Complete question:
A parallel-plate capacitor has plates with an area of 405 cm² and an air-filled gap between the plates that is 2.25 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery. (a) How much energy is stored in the capacitor? (b) The separation between the plates is now increased to 4.50 mm. How much energy is stored in the capacitor now?
Answer:
The energy stored in the capacitor when the plates is increased to 4.50 mm is 1.32 X 10⁻⁵ J
Explanation:
Given:
Area of the plates = 405 cm² = 405 X (10⁻²)² m²= 405 X 10⁻⁴m² = 0.0405m²
Energy stored in a capacitor = CV²/2
Where;
V is the voltage across the plates = 575 V
C is the capacitor =?
C = Kε(A/d)
K is constant = 1.0
ε is permittivity of free space = 8.885 X 10⁻¹²
d is the diameter of the two plates = 2.25 mm = 0.00225m
C = 1.0 x 8.885 X 10⁻¹² x (0.0405/0.00225)
C = 1.5993 X 10⁻¹⁰ F
(a) Energy stored in a capacitor = 0.5 X 1.5993 X 10⁻⁹ X 575²
= 2.64 X 10⁻⁵ J
(b) The separation between the plates is now increased to 4.50 mm.
C = Kε(A/d)
New diameter, d = 4.5 mm = 0.0045 m
C = 1.0 x 8.885 X 10⁻¹² x (0.0405/0.0045)
C = 7.9965 X 10 ⁻¹¹ F
Energy stored in a capacitor = 0.5 X 7.9965 X 10 ⁻¹¹ X 575²
= 1.32 X 10⁻⁵ J
Therefore, the energy stored in the capacitor when the plates is increased to 4.50 mm is 1.32 X 10⁻⁵ J
Astronomers have not yet reported any Earth-like planets orbiting other stars because (a) there are none; (b) they are not detectable with current technology; (c) no nearby stars are of the type expected to have Earth-like planets; (d) the government is preventing us from reporting their discovery.
Answer:
(b) they are not detectable with current technology
Explanation:
The improvement in exoplanet detection methods in recent years, thanks to spaces telescopes such as the Kepler, has led to a revolution in the field of astronomy. The findings have gone from focusing on Jupiters hot to super-earth and, ultimately, to Earth-like planets.
Calculate the energy separations in joules, kilojoules per mole, electronvolts, and reciprocal centimeters (cm^-1) between the levels:
a) n = 1 and n = 2, and
b) n = 5 and n = 6
Assume that the length of the box is 1.0 nm and the particle you are dealing with is an electron with m_c = 9.109 times 10^-31 kg.
This question asks for calculations of energy separations between specific quantum states (n) of an electron in a one-dimensional box, using Quantum Physics principles. The energy difference is calculated using a standard formula, and the values are converted into different units. This demonstrates the important foundational concepts of Quantum Physics.
Explanation:In Quantum Physics, the energy difference between specific states (n) can be calculated using the formula for the potential energy inside a one-dimensional box, which is defined as E = n^2h^2 / (8mL^2), where h is the Planck constant, m is the mass of the particle (electron in this case), and L is the length of the box.
The energy separation between n = 1 and n = 2 can be found by simply finding the difference in energy levels. This can also be done for n = 5 and n = 6. Thus, a straightforward calculation will yield the energy separations in joules (J). To convert to alternative units, use convenient relations: 1 J = 0.239006 kJ/mol, 1 J = 6.242×10^18 eV, and 1 cm^-1 = 1.98644582 x 10^-23 J.
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If A > B, under what condition is |A-BI=|A|- IB|? a. Vectors A and B are in opposite directions b. Vectors A and B are in the same direction. c. The statement is never true. d. Vectors A and B are in perpendicular directions. e. The statement is always true.
Answer:
b) Vectors A and B are in the same direction.
Explanation:
To understand this problem we will say that vector A has a magnitude of 5 units and vector B a magnitude of 3 units. In the subtraction of vectors the initial parts of vectors always bind together. And the vector resulting from the subtraction is traced from the end of the second vector (B) to the end of the first vector (A).
The length of the resultant vector will be 5 - 3 = 2
In the attached image, we analyze case a), b), and d)
For a)
As we can see in the attached image the resultant vector has a length of 8 units.
For d)
As we can see in the attached image the resultant vector has a length of 5.83 units.
For b)
The resultant vector has a length of 2 units.
Therefore the case given in b) is true
Final answer:
The condition required for |A-B| to equal |A| - |B| is when vectors A and B are in the same direction.
Explanation:
The correct answer is b. Vectors A and B are in the same direction. For two vectors A and B, the operation A - B is equivalent to adding -B to A, where -B is a vector of the same magnitude as B but in the opposite direction. When A and B are aligned and pointing in the same direction, the magnitude of their difference is equal to the difference of their magnitudes because the subtraction does not involve any trigonometric component due to the angle between them, since there is none. In mathematical symbols, |A - B| = |A| - |B|.
A skydiver leaves a helicopter with zero velocity and falls for 10.0 seconds before she opens her parachute. During the fall, the air resistance, F_DF D , is given by the formula F_D= - bvF D =−bv, where b=0.75b=0.75 kg/s and vv is the velocity. The mass of the skydiver with all the gear is 82.0 kg. Set up differential equations for the velocity and the position, and then find the distance fallen before the parachute opens.
a. 485 m
b. 458 m
c. 490 m
d. 257 m
e. 539 m
The distance fallen before the parachute opens is 485 m. The correct option is (a).
What is velocity?Velocity is a vector quantity that describes the rate of change of an object's position with respect to time. It specifies the object's speed and the direction of its motion. In other words, velocity is a measure of how fast an object is moving in a particular direction. The SI unit for velocity is meters per second (m/s). Velocity can be positive or negative depending on the direction of motion. For example, a car traveling northward has a positive velocity, while a car traveling southward has a negative velocity.
Here in the Question,
We can use the following equations of motion to set up the differential equations for the velocity and position of the skydiver:
F = ma
v = dx/dt
The net force on the skydiver is given by:
F_net = F_g - F_D
where F_g is the force due to gravity, F_D is the force due to air resistance, and the negative sign indicates that F_D is opposite to the direction of motion.
So we have:
F_g - F_D = ma
where a is the acceleration of the skydiver.
Substituting F_D = -bv into the above equation, we get:
mg - bv = ma
Dividing both sides by m, we get:
g - (b/m)v = a
Now we can set up the differential equations as follows:
dv/dt = g - (b/m)v
dx/dt = v
To solve these differential equations, we can use separation of variables:
dv/(g - (b/m)v) = dt
dx/v = dt
Integrating both sides with respect to t, we get:
-(m/b) ln(g - (b/m)v) = t + C1
ln(v) = t + C2
where C1 and C2 are constants of integration.
Solving for v and simplifying, we get:
v = (g*m/b) - Ce^(-bt/m)
where C = (g*m/b - v0) is another constant of integration, and v0 is the initial velocity (which is zero).
Using the initial condition v(0) = 0, we get:
C = g*m/b
So we have:
v = (gm/b) - (gm/b)e^(-bt/m)
To find the distance fallen before the parachute opens, we need to integrate v with respect to time from t = 0 to t = 10 seconds (when the parachute opens):
x = ∫[0 to 10] v dt
Substituting v, we get:
x = ∫[0 to 10] (gm/b) - (gm/b)e^(-bt/m) dt
x = (g*m/b) t + (m^2/b^2) (e^(-bt/m) - 1)
Substituting g = 9.81 m/s^2, m = 82.0 kg, and b = 0.75 kg/s, we get:
x = 485.1 m
Therefore, the distance fallen before the parachute opens is approximately 485 m. The answer is (a).
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During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the earth's surface and is to reach a maximum height of 940 m above the earth's surface. The rocket's engines give the rocket an upward acceleration so it moves with acceleration of 16.0 m/s2 during the time T that they fire. After the engines shut off, the rocket is in free fall. Ignore air resistance. Assume that the acceleration due to gravity does not change with the height of the rocket.
Answer:
solution:
when the engine are fired the rocket has a linear constant acceleration motion:
[tex]V_{1} ^{2} =v_{0} ^{2} +2a_{1} (y_{1} -y_{0} )t=(0)^2+2(16)(y_{1}-0)\\V_{1} ^{2}=32y_{1}................. eq(1)[/tex]
[tex]V_{1}[/tex] is the final velocity of the rocket
when the engines are fired it become equal to the initial velocity of the rocket,
when the engines are shut off
[tex]V_{2} ^{2} =v_{1} ^{2} +2a_{2} (y_{2} -y_{1} )t=>v_{1} ^{2}-2(9.8)(960-y_{1})\\V_{2} ^{2} =v_{1} ^{2}-18816+19.6y_{1}...................eq(2)[/tex]
solve eq(1) and eq(2) we find
[tex](0)^2=(32y_{1} )-18816+19.6y_{1}[/tex]
solving for [tex]y_{1}[/tex]=364.65 m
Where [tex]y_{1}[/tex] is the distance travelled by the rockets for shutting off the engine
when the engines are fired:
[tex]y_{1} =y_{o} + v_{0}t_{1} +\frac{1}{2}at^{2} =>0+(0)T+\frac{1}{2}(16)t^{2\\\\\\364.65=8T^{2} -->T=6.75s[/tex]
NOTE:
DIAGRAM IS ATTACHED
In order to work well, a square antenna must intercept a flux of at least 0.040 N⋅m2/C when it is perpendicular to a uniform electric field of magnitude 5.0 N/C.
Answer:
L > 0.08944 m or L > 8.9 cm
Explanation:
Given:
- Flux intercepted by antenna Ф = 0.04 N.m^2 / C
- The uniform electric field E = 5.0 N/C
Find:
- What is the minimum side length of the antenna L ?
Solution:
- We can apply Gauss Law on the antenna surface as follows:
Ф = [tex]\int\limits^S {E} \, dA[/tex]
- Since electric field is constant we can pull it out of integral. The surface at hand is a square. Hence,
Ф = E.(L)^2
L = sqrt (Ф / E)
L > sqrt (0.04 / 5.0)
L > 0.08944 m
The area of a square antenna needed to intercept a flux of 0.040 N⋅m2/C in a uniform electric field of magnitude 5.0 N/C is 0.008 m². Consequently, each side of the antenna must be about 0.089 meters (or 8.9 cm) long.
Explanation:The question pertains to the relationship between electric field and flux. The electric flux through an area is defined as the electric field multiplied by the area through which it passes, oriented perpendicularly to the field.
We are given that the electric field (E) is 5.0 N/C and the flux Φ must be 0.040 N⋅m2/C.
Hence, to intercept this amount of flux, the antenna must have an area (A) such that A = Φ / E.
That is, A = 0.040 N⋅m2/C / 5.0 N/C = 0.008 m².
Since the antenna is square, each side will have a length of √(0.008) ≈ 0.089 meters (or 8.9 cm).
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In an experiment to measure the acceleration of g due to gravity, two values, 9.96m/s (s is squared) and 9.72m/s (s is squared), are determined. Find (1) the percent difference of the measurements, (2) the percent error of each measurement, and (3) the percent error of their mean. (Accepted value:g=9.80m/s (s is squared))
Answer:
a)2.46 %
b)For 1 :101.52 %
For 2 : 99.08 %
c)100..4 %
Explanation:
Given that
g₁ = 9.96 m/s²
g₂ = 9.72 m/s²
The actual value of g = 9.8 m/s²
a)
The difference Δ g = 9.96 -9.72 =0.24 m/s²
[tex]The\ percentage\ difference=\dfrac{0.24}{9.72}\times 100=2.46\ percentage\\[/tex]
b)
For first one :
[tex]Error\ in\ the\ percentage =\dfrac{9.96}{9.81}\times 100 =101.52\ perncetage[/tex]
For second :
[tex]Error\ in\ the\ percentage =\dfrac{9.72}{9.81}\times 100 =99.08\ perncetage[/tex]
c)
The mean g(mean )
[tex]g(mean )=\dfrac{9.96+9.72}{2}\ m/s^2\\g(mean)=9.84\ m/s^2[/tex]
[tex]The\ percentage=\dfrac{9.84}{9.8}\times 100=100.40\ percentage[/tex]
a)2.46 %
b)For 1 :101.52 %
For 2 : 99.08 %
c)100..4 %
The percent difference between the two measurements is 2.44%. The percent error for the first measurement is 1.63%, and for the second measurement is 0.82%. The percent error of their mean is 0.41%.
Explanation:In physics, the percent difference is calculated by subtracting the two values, taking the absolute value, dividing by the average of the two values, and then multiplying by 100. Therefore, the percent difference between the two measurements 9.96m/s² and 9.72m/s² is:
|(9.96-9.72)|/((9.96+9.72)/2)*100 = 2.44%
The percent error involves taking the absolute difference between the experimental value and the accepted value, divided by the accepted value, then multiplied by 100. So, the percent error for the two measurements with accepted value of 9.80m/s² are:
For 9.96m/s²: |(9.96-9.80)|/9.80*100 = 1.63%
For 9.72m/s²: |(9.72-9.8)|/9.8*100 = 0.82%
The percent error of the mean involves doing the above but using the mean of the experimental measurements:
|(Mean of measurements - Accepted value)|/Accepted value * 100 |(9.96+9.72)/2-9.8|/9.8*100 = 0.41%
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A student adds 75.0 g of hot water at 80.0 0C into a calorimeter containing 100.0 g cold water at 20.0 0C. The final temperature is 42.5 0C. The heat capacity of water is 4.186 J/gK. What is the heat capacity of the calorimeter?
Answer:
The heat capacity of the calorimeter is 104.65 J/K
Explanation:
Heat lost by the hot water = Heat gained by cold water + Heat gained by the calorimeter
Heat lost by hot water = mCΔT
m = 75 g, C = 4.186 J/g.K, ΔT = (80 - 42.5) = 37.5 K
Heat lost by hot water = 75 × 4.186 × 37.5 = 11773.125 J
Heat gained by cold water = mCΔT
m = 100 g, C = 4.186 J/g.K, ΔT = (42.5 - 20) = 22.5 K
Heat gained by cold water = 100 × 4.186 × 22.5 = 9418.5 J
Heat gained by the calorimeter = Heat lost by hot water - Heat gained by cold water
Heat gained by the calorimeter = 11773.125 - 9418.5 = 2354.625 J
Heat gained by the calorimeter = Heat capacity of the calorimeter × ΔT
ΔT = (42.5 - 20) = 22.5 K
Heat capacity of the calorimeter = (Heat gained by the calorimeter)/(ΔT) = 2351.25/22.5 = 104.65 J/K
A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing at a constant rate of 4 m/s2. What is the magnitude of its total acceleration at this instant?
Answer:
5 m/s2
Explanation:
The total acceleration of the circular motion is made of 2 components: centripetal acceleration and linear acceleration of 4 m/s2. They are perpendicular to each other.
The centripetal acceleration is the ratio of instant velocity squared and the radius of the circle
[tex]a_c = \frac{v^2}{r} = \frac{30^2}{300} = \frac{900}{300} = 3 m/s^2[/tex]
So the magnitude of the total acceleration is
[tex]a = \sqrt{a_c^2 + a_l^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 m/s^2[/tex]
Answer:
truu dat
Explanation:
df
In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal. The shot leaves her hand at a height of 1.80 m above the ground.
A. How far does the shot travel?
B. Repeat the calculation of the first part for angle 42.5? .
C. Repeat the calculation of the first part for angle 45 ? .
D. Repeat the calculation of the first part for angle 47.5? .
E. At what angle of release does she throw the farthest?
A) Horizontal range: 16.34 m
B) Horizontal range: 16.38 m
C) Horizontal range: 16.34 m
D) Horizontal range: 16.07 m
E) The angle that gives the maximum range is [tex]41.9^{\circ}[/tex]
Explanation:
A)
The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.
The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:
[tex]s=u_y t + \frac{1}{2}at^2[/tex] (1)
where
s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)
[tex]u_y = u sin \theta[/tex] is the initial vertical velocity, where
u = 12.0 m/s is the initial speed
[tex]\theta=40.0^{\circ}[/tex] is the angle of projection
So
[tex]u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s[/tex]
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration due to gravity (downward)
Substituting the numbers, we get
[tex]-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0[/tex]
which has two solutions:
t = -0.21 s (negative, we ignore it)
t = 1.778 s (this is the time of flight)
The horizontal motion is instead uniform, so the horizontal range is given by
[tex]d=u_x t[/tex]
where
[tex]u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s[/tex] is the horizontal velocity
t = 1.778 s is the time of flight
Solving, we find
[tex]d=(9.19)(1.778)=16.34 m[/tex]
B)
In this second case,
[tex]\theta=42.5^{\circ}[/tex]
So the vertical velocity is
[tex]u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s[/tex]
So the equation for the vertical motion becomes
[tex]4.9t^2-8.1t-1.80=0[/tex]
Solving for t, we find that the time of flight is
t = 1.851 s
The horizontal velocity is
[tex]u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s[/tex]
So, the range of the shot is
[tex]d=u_x t = (8.85)(1.851)=16.38 m[/tex]
C)
In this third case,
[tex]\theta=45^{\circ}[/tex]
So the vertical velocity is
[tex]u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s[/tex]
So the equation for the vertical motion becomes
[tex]4.9t^2-8.5t-1.80=0[/tex]
Solving for t, we find that the time of flight is
t = 1.925 s
The horizontal velocity is
[tex]u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s[/tex]
So, the range of the shot is
[tex]d=u_x t = (8.49)(1.925)=16.34[/tex] m
D)
In this 4th case,
[tex]\theta=47.5^{\circ}[/tex]
So the vertical velocity is
[tex]u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s[/tex]
So the equation for the vertical motion becomes
[tex]4.9t^2-8.8t-1.80=0[/tex]
Solving for t, we find that the time of flight is
t = 1.981 s
The horizontal velocity is
[tex]u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s[/tex]
So, the range of the shot is
[tex]d=u_x t = (8.11)(1.981)=16.07 m[/tex]
E)
From the previous parts, we see that the maximum range is obtained when the angle of releases is [tex]\theta=42.5^{\circ}[/tex].
The actual angle of release which corresponds to the maximum range can be obtained as follows:
The equation for the vertical motion can be rewritten as
[tex]s-u sin \theta t + \frac{1}{2}gt^2=0[/tex]
The solutions of this quadratic equation are
[tex]t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}[/tex]
This is the time of flight: so, the horizontal range is
[tex]d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta[/tex]
It can be found that the maximum of this function is obtained when the angle is
[tex]\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})[/tex]
Therefore in this problem, the angle which leads to the maximum range is
[tex]\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}[/tex]
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A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a car will not have to rely on friction to round the curve without skidding. In other words, a car of mass m moving at the designated speed can negotiate the curve even when the road is covered with ice. Such a road is usually banked, which means that the roadway is tilted toward the inside of the curve. Suppose the designated speed for the road is to be v = 12.8 m/s (28.6 mi/h) and the radius of the curve is r = 37.0 m. At what angle should the curve be banked?
Answer:
24.3 degrees
Explanation:
A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:
[tex]a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2[/tex]
Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.
Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.
So [tex] gsin\alpha = 4.43cos\alpha[/tex]
[tex]\frac{sin\alpha}{cos\alpha} = \frac{4.43}{g} = \frac{4.43}{9.81} = 0.451[/tex]
[tex]tan\alpha = 0.451[/tex]
[tex]\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0[/tex]
Using Newton's second law and free-body diagram the angle with which the curve is to be banked is obtained as [tex]24.31^\circ[/tex].
Newton's Second Law of MotionThis problem can be analysed using a free-body diagram.
The acceleration in the horizontal direction (radial diretion) is the centripetal acceleration.
Applying Newton's second law of motion in the x-direction, we get;
[tex]\sum F = ma[/tex]
[tex]\implies N\,sin\, \theta =\frac{mv^2}{r}[/tex]
Now, applying Newton's second law of motion in the y-direction, we get;
[tex]\implies N\,cos \, \theta=mg[/tex]
Dividing both the equations, we get;
[tex]\frac{N\,sin\, \theta}{N\,cos\, \theta} =\frac{mv^2}{r\, mg}[/tex]
[tex]\implies tan\theta=\frac{v^2}{rg}=\frac{12.8^2}{37\times9.8} =0.4518[/tex]
[tex]\implies \theta=tan^{-1}(0.4518)=24.31^\circ[/tex]
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Charges of 3.5 µC and −7.6 µC are placed at two corners of an equilateral triangle with sides of 0.1 m. At the third corner, what is the electric field magnitude created by these two charges? (ke = 8.99 × 109 N·m2/C2)
Answer:
E = -3.6859 x 10∧6 N/C
Explanation:
Let q1 = 3.5 μF = 3.5 x 10∧-6 F and q2 =-7.6 μF = -7.6 x 10∧-6 F
are the charges placed at two corner of equilateral triangle. Electric Field Magnitude "E" on the third corner will be equal to the sum of Electric Filed Magnitude generated by q1 and q2.
E = E1 + E2
E = Ke × q1 / ([tex]d^{2}[/tex]) + Ke × q2 / ([tex]d^{2}[/tex])
E = Ke/[tex]d^{2}[/tex] (q1 + q2) (taking common)
(Ke 8.99 × 10∧9 N[tex]m^{2}[/tex]/[tex]C^{2}[/tex] and distance d= 0.1 m)
E = 8.99 × 10∧9 N[tex]m^{2}[/tex]/[tex]C^{2}[/tex] /[tex](0.1 m)^{2}[/tex] (3.5 x 10∧-6 F - 7.6 x 10∧-6 F)
E = -3685.9 x 10∧3 N/C
E = -3.6859 x 10∧6 N/C
An electric dipole with dipole moment p⃗ is in a uniform electric field E⃗ . A. Find all the orientation angles of the dipole measured counterclockwise from the electric field direction for which the torque on the dipole is zero. B. Which part of orientation in part (a) is stable and which is unstable?
A. To find the orientation angles of the dipole for which the torque is zero, we need to consider the torque equation for an electric dipole in an external electric field.
The torque ([tex]\(\tau\)[/tex]) acting on an electric dipole (p) in an electric field (E) is given by:
[tex]\[ \tau = p \cdot E \cdot \sin(\theta) \][/tex]
Where:
p is the magnitude of the electric dipole moment,
E is the magnitude of the electric field,
[tex]\(\theta\)[/tex] is the angle between the dipole moment (p) and the electric field (E).
For the torque to be zero, [tex]\(\sin(\theta)\)[/tex] must be zero. This happens when [tex]\(\theta = 0\)[/tex] or [tex]\(\theta = \pi\) (180 degrees)[/tex], as [tex]\(\sin(0) = 0\)[/tex] and [tex]\(\sin(\pi) = 0\)[/tex].
So, the two possible orientation angles for which the torque is zero are:
[tex]\(\theta = 0\)[/tex] degrees (aligned with the electric field)
[tex]\(\theta = 180\)[/tex] degrees (opposite to the electric field)
B. Now, let's analyze the stability of these orientations:
1. [tex]\(\theta = 0\)[/tex] degrees (aligned with the electric field):
In this orientation, the dipole moment is aligned with the electric field.
Any slight deviation from this orientation will result in a torque that tends to restore the dipole to its original position. Therefore, this orientation is stable.
2. [tex]\(\theta = 180\)[/tex] degrees (opposite to the electric field):
In this orientation, the dipole moment is opposite to the electric field. Similar to the previous case, any slight deviation from this orientation will result in a torque that tends to restore the dipole to its original position. Therefore, this orientation is also stable.
Both orientations are stable because they represent energy minima. Any deviation from these orientations results in a restoring torque that tries to bring the dipole back to its stable position.
Thus, both orientations ([tex]\(\theta = 0\) degrees and \(\theta = 180\)[/tex] degrees) are stable orientations for the electric dipole in a uniform electric field.
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Final answer:
The torque on an electric dipole in a uniform electric field is zero when the dipole is parallel to the field. This orientation is stable.
Explanation:
In general, the torque vector on an electric dipole, p, from an electric field, E, is given by the equation T = p x E, where the cross product represents the direction of the torque. To find all the orientation angles of the dipole for which the torque is zero, we set the cross product equal to zero:
p x E = 0
The torque is zero when the dipole and electric field vectors are parallel. Therefore, the dipole will experience a torque that tends to align it with the electric field vector. This means that the dipole is in a stable equilibrium when it is parallel to the electric field.
Therefore, the orientation angles that result in a zero torque are all angles that make the dipole parallel to the electric field.
Azurite is a mineral that contains 55.1% of copper. How many meter of copper wire with diameter of 0.0113 in can be produced from 3.25 lb of azurite?
Answer:
1402.73 m
Explanation:
Mass of Azurite=3.25 lb
Percent of copper in AZurite mineral=55.1%
Diameter of copper wire,d=0.0113 in
Radius of copper wire=[tex]r=\frac{d}{2}=\frac{0.0113}{2}=0.00565 in=\frac{565}{100000}=\frac{565}{100}\times \frac{1}{1000}=5.65\times 10^{-3}in[/tex]
[tex]\frac{1}{1000}=10^{-3}[/tex]
Density of copper=[tex]\rho=8.96g/cm^3[/tex]
1 lb=454 g
3.25 lb=[tex]3.25\times 454=1475.5 g[/tex]
Mass of Azurite=[tex]1475.5 g[/tex]
Mass of copper=[tex]\frac{55.1}{100}\times 1475.5=813 g[/tex]
Density=[tex]\frac{Mass}{volume}[/tex]
Using the formula
[tex]8.96=\frac{813}{volume\;of\;copper}[/tex]
Volume of copper wire=[tex]\frac{813}{8.96}=90.7cm^3[/tex]
Radius of copper wire=[tex]5.65\times 10^{-3}\times 2.54=14.35\times 10^{-3} cm[/tex]
1 in=2.54 cm
Volume of copper wire=[tex]\pi r^2 h[/tex]
[tex]\pi=3.14[/tex]
Using the formula
[tex]90.7=3.14\times (14.35\times 10^{-3})^2\times h[/tex]
[tex]h=\frac{90.7}{3.14\times (14.35\times 10^{-3})^2}[/tex]
[tex]h=140273 cm[/tex]
1 m=100 cm
[tex]h=\frac{140273}{100}=1402.73 m[/tex]
Hence, the length of copper wire required=1402.73 m
Calculate the approximate mass of the Milky Way Galaxy from the fact that the Sun orbits the galactic center every 230 million years at a distance of 27,000 light-years. (As dis-cussed in Chapter 19, this calculation tells us only the mass of the galaxy within the Sun’s orbit.)
To calculate the approximate mass of the Milky Way Galaxy within the Sun's orbit, we use circular motion and gravity principles. The mass is approximately 5.8 trillion solar masses.
To calculate the approximate mass of the Milky Way Galaxy within the Sun's orbit, we can use the principles of circular motion and gravity. The mass of the Milky Way Galaxy can be determined using the formula gravitational force = centripetal force. By rearranging the equation and plugging in the given values, we can solve for the mass. The approximate mass of the Milky Way Galaxy within the Sun's orbit is approximately 5.8 trillion solar masses.
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The mass of the Milky Way Galaxy within the Sun's orbit is approximately 1.0 x 10⁴¹ kg or about 5 x 10¹¹ solar masses.
We know the Sun orbits the galactic center every 230 million years (2.3 x 10⁸ years) and is approximately 27,000 light-years (2.54 x 10¹⁷ meters) away from the center.
1. First, we convert the orbital period to seconds:
(2.3 x 10⁸ years) x (3.15 x 10⁷ seconds/year) ≈ 7.25 x 10¹⁵ seconds2. Next, we apply Newton's form of Kepler's third law, which states:
T2 = (4π²r³) / (GM)where
T is the orbital period r is the radius of the orbit G is the gravitational constant (6.67 x 10⁻¹¹ N(m²/kg²))M is the mass of the galaxy within the Sun's orbit.3. Rearranging for M gives us:
M = (4π²r³) / (G T²)Substituting known values:
M = (4 x π² x (2.54 x 10¹⁷ m)³) / (6.67 x 10⁻¹¹ N(m²/kg²) x (7.25 x 10¹⁵ s)²)M ≈ 1.0 x 10⁴¹ kgThis mass is approximately 5 x 10¹¹ solar masses, considering one solar mass is about 2 x 10³⁰ kg.
A projectile is launched from ground level to the top of a cliff which is 195 m away and 155 m high. If the projectile lands on top of the cliff 7.6 s after it is fired, find the initial velocity of the projectile. Neglect air resistance.
Answer:
66.02m/s
Explanation:
the equation describing the distance covered in the horizontal direction is
[tex]x=ucos\alpha t-(1/2)gt^{2}[/tex] but the acceleration in the horizontal path is zero, hence we have
[tex]x=ucos\alpha t[/tex]
Since the horizontal distance covered is 155m at 7.6secs, we have [tex]ucos\alpha =\frac{155}{7.6} \\ucos\alpha =20.38.............equation 1[/tex]
Also from the vertical path, the distance covered is expressed as
[tex]y=usin\alpha t-(1/2)gt^{2}[/tex]
since the horizontal distance covered in 7.6secs is 195m, then we have
[tex]y=usin\alpha t-(1/2)gt^{2}\\y=7.6usin\alpha -4.9(7.6)^{2}\\478.02=7.6usin\alpha \\usin\alpha =62.9...........equation 2[/tex]
Hence if we divide both equation 1 and 2 we arrive at
[tex]\frac{usin\alpha }{ucos\alpha } =\frac{62.9}{20.38} \\tan\alpha =3.08\\\alpha =tan^{-1}(3.08)\\\alpha =72.02^{0}\\[/tex]
Hence if we substitute the angle into the equation 1 we have
[tex]ucos72.02=20.38\\u=66.02m/s[/tex]
Hence the initial velocity is 66.02m/s
A parallel-plate capacitor is charged and then disconnected from the battery, so that the charge Q on its plates cant change. Originally the separation between the plates of the capacitor is d and the electrical field between the plates of the capacitor is E = 6.0 × 10^4 N/C If the plates are moved closer together, so that their separation is halved and becomes d/2, what then is the electrical field between the plates of the capacitor? What if the battery is left connected?
Answer:
[tex] E_f = \frac{V}{\frac{d}{2}}= 2 \frac{V}{d}= 2* 6x10^4 \frac{N}{C} = 1.2x10^5 \frac{N}{C}[/tex]
Explanation:
For this case we know that the electric field is given by:
[tex] E= 6 x 10^4 \frac{N}{C}[/tex]
And we want to find the final electric field assuming that the separation is halved and becomes d/2.
For this case we can use the following two equations:
[tex] C = \epsilon_o \frac{A}{d} = \frac{Q}{V}[/tex] (1)
[tex] E = \frac{\sigma}{\epsilon_o}[/tex] (2)
Where Q represent the charge, V the voltage, d the distance, A the area.
We can rewrite the equation (2) like this:
[tex] E = \frac{\sigma}{\epsilon_o} = \frac{Q}{A \epsilon_o}[/tex] (3)
And we can solve for Q from equation (1)like this:
[tex] Q = \frac{\epsilon_o A V}{d}[/tex]
And if we replace into equation (3) the previous result we got:
[tex] E = \frac{\epsilon_o A V}{A d \epsilon_o} = \frac{V}{d}[/tex]
And since the the electric field not change and the distance would be the half we have that the final electric field would be given by:
[tex] E_f = \frac{V}{\frac{d}{2}}= 2 \frac{V}{d}= 2* 6x10^4 \frac{N}{C} = 1.2x10^5 \frac{N}{C}[/tex]
What is a blackbody? Describe the radiation it emits.
Answer:
A black body or blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. (It does not only absorb radiation, but can also emit radiation. The name "black body" is given because it absorbs radiation in all frequencies, not because it only absorbs.) A white body is one with a "rough surface that reflects all incident rays completely and uniformly in all directions."
A block with mass m1 = 8.8 kg is on an incline with an angle theta = 40 degree with respect to the horizontal. For the first question there is no friction, but for the rest of this problem the coefficients of friction are: mu_k = 0.38 and mu_s = 0.418. 1) 1) When there is no friction, what is the magnitude of the acceleration of the block? m/s^2 2) Now with friction, what is the magnitude of the acceleration of the block after it begins to slide down the plane? m/s^2 3) 3) To keep the mass from accelerating, a spring is attached. What is the minimum spring constant of the spring to keep the block from sliding if it extends x = 0.13 m from its unstretched length. N/m 4) Now a new block with mass m2 = 16.3 kg is attached to the first block. The new block is made of a different material and has a greater coefficient of static friction. What minimum value for the coefficient of static friction is needed between the new block and the plane to keep the system from accelerating?
To solve the problem, we analyze a block on an incline to determine its acceleration with and without friction, calculate the minimum spring constant to prevent motion, and find the required coefficient of static friction for a new block added to the system.
Explanation:This problem involves concepts from Physics, specifically dynamics and statics, to understand the motion of a block on an incline and the effects of friction and spring force. To solve such problems, we use Newton's second law of motion, the equation for frictional force, and Hooke's law for the spring force. These principles help us calculate the acceleration of the block, the conditions required to prevent its motion, and the properties of friction between surfaces in contact.
For the first part, without friction, the acceleration can be found using the component of gravitational force along the incline. With friction, the net force includes both the component of gravitational force along the incline and the frictional force, affecting the acceleration. To find the minimum spring constant, Hooke's law is applied considering the balance of forces to prevent the block's motion. Lastly, calculating the minimum coefficient of static friction for a new block involves considering the combined system's weight and ensuring the frictional force is sufficient to prevent motion.
A certain volcano on earth can eject rocks vertically to a maximum height H. (a) How high (in terms of H) would these rocks go if a volcano on Mars ejected them with the same initial velocity? The acceleration due to gravity on Mars is 3.71 m/s2; ignore air resistance on both planets. (b) If the rocks are in the air for a time T on earth, for how long (in terms of T) would they be in the air on Mars?
Explanation:
a) By conservation of energy we can write
mgh on earth = mgh on mars.
[tex]mg_Eh_E=mg_Mh_M[/tex]
M,E are earth and mars respectively.
[tex]h_m =\frac{g_E}{g_M}\times h_E[/tex]
[tex]h_m=\frac{9.81}{3.71}\times h_E[/tex]
h_m= 2.64 h_E
b) Consider the time taken for the rock to reach the top of its trajectory. By symmetry, this is T/2. Inserting this into the kinematics equation v = u+at, we get the following two sets of equations:
final velocities will be zero v= 0
[tex]0= v- g_E\frac{T}{2}[/tex]
[tex]0=v- g_M\frac{T_M}{2}[/tex]
This gives [tex]2v= g_ET_E=g_mT_m[/tex] and
therefore,
[tex]T_M= 2.64T_E[/tex]
The rocks ejected by a volcano on Mars with the same initial velocity as one on Earth would reach a maximum height about 2.69 times greater, and stay aloft approximately 2.69 times longer.
Explanation:The maximum height a rock reaches is given by the formula H = (v^2) / (2g), where v is the initial velocity and g is the acceleration due to gravity. If we ignore potential differences in air resistance and other factors, and the initial velocity of the rocks is the same on both planets, then the change in maximum height is directly proportional to the change in gravity because the initial velocity is constant.
(a) Mars has 37.1% of Earth's gravitational force. Thus, if a volcano on Mars ejected rocks with the same initial velocity as a volcano on Earth, they would reach approximately 2.69 times the height H.
(b) The time a rock stays in the air is given by the formula T = 2v / g. So the rock will stay aloft on Mars approximately 2.69 times as long as T, the time that would stay on Earth given the same initial velocity.
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Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infinity as your reference point. Compute the gradient of V in each region, and check that it yields the correct field. Sketch V(r).
Answer:
Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:
[tex]E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}[/tex]
lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:
[tex]E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}[/tex]
To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):
[tex]V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r<R)=- \int\limits^r_\infty{E.dl\\\\= -\int\limits^R_\infty{\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } -\int\limits^r_R{\frac{1}{4\pi(e_{0)} }\frac{Q}{R^2}\frac{r}{R} dr\\[/tex]
=[tex]\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ][/tex]
∴NOTE: Graph is attached
The potential inside the uniformly charged solid sphere is given by a specific equation, while the potential outside the sphere is given by a different equation. The gradients of these potentials can also be calculated. A sketch of V(r) shows how the potential changes within and outside the sphere.
Explanation:Inside the uniformly charged solid sphere:
The potential is given by the equation:
V(r) = k(q / R)((3R - r^2) / (2R^3))
The gradient of V inside the sphere is given by:
∇V(r) = -(kqr) / (R^3)
Outside the sphere:
The potential is given by the equation:
V(r) = (kq) / (r)
The gradient of V outside the sphere is given by:
∇V(r) = -(kq) / (r^2)
Sketch of V(r):
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The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizontal, some of the tiny critters have reached a maximum height of 58.7 cm above the level ground.
(a) What was the takeoff speed for such a leap?
(b) What horizontal distance did the froghopper cover for this world-record leap?
Answer:
0.528m
Explanation:
a)58.7 cm = 0.587 m
Let g = 9.8m/s2. When the frog jumps from ground to the highest point its kinetic energy is converted to potential energy:
[tex]E_p = E_k[/tex]
[tex]mgh = mv^2/2[/tex]
where m is the frog mass and h is the vertical distance traveled, v is the frog velocity at take-off
[tex]v^2 = 2gh = 2*9.8*0.587 = 11.5[/tex]
[tex]v = \sqrt{11.5} = 3.4 m/s[/tex]
b) Vertical and horizontal components of the velocity are
[tex]v_v = vsin(\alpha) = 3.4sin(58^0) = 2.877 m/s[/tex]
[tex]v_h = vcos(\alpha) = 3.4cos(58^0) = 1.8 m/s[/tex]
The time it takes for the vertical speed to reach 0 (highest point) under gravitational acceleration g = -9.8m/s2 is
[tex]\Delta t = \Delta v / g = \frac{0 - 2.877}{-9.8} = 0.293s[/tex]
This is also the time it takes to travel horizontally, we can multiply this with the horizontal speed to get the horizontal distance it travels
[tex]s_h = v_ht = 1.8*0.293 = 0.528 m[/tex]
How much heat has to be added to 508 g of copper at 22.3°C to raise the temperature of the copper to 49.8°C? (The specific heat of copper is 0.377 J/g·°C.)
Answer:
Q = 5267J
Explanation:
Specific heat capacity of copper (S) = 0.377 J/g·°C.
Q = MSΔT
ΔT = T2 - T1
ΔT=49.8 - 22.3 = 27.5C
Q = change in energy = ?
M = mass of substance =508g
Q = (508g) * (0.377 J/g·°C) * (27.5C)
Q= 5266.69J
Approximately, Q = 5267J
Final answer:
To raise the temperature of 508 g of copper from 22.3°C to 49.8°C, 5096.925 J of heat needs to be added, using the specific heat capacity of copper.
Explanation:
To calculate the amount of heat (Q) needed to raise the temperature of a substance, we use the formula Q = mcΔT, where m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. For 508 g of copper with a specific heat of 0.377 J/g°C, needing to increase from 22.3°C to 49.8°C, the change in temperature (ΔT) is 49.8°C - 22.3°C = 27.5°C.
Therefore, Q = (508 g)(0.377 J/g°C)(27.5°C) = 5096.925 J. Hence, 5096.925 J of heat needs to be added to the copper to achieve the desired temperature increase.
Is it possible for an object to be (a) slowing down while its acceleration is increasing in magnitude; (b) speeding up while its acceleration is decreasing? In both cases, explain your reasoning.
a) Yes, if acceleration and velocity have opposite directions
b) Yes, if acceleration and velocity have same direction
Explanation:
a)
In order to answer this question, we have to keep in mind that both velocity and acceleration are vector quantities, so they have also a direction.
Acceleration is defined as the rate of change in velocity:
[tex]a=\frac{v-u}{t}[/tex]
where
u is the initial velocity
v is the final velocity
t is the time elapsed
In this problem, the object is slowing down: this means that the magnitude of its velocity is decreasing, so
[tex]|v|<|u|[/tex]
This means that the direction of the acceleration is opposite to the direction of the velocity. Then, the magnitude of the acceleration can be increasing (this will not affect the fact that the object will slow down, but it will affect only the rate at which the object is slowing down).
b)
In this case, the object is speeding up. This means that the magnitude of its velocity is increasing, so we have
[tex]|v|>|u|[/tex]
In order for this to happen, it must be that the direction of the acceleration is the same as the direction of the velocity: therefore, this way, the magnitude of the velocity will be increasing (either in the positive or in the negative direction).
Then, the magnitude of the acceleration can be decreasing (this will not affect the fact that the object will speed up, but it will only affect the rate at which the object is speeding up).
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What can you conclude about the relative magnitudes of the lattice energy of lithium iodide and its heat of hydration?
Complete question:
When lithium iodide is dissolved in water, the solution becomes hotter.
What can you conclude about the relative magnitudes of the lattice energy of lithium iodide and its heat of hydration?
Answer:
The heat of hydration is greater in magnitude than the lattice energy and lattice energy is smaller in magnitude that the heat of hydration.
Explanation:
When the solution becomes hotter on addition of lithium iodide, it shows exothermic reaction and it means that the heat of hydration is greater than the lattice energy and lattice energy is smaller in magnitude that the heat of hydration.
This can also be observed in the formula below;
[tex]H_{solution} = H_{hydration} + H_{lattice. energy}[/tex]
Heat of the hydration is thus greater in magnitude than that of the lattice energy and the lattice energy is smaller in the magnitude than the heat of hydration.
What are the lattice energy and the heat of hydration ?.The lattice energy is defined as the energy needed to separate the mole of the icon into a slid gaseous ion. It can be measured empirically and is calculated by use of electrostatics.
The heat of hydration is the great energy generated when the water reacts with the contact of cement powder. This leas to high temperatures and cause thermal cracking and the reduction of mechanical properties.
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A voltage of 169 V is applied across a 199 μF capacitor. Calculate the charge stored on the capacitor.
Answer:
Q = 3.363 x 10⁻² C
Explanation:
given,
Voltage, V= 169 V
Capacitance of the capacitor, C = 199 μF
Charge in the capacitor = ?
We know,
Q = CV
Q = 169 x 199 x 10⁻⁶
Q = 3.363 x 10⁻² C
Hence, the Charge stored in the capacitor is equal to Q = 3.363 x 10⁻² C
The cheetah is considered the fastest running animal in the world. Cheetahs can accelerate to a speed of 20.0 m/s in 2.50 s and can continue to accelerate to reach a top speed of 28.0 m/s. Assume the acceleration is constant until the top speed is reached and is zero thereafter. 1) Express the cheetah's top speed in mi/h. (Express your answer to three significant figures.)
Answer:
62.6m
Explanation:
We are given that
Speed of Cheetah,v=20 m/s
Time=2.5 s
Top speed=[tex]v'=28m/s[/tex]
We have to express Cheetah's top speed in mi/h
1 mile=1609.34 m
1 m=[tex]\frac{1}{1609.34}miles[/tex]
1 hour=3600 s
By using these values
[tex]v'=\frac{\frac{28}{1609.34}}{\frac{1}{3600}}[/tex]mi/h
[tex]v'=\frac{28}{1609.34}\times 3600[/tex]mi/h
Top speed of Cheetah=62.6mph
An object whose weight is 100lbf( pound force) experiences a decrease i kinetic energy of 500ft-lb, and an increase in potential energy of 1500ft-lbf. The initial velocity and elevation of the object, each relative to the surface of the Earth are 40ft/s and 30ft.
(a). Find final velocity in ft/s . (b) find final elevation.
The final velocity is [tex]v=35.75\ ft/s[/tex] and the final elevation is [tex]45ft[/tex].
The potential energy is the energy possessed by virtue of configuration and the kinetic energy is the energy possessed due to motion.
Given:
Initial velocity, [tex]u=40 ft/s[/tex]
Initial elevation, [tex]h_i=30 ft[/tex]
Weight of the object, [tex]w=100lbf[/tex]
Increase in potential energy, [tex]\Delta PE=1500 ft{-}lbf[/tex]
Decrease in kinetic energy, [tex]\Delta KE=-500 ft{-}lbf[/tex]
(a)
The final velocity of the object is computed as:
[tex]\frac{1}{2} \frac{w}{g}v^2-\frac{1}{2} \frac{w}{g}u^2= \Delta KE\\v^2=40^2 + 2 \times \frac{32.17}{100} \times (-500)\\v= \sqrt{1278.3}\\v=35.75 \ ft/s[/tex]
(b)
The final elevation of the object is computed as:
[tex]mgh_f-mgh_i= \Delta PE\\h_f= h_i+ \frac {\Delta PE}{mg}\\h_f= 30 + \frac{1500}{100}\\h_f=45 ft[/tex]
Therefore, the final velocity is [tex]v=35.75\ ft/s[/tex] and the final elevation is [tex]45ft[/tex].
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A water pump increases the water pressure from 15 psia to 70 psia. Determine the power input required, in hp, to pump 1.1 ft3/s of water. Does the water temperature at the inlet have any significant effect on the required flow power?
Answer:
[tex]P= 60.5 \frac{psia ft^3}{s} *\frac{1 Btu}{5.404 psia ft^3} *\frac{1 hp}{0.7068 Btu/s}= 15.839 hp[/tex]
Explanation:
Notation
For this case we have the following pressures:
[tex] p_1 = 15 psia[/tex] initial pressure
[tex]p_2 = 70 psia[/tex] final pressure
[tex] V^{*} = 1.1 ft^3/s[/tex] represent the volumetric flow
[tex] rho[/tex] represent the density
[tex]m^{*}[/tex] represent the mass flow
Solution to the problem
From the definition of mass flow we have the following formula:
[tex] m^{*} = \rho V^{*}[/tex]
For this case we can calculate the total change is the sytem like this:
[tex] \Delta E= \frac{p_2 -p_1}{\rho}[/tex]
Since we just have a change of pressure and we assume that all the other energies are constant.
The power is defined as:
[tex] P = m^* \Delta E[/tex]
And replacing the formula for the change of energy we got:
[tex]P = m V^* \frac{p_2 -p_1}{\rho} = V^* (p_2 -p_1)[/tex]
And replacing we have this:
[tex] P= (70-15) psia * 1.1 \frac{ft^3}{s} =60.5 \frac{psia ft^3}{s}[/tex]
And we can convert this into horsepower like this:
[tex]P= 60.5 \frac{psia ft^3}{s} *\frac{1 Btu}{5.404 psia ft^3} *\frac{1 hp}{0.7068 Btu/s}= 15.839 hp[/tex]
List the following items in order of decreasing speed, from greatest to least: (A) A wind-up toy car that moves 0.10 m in 2.5 s . (B) A soccer ball that rolls 2.5 m in 0.55 s . (C) A bicycle that travels 0.60 m in 7.5×10−2 s .(D) A cat that runs 8.0 m in 2.5 s . Enter your answer as four letters separated with commas.
Answer:
bicycle>soccer ball>cat>toy car
Explanation:
s = Distance
t = Time
Speed is given by
[tex]v=\dfrac{s}{t}[/tex]
For toy car
[tex]v=\dfrac{0.1}{2.5}=0.04\ m/s[/tex]
For soccer ball
[tex]v=\dfrac{2.5}{0.55}=4.54\ m/s[/tex]
For bicycle
[tex]v=\dfrac{0.6}{7.5\times 10^{-2}}=8\ m/s[/tex]
For cat
[tex]v=\dfrac{8}{2.5}=3.2\ m/s[/tex]
8>4.54>3.2>0.04
bicycle>soccer ball>cat>toy car