Answer:
a) P = 2319.6[kPa]; b) 2.6%
Explanation:
Since the problem data is not complete, the following information is entered:
A 1.78-m3 rigid tank contains steam at 220°C. One-third of the volume is in the liquid phase and the rest is in the vapor form. Determine (a) the pressure of the steam, and (b) the quality of the saturated mixture.
From the information provided in the problem we can say that you have a mixture of liquid and steam.
a) Using the steam tables we can see (attached image) that the saturation pressure at 220 °C is equal to:
[tex]P_{sat} =2319.6[kPa][/tex]
[tex]v_{f}=0.001190[m^{3}/hr]\\v_{g}=0.08609[m^{3}/hr]\\[/tex]
b) Since the specific volume of the gas and liquid is known, we can find the mass of each phase using the following equation:
[tex]m_{f}=\frac{V_{f} }{v_{f} } \\m_{g}=\frac{V_{g} }{v_{g} } \\where:\\V_{f}=volume of the fluid[m^3]\\v_{f}=specific volume of the fluid [m^3/kg]\\[/tex]
We know that the volume of the fluid is equal to:
[tex]V_{f}=1/3*V_{total} \\V_{total}=1.78[m^3]\\[/tex]
Now we can find the mass of the gas and the liquid.
[tex]m_{f}=\frac{1/3*1.78}{0.001190} \\m_{f}=498.6[kg]\\m_{g}=\frac{2/3*1.78}{0.08609}\\m_{g}=\ 13.78[kg][/tex]
The total mass is the sum of both
[tex]m_{total} =m_{g} + m_{fluid} \\m_{total} = 498.6 + 13.78\\m_{total} = 512.38[kg][/tex]
The quality will be equal to:
[tex]x = \frac{m_{g} }{m_{T} }\\ x= \frac{13.78}{512.38} \\x = 0.026 = 2.6%[/tex]
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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Tim and Rick both can run at speed v_r and walk at speed v_w, with v_r > v_w They set off together on a journey of distance D. Rick walks half of the distance and runs the other half. Tim walks half of the time and runs the other half.
How long does it take Rick to cover the distance D?
Express the time taken by Rick in terms of v_r, v_w, and D.
Find Rick's average speed for covering the distance D.
Express Rick's average speed in terms of v_r and v_w.
How long does it take Tim to cover the distance?
Express the time taken by Tim in terms of v_r, v_w, and D.
Who covers the distance D more quickly?
In terms of given quantities, by what amount of time, Delta t, does Tim beat Rick?
It will help you check your answer if you simplify it algebraically and check the special case v_r = v_w
Express the difference in time, Delta t in terms of v_r, v_w, and D.
In the special case that v_r = v_w, what would be Tim's margin of victory Delta t(v_r = v_w)?
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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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 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]
Raindrops hitting the side windows of a car in motion often leave diagonal streaks even if there is no wind. Why? Is the explanation the same or different for diagonal streaks on the windshield?
Answer:
because of the raindrop velocity relative of the car has a vertical and horizontal component
Explanation:
The car moves in a horizontal direction relative to the ground. The raindrops fall in the vertical direction relative to the ground. Their velocity relative to the moving car has both vertical and horizontal components and this is the reason for the diagonal streaks on the side window. The diagonal streaks on the windshield arise from a different reason. The drops are pushed off to one side of the windshield because of air resistance.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
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
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 the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.100 mm length of the axon?
Answer:
Charge enter a 0.100 mm length of the axon is [tex]8.98\times 10^{-12} C[/tex]
Explanation:
Electric field E at a point due to a point charge is given by
[tex]E=k \frac{q}{r^2}[/tex]
where [tex]k[/tex] is the constant =[tex]9.0 \times 10^9 Nm^2 / C^2[/tex]
[tex]q[/tex] is the magnitude of point charge and [tex]r[/tex] is the distance from the point charge
Charges entering one meter of axon is [tex]5.\times 10^{11} \times (+e)[/tex]
Charges entering 0.100 mm of axon is [tex]5.\times 10^{11} \times (+e) \times (0.1 \times 10^{-3}[/tex]
substituting the value of [tex]+e=1.6\times 10^{-19} C[/tex] in above equation, we get charge enter a 0.100 mm length of the axon is
[tex]q=5.\times 10^{11} \times1.6\times 10^{-19} \times (0.1 \times 10^{-3}\\q=8.98\times 10^{-12} C[/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
The nose of an ultralight plane is pointed south, and its airspeed indicator shows 39 m/s m/s . The plane is in a 17 m/s m/s wind blowing toward the southwest relative to the earth. For help with math skills, you may want to review: Vector Addition Resolving Vector Components For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Flying in a crosswind. Part A Letting x be east and y be north, find the components of v ⃗ P/E v→P/E (the velocity of the plane relative to the earth). Express your answers in meters per second separated by a comma. View Available Hint(s) v x vx , v y vy = nothing m/s m/s Submit Part B Find the magnitude of v ⃗ P/E v→P/E . Express your answer in meters per second. View Available Hint(s) v P/E vP/E = nothing m/s m/s Submit Part C Find the direction of v ⃗ P/E v→P/E . Express your answer in degrees. View Available Hint(s) ϕ ϕ = nothing ∘ ∘ south of west Submit Provide Feedback Next
Answer:
A. [tex]|\vec v_t|=52.42\ m/s[/tex]
B. [tex]\theta=256.74^o[/tex]
Explanation:
Velocity Vector
The velocity vector has two components. Depending on the reference system they could be magnitude and direction in the polar coordinates or x-component and y-component in the rectangular coordinates system.
We are given two velocities in the form of magnitude-direction. The plane's velocity goes south at 39 m/s. The zero reference for angles is pointed East, so the south direction has a 270° angle respect to the reference. If the polar coordinates are known, the rectangular coordinates are computed as
[tex]v_{xp}=|v_p|cos\alpha_p[/tex]
[tex]v_{yp}=|v_p|sin\alpha_p[/tex]
[tex]Since\ |v_p|=39 m/s,\ \alpha_p=270^o,[/tex]
[tex]v_{xp}=39cos\ 270^o=0[/tex]
[tex]v_{yp}=39sin\ 270^o=-39[/tex]
Thus, the velocity of the plane is
[tex]\vec v_p=0\hat i-39\hat j[/tex]
The wind is blowing toward the southwest. It means its angle is 225° (3rd quadrant):
[tex]v_{xw}=|v_w|cos\alpha_w[/tex]
[tex]v_{yw}=|v_w|sin\alpha_w[/tex]
[tex]v_{xw}=17cos\ 215^o=-12.02[/tex]
[tex]v_{yw}=17sin\ 215^o=-12.02[/tex]
Thus, the velocity of the wind is
[tex]\vec v_w=-12.02\hat i-12.02\hat j[/tex]
Now we perform the vector addition to compute the plane's final speed
[tex]\vec v_t=\vec v_p+\vec v_w[/tex]
[tex]\vec v_t=0\hat i-39\hat j-12.02\hat i-12.02\hat j[/tex]
[tex]\vec v_t=-12.02\hat i-51.02\hat j[/tex]
A) The magnitude of the total velocity is
[tex]|\vec v_t|=\sqrt{(-12.02)^2+(-51.02)^2}[/tex]
[tex]\boxed{|\vec v_t|=52.42\ m/s}[/tex]
B) The direction angle is given by
[tex]\displaystyle \tan\theta=\frac{-51.02}{-12.02}=4.24[/tex]
[tex]\theta=arctan\ 4.24[/tex]
[tex]\theta=76.74^o[/tex]
This angle is west of south, we must add 180° to express it in due reference, thus
[tex]\boxed{\theta=256.74^o}[/tex]
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.
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 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
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
What is the final volume of a balloon that was initially 500.0 mL at 25°C and was then heated to 50°C?
Answer:
V₂ =541.94 m L.
Explanation:
Given that
V₁ = 500 mL
T₁ = 25°C = 273 + 25 = 298 K
T₂ = 5°C = 273+50 =323 K
The final volume = V₂
We know that ,the ideal gas equation
If the pressure of the gas is constant ,then we can say that
[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]
[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]
Now by putting the values in the above equation we get
[tex]V_2=500\times \dfrac{323}{298}\ mL\\V_2=541.94\ mL[/tex]
The final volume of the balloon will be 541.94 m L.
V₂ =541.94 m L.
The final volume of the balloon will be "541.94 mL".
Given:
Volume,
[tex]V_1 = 500 \ mL[/tex]Temperature,
[tex]T_1 = 25^{\circ} C[/tex][tex]= 273+25[/tex]
[tex]= 298 \ K[/tex]
[tex]T_2 = 5^{\circ} C[/tex][tex]=273+50[/tex]
[tex]=323 \ K[/tex]
By using the Ideal gas equation, we get
→ [tex]\frac{V_2}{V_1} = \frac{T_2}{T_1}[/tex]
or,
→ [tex]V_2 = \frac{T_2\times V_1}{T_1}[/tex]
[tex]= \frac{500\times 323}{298}[/tex]
[tex]= 541.94 \ mL[/tex]
Thus the above approach is correct.
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A cliff diver positions herself on a cliff that angles downwards towards the edge. The length of the top of the cliff is 50.0 m and the angle of the cliff is θ = 21.0° below the horizontal. The cliff diver runs towards the edge of the cliff with a constant speed, and reaches the edge of the cliff in a time of 6.10 s. After running straight off the edge of the cliff (without jumping up), the diver falls h = 30.0 m before hitting the water.
After leaving the edge of the cliff how much time does the diver take to get to the water?
How far horizontally does the diver travel from the cliff face before hitting the water?
Remember that the angle is at a downward slope to the right.
Final answer:
The diver takes approximately 2.18 seconds to reach the water after leaving the edge of the cliff. The diver travels approximately 3.21 meters horizontally from the cliff face before hitting the water.
Explanation:
To find the time it takes for the diver to reach the water after leaving the edge of the cliff, we can use the equation of motion:[tex]h = 1/2 * g * t^2,[/tex]where h is the height of the cliff and g is the acceleration due to gravity. Rearranging the equation to solve for t, we get t = sqrt(2h/g). Plugging in the values given, we have t = sqrt(2*30/9.8) ≈ 2.18 s.
To find the horizontal distance the diver travels, we can use the equation s = v * t, where s is the distance, v is the horizontal velocity, and t is the time. Rearranging the equation to solve for v, we get v = s/t. Plugging in the values given, we have v = 7/2.18 ≈ 3.21 m/s.
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.
A gun is fired with angle of elevation . What is the muzzle speed if the maximum height of the shell is 500 m?
Answer:
The muzzle speed u =200 m/s
Explanation:
Let u be the muzzle speed.
maximum height s= 500 m
Assuming angle of elevation to be 30°
then initial vertical speed = usin30° = u/2
moreover, when the bullet reaches the maximum height its vertical velocity will be zero.
then using
[tex]v^2=u^2-2as[/tex]
a= 10 m/s^2
u= u/2
v= 0
s=500
we get
[tex]0^2=u^2/4-2\times10\times500[/tex]
u= 200 m/s.
Therefore, muzzle speed = 200 m/s
Answer:
The muzzle speed is 197.9 m/s.
Explanation:
Given that,
Maximum height = 500 m
Suppose the angle is 30°.
We need to calculate the muzzle speed
Using formula of maximum height
[tex]h_{max}=\dfrac{u^2}{2g}[/tex]
Where, h = maximum height
g = acceleration due to gravity
u = initial vertical velocity
Put the value into the formula
[tex]500=\dfrac{u^2\sin^{2}30}{2\times9.8}[/tex]
[tex]u^2=8\times500\times9.8[/tex]
[tex]u=\sqrt{4\times500\times9.8}[/tex]
[tex]u=197.9\ m/s[/tex]
Hence, The muzzle speed is 197.9 m/s.
Electrons are ejected from a metallic surface with speeds of up to 4.60 3 105 m/s when light with a wavelength of 625 nm is used. (a) What is the work function of the surface? (b) What is the cutoff f
Complete question:
Electrons are ejected from a metallic surface with speeds ranging up to 4.60 x10⁵ m/s when light with a wavelength of 625nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface?
Answer:
Part(a) The work function of the surface is 22.177 x 10⁻²⁰ J = 1.384 eV
Part(b) The cutoff frequency for this surface is 3.347 x 10¹⁴ Hz
Explanation:
The kinetic energy (KE) of the emitted photon:
KE = 0.5mv²
m is mass of electron = 9.1 X 10⁻³¹ kg
KE = 0.5 * 9.1 X 10⁻³¹ * (460000)² = 9.628 X 10⁻²⁰ J
in eV = 9.628 X 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 0.601 eV
The photon energy of the incoming radiation:
E = hf = hc/λ
c is speed of light (photon) = 3 x 10⁸
h is Planck's constant = 6.626 × 10⁻³⁴ J.s
E = (6.626 × 10⁻³⁴ *3 x 10⁸) /(625 X 10⁻⁹)
E = 31.805 X 10⁻²⁰ J
in eV = 31.805 X 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 1.985 eV
Part (a) the work function of the surface
KE = hf - W
where;
W is work function
W = hf - KE
W = 31.805 X 10⁻²⁰ J - 9.628 X 10⁻²⁰ J = 22.177 x 10⁻²⁰ J
in eV = 22.177 x 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 1.384 eV
Part(b) the cutoff frequency for this surface
W =hf
f = W/h
f = (22.177 x 10⁻²⁰ J)/(6.626 × 10⁻³⁴ J.s)
f = 3.347 x 10¹⁴ Hz
The work function is the minimum energy required to eject an electron from a metallic surface, which can be calculated using the maximum kinetic energy of ejected electrons and the energy of the incident light. The cutoff frequency is the minimum frequency of light required to eject an electron.
Explanation:The question is asking to calculate the work function of the surface and the cutoff frequency in a context related to the photoelectric effect. The photoelectric effect is a phenomenon in which electrons are ejected from a metallic surface when it is exposed to light of a certain frequency, in this case, light with a wavelength of 625 nm.
The energy of the incident light is used to expel electrons from the surface of the metal. Any remaining energy contributes toward the ejected electrons' kinetic energy. This can be modeled by the equation KE_maximum = hf - Φ, where KE_maximum is the electron's maximum kinetic energy, h is Planck's constant, f is the frequency of the light, and Φ is the work function of the material. The work function Φ is the minimum amount of energy required to eject an electron from the material's surface.
The cutoff frequency or threshold frequency is the minimum frequency of the incident light required to eject electrons. If the frequency of the light is less than this value, no electrons are ejected
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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.
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|.
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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Temperature change time base problem. Suppose V = 24 V, I = 0.1 A, for water: mw = 51 gm, cw = 4.18 J/gm ∘K-1, for resistor: mr = 8 gm, and cr = 3.7 J/gm ∘K-1. If the water is initially at room temperature, how long will it take for the water to heat up 5∘K? (Hint: dT/dt is approximately equal to Δ T / Δt .)
Answer:
[tex]t=5057.9167\ s[/tex]
Explanation:
Given:
Voltage supply to the resistor, [tex]V=24\ V[/tex]current supply to the resistor, [tex]I=0.1\ A[/tex]mass of water, [tex]m_w= 51\ g[/tex]specific heat of water, [tex]c_w=4180\ J.kg^{-1}.K^{-1}[/tex]specific heat of resistor, [tex]c_r=3700\ J.kg^{-1}.K^{-1}[/tex]mass of resistor, [tex]m_r=0.008\ kg[/tex]change in temperature, [tex]\Delta T=50\ K[/tex]Now the amount of heat required to heat the water by 50 K:
[tex]Q_w=m_w.c_w.\Delta T[/tex]
[tex]Q_w=0.051\times 4180\times 50[/tex]
[tex]Q_w=10659\ J[/tex]
Now the amount of heat required to heat the resistor by 50 K:
[tex]Q_r=m_r.c_r.\Delta T[/tex]
[tex]Q_r=0.008\times 3700\times 50[/tex]
[tex]Q_r=1480\ J[/tex]
Now the total heat to converted from the electrical energy:
[tex]Q=Q_w+Q_r[/tex]
[tex]Q=12139\ J[/tex]
Now Using Joule's law of heating:
[tex]Q=V.I.t[/tex]
[tex]12139=24\times 0.1\times t[/tex]
[tex]t=5057.9167\ s[/tex]
Final answer:
To determine the time needed to heat water by 5°K, calculate the total energy needed using the specific heat capacity of water, then use the power equation P = V × I to find the time. It will take approximately 444.125 seconds for the water to reach the desired temperature.
Explanation:
To calculate the time it will take for the water to heat up by 5°K using the provided values, first determine the total energy required to raise the temperature of the water by using the equation E = mw × cw × ΔT. Here, E is the energy in joules (J), mw is the mass of water, cw is the specific heat capacity of water, and ΔT is the change in temperature.
Using the given figures:
E = 51 g × 4.18 J/g°K × 5°K = 1065.9 J
Since power P is the rate at which energy is used and P = V × I where V is the voltage and I is the current, we can rearrange the equation to find time: t = E / P.
Substitute the power using the given voltage and current:
P = 24 V × 0.1 A = 2.4 W
The time in seconds to heat the water 5°K is thus:
t = 1065.9 J / 2.4 W = 444.125 seconds
Therefore, it will take approximately 444.125 seconds to heat up the water by 5°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
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 large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. (a) At what time after being ejected is the boulder moving at 20.0 m/s upward? (b) At what time is it moving at 20.0 m/s downward? (c) When is the displacement of the boulder from its initial position zero? (d) When is the velocity of the boulder zero? (e) What are the magnitude and direction of the acceleration while the boulder is (i) moving upward? (ii) Moving downward? (iii) At the highest point? (f) Sketch ay-t, vy-t, and y-t graphs for the motion.
a) Time at which velocity is +20.0 m/s: 2.04 s
b) Time at which velocity is -20.0 m/s: 6.12 s
c) Time at which the displacement is zero: t = 0 and t = 8.16 s
d) Time at which the velocity is zero: t = 4.08 s
e) i) ii) iii) The acceleration of the boulder is always [tex]9.8 m/s^2[/tex] downward
f) See graphs in attachment
Explanation:
a)
The motion of the boulder is a uniformly accelerated motion, with constant acceleration
[tex]a=g=-9.8 m/s^2[/tex]
downward (acceleration due to gravity). So, we can use the following suvat equation:
[tex]v=u+at[/tex]
where:
v is the velocity at time t
u = 40.0 m/s is the initial velocity
a=g=-9.8 m/s^2 is the acceleration
We want to find the time t at which the velocity is
v = 20.0 m/s
Therefore,
[tex]t=\frac{v-u}{a}=\frac{20-40}{-9.8}=2.04 s[/tex]
b)
In this case, we want to find the time t at which the boulder is moving at 20.0 m/s downward, so when
v = -20.0 m/s
(the negative sign means downward)
We use again the suvat equation
[tex]v=u+at[/tex]
And substituting
u = +40.0 m/s
a=g=-9.8 m/s^2
We find the corresponding time t:
[tex]t=\frac{v-u}{a}=\frac{-20-(+40)}{-9.8}=6.12 s[/tex]
c)
To solve this part, we can use the following suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s is the displacement
u = +40.0 m/s is the initial velocity
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration
t is the time
We want to find the time t at which the displacement is zero, so when
s = 0
SUbstituting into the equation and solving for t,
[tex]0=ut+\frac{1}{2}at^2\\t(u+\frac{1}{2}a)=0[/tex]
which gives two solutions:
t = 0 (initial instant)
[tex]u+\frac{1}{2}at=0\\t=-\frac{2u}{a}=-\frac{2(40)}{-9.8}=8.16 s[/tex]
which is the instant at which the boulder passes again through the initial position, but moving downward.
d)
To solve this part, we can use again the suvat equation
[tex]v=u+at[/tex]
where
u = +40.0 m/s is the initial velocity
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration
We want to find the time t at which the velocity is zero, so when
v = 0
Substituting and solving for t, we find:
[tex]t=\frac{v-u}{a}=\frac{0-(40)}{-9.8}=4.08 s[/tex]
e)
In order to evaluate the acceleration of the boulder, let's consider the forces acting on it.
If we neglect air resistance, there is only one force acting on the boulder: the force of gravity, acting downward, with magnitude
[tex]F=mg[/tex]
where m is the mass of the boulder and [tex]g[/tex] the acceleration of gravity.
According to Newton's second law, the net force on the boulder is equal to the product between its mass and its acceleration:
[tex]F=ma[/tex]
Combining the two equations, we get
[tex]ma=mg\\a=g[/tex]
So, the acceleration of the boulder is [tex]g=9.8 m/s^2[/tex] downward at any point of the motion, no matter where the boulder is (because the force of gravity is constant during the motion).
f)
Find the three graphs in attachment:
- Position-time graph: the position of the boulder initially increases as the boulder goes upward; however, the slope of the curve decreases as the boulder goes higher (because the velocity decreases). The boulder reaches its maximum height at t = 4.08 s (when velocity is zero), then it starts going downward, until reaching its initial position at t = 8.16 s
- Velocity-time graph: the initial velocity is +40 m/s; then it decreases linearly (because the acceleration is constant), and becomes zero when t = 4.08 s. Then the velocity becomes negative (because the boulder is now moving downward) and its magnitude increases.
- Acceleration-time graph: the acceleration is constant and it is [tex]-9.8 m/s^2[/tex], so this graph is a straight horizontal line.
Learn more about accelerated motion:
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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]