Answer:
(a). The shear stress on the pipe wall is 0.2062 lb/ft²
(b). The shear stress at the distance 0.3 is 0.12375 lb/ft²
(c). The shear stress at the distance 0.5 in away from the pipe wall is zero.
Explanation:
Given that,
Diameter = 1 in
Pressure = 0.55 psi
Length = 8 ft
We need to calculate the radius of the pipe
Using formula of radius
[tex]r=\dfrac{D}{2}[/tex]
Put the value into the formula
[tex]r=\dfrac{1}{2}[/tex]
[tex]r=0.5\ in[/tex]
(a). We need to calculate the shear stress on the pipe wall
Using formula of shear stress
[tex]\dfrac{\Delta p}{L}=\dfrac{2\tau}{r}[/tex]
[tex]\tau=\dfrac{\Delta p\times r}{2L}[/tex]
Put the value into the formula
[tex]\tau=\dfrac{0.55\times144\times0.5}{2\times8\times12}[/tex]
[tex]\tau=0.2062\ lb/ft^2[/tex]
(b). We need to calculate the shear stress at the distance 0.3 in
Using formula of shear stress
[tex]\tau=\dfrac{\Delta p\times r}{2L}[/tex]
Put the value into the formula
[tex]\tau=\dfrac{0.55\times144\times0.3}{2\times8\times12}[/tex]
[tex]\tau=0.12375\ lb/ft^2[/tex]
(c). We need to calculate the shear stress at the distance 0.5 in away from the pipe wall
r = 0.5-0.5 = 0
Using formula of shear stress
[tex]\tau=\dfrac{\Delta p\times r}{2L}[/tex]
Put the value into the formula
[tex]\tau=\dfrac{0.55\times144\times0}{2\times8\times12}[/tex]
[tex]\tau=0[/tex]
Hence, (a). The shear stress on the pipe wall is 0.2062 lb/ft²
(b). The shear stress at the distance 0.3 is 0.12375 lb/ft²
(c). The shear stress at the distance 0.5 in away from the pipe wall is zero.
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]
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 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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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
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 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
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
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
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
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
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.
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.
What is the temperature of a sample of gas when the average translational kinetic energy of a molecule in the sample is 8.37 × 10 − 21 J ?
Answer:
404K
Explanation:
Data given, Kinetic Energy.K.E=8.37*10^-21J
Note: as the temperature of a is increase, the rate of random movement will increase, hence leading to more collision per unit time. Hence we can say that the relationship between the kinetic energy and the temperature is a direct variation.
This relationship can be expressed as
[tex]K.E=\frac{3}{2}KT[/tex]
where K is a constant of value 1.38*10^-23
Hence if we substitute the values, we arrive at
[tex]T=\frac{2/3(8.37*10^{-21})}{1.38*10^-23}\\ T=404K[/tex]
converting to degree we have [tex]131^{0}C[/tex]
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
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]
Two tuning forks sounding together result in a beat frequency of 3 Hz. If the frequency of one of the forks is 256 Hz, what is the frequency of the other?
Answer:
The other frequency is either 253Hz (the altered tuning frequency vibrates lower) or 259Hz(if the altered tuning frequency vibrates higher):
Explanation:
We have two frequencies: the tuning fork frequency and the altered tuning fork frequency
F1 -----tuning fork frequency
F2-----Altered tuning fork frequency
Fbeat=3
F1=256Hz
To find the alterd tuning fork frequency:
The beat frequency of any two waves is:
Fbeat =|f1 – f2|
i.e we have two choices
for frequency tied to length increase (wavelength increases as length increases, therefore frequency decrease (the altered tuning frequency vibrates lower): Fbeat = f2-f1
F2=F1-Fbeat
F2= 256-3
=253Hz
For frequency tied to length decrease wavelength increases as length increases, therefore frequency also increases(the altered tuning frequency vibrates higher) : Fbeat = -(f2-f1)
F2= Fbeat+F1
= 256+3
= 259Hz
Note that a wavelength increase means a decrease of frequency because v = fλ
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|.
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
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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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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A sample of gas occupies a volume of 57.8 mL 57.8 mL . As it expands, it does 110.8 J 110.8 J of work on its surroundings at a constant pressure of 783 Torr 783 Torr . What is the final volume of the gas?
Answer: The final volume of the gas is 1.12 L
Explanation:
W = work done on or by the system
w = work done by the system = [tex]P\Delta V[/tex]
P = pressure = 783 torr = 1.03 atm (1atm=760 torr)
w= 110.8 J = 1.094 J (1Latm=101.3J)
[tex]V_1[/tex] = initial volume = 57.8 ml = 0.0578L (1L=1000ml)
[tex]V_2[/tex] = final volume = ?
[tex]1.094Latm=1.03atm\times (V_2-0.0578)L[/tex]
[tex](V_2-0.0578)=1.06[/tex]
[tex]V_2=1.12L[/tex]
The final volume of the gas is 1.12 L
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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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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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."
Problem 4 A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true anomaly of 150 deg. If the speed of the meteoroid at that time is 2.23 km/s, calculate: (a) the eccentricity of the trajectory, (b) the altitude at closest approach, (c) the speed at the closest approach, (d) the aiming radius and turn angle, and (e) the C3 of the meteoroid. Sketch the orbit.
a: The eccentricity is 1.086.
b: The altitude at closest approach is 5088 km
c: The velocity at perigee is 8.516 km/s
d: The turn angle is 134.08 while the aiming radius is 5641.28 km
Part a
Specific energy is given by
[tex]\epsilon=(v^2)/(2)-(\mu)/(r)[/tex]
Here
ε is the specific energy
v is the velocity which is given as 2.23 km/s
μ is the gravitational constant whose value is 398600
r is the distance between earth and the meteorite which is 402,000 km
[tex]\epsilon=(v^2)/(2)-(\mu)/(r)\n\epsilon=(2.2^2)/(2)-(398600)/(402,000)\n\epsilon=1.495 km^2/s^2[/tex]
Value of specific energy is also given as
[tex]\epsilon=(\mu)/(2a)\na=(\mu)/(2\epsilon)\na=(398600)/(2* 1.495)\na=13319 km[/tex]
Orbit formula is given as
[tex]r=a((e^2-1)/(1+ecos \theta))\nae^2-recos\theta-(a+r)=0[/tex]
Putting values in this equation and solving for e via the quadratic formula gives
[tex]ae^2-recos\theta-(a+r)=0\n(133319)e^2-(402000)(cos 150) e-(133319+402000)[/tex]
[tex]=0\n133319e^2+348142.21 e-535319=0\n\ne[/tex]
[tex]=(-348142.21 \pm \sqrt{(348142.21^2-4(133319)(535319)))}/(2 (133319))\n\ne[/tex]
=1.086 , or , -3.69
As the value of eccentricity cannot be negative so the eccentricity is 1.086.
Part b
The radius of trajectory at perigee is given as
[tex]r_p=a(e-1)\n[/tex]
Substituting values gives
[tex]r_p=133319 (1.086-1)\nr_p=11465.4 km[/tex]
Now for estimation of altitude z above earth is given as
[tex]z=r_p-R_E\nz=11465.4-6378\nz=5087.434\nz\approx 5088 km[/tex]
So the altitude at closest approach is 5088 km
Part c
radius of perigee is also given as
[tex]r_p=(h^2)/(\mu)(1)/(1+e)[/tex]
Rearranging this equation gives
[tex]h=√(r_p\mu(1+e))\nh=√(11465.4 * 3986000 * (1+1.086))\nh=97638.489 km^2/s[/tex]
Now the velocity at perigee is given as
[tex]v_p=(h)/(r_p)\nv_p=(97638.489)/(11465.4)\nv_p=8.516 km/s\n[/tex]
So the velocity at perigee is 8.516 km/s
Part d
Turn angle is given as
[tex]\delta =2 sin^{(-1)} ((1)/(e))[/tex]
Substituting value in the equation gives
[tex]\delta =2 sin^{(-1)} ((1)/(e))\n\delta =2 sin^{(-1)} ((1)/(1.086))\n\delta =134.08[/tex]
Aiming radius is given as
[tex]\Delta =a \sqrt{(e^2-1)[/tex]
Substituting value in the equation gives
[tex]\Delta =a \sqrt{(e^2-1)\n\Delta} =13319 \sqrt{(1.086^2-1)\n\Delta}=5641.28 km[/tex]
So the turn angle is 134.08 while the aiming radius is 5641.28 km
Therefore, a: The eccentricity is 1.086.
b: The altitude at closest approach is 5088 km
c: The velocity at perigee is 8.516 km/s
d: The turn angle is 134.08 while the aiming radius is 5641.28 km
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.
Find out more information about the relative magnitudes.
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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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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 would be the frequency of an electromagnetic wave having a wavelength equal to Earth’s diameter (12,800 km)? In what part of the electromagnetic spectrum would such a wave lie?
Answer:
f = 23.43 Hz for radio waves.
Explanation:
In this case, we need to find the frequency of an electromagnetic wave having a wavelength equal to Earth’s diameter i.e. 12,800 km. The relation between the frequency, wavelength and the speed is given by :
[tex]v=f\lambda[/tex]
Electromagnetic wave moves with a speed of light
[tex]c=f\lambda[/tex]
[tex]f=\dfrac{c}{\lambda}[/tex]
[tex]f=\dfrac{3\times 10^8}{12800\times 10^3}[/tex]
f = 23.43 Hz
So, the frequency of an electromagnetic wave is 23.43 Hz. It belongs to radio waves.
Final answer:
The frequency of an electromagnetic wave with a wavelength of Earth's diameter (12,800 km) would be approximately 23.4 Hz and would lie in the extremely low frequency (ELF) portion of the electromagnetic spectrum, typically used for submarine communication.
Explanation:
The frequency of an electromagnetic wave can be calculated using the formula c = λf, where c is the speed of light (approximately 3×108 m/s), λ is the wavelength, and f is the frequency. Given the Earth’s diameter as the wavelength (12,800 km, which is 12,800,000 meters), we can rearrange the formula to solve for frequency: f = c / λ.
Plugging in the values, we get:
f = 3×108 m/s / 12,800,000 m
This results in a frequency of approximately 23.4 Hz. Electromagnetic waves with this frequency would fall in the extremely low frequency (ELF) range of the electromagnetic spectrum, which is commonly used for submarine communication due to its ability to penetrate deep into the ocean.