle on the medium from its rest position. It is the distance from rest to crest.
Out of three graphs, the amplitude is greatest in graph 3 as the distance from rest is crest in this case is maximum. Hence, the correct option is (c).
Which is the most difficult subject?
Answer:
Quantum Mechanics
Explanation:
Well, that's what I think personally.
A 49.5-turn circular coil of radius 5.10 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.535 T. If the coil carries a current of 26.5 mA, find the magnitude of the maximum possible torque exerted on the coil.
Answer:
The magnitude of the maximum possible torque exerted on the coil is 5.73 x 10⁻³ Nm
Explanation:
Given;
number of turns of the circular coil, N = 49.5 turns
radius of the coil, r = 5.10 cm = 0.051 m
magnitude of the magnetic field, B = 0.535 T
current in the coil, I = 26.5 mA = 0.0265 A
The magnitude of the maximum possible torque exerted on the coil is calculated as;
τ = NIAB
where;
A is the area of the coil
A = πr² = π(0.051)² = 0.00817 m²
Substitute the given values and solve for the maximum torque
τ = (49.5) x (0.0265) x (0.00817) x (0.535)
τ = 0.00573 Nm
τ = 5.73 x 10⁻³ Nm
The north pole of magnet A will __?____ the south pole of magnet B
Answer:
A will attract
B will repare
7. The gravitational potential energy of a body depends on its A speed and position B. mass and volume. C. weight and position D.speed and mass
Answer:
Option "D" is the correct answer to the following question.
Explanation:
The gravitational potential energy of an item is determined by its mass, elevation, and gravitational acceleration. As a result, angular momentum and energy are preserved. The gravitational potential energy, on the other hand, varies with distance. When a consequence, kinetic energy varies during each orbit, resulting in a faster speed as a planet approaches the Sun.
Answer:
SPEED AND MASS
Explanation:
TOOK THE TEST
A 3.10 mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1550 m3 to 0.742 m3 . Initially the pressure was 1.00 atm.(a) Determine the initial and final temperatures.initial Kfinal K(b) Determine the change in internal energy. J(c) Determine the heat lost by the gas. J(d) Determine the work done on the gas. J
Answer:
a) Initial Temperature = 609.4 K and Final Temperature = 325.7 K
b) the change in internal energy is -18279.78 J
c) heat lost by the gas is zero or 0
d) the work done on the gas is -18279.78 J
Explanation:
Given the data in the question;
P[tex]_i[/tex] = 1 atm = 101325 pascal
P[tex]_f[/tex] = ?
V[tex]_i[/tex] = 0.1550 m³
V[tex]_f[/tex] = 0.742 m³
we know that for an adiabatic process γ = 1.4
P[tex]_i[/tex]V[tex]_i^Y[/tex] = P[tex]_f[/tex]V[tex]_f^Y[/tex]
P[tex]_f[/tex] = P[tex]_i[/tex][tex]([/tex] V[tex]_i[/tex] / V[tex]_f[/tex] [tex])^Y[/tex]
we substitute
P[tex]_f[/tex] = 1 × [tex]([/tex] 0.1550 / 0.742 [tex])^{1.4[/tex]
= [tex]([/tex] 0.2088948787 [tex])^{1.4[/tex]
= 0.11166 atm
a) the initial and final temperatures
Initial temperature
T[tex]_i[/tex] = P[tex]_i[/tex]V[tex]_i[/tex] / nR
given that n = 3.10 mol
= ( 101325 × 0.1550 ) / ( 3.10 × 8.314 )
= 15705.375 / 25.7734
T[tex]_i[/tex] = 609.4 K
Final temperature
T[tex]_f[/tex] = P[tex]_f[/tex]V[tex]_f[/tex] / nR
= ( 0.11166 × 101325 × 0.742 ) / ( 3.10 × 8.314 )
= 8394.95 / 25.7734
= 325.7 K
Therefore, Initial Temperature = 609.4 K and Final Temperature = 325.7 K
b) the change in internal energy
ΔE[tex]_{int[/tex] = nC[tex]_v[/tex]ΔT
here, C[tex]_v[/tex] = ( 5/2 )R
ΔE[tex]_{int[/tex] = 3.10 × ( 5/2 )8.314 × ( 325.7 - 609.4 )
= -18279.78 J
Therefore, the change in internal energy is -18279.78 J
c) the heat lost by the gas
Since its an adiabatic process,
Q = 0
Therefore, heat lost by the gas is zero or 0
d) the work done on the gas
W = ΔE[tex]_{int[/tex] - Q
= -18279.78 J - 0
W = -18279.78 J
Therefore, the work done on the gas is -18279.78 J
a) The Initial Temperature and Final Temperature of gas are 601.68 K and 321.61 K respectively.
b) The change in internal energy is -18279.78 J.
c) The heat lost by the gas is zero.
d) The work done on the gas is -18279.78 J.
Given data:
The moles of sample is, n = 3.10 mol.
The initial volume of sample is, [tex]V_{1}=0.1550 \;\rm m^{3}[/tex].
The final volume of sample is, [tex]V_{2}=0.742 \;\rm m^{3}[/tex].
The initial pressure of the sample is, [tex]P_{1}=1.00 \;\rm atm[/tex].
(a)
We know that the relation between the pressure and volume for an adiabatic process is as follows,
[tex]P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}[/tex]
Here, [tex]\gamma[/tex] is a adiabatic index. And for air, its value is 1.41.
Solving as,
[tex]P_{2}=P_{1} \times\dfrac{V_{1}^{\gamma}}{V_{2}^{\gamma}}\\\\\\P_{2}=1.00 \times\dfrac{0.1550^{1.41}}{0.742^{1.41}}\\\\\\P_{2} = 0.11166 \;\rm atm[/tex]
Now, calculate the final temperature using the ideal gas equation as,
[tex]P_{2}V_{2}=nRT_{2}\\\\T_{2}= \dfrac{P_{2} \times V_{2}}{nR}\\\\T_{2}= \dfrac{0.11166 \times 10^{5}\times 0.742}{3.10 \times 8.31}\\\\T_{2}=321.61 \;\rm K[/tex]
Similarly, calculate the initial temperature as,
[tex]P_{1}V_{1}=nRT_{1}\\\\T_{1}= \dfrac{P_{1} \times V_{1}}{nR}\\\\T_{1}= \dfrac{1 \times 10^{5}\times 0.1550}{3.10 \times 8.31}\\\\T_{1}=601.68 \;\rm K[/tex]
Thus, we can conclude that the initial and final temperature of the gas is 601.68 K and 321.61 K respectively.
(b)
The change in internal energy is given as,
ΔE = nCΔT
here, C = ( 5/2 )R
ΔE = 3.10 × ( 5/2 )8.314 × ( 325.7 - 609.4 )
= -18279.78 J
Therefore, the change in internal energy is -18279.78 J.
c)
The heat lost by the gas . Since its an adiabatic process, so there will be no heat interaction.
Q = 0
Therefore, heat lost by the gas is zero or 0
d)
The work done on the gas
W = ΔE - Q
W = -18279.78 J - 0
W = -18279.78 J
Therefore, the work done on the gas is -18279.78 J.
Learn more about the adiabatic process here:
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which has higher eneergy electron r proton
Answer:
proton have higher energy than electron
Explanation:
tag me brainliest
Answer:
proton
Explanation:
proton is higher energy than the electron
Warm air rises because faster moving molecules tend to move to regions of less
A) density.
B) pressure.
C) both of these
D) none of the above
Answer:
76rsfy7zfyuutfzufyztudzutdT7dFy9y8fr6s
Explanation:
rshyyjfshfsgfshfsyhrsyhuydtufhr6ra6yris7toe7r9w7rr6w996ryrowosotusuogsuoufsutot
i.Name two commonly used thermometric liquids.
ii.State two advantages each of the thermometric liquids mentioned above
Answer:
mercury and alcohol
ii) used to test temperatures
i) It is a good conductor of heat and therefore the whole liquid reaches the temperature of the surroundings quickly.
ii) It does not wet (cling to the sides of) the tube.
Alcohol:
i) Alcohol has greater value of temperature coefficient of expansion than mercury.
ii) it's freezing point is below –100°C.
Topic: Chapter 19: Some wiggle room
A hummingbird flaps its wings up to 70 times per second, producing a 70 Hz
hum as it flies by. If the speed of sound is 340 m/s, how far does the sound
travel between wing flaps?
= 4.86 m
= 58.9 m
= 0.206 m
= 23,800 m
Answer:
4.86 m
Explanation:
Given that,
The frequency produced by a humming bird, f = 70 Hz
The speed of sound, v = 340 m/s
We need to find how far does the sound travel between wing flaps. Let the distance is equal to its wavelength. So,
[tex]v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{340}{70}\\\\\lambda=4.86\ m[/tex]
So, the sound travel 4.86 m between wings flaps.
why do you like the full moon ?
Answer:
The Moon brings perspective. Observing the Moon, and I mean really looking – sitting comfortably, or lying down on a patch of grass and letting her light fill your eyes, it's easy to be reminded of how ancient and everlasting the celestial bodies are. When I do this, it always puts my life into perspective.Answer:
because it look more impressive than empty dark sky .
Click Stop Using the slider set the following: coeff of restitution to 1.00 A velocity (m/s) to 6.0 A mass (kg) to 6.0 B velocity (m/s) to 0.0 Calculate what range can the mass of B be to cause mass A to bounce off after the collision. Calculate what range can the mass of B be to cause mass A to continue forward after the collision. Check your calculations with the simulation. What are the ranges of B mass (kg)
Answer:
[tex]M_b=6kg[/tex]
Explanation:
From the question we are told that:
Coefficient of restitution [tex]\mu=1.00[/tex]
Mass A [tex]M_a=6kg[/tex]
Initial Velocity of A [tex]U_a=6m/s[/tex]
Initial Velocity of B [tex]U_b=0m/s[/tex]
Generally the equation for Coefficient of restitution is mathematically given by
[tex]\mu=\frac{V_b-V_a}{U_a-U_b}[/tex]
[tex]1=\frac{v_B}{6}[/tex]
[tex]V_b=6*1[/tex]
[tex]V_b=6m/s[/tex]
Generally the equation for conservation of linear momentum is mathematically given by
[tex]M_aU_a+M_bU_b=M_aV_a+M_bV_b[/tex]
[tex]6*6+=M_b*6[/tex]
[tex]M_b=6kg[/tex]
Two forces act on the screw eye. If F = 600 N, determine the magnitude of the resultant force and the angle θ if the resultant force is directed vertically upward.
Answer:
how to solve this problem ???????
The magnitude of the resultant force is 919.6 N and the value of angle θ is 36.87⁰.
Resultant of the two forces
The resultant of the two forces is determined by resolving the force into x and y component as shown below;
[tex]F_1_x + F_2x_x = F_R_x \ --- (1) \\\\F_1_y + F_2_y = F_R_y\ ---(2)[/tex]
where;
F1 = 500 NF2 = 600 NValue of Angle θThe value of Angle θ is determined from equation (1)
-500sinθ + 600sin(30) = 0
500sinθ = 600sin(30)
500sinθ = 300
sinθ = 3/5
θ = 36.87⁰
Resultant of the two forcesThe resultant of the forces is determined using the second equation;
500cosθ + 600cos(30) = R
500 x cos(36.87) + 600 x cos(30) = R
919.6 N = R
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A
Fluids in which the shear stress must reach
certain minimum value(yield stress)
before flow commences are called
Answer:
Plastic
Explanation:
Shear Modulus can be defined as the ratio of shear stress to shear strain with respect to a physical object.
This ultimately implies that, Shear Modulus arises as a result of the application of a shear force on an object or body which eventually leads to its deformation. Thus, this phenomenon is simply used by scientists to measure or determine the rigidity of an object or body.
Fluids in which the shear stress must reach certain minimum value (yield stress) before flow commences are called plastic. Thus, a plastic would only begin to flow when its shear stress attain a certain minimum value (yield stress). The unit of measurement of yield stress is usually mega pascal (MPa).
The cannon on a battleship can fire a shell a maximum distance of 33.0 km.
(a) Calculate the initial velocity of the shell.
Answer:
v = 804.23 m/s
Explanation:
Given that,
The maximum distance covered by a cannon, d = 33 km = 33000 m
We need to find the initial velocity of the shell. Let it is v. It can be calculated using the conservation of energy such that,
[tex]v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 33000} \\\\v=804.23\ m/s[/tex]
So, the initial velocity of the shell is 804.23 m/s.
1.a machine gun fires a ball with an initial velocity of 600m/s with an elevation of 30° with respect to the ground neglecting air resistance calculate:
a.the maximum height that can be reached?
b.the time of flight of the bullet?
c.the maximum horizontal displacement of the ired bullet?
Answer:
See explanation
Explanation:
a) maximum height of a projectile = u sin^2θ/2g
H= 600 × (sin 30)^2/2 × 10
H= 7.5 m
b) Time of flight
t= 2u sinθ/g
t= 2 × 600 sin 30/10
t= 60 seconds
Range
R= u^2sin2θ/g
R= (600)^2 × sin2(30)/10
R= 31.2 m
How do you find the product of gamma decay?
Answer:
The mass and atomic numbers don't change
Explanation:
An excited atom relaxes to the ground state emitting a photon...called a gamma ray.
The answer is that the mass and atomic numbers don't change.
In gamma decay, the product refers to the nucleus resulting from the emission of a gamma ray. Gamma decay occurs when an excited atomic nucleus releases excess energy in the form of a high-energy photon called a gamma ray.
To find the product of gamma decay, you need to identify the nucleus before and after the decay process. The product nucleus is determined by the parent nucleus that undergoes gamma decay.
During gamma decay, the number of protons and neutrons in the nucleus remains unchanged. Therefore, the identity of the element remains the same, but the energy state of the nucleus is altered.
The product nucleus is typically represented by the same chemical symbol as the parent nucleus, followed by a superscript indicating the mass number (total number of protons and neutrons) and a subscript indicating the atomic number (number of protons).
For example, if a parent nucleus with an atomic number of Z and a mass number of A undergoes gamma decay, the product nucleus will have the same atomic number Z and mass number A.
It's important to note that gamma decay does not involve the emission or absorption of any particles, only the release of electromagnetic radiation (gamma ray).
Thus, the product nucleus remains unchanged in terms of atomic number and mass number.
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Find the force on a negative charge that is placed midway between two equal positive charges. All charges have the same magnitude.
Answer: The force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.
Explanation:
Let us assume that
[tex]q_{1} = q_{2} = +q[/tex]
[tex]q_{3} = -q[/tex]
As [tex]q_{3}[/tex] is the negative charge and placed midway between two equal positive charges ([tex]q_{1}[/tex] and [tex]q_{2}[/tex]).
Total distance between [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is 2r. This means that the distance between [tex]q_{1}[/tex] and [tex]q_{3}[/tex], [tex]q_{2}[/tex] and [tex]q_{3}[/tex] = d = r
Now, force action on charge [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.
[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})[/tex]
where,
k = electrostatic constant = [tex]9 \times 10^{9} Nm^{2}/C^{2}[/tex]
Substitute the values into above formula as follows.
[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})\\= 9 \times 10^{9} (\frac{q \times (-q)}{r^{2}})\\= - 9 \times 10^{9} (\frac{q^{2}}{r^{2}})[/tex] ... (1)
Similarly, force acting on [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.
[tex]F_{32} = k \frac{q_{2}q_{3}}{d^{2}}\\= -9 \times 10^{9} \frac{q^{2}}{r^{2}}\\[/tex] ... (2)
As both the forces represented in equation (1) and (2) are same and equal in magnitude. This means that the net force acting on charge [tex]q_{3}[/tex] is zero.
Thus, we can conclude that the force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.
What is the effect on range and maximum height of a projectile as the launch height, launch speed, and launch angle are increased?
Answer:
The highest point in the trajectory occurs at the midpoint of the path. This highest point increases as the angle increases. At a 75° launch angle, the maximum height is approximately 76 meters. However, a further increase in launch angle beyond this 75° angle will increase the peak height even more.
A grade 12 Physics student shoots a basketball
from the ground at a hoop which is 2.0 m above
her release. The shot was at a velocity of 10 m/s
and at an angle of 80° to the ground.
a. Determine the vertical velocity of the ball
when it is at the level of the net. You
should get two answers.
Please show ALL steps
Answer:
7.84 m/s
Explanation:
Height, h = 2 m
Initial velocity, u = 10 m/s
Angle, A = 80°
(a) Let the time taken to go to the net is t.
Use second equation of motion
[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]
Time cannot be negative.
So, t = 0.2 s
The vertical velocity at t = 0.2 s is
v = u + at
v = 10 sin 80 - 9.8 x0.2
v = 9.8 - 1.96 = 7.84 m/s
Olympus Mons on Mars is the largest volcano in the solar system, at a height of 25 km and with a radius of 309 km. If you are standing on the summit, with what initial velocity would you have to fire a projectile from a cannon horizontally to clear the volcano and land on the surface of Mars
Answer:
The velocity is 2661.5 m/s.
Explanation:
Radius, horizontal distance, d = 309 km
height, h = 25 km
acceleration due to gravity on moon, g =3.71 m/s^2
Let the time taken is t and the horizontal velocity is u.
horizontal distance = horizontal velocity x time
309 x 1000 = u t .... (1)
Use second equation of motion in vertical direction.
[tex]h = u_yt +0.5 gt^2\\\\25000 = 0 + 0.5\times 3.71\times t^2\\\\t =116.1 s[/tex]
So, put in (1)
309 x 1000 = u x 116.1
u = 2661.5 m/s
A student has to work the following problem: A block is being pulled along at constant speed on a horizontal surface a distance d by a rope supplying a force F at an angle of elevation q. The surface has a frictional force acting during this motion. How much work was done by friction during this motion? The student calculates the value to be –Fd sinq. How does this value compare to the correct value?
a. It is the correct value.
b. It is too high.
c. It is too low.
d. The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.
Answer:
D
The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.
Explanation:
Since block moves with constant speed
So, frictional force
f = FCosq
Work done by friction
W = - fd
W = - fd Cos q
The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.
A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 60.0-g suspended mass. The suspended mass is given an initial downward speed of 1.60 m/s .
How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)
Express your answer using two significant figures.
Answer:
0.5
Explanation:
because the block is attached to the pulley of the string
A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the average intensity of the light from this bulb at a distance of 0.400 m from the bulb
Answer: [tex]29.85\ W/m^2[/tex]
Explanation:
Given
Power [tex]P=60\ W[/tex]
Distance from the light source [tex]r=0.4\ m[/tex]
Intensity is given by
[tex]I=\dfrac{P}{4\pi r^2}[/tex]
Inserting values
[tex]\Rightarrow I=\dfrac{60}{4\pi (0.4)^2}\\\\\Rightarrow I=\dfrac{60}{2.010}\\\\\Rightarrow I=29.85\ W/m^2[/tex]
Answer:
29.85 W/ m^2
Explanation:
A double-slit experiment is performed with light of wavelength 550 nm. The bright interference fringes are spaced 2.3 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 360 nm?
Answer:
[tex]d_2=1.5*10^-3m[/tex]
Explanation:
From the question we are told that:
Initial Wavelength [tex]\lambda_1=550nm=550*10^{-9}[/tex]
Space 1 [tex]d_1=2.3*10^{-3}[/tex]
Final wavelength [tex]\lambda_2=360*10^{-9}[/tex]
Generally the equation for Fringe space at [tex]\lambda _2[/tex] is mathematically given by
[tex]d_2=\frac{d_1}{\lambdaI_1}*\lambda_2[/tex]
[tex]d_2=\frac{2.3*10^{-3}}{550*10^{-9}}*360*10^{-9}[/tex]
[tex]d_2=1.5*10^-3m[/tex]
When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C where C is a constant. Suppose that at a certain instant the volume is 420 cubic centimeters and the pressure is 99 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?
Answer:
[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]
Explanation:
We are given that
[tex]PV^{1.4}=C[/tex]
Where C=Constant
[tex]\frac{dP}{dt}=-7KPa/minute[/tex]
V=420 cubic cm and P=99KPa
We have to find the rate at which the volume increasing at this instant.
Differentiate w.r.t t
[tex]V^{1.4}\frac{dP}{dt}+1.4V^{0.4}P\frac{dV}{dt}=0[/tex]
Substitute the values
[tex](420)^{1.4}\times (-7)+1.4(420)^{0.4}(99)\frac{dV}{dt}=0[/tex]
[tex]1.4(420)^{0.4}(99)\frac{dV}{dt}=(420)^{1.4}\times (7)[/tex]
[tex]\frac{dV}{dt}=\frac{(420)^{1.4}\times (7)}{1.4(420)^{0.4}(99)}[/tex]
[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]
Answer:
[tex]\dot V=2786.52~cm^3/min[/tex]
Explanation:
Given:
initial pressure during adiabatic expansion of air, [tex]P_1=99~kPa[/tex]
initial volume during the process, [tex]V_1=420~cm^3[/tex]
The adiabatic process is governed by the relation [tex]PV^{1.4}=C[/tex] ; where C is a constant.
Rate of decrease in pressure, [tex]\dot P=7~kPa/min[/tex]
Then the rate of change in volume, [tex]\dot V[/tex] can be determined as:
[tex]P_1.V_1^{1.4}=\dot P.\dot V^{1.4}[/tex]
[tex]99\times 420^{1.4}=7\times V^{1.4}[/tex]
[tex]\dot V=2786.52~cm^3/min[/tex]
[tex]\because P\propto\frac{1}{V}[/tex]
[tex]\therefore[/tex] The rate of change in volume will be increasing.
What is the length of the x-component of the vector shown below?
у
6
28°
Answer:
Explanation:
6cos28
=5.3 N
