Complete question:
Two cars start moving from the same point. One travels south at 28 mi/h and the other travels west at 21 mi/h. At what rate is the distance between the cars increasing four hours later.
Answer:
The rate at which the distance between the cars is increasing four hours later is 35 mi/h.
Explanation:
Given;
speed of one car, dx/dt = 28 mi/h South
speed of the second car, dy/dt = 21 mi/h West
The distance between the cars is the line joining west to south, which forms a right angled triangle with the two positions.
Apply Pythagoras theorem to evaluate this distance;
let the distance between the cars = z
x² + y² = z² -------- equation (1)
Differentiate with respect to time (t)
[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}[/tex] ----- equation (2)
Since the speed of the cars is constant, after 4 hours their different distance will be;
x: 28(4) = 112 mi
y: 21(4) = 84 mi
[tex]z = \sqrt{x^2 + y^2} \\\\z = \sqrt{112^2 + 84^2} \\\\z = 140 \ mi[/tex]
Substitute in the value of x, y, z, dx/dt, dy /dt into equation (2) and solve for dz/dt
[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \\\\2(112)(28) + 2(84)(21) = 2(140)\frac{dz}{dt} \\\\9800 = 280\frac{dz}{dt} \\\\\frac{dz}{dt} = \frac{9800}{280} \\\\\frac{dz}{dt} = 35 \ mi/h[/tex]
Therefore, the rate at which the distance between the cars is increasing four hours later is 35 mi/h
A cowboy fires a silver bullet with a muzzle speed of 200 m/s into the pine wall of a saloon. Assume all the internal energy generated by the impact remains with the bullet. What is the temperature change of the bullet?
Explanation:
KE = q
½ mv² = mCΔT
ΔT = v² / (2C)
ΔT = (200 m/s)² / (2 × 236 J/kg/°C)
ΔT = 84.7°C
This question involves the concepts of the law of conservation of energy.
The temperature change of the bullet is "84.38°C".
What is the Law of Conservation of Energy?According to the law of conservation of energy, total energy of the system must remain constant. Therefore, in this situation.
[tex]Kinetic\ energy\ of\ bullet\ before\ impact=heat\ absorbed\ in\ bullet\\\\\frac{1}{2}mv^2=mC\Delta T\\\\\Delta T = \frac{v^2}{2C}[/tex]
where,
ΔT = change in temperature of the bullet = ?C = specific heat capacity of silver = 237 J/kg°Cv = speed of bullet = 200 m/sTherefore,
[tex]\Delta T = \frac{(200\ m/s)^2}{2(237\ J/kg.^oC)}[/tex]
ΔT = 84.38°C
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A circular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 A. It is placed in a magnetic field of 0.70 T with the plane of the coil making an angle of 30° with the magnetic field. What is the magnetic torque on the coil?
Answer:
0.087976 Nm
Explanation:
The magnetic torque (τ) on a current-carrying loop in a magnetic field is given by;
τ = NIAB sinθ --------- (i)
Where;
N = number of turns of the loop
I = current in the loop
A = area of each of the turns
B = magnetic field
θ = angle the loop makes with the magnetic field
From the question;
N = 200
I = 4.0A
B = 0.70T
θ = 30°
A = π d² / 4 [d = diameter of the coil = 2.0cm = 0.02m]
A = π x 0.02² / 4 = 0.0003142m² [taking π = 3.142]
Substitute these values into equation (i) as follows;
τ = 200 x 4.0 x 0.0003142 x 0.70 sin30°
τ = 200 x 4.0 x 0.0003142 x 0.70 x 0.5
τ = 200 x 4.0 x 0.0003142 x 0.70
τ = 0.087976 Nm
Therefore, the torque on the coil is 0.087976 Nm
In a double-slit interference experiment you are asked to use laser light of different wavelengths and determine the separation between adjacent maxima. You observe that this separation is greatest when you illuminate the double slit with In a double-slit interference experiment you are asked to use laser light of different wavelengths and determine the separation between adjacent maxima. You observe that this separation is greatest when you illuminate the double slit with:_________.
1. yellow light.
2. red light.
3. blue light.
4. green light.
5. The separation is the same for all wavelengths.
Answer:
Red light
Explanation:
This because All interference or diffraction patterns depend upon the wavelength of the light (or whatever wave) involved. Red light has the longest wavelength (about 700 nm)
Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to:_______
a. one-half.
b. double.
c. reduce to one-fourth.
d. quadruple.
Answer:
D. quadrupleExplanation:
The stored energy varies with the square of the electric charge stored in the capacitor. If you double the charge, the stored energy in the capacitor will quadruple or increase by a factor of 4.
Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to :
D. Quadruple
"Energy"Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to Quadruple.
The stored energy shifts with the square of the electric charge put away within the capacitor.
In case you twofold the charge, the put away vitality within the capacitor will fourfold or increment by a calculate of 4.
Thus, the correct answer is D.
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5) What is the weight of a body in earth. if its weight is 5Newton
in moon?
Answer:
8.167
Explanation:
An experimenter finds that standing waves on a string fixed at both ends occur at 24 Hz and 32 Hz , but at no frequencies in between. Part A What is the fundamental frequency
Answer:
8 Hz
Explanation:
Given that
Standing wave at one end is 24 Hz
Standing wave at the other end is 32 Hz.
Then the frequency of the standing wave mode of a string having a length, l, is usually given as
f(m) = m(v/2L), where in this case, m could be 1. 2. 3. 4 etc
Also, another formula is given as
f(m) = m.f(1), where f(1) is the fundamental frequency..
Thus, we could say that
f(m+1) - f(m) = (m + 1).f(1) - m.f(1) = f(1)
And as such,
f(1) = 32 - 24
f(1) = 8 Hz
Then, the fundamental frequency needed is 8 Hz
what are the rays that come of the sun called? A. Ultraviolent rays B. Gamma rays C. soundwaves D. sonic rays
Answer:
The answer is ultraviolet rays...
Explanation:
...because the ozone layer protects us from the UV rays of the sun.
Problem 2: A 21-W horizontal beam of light of wavelength 430 nm, travelling at speed c, passes through a rectangular opening of width 0.0048 m and height 0.011 m. The light then strikes a screen at a distance 0.36 m behind the opening.
a) E = 4.62E-19 Joules
b) N = 4.543E+19 # of photons emitted
c) If the beam of light emitted by the source has a constant circular cross section whose radius is twice the height of the opening the beam is approaching, find the flow density of photons as the number of photons passing through a square meter of cross-sectional area per second.
d) Calculate the number of photons that pass through the rectangular opening per second.
e) The quantity NO that you found in part (d) gives the rate at which photons enter the region between the opening and the screen. Assuming no reflection from the screen, find the number of photons in that region at any time.
f) How would the number of photons in the region between the opening and the screen change, if the photons traveled more slowly? Assume no change in any other quantity, including the speed of light.
Answer:
a. E = 4.62 × 10⁻¹⁹J
b. n = 4.54 × 10¹⁹photons
c. 2.99 × 10²²photons/m²
d. 1.58 ×10¹⁸photons/seconds
e. n = 1.896 × 10 ⁹ photons
f. the number of photons will be more if it travels slowly
Explanation:
How much heat does it take to raise the temperature of 7.0 kg of water from
25-C to 46-C? The specific heat of water is 4.18 kJ/(kg.-C).
Use Q = mcTr-T)
A. 148 kJ
B. 176 kJ
C. 610 kJ
D. 320 kJ
Answer:
non of the above
Explanation:
Quantity of heat = mass× specific heat× change in temperature
m= 7kg c= 4.18 temp= 46-25=21°
.......H= 7×4.18×21= 614.46kJ
Answer:610 KJ
Explanation:A P E X answers
Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits
Answer:
The maximum number of bright spot is [tex]n_{max} =5001[/tex]
Explanation:
From the question we are told that
The slit distance is [tex]d = 1 \ mm = 0.001 \ m[/tex]
The wavelength is [tex]\lambda = 400 \ nm = 400*10^{-9 } \ m[/tex]
Generally the condition for interference is
[tex]n * \lambda = d * sin \theta[/tex]
Where n is the number of fringe(bright spots) for the number of bright spots to be maximum [tex]\theta = 90[/tex]
=> [tex]sin( 90 )= 1[/tex]
So
[tex]n = \frac{d }{\lambda }[/tex]
substituting values
[tex]n = \frac{ 1 *10^{-3} }{ 400 *10^{-9} }[/tex]
[tex]n = 2500[/tex]
given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as
[tex]n_{max} = 2 * n + 1[/tex]
The 1 here represented the central bright spot
So
[tex]n_{max} = 2 * 2500 + 1[/tex]
[tex]n_{max} =5001[/tex]
A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to the incline, which makes an angle of 20.2° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.10 m.A) How much work is done by the gravitational force on thecrate?
B) Determine the increase in internal energy of the crate-inclinesystem owing to friction.
C) How much work is done by the 100N force on the crate?
D) What is the change in kinetic energy of the crate?
E) What is the speed of the crate after being pulled 5.00m?
Given that,
Mass = 9.2 kg
Force = 110 N
Angle = 20.2°
Distance = 5.10 m
Speed = 1.58 m/s
(A). We need to calculate the work done by the gravitational force
Using formula of work done
[tex]W_{g}=mgd\sin\theta[/tex]
Where, w = work
m = mass
g = acceleration due to gravity
d = distance
Put the value into the formula
[tex]W_{g}=9.2\times(-9.8)\times5.10\sin20.2[/tex]
[tex]W_{g}=-158.8\ J[/tex]
(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction
Using formula of potential energy
[tex]\Delta U=-W[/tex]
Put the value into the formula
[tex]\Delta U=-(-158.8)\ J[/tex]
[tex]\Delta U=158.8\ J[/tex]
(C). We need to calculate the work done by 100 N force on the crate
Using formula of work done
[tex]W=F\times d[/tex]
Put the value into the formula
[tex]W=100\times5.10[/tex]
[tex]W=510\ J[/tex]
We need to calculate the work done by frictional force
Using formula of work done
[tex]W=-f\times d[/tex]
[tex]W=-\mu mg\cos\theta\times d[/tex]
Put the value into the formula
[tex]W=-0.4\times9.2\times9.8\cos20.2\times5.10[/tex]
[tex]W=-172.5\ J[/tex]
We need to calculate the change in kinetic energy of the crate
Using formula for change in kinetic energy
[tex]\Delta k=W_{g}+W_{f}+W_{F}[/tex]
Put the value into the formula
[tex]\Delta k=-158.8-172.5+510[/tex]
[tex]\Delta k=178.7\ J[/tex]
(E). We need to calculate the speed of the crate after being pulled 5.00m
Using formula of change in kinetic energy
[tex]\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})[/tex]
[tex]v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2[/tex]
Put the value into the formula
[tex]v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58[/tex]
[tex]v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}[/tex]
[tex]v_{2}=6.35\ m/s[/tex]
Hence, (A). The work done by the gravitational force is -158.8 J.
(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.
(C). The work done by 100 N force on the crate is 510 J.
(D). The change in kinetic energy of the crate is 178.7 J.
(E). The speed of the crate after being pulled 5.00m is 6.35 m/s
A 2.5-kg object falls vertically downward in a viscous medium at a constant speed of 2.5 m/s. How much work is done by the force the viscous medium exerts on the object as it falls 80 cm?
Answer:
The workdone is [tex]W_v = - 20 \ J[/tex]
Explanation:
From the question we are told that
The mass of the object is [tex]m = 2.5 \ kg[/tex]
The speed of fall is [tex]v = 2.5 \ m/s[/tex]
The depth of fall is [tex]d = 80\ cm = 0.8 \ m[/tex]
Generally according to the work energy theorem
[tex]W = \frac{1}{2} mv_2^2 - \frac{1}{2} mv_1^2[/tex]
Now here given the that the velocity is constant i.e [tex]v_1 = v_2 = v[/tex] then
We have that
[tex]W = \frac{1}{2} mv^2 - \frac{1}{2} mv^2 = 0 \ J[/tex]
So in terms of workdone by the potential energy of the object and that of the viscous liquid we have
[tex]W = W_v - W_p[/tex]
Where [tex]W_v[/tex] is workdone by viscous liquid
[tex]W_p[/tex] is the workdone by the object which is mathematically represented as
[tex]W_p = mgd[/tex]
So
[tex]0 = W_v + mgd[/tex]
=> [tex]W_v = - m * g * d[/tex]
substituting values
[tex]W_v = - (2.5 * 9.8 * 0.8)[/tex]
[tex]W_v = - 20 \ J[/tex]
Three sleds (30kg sled connected by tension rope B to 20kg sled connected by tension rope A to 10kg sled) are being pulled horizontally on frictionless horizontal ice using horizontal ropes. The pull is horizontal and of magnitude 143N . Required:a. Find the acceleration of the system. b. Find the tension in rope A. c. Find the tension in rope B.
Answer:
a) a = 2.383 m / s², b) T₂ = 120,617 N , c) T₃ = 72,957 N
Explanation:
This is an exercise of Newton's second law let's fix a horizontal frame of reference
in this case the mass of the sleds is 30, 20 10 kg from the last to the first, in the first the horizontal force is applied.
a) request the acceleration of the system
we can take the sledges together and write Newton's second law
T = (m₁ + m₂ + m₃) a
a = T / (m₁ + m₂ + m₃)
a = 143 / (10 +20 +30)
a = 2.383 m / s²
b) the tension of the cables we think through cable A between the sledges of 1 and 20 kg
on the sled of m₁ = 10 kg
T - T₂ = m₁ a
in this case T₂ is the cable tension
T₂ = T - m₁ a
T₂ = 143 - 10 2,383
T₂ = 120,617 N
c) The cable tension between the masses of 20 and 30 kg
T₂ - T₃ = m₂ a
T₃ = T₂ -m₂ a
T₃ = 120,617 - 20 2,383
T₃ = 72,957 N
How do I find an apparent weight in N for a metal connected to a string submerged in water if a scale shows the mass 29.52 g when it is submerged ? Also how do I measure its density
The Tension of the string is going to be less when submerged in water by a value called the buoyancy force, so below in the attached file is explanation on how to calculate the apparent weight and density of the submerged object
An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.500 mm .
a. If the slits are very narrow, what would be the angular position of the second- order, two-slit interference maxima?
b. Let the slits have a width 0.300 mm. In terms of the intensity lo at the center of the central maximum, what is the intensity at the angular position in part "a"?
Answer:
a
[tex]\theta = 0.0022 rad[/tex]
b
[tex]I = 0.000304 I_o[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 550 \ nm = 550 *10^{-9} \ m[/tex]
The distance of the slit separation is [tex]d = 0.500 \ mm = 5.0 *10^{-4} \ m[/tex]
Generally the condition for two slit interference is
[tex]dsin \theta = m \lambda[/tex]
Where m is the order which is given from the question as m = 2
=> [tex]\theta = sin ^{-1} [\frac{m \lambda}{d} ][/tex]
substituting values
[tex]\theta = 0.0022 rad[/tex]
Now on the second question
The distance of separation of the slit is
[tex]d = 0.300 \ mm = 3.0 *10^{-4} \ m[/tex]
The intensity at the the angular position in part "a" is mathematically evaluated as
[tex]I = I_o [\frac{sin \beta}{\beta} ]^2[/tex]
Where [tex]\beta[/tex] is mathematically evaluated as
[tex]\beta = \frac{\pi * d * sin(\theta )}{\lambda }[/tex]
substituting values
[tex]\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }[/tex]
[tex]\beta = 0.06581[/tex]
So the intensity is
[tex]I = I_o [\frac{sin (0.06581)}{0.06581} ]^2[/tex]
[tex]I = 0.000304 I_o[/tex]
Two small identical speakers are connected (in phase) to the same source. The speakers are 3 m apart and at ear level. An observer stands at X, 4 m in front of one speaker. If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is:
a. 1 m
b. 2 m
c. 3 m
d. 4 m
e. 5 m
Answer:
b. 2 m
Explanation:
Given that:
the identical speakers are connected in phases ;
Let assume ; we have speaker A and speaker B which are = 3 meter apart
An observer stands at X = 4m in front of one speaker.
If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is:
From above; the distance between speaker A and speaker B can be expressed as:
[tex]\sqrt{3^2 + 4^2 } \\ \\ = \sqrt{9+16 } \\ \\ = \sqrt{25} \\ \\ = 5 \ m[/tex]
The path length difference will now be:
= 5 m - 4 m
= 1 m
Since , we are to determine the least intense sound; the destructive interference for that path length will be half the wavelength; which is
= [tex]\dfrac{1}{2}*4 \ m[/tex]
= 2 m
The sound will be heard with least intensity if the wavelength is 2 m. Hence, option (b) is correct.
Given data:
The distance between the speakers is, d = 3 m.
The distance between the observer and speaker is, s = 4 m.
The amplitude of sound wave is the vertical distance from the base to peak of wave. Since sound amplitudes are not changed in the given problem. Then the distance between speaker A and speaker B can be expressed as:
[tex]=\sqrt{3^{2}+4^{2}}\\\\=\sqrt{25}\\\\=5\;\rm m[/tex]
And the path length difference is,
= 5 m - 4 m
= 1 m
Since , we are to determine the least intense sound; the destructive
interference for that path length will be half the wavelength; which is
[tex]=\dfrac{1}{2} \times s\\\\=\dfrac{1}{2} \times 4[/tex]
= 2 m
Thus, we can conclude that the sound will be heard with least intensity if the wavelength is 2 m.
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The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin/eo, with Qin/e, where ε is the permittivity of the material. (Technically, Eo is called the vacuum permittivity.) Suppose that a 70 nC point charge is surrounded by a thin, 32-cm-diameter spherical rubber shell and that the electric field strength inside the rubber shell is 2500 N/C.
What is the permittivity of rubber?
Answer:
The permittivity of rubber is [tex]\epsilon = 8.703 *10^{-11}[/tex]
Explanation:
From the question we are told that
The magnitude of the point charge is [tex]q_1 = 70 \ nC = 70 *10^{-9} \ C[/tex]
The diameter of the rubber shell is [tex]d = 32 \ cm = 0.32 \ m[/tex]
The Electric field inside the rubber shell is [tex]E = 2500 \ N/ C[/tex]
The radius of the rubber is mathematically evaluated as
[tex]r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m[/tex]
Generally the electric field for a point is in an insulator(rubber) is mathematically represented as
[tex]E = \frac{Q}{ \epsilon } * \frac{1}{4 * \pi r^2}[/tex]
Where [tex]\epsilon[/tex] is the permittivity of rubber
=> [tex]E * \epsilon * 4 * \pi * r^2 = Q[/tex]
=> [tex]\epsilon = \frac{Q}{E * 4 * \pi * r^2}[/tex]
substituting values
[tex]\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}[/tex]
[tex]\epsilon = 8.703 *10^{-11}[/tex]
A car moving at a speed of 25 m/s enters a curve that traces a circular quarter turn of radius 129 m. The driver gently applies the brakes, slowing the car with a constant tangential acceleration of magnitude 1.2 m/s2.a) Just before emerging from the turn, what is the magnitudeof the car's acceleration?
b) At that same moment, what is the angle q between the velocity vector and theacceleration vector?
I am having trouble because this problem seems to have bothradial and tangential accleration. I tried finding the velocityusing V^2/R, but then that didnt take into account thedeceleration. Any help would be great.
Answer:
8.7 m/s^2
82.15°
Explanation:
Given:-
- The initial speed of the car, vi = 25 m/s
- The radius of track, r = 129 m
- Car makes a circular " quarter turn "
- The constant tangential acceleration, at = 1.2 m/s^2
Solution:-
- We will solve the problem using rotational kinematics. Determine the initial angular velocity of car ( wi ) as follows:
[tex]w_i = \frac{v_i}{r} \\\\w_i = \frac{25}{129}\\\\w_i = 0.19379 \frac{rad}{s}[/tex]
- Now use the constant tangential acceleration ( at ) and determine the constant angular acceleration ( α ) for the rotational motion as follows:
at = r*α
α = ( 1.2 / 129 )
α = 0.00930 rad/s^2
- We know that the angular displacement from the initial entry to the exit of the turn is quarter of a turn. The angular displacement would be ( θ = π/2 ).
- Now we will use the third rotational kinematic equation of motion to determine the angular velocity at the exit of the turn (wf) as follows:
[tex]w_f^2 = w_i^2 + 2\alpha*theta\\\\w_f = \sqrt{0.19379^2 + 0.00930\pi } \\\\w_f = 0.25840 \frac{rad}{s}[/tex]
- We will use the evaluated final velocity ( wf ) and determine the corresponding velocity ( vf ) as follows:
[tex]v_f = r*w_f\\\\v_f = 129*0.2584\\\\v_f = 33.33380 \frac{x}{y}[/tex]
- Now use the formulation to determine the centripetal acceleration ( ac ) at this point as follows:
[tex]a_c = \frac{v_f^2}{r} \\\\a_c = \frac{33.3338^2}{129} \\\\a_c = 8.6135 \frac{m}{s^2}[/tex]
- To determine the magnitude of acceleration we will use find the resultant of the constant tangential acceleration ( at ) and the calculated centripetal acceleration at the exit of turn ( ac ) as follows:
[tex]|a| = \sqrt{a^2_t + a_c^2} \\\\|a| = \sqrt{1.2^2 + 8.6135^2} \\\\|a| = 8.7 \frac{m}{s^2}[/tex]
- To determine the angle between the velocity vector and the acceleration vector. We need to recall that the velocity vector only has one component and always tangential to the curved path. Hence, the velocity vector is parallel to the tangential acceleration vector ( at ). We can use the tangential acceleration ( at ) component of acceleration ( a ) and the centripetal acceleration ( ac ) component of the acceleration and apply trigonometric ratio as follows:
[tex]q = arctan \frac{a_c}{a_t} = arctan \frac{8.7}{1.2} \\\\q = 82.15 ^.[/tex]
Answer: The angle ( q ) between acceleration vector ( a ) and the velocity vector ( v ) at the exit of the turn is 82.15° .
Charge of uniform surface density (0.20 nC/m2) is distributed over the entire xy plane. Determine the magnitude of the electric field at any point having z
The question is not complete, the value of z is not given.
Assuming the value of z = 4.0m
Answer:
the magnitude of the electric field at any point having z(4.0 m) =
E = 5.65 N/C
Explanation:
given
σ(surface density) = 0.20 nC/m² = 0.20 × 10⁻⁹C/m²
z = 4.0 m
Recall
E =F/q (coulumb's law)
E = kQ/r²
σ = Q/A
A = 4πr²
∴ The electric field at point z =
E = σ/zε₀
E = 0.20 × 10⁻⁹C/m²/(4 × 8.85 × 10⁻¹²C²/N.m²)
E = 5.65 N/C
A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacement of 4.9 rad. What is its average angular acceleration
Answer:
The average angular acceleration is 0.05 radians per square second.
Explanation:
Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:
[tex]\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})[/tex]
Where:
[tex]\omega_{o}[/tex], [tex]\omega[/tex] - Initial and final angular velocities, measured in radians per second.
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
[tex]\theta_{o}[/tex], [tex]\theta[/tex] - Initial and final angular position, measured in radians.
Then,
[tex]\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}[/tex]
Given that [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega = 0.70\,\frac{rad}{s}[/tex] and [tex]\theta-\theta_{o} = 4.9\,rad[/tex], the angular acceleration is:
[tex]\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}[/tex]
[tex]\alpha = 0.05\,\frac{rad}{s^{2}}[/tex]
Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:
[tex]\omega = \omega_{o} + \alpha \cdot t[/tex]
Where [tex]t[/tex] is the time measured in seconds.
The time is cleared and obtain after replacing every value:
[tex]t = \frac{\omega-\omega_{o}}{\alpha}[/tex]
If [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega = 0.70\,\frac{rad}{s}[/tex] and [tex]\alpha = 0.05\,\frac{rad}{s^{2}}[/tex], the required time is:
[tex]t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }[/tex]
[tex]t = 14\,s[/tex]
Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:
[tex]\bar \alpha = \frac{\omega-\omega_{o}}{t}[/tex]
If [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega = 0.70\,\frac{rad}{s}[/tex] and [tex]t = 14\,s[/tex], the average angular acceleration is:
[tex]\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}[/tex]
[tex]\bar \alpha = 0.05\,\frac{rad}{s^{2}}[/tex]
The average angular acceleration is 0.05 radians per square second.
Consider the Earth and the Moon as a two-particle system.
Find an expression for the gravitational field g of this two-particle system as a function of the distance r from the center of the Earth. (Do not worry about points inside either the Earth or the Moon. Assume the Moon lies on the +r-axis. Give the scalar component of the gravitational field. Do not substitute numerical values; use variables only. Use the following as necessary: G, Mm, Me, r, and d for the distance from the center of Earth to the center of the Moon.)"
sorry but I don't understand
200 J of heat is added to two gases, each in a sealed container. Gas 1 is in a rigid container that does not change volume. Gas 2 expands as it is heated, pushing out a piston that lifts a small weight. Which gas has the greater increase in its thermal energy?Which gas has the greater increase in its thermal energy?Gas 1Gas 2Both gases have the same increase in thermal energy.
Answer:
Gas 1
Explanation:
The reason for this is that for gases attached to both gases or containers, with a heat of 200 J, the change in volume is only observed in gas 2, whereas the volume of gas 1 is the same as that of gas. Therefore, the internal energy (heat) or thermal energy of the system is not utilized for Gas 1 and hence the absorption and transfer of energy is the same, whereas Gas 2 is propagated by the use of additional heat of heat. Thus there is a large increase in the thermal energy of Gas1.A parallel-plate capacitor with circular plates of radius R is being discharged. The displacement current through a central circular area, parallel to the plates and with radius R/2, is 9.2 A. What is the discharging current?
Answer:
The discharging current is [tex]I_d = 36.8 \ A[/tex]
Explanation:
From the question we are told that
The radius of each circular plates is R
The displacement current is [tex]I = 9.2 \ A[/tex]
The radius of the central circular area is [tex]\frac{R}{2}[/tex]
The discharging current is mathematically represented as
[tex]I_d = \frac{A}{k} * I[/tex]
where A is the area of each plate which is mathematically represented as
[tex]A = \pi R ^2[/tex]
and k is central circular area which is mathematically represented as
[tex]k = \pi [\frac{R}{2} ]^2[/tex]
So
[tex]I_d = \frac{\pi R^2 }{\pi * [ \frac{R}{2}]^2 } * I[/tex]
[tex]I_d = \frac{\pi R^2 }{\pi * \frac{R^2}{4} } * I[/tex]
[tex]I_d = 4 * I[/tex]
substituting values
[tex]I_d = 4 * 9.2[/tex]
[tex]I_d = 36.8 \ A[/tex]
Calculate the angular momentum of a solid uniform sphere with a radius of 0.150 m and a mass of 13.0 kg if it is rotating at 5.70 rad/s about an axis through its center.
Answer:
The angular momentum of the solid sphere is 0.667 kgm²/s
Explanation:
Given;
radius of the solid sphere, r = 0.15 m
mass of the sphere, m = 13 kg
angular speed of the sphere, ω = 5.70 rad/s
The angular momentum of the solid sphere is given;
L = Iω
Where;
I is the moment of inertia of the solid sphere
ω is the angular speed of the solid sphere
The moment of inertia of solid sphere is given by;
I = ²/₅mr²
I = ²/₅ x (13 x 0.15²)
I = 0.117 kg.m²
The angular momentum of the solid sphere is calculated as;
L = Iω
L = 0.117 x 5.7
L = 0.667 kgm²/s
Therefore, the angular momentum of the solid sphere is 0.667 kgm²/s
Lightbulbs are typically rated by their power dissipation when operated at a given voltage. Which of the following lightbulbs has the largest resistance when operated at the voltage for which it's rated?
A. 0.8 W, 1.5 V
B. 6 W 3 V
C. 4 W, 4.5 V
D. 8 W, 6 V
Answer:
The arrangement with the greatest resistance is the light bulb of option C. 4 W, 4.5 V
Explanation:
The equation for electric power is
power P = IV
also, I = V/R,
substituting into the equation, we have
[tex]P = \frac{V^{2} }{R}[/tex]
[tex]R = \frac{V^{2} }{P}[/tex]
a) [tex]R = \frac{1.5^{2} }{0.8}[/tex] = 2.8 Ω
b) [tex]R = \frac{3^{2} }{6}[/tex] = 1.5 Ω
c) [tex]R = \frac{4.5^{2} }{4}[/tex] 5.06 Ω
d) [tex]R = \frac{6^{2} }{8}[/tex] = 4.5 Ω
from the calculations, one can see that the lightbulb with te greates resistance is
C. 4 W, 4.5 V
at the temperature at which we live, earth's core is solid or liquid?
Explanation:
The Earth has a solid inner core
Use Kepler's third law to determine how many days it takes a spacecraft to travel in an elliptical orbit from a point 6 590 km from the Earth's center to the Moon, 385 000 km from the Earth's center.
Answer:
1.363×10^15 seconds
Explanation:
The spaceship travels an elliptical orbit from a point of 6590km from the earth center to the moon and 38500km from the earth center.
To calculate the time taken from Kepler's third Law :
T^2 = ( 4π^2/GMe ) r^3
Where Me is the mass of the earth
r is the average distance travel
G is the universal gravitational constant. = 6.67×10-11 m3 kg-1 s-2
π = 3.14
Me = mass of earth = 5.972×10^24kg
r =( r minimum + r maximum)/2 ......1
rmin = 6590km
rmax = 385000km
From equation 1
r = (6590+385000)/2
r = 391590/2
r = 195795km
From T^2 = ( 4π^2/GMe ) r^3
T^2 = (4 × 3.14^2/ 6.67×10-11 × 5.972×10^24) × 195795^3
= ( 4×9.8596/ 3.983×10^14 ) × 7.5059×10^15
= 39.4384/ 3.983×10^14 ) × 7.5059×10^15
= (9.901×10^14) × 7.5059×10^15
T^2 = 7.4321× 10^30
T =√7.4321× 10^30
T = 2.726×10^15 seconds
The time for one way trip from Earth to the moon is :
∆T = T/2
= 2.726×10^15 /2
= 1.363×10^15 secs
Two cannonballs are dropped from a second-floor physics lab at height h above the ground. Ball B has four times the mass of ball A. When the balls pass the bottom of a first-floor window at height above the ground, the relation between their kinetic energies, KA and KB, is
Answer:
1:4
Explanation:
The formula for calculating kinetic energy is:
[tex]KE=\dfrac{1}{2}mv^2[/tex]
If the mass is multiplied by 4, then, the kinetic energy must be increased by 4 as well. Since they will be travelling at the same speed when they are at the same point, the relation between KA and KB must be 1:4 or 1/4. Hope this helps!
The relation between the kinetic energies of the freely falling balls A and B is obtained as [tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex].
Kinetic EnergyThe kinetic energy of an object depends on the mass and velocity with which it moves.
While under free-fall, the mass of an object does not affect the velocity with which it falls.
So, the velocities of both the balls are the same.
Let the mass of ball A is 'm'
So, the mass of ball B is '4m'
The kinetic energy of ball A is given by;
[tex]KE_{A}=\frac{1}{2} mv^2[/tex]
The kinetic energy of ball B is given by;
[tex]KE_{B}=\frac{1}{2} 4mv^2 = 2mv^2[/tex]
Therefore, the ratio of kinetic energies of A and B is,
[tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex]
Learn more about kinetic energy here:
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A circle has a radius of 13m Find the length of the arc intercepted by a central angle of .9 radians. Do not round any intermediate computations, and round your answer to the nearest tenth.
Answer:
11.7 m
Explanation:
The radius of the circle is 13 m.
The central angle of the arc is 0.9 radians
The length of an arc is given as:
L = r θ
where θ = central angle in radians = 0.9
=> L = 0.9 * 13 = 11.7 m
Length of the arc will be 11.7 m ≈ 10 m
What is an arc length?
Arc length refers to the distance between two points along a curve’s section.
Arc length = radius * theta
where
Arc length = ? to find
given :
radius = 13 m
theta ( central angle) = 0.9 radians
Arc length = 13 m * 0.9 radians
= 11.7 m ≈ 10 m
length of the arc will be 11.7 m ≈ 10 m
Learn more about Arc length
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The average density of the body of a fish is 1080kg/m^3 . To keep from sinking, the fish increases its volume by inflating an internal air bladder, known as a swim bladder, with air.
By what percent must the fish increase its volume to be neutrally buoyant in fresh water? Use 1.28kg/m^3 for the density of air at 20 degrees Celsius. (change in V/V)
Answer:
Increase of volume (F) = 8.01%
Explanation:
Given:
Density of fish = 1,080 kg/m³
Density of water = 1,000 kg/m³
density of air = 1.28 kg/m³
Find:
Increase of volume (F)
Computation:
1,080 kg/m³ + [F × 1.28 kg/m³ ] = (1+F) × 1,000 kg/m³
1,080 + 1.28 F =1,000 F + 1,000
80 = 998.72 F
F = 0.0801 (Approx)
F = 8.01% (Approx)