To solve this problem it is necessary to use the concepts given through the ideal gas equation.
For this it is defined that
[tex]PV = nRT[/tex]
Where,
P = Pressure
V = Volume
R = Gas ideal constant
T = Temperature
n = number of moles.
In this problem we have two states in which the previous equation can be applied, so
[tex]1) P_1V_1 = n_1RT_1[/tex]
[tex]2) P_2V_2 = n_2RT_2[/tex]
From the first state we can calculate the Volume
[tex]V_1 = \frac{n_1RT_1}{P_1}[/tex]
Replacing
[tex]V_1 = \frac{5.1*8.314*300.15}{3.1*10^5}[/tex]
[tex]V_1 = 0.041m^3[/tex]
From the state two we can calculate now the number of the moles, considering that there is a change of 0.8 from Volume 1, then
[tex]n_2 = \frac{P_2(0.8*V_2)}{RT_2}[/tex]
[tex]n_2 = \frac{2.6*10^5(0.8*0.041)}{8.314*310.15}[/tex]
[tex]n_2 = 3.3moles[/tex]
Therefore there are 3.3moles of air left in the tire.
Final answer:
To find the number of moles of air left in the tire, we can use the ideal gas law. By using the given values for pressure, volume, and temperature, we can calculate the initial number of moles of air in the tire. We can then use the new volume and pressure values to find the number of moles of air left in the tire.
Explanation:
To find the number of moles of air left in the tire, we can use the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to convert the temperatures from Celsius to Kelvin. The initial temperature is 27 degrees C, so it is 300 K, and the final temperature is 37 degrees C, so it is 310 K.
Next, we use the ideal gas law to find the initial number of moles of air in the tire:
5.1 moles = (2.1 x 10^5 N/m^2) x V / ((8.314 J/mol*K) x 300 K)
Solving for V gives us V = (5.1 moles x 8.314 J/mol*K x 300 K) / (2.1 x 10^5 N/m^2) = 0.614 m^3.
Finally, we can use the new volume and pressure values to find the number of moles of air left in the tire:
n = (1.6 x 10^5 N/m^2) x (0.8 x 0.614 m^3) / (8.314 J/mol*K x 310 K) = 2.23 moles.
A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction is mu_k. During time interval t, the sled moves a distance s, starting from rest.Find the average velocity vavg of the sled during that time interval.Express your answer in terms of the given quantities and, if necessary, appropriate constants. You may or may not use all of the given quantities.
Answer:
The average velocity of the sled is vavg = s/t.
Explanation:
Hi there!
The average velocity is calculated as the traveled distance over time:
vavg = Δx/Δt
Where:
vavg = average velocity.
Δx = traveled distance.
Δt = elapsed time.
We already know the traveled distance (s) and also know the time it takes the sled to travel that distance (t). Then, the average velocity can be calculated as follows:
vavg = s/t
Have a nice day!
The average velocity of the sled, given by the formula v_avg = s / t, evaluates the distance covered over a certain duration of time. The sled has moved a distance 's' during a time interval 't', and so its average velocity over this interval is simply 's' divided by 't'. The presence of friction does not affect the calculation of average velocity.
Explanation:The average velocity of the sled is simply the distance traveled over a certain time interval. In this case, the sled starts from rest and is being pulled by a horizontal force F against a frictional force. Average Velocity, v_avg, is given by the formula:
v_avg = s / t
In physics, velocity is a measure of the rate of change of position concerning time, so the average velocity is simply the total distance (or displacement) divided by the total time. Given the fact that the sled has moved a distance, s, during the time interval t, the average velocity is simply s divided by t.
With this definition, the sled's average velocity can be determined without needing to tabulate its speed or direction throughout the time interval even in the presence of friction.
Learn more about Average Velocity here:https://brainly.com/question/28512079
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Which of the following statements concerning the electric field inside a conductor is true?
A) The electric field inside a conductor is never zero.
B) The electric field inside a conductor is always zero.
C) The electric field inside a conductor is always zero if charges inside the conductor are not moving.
D) The electric field inside a conductor is always zero unless there are excess charges inside the conductor.
Answer:
C
Explanation:
The electric field inside a conductor is always zero if the charges inside the conductor are not moving.
Since the electron are not moving then they must be in electrostatic equilibrium which means the electric field inside the conductor is zero. if the electric field existed inside the conductor then there will be net force on all the electrons and the electrons will accelerate.
The electric field inside a conductor is always zero if charges inside the conductor are not moving. Under electrostatic conditions, free electrons in the conductor rearrange to cancel any internal electric fields. Thus, the correct answer is Option C.
1. In a conductor in electrostatic equilibrium, the electric field inside the conductor is zero. This is because any free electrons within the conductor will move in response to any electric field until they reach a state where there is no net force acting on them. This movement of electrons cancels out any existing electric field.
2. Properties of conductors in electrostatic equilibrium include that any excess charge resides on the surface of the conductor, and the electric field just outside the surface is perpendicular to the surface.
3. Therefore, in the absence of moving charges (static conditions), the electric field inside a conductor must be zero.
Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as the earth, as having a density that decreases linearly with distance from the center. Let the density be 1.60 x 10^4 kg/m^3 at the center and 2100 kg/m^3 at the surface.
What is the acceleration due to gravity at the surface of this planet?
Answer:
a = 9.94 m/s²
Explanation:
given,
density at center= 1.6 x 10⁴ kg/m³
density at the surface = 2100 Kg/m³
volume mass density as function of distance
[tex]\rho(r) = ar^2 - br^3[/tex]
r is the radius of the spherical shell
dr is the thickness
volume of shell
[tex]dV = 4 \pi r^2 dr[/tex]
mass of shell
[tex]dM = \rho(r)dV[/tex]
[tex]\rho = \rho_0 - br[/tex]
now,
[tex]dM = (\rho_0 - br)(4 \pi r^2)dr[/tex]
integrating both side
[tex]M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr[/tex]
[tex]M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})[/tex]
[tex]M = \pi R^3(\dfrac{\rho_0}{3}+\rho)[/tex]
we know,
[tex]a = \dfrac{GM}{R^2}[/tex]
[tex]a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}[/tex]
[tex]a =\pi RG(\dfrac{\rho_0}{3}+\rho)[/tex]
[tex]a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)[/tex]
a = 9.94 m/s²
In an electricity experiment, a 1.20 g plastic ball is suspended on a 59.0 cm long string and given an electric charge. A charged rod brought near the ball exerts a horizontal electrical force F⃗ elec on it, causing the ball to swing out to a 24.0 ∘ angle and remain there.
a. What is the magnitude of Felec ?
b.What is the tension in the string?
Answer:
a.) 5.24 10⁻³ N . b) 0.013 N
Explanation:
a) In absence of other forces, the plastic ball is only subject to the force of gravity (downward) , and to the tension in the string, which are equal each other.
We are told that there exists an horizontal force , of an electric origin, that causes the ball to swing out to a 24º angle (respect the normal) and remain there, so there exists a new equilibrium condition.
In this situation, both the vertical and horizontal components of the external forces acting on the ball (gravity, tension and the electrical force) must be equal to 0.
The only force that has horizontal and vertical components, is the tension in the string.
We can apply Newton's 2nd Law to both directions, as follows:
T cos 24º - mg = 0
-T sin 24º + Fe = 0
where T= Tension in the string.
Fe = Electrical Force
mg = Fg = gravity force
⇒ T = mg/ cos 24º
Replacing in the horizontal forces equation:
-mg/cos 24º . sin 24º = -Fe
∴ Fe = mg. tg 24º = 0.0012 kg. 9.8 m/s². tg 24º = 5.24 10⁻³ N
b) In order to get the value of T, we can simply solve for T the vertical forces component equation , as follows:
T = mg/ cos 24º = 0.0012 kg. 9.8 m/s² / 0.914 = 0.013 N
Final answer:
To find the magnitude of the electrical force (F⃗ elec) exerted on the plastic ball, we can use the fact that the ball is in equilibrium, meaning the net force acting on it is zero. The electrical force is the only horizontal force acting on the ball, so it must be balanced by the horizontal component of the tension in the string.
Explanation:
The horizontal component of the tension (T) in the string can be found using trigonometry. The angle between the string and the vertical is 24.0 degrees, so the horizontal component of the tension is:
T_horizontal = T * cos(24.0 degrees)
Since the ball is in equilibrium, the magnitude of the electrical force is equal to the horizontal component of the tension:
|F⃗ elec| = T_horizontal
To find the tension in the string, we can use the fact that the ball is in equilibrium, meaning the net force acting on it is zero. The gravitational force (F⃗ g) acting on the ball is balanced by the vertical component of the tension in the string.
The vertical component of the tension (T) in the string can be found using trigonometry. The angle between the string and the vertical is 24.0 degrees, so the vertical component of the tension is:
T_vertical = T * sin(24.0 degrees)
Since the ball is in equilibrium, the magnitude of the gravitational force is equal to the vertical component of the tension:
|F⃗ g| = T_vertical
The gravitational force can be calculated using the mass of the ball (m) and the acceleration due to gravity (g):
|F⃗ g| = m * g
Substituting the expression for the vertical component of the tension into the equation for the gravitational force, we get:
m * g = T * sin(24.0 degrees)
Solving for the tension (T), we get:
T = (m * g) / sin(24.0 degrees)
Substituting the given values for the mass of the ball (m) and the acceleration due to gravity (g), we get:
T = (1.20 g * 9.8 m/s^2) / sin(24.0 degrees)
T ≈ 5.10 N
b. The tension in the string is approximately 5.10 N.
Two waves traveling on a string in the same direction both have a frequency of 135 Hz, a wavelength of 2 cm, and an amplitude of 0.04 m. What is the amplitude of the resultant wave if the original waves differ in phase by each of the following values?
(a) p/6 cm(b) p/3 cm
Answer:
The amplitude of the resultant wave are
(a). 0.0772 m
(b). 0.0692 m
Explanation:
Given that,
Frequency = 135 Hz
Wavelength = 2 cm
Amplitude = 0.04 m
We need to calculate the angular frequency
[tex]\omega=2\pi f[/tex]
[tex]\omega=2\times\pi\times135[/tex]
[tex]\omega=848.23\ rad/s[/tex]
As the two waves are identical except in their phase,
The amplitude of the resultant wave is given by
[tex]y+y=A\sin(kx-\omega t)+Asin(kx-\omega t+\phi)[/tex]
[tex]y+y=A[2\sin(kx-\omega t+\dfrac{\phi}{2})\cos\phi\dfrac{\phi}{2}[/tex]
[tex]y'=2A\cos(\dfrac{\phi}{2})\sin(kx-\omega t+\dfrac{\phi}{2})[/tex]
(a). We need to calculate the amplitude of the resultant wave
For [tex]\phi =\dfrac{\pi}{6}[/tex]
The amplitude of the resultant wave is
[tex]A'=2A\cos(\dfrac{\phi}{2})[/tex]
Put the value into the formula
[tex]A'=2\times0.04\cos(\dfrac{\pi}{12})[/tex]
[tex]A'=0.0772\ m[/tex]
(b), We need to calculate the amplitude of the resultant wave
For [tex]\phi =\dfrac{\pi}{3}[/tex]
[tex]A'=2\times0.04\cos(\dfrac{\pi}{6})[/tex]
[tex]A'=0.0692\ m[/tex]
Hence, The amplitude of the resultant wave are
(a). 0.0772 m
(b). 0.0692 m
Selena uses a garden hose to fill a bucket of water. The water enters the hose through a faucet with a 6.0 cm diameter.
The speed of the water at the faucet is 5/ms.
If the faucet and nozzlw are at the same height, and the water leaves the nozzle with a speed of 20 m/s, what is the diameter of the nozzle?
Answer:
3 cm
Explanation:
To answer this question the equation of continuity can be used
For us to use the equaition of continuity, we will make a few assumptions:
That the temperature of the water does not change, therefore there is not expansion or contraction (change in volume)That the flow is non-viscousThat there is a single entry and a single exitThat the water is incompressible[tex]A_{1}v_{1} = A_{2} v_{2} \\\pi (3cm)^{2} * 5m/s = \pi r^{2} *20m/s\\\\r = 1.5 cm\\D = 2r = 2*1.5cm=3 cm[/tex]
A soup can is in the shape of a right cylinder. The can has a volume of 16 fluid ounces. The height is three times its radius. The metal used to make the lateral surface of the can costs $0.01 per square inch. The metal used to make the top and bottom costs $0.02 per square inch. If one fluid ounce is approximately 1.805 cubic inches, what is the total cost to make one empty soup can? Use 3.14 for straight pi
Answer:
$0.662772
Explanation:
v = Volume of can = 16 fl oz.
[tex]1\ floz.=1.805\ in^3[/tex]
r = Radius of can
h = Height of can = 3r
Volume of cylinder is given by
[tex]\pi r^2h=16\times 1.805\\\Rightarrow \pi r^23r=16\times 1.805\\\Rightarrow 3\pi r^3=16\times 1.805\\\Rightarrow r=\left(\frac{16\times 1.805}{3\times 3.14}\right)^{\frac{1}{3}}\\\Rightarrow r=1.45247\ in[/tex]
h=3r\\\Rightarrow h=3\times 1.45247\\\Rightarrow h=4.35741\ in[/tex]
Surface area of sides is given by
[tex]2\pi rh\\ =2\times 3.14\times 1.45247\times 4.35741\\ =39.76632\ in^2[/tex]
Surface area of top and bottom is given by
[tex]2\pi r^2\\ =2\times 3.14\times 1.45247^2\\ =13.25544\ in^2[/tex]
Cost of making the can will be
[tex]39.76632\times 0.01+13.25544\times 0.02=\$0.662772[/tex]
The cost to make the can is $0.662772
Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg), who is sitting at the very bottom end, and whom he holds onto when he arrives. Laughing, John & William leave the slide horizontally and land in the muddy ground near the foot of the slide. (A) If John starts out 1.8 m above William, and the slide is essentially frictionless, how fast are they going when they leave the slide? (B) Thanks to the mud he acquired, John will now experience an average frictional force of 105 N as he slides down. How much slower is he going when he reaches the bottom than when friction was absent?
Answer:
[tex]v=3.564\ m.s^{-1}[/tex]
[tex]\Delta v =2.16\ m.s^{-1}[/tex]
Explanation:
Given:
mass of John, [tex]m_J=30\ kg[/tex]mass of William, [tex]m_W=30\ kg[/tex]length of slide, [tex]l=3\ m[/tex](A)
height between John and William, [tex]h=1.8\ m[/tex]
Using the equation of motion:
[tex]v_J^2=u_J^2+2 (g.sin\theta).l[/tex]
where:
v_J = final velocity of John at the end of the slide
u_J = initial velocity of John at the top of the slide = 0
Now putting respective :
[tex]v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3[/tex]
[tex]v_J=5.94\ m.s^{-1}[/tex]
Now using the law of conservation of momentum at the bottom of the slide:
Sum of initial momentum of kids before & after collision must be equal.
[tex]m_J.v_J+m_w.v_w=(m_J+m_w).v[/tex]
where: v = velocity with which they move together after collision
[tex]30\times 5.94+0=(30+20)v[/tex]
[tex]v=3.564\ m.s^{-1}[/tex] is the velocity with which they leave the slide.
(B)
frictional force due to mud, [tex]f=105\ N[/tex]Now we find the force along the slide due to the body weight:
[tex]F=m_J.g.sin\theta[/tex]
[tex]F=30\times 9.8\times \frac{1.8}{3}[/tex]
[tex]F=176.4\ N[/tex]
Hence the net force along the slide:
[tex]F_R=71.4\ N[/tex]
Now the acceleration of John:
[tex]a_j=\frac{F_R}{m_J}[/tex]
[tex]a_j=\frac{71.4}{30}[/tex]
[tex]a_j=2.38\ m.s^{-2}[/tex]
Now the new velocity:
[tex]v_J_n^2=u_J^2+2.(a_j).l[/tex]
[tex]v_J_n^2=0^2+2\times 2.38\times 3[/tex]
[tex]v_J_n=3.78\ m.s^{-1}[/tex]
Hence the new velocity is slower by
[tex]\Delta v =(v_J-v_J_n)[/tex]
[tex]\Delta v =5.94-3.78= 2.16\ m.s^{-1}[/tex]
You’ve been given the challenge of balancing a uniform, rigid meter-stick with mass M = 95 g on a pivot. Stacked on the 0-cm end of the meter stick are n identical coins, each with mass m = 3.1 g, so that the center of mass of the coins is directly over the end of the meter stick. The pivot point is a distance d from the 0-cm end of the meter stick.
Part (a): Determine the distance d = d1, in centimeters, if there is only one coin o the 0 end of the meter stick and the system is in static equilibrium
The distance [tex]\( d_1 \)[/tex] from the pivot point to the center of mass of the meter stick with one coin on the 0-cm end, maintaining static equilibrium, is approximately 46.8 cm.
1. To achieve static equilibrium, the torques on both sides of the pivot point must balance out.
2. The torque due to the meter stick with mass ( M ) is [tex]\( M \times g \times \frac{L}{2} \)[/tex], where ( L ) is the length of the meter stick (100 cm) and \( g \) is the acceleration due to gravity (approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex]).
3. The torque due to the coin on the 0-cm end is [tex]\( m \times g \times d_1 \)[/tex], where [tex]\( m \)[/tex] is the mass of the coin.
4. Since the torques balance out, we have the equation: [tex]\( M \times g \times \frac{L}{2} = m \times g \times d_1 \).[/tex]
5. Rearrange the equation to solve for [tex]\( d_1 \): \( d_1 = \frac{M \times \frac{L}{2}}{m} \).[/tex]
6. Substitute the given values: [tex]\( d_1 = \frac{95 \, \text{g} \times \frac{100}{2} \, \text{cm}}{3.1 \, \text{g}} \).[/tex]
7. Calculate [tex]\( d_1 \)[/tex]to find the distance from the pivot point to the center of mass of the meter stick with one coin on the 0-cm end, which is approximately 46.8 cm.
A 3 GHz line-of-sight microwave communication link consists of two lossless parabolic dish antennas, each1 m in diameter. If the receive antenna requires 10 nW of receive power for good reception and the distance between the antennas is 40 km, how much power should be transmitted?
Answer:
0.25938 W
Explanation:
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
[tex]\nu[/tex] = Frequency = 3 GHz
d = Diameter of lossless antenna = 1 m
r = Radius = [tex]\frac{d}{2}=\frac{1}{2}=0.5\ m[/tex]
[tex]A_t[/tex] = Area of transmitter
[tex]A_r[/tex] = Area of receiver
R = Distance between the antennae = 40 km
[tex]P_r[/tex] = Power of receiver = [tex]10\times 10^{-9}\ W[/tex]
[tex]P_t[/tex] = Power of Transmitter
Wavelength
[tex]\lambda=\frac{c}{\nu}\\\Rightarrow \lambda=\frac{3\times 10^8}{3\times 10^9}\\\Rightarrow \lambda=0.1\ m[/tex]
From Friis transmission formula we have
[tex]\frac{P_t}{P_r}=\frac{\lambda^2R^2}{A_tA_r}\\\Rightarrow P_t=P_r\frac{\lambda^2R^2}{A_tA_r}\\\Rightarrow P_t=10\times 10^{-9}\frac{0.1^2\times (40\times 10^3)^2}{\pi 0.5^2\times \pi 0.5^2}\\\Rightarrow P_t=0.25938\ W[/tex]
The power that should be transmitted is 0.25938 W
Compute the longitudinal strength of an aligned carbon fiber-epoxy matrix composite having a 0.25 volume fraction of fibers, assuming the following: (1) an average fiber diameter of 10 10-3 mm, (2) an average fiber length of 1 mm, (3) a fiber fracture strength of 2.5 GPa, (4) a fiber-matrix bond strength of 10 MPa, (5) a matrix stress at fiber failure of 10.0 MPa, and (6) a matrix tensile strength of 75 MPa.
Answer:
632.5 MPa
Explanation:
[tex]\sigma_{m}[/tex] = Matrix stress at fiber failure = 10 MPa
[tex]V_f[/tex] = Volume fraction of fiber = 0.25
[tex]\sigma_f[/tex] = Fiber fracture strength = 2.5 GPa
The longitudinal strength of a composite is given by
[tex]\sigma_{cl}=\sigma_{m}(1-V_f)+\sigma_fV_f\\\Rightarrow \sigma_{cl}=10(1-0.25)+(2.5\times 10^3)\times 0.25\\\Rightarrow \sigma_{cl}=632.5\ MPa[/tex]
The longitudinal strength of the aligned carbon fiber-epoxy matrix composite is 632.5 MPa
Final answer:
The given question seeks to calculate the longitudinal strength of a carbon fiber-epoxy composite, but lacks sufficient detail or formulae for a complete answer. Typically, this computation would involve using materials science models that consider fiber orientation and other stress-strain interactions between components.
Explanation:
To compute the longitudinal strength of an aligned carbon fiber-epoxy matrix composite with a 0.25 volume fraction of fibers, we'd need to consider the following given properties: the average fiber diameter, average fiber length, fiber fracture strength, fiber-matrix bond strength, matrix stress at fiber failure, and matrix tensile strength.
Although the exact method for calculating the longitudinal strength would typically involve applying principles from materials science, such as the rule of mixtures, in combination with the given data points, the actual question does not provide enough information or a specific formula to complete the calculation. For a real-life carbon fiber-epoxy composite, the longitudinal strength could be substantially influenced by the alignment of the fibers, bond quality between the fibers and matrix, and the interaction between the stress and strain of the components.
If we had a suitable model or empirical formula, we would proceed by plugging in the given values to determine the longitudinal strength. However, as the provided data from the question is incomplete for this calculation, it is recommended to refer to a textbook or comprehensive resource on composite material mechanics for the detailed step-by-step methodology and equations.
A proton has a speed of 3.50 Ã 105 m/s when at a point where the potential is +100 V. Later, itâs at a point where the potential is â150 V. What is the change in the protonâs electric potential? What is the change in the potential energy of the proton? What is the work done on the proton?
Answer:
(a). The change in the protons electric potential is 0.639 kV.
(b). The change in the potential energy of the proton is [tex]1.022\times10^{-16}\ J[/tex]
(c). The work done on the proton is [tex]-8\times10^{-18}\ J[/tex].
Explanation:
Given that,
Speed [tex]v= 3.50\times10^{5}\ m/s[/tex]
Initial potential V=100 V
Final potential = 150 V
(a). We need to calculate the change in the protons electric potential
Potential energy of the proton is
[tex]U=qV=eV[/tex]
Using conservation of energy
[tex]K_{i}+U_{i}=K_{f}+U_{f}[/tex]
[tex]\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}[/tex]
[tex]]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})[/tex]
[tex]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V[/tex]
[tex]\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}[/tex]
Put the value into the formula
[tex]\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}[/tex]
[tex]\Delta V=639.2=0.639\ kV[/tex]
(b). We need to calculate the change in the potential energy of the proton
Using formula of potential energy
[tex]\Delta U=q\Delta V[/tex]
Put the value into the formula
[tex]\Delta U=1.6\times10^{-19}\times639.2[/tex]
[tex]\Delta U=1.022\times10^{-16}\ J[/tex]
(c). We need to calculate the work done on the proton
Using formula of work done
[tex]\Delta U=-W[/tex]
[tex]W=q(V_{2}-V_{1})[/tex]
[tex]W=-1.6\times10^{-19}(150-100)[/tex]
[tex]W=-8\times10^{-18}\ J[/tex]
Hence, (a). The change in the protons electric potential is 0.639 kV.
(b). The change in the potential energy of the proton is [tex]1.022\times10^{-16}\ J[/tex]
(c). The work done on the proton is [tex]-8\times10^{-18}\ J[/tex].
Final answer:
The change in electric potential and potential energy of the proton can be calculated based on the provided potentials. The work done on the proton equals the change in potential energy.
Explanation:
The change in the proton's electric potential: The change in electric potential is the final potential minus the initial potential, thus the change is -150 V - 100 V = -250 V.
The change in potential energy of the proton: The potential energy change equals the charge of the proton times the change in potential, giving -proton charge x change in potential.
The work done on the proton: The work done is equal to the change in the potential energy of the proton.
The radius of the aorta is «10 mm and the blood flowing through it has a speed of about 300 mm/s. A capillary has a radius of about 4ˆ10´3 mm but there are literally billions of them. The average speed of blood through the capillaries is about 5ˆ10´4 m/s. (i) Calculate the effective cross sectional area of the capillaries and (ii) the approximate number of capillaries.
Answer:
(I). The effective cross sectional area of the capillaries is 0.188 m².
(II). The approximate number of capillaries is [tex]3.74\times10^{9}[/tex]
Explanation:
Given that,
Radius of aorta = 10 mm
Speed = 300 mm/s
Radius of capillary [tex]r=4\times10^{-3}\ mm[/tex]
Speed of blood [tex]v=5\times10^{-4}\ m/s[/tex]
(I). We need to calculate the effective cross sectional area of the capillaries
Using continuity equation
[tex]A_{1}v_{1}=A_{2}v_{2}[/tex]
Where. v₁ = speed of blood in capillarity
A₂ = area of cross section of aorta
v₂ =speed of blood in aorta
Put the value into the formula
[tex]A_{1}=A_{2}\times\dfrac{v_{2}}{v_{1}}[/tex]
[tex]A_{1}=\pi\times(10\times10^{-3})^2\times\dfrac{300\times10^{-3}}{5\times10^{-4}}[/tex]
[tex]A_{1}=0.188\ m^2[/tex]
(II). We need to calculate the approximate number of capillaries
Using formula of area of cross section
[tex]A_{1}=N\pi r_{c}^2[/tex]
[tex]N=\dfrac{A_{1}}{\pi\times r_{c}^2}[/tex]
Put the value into the formula
[tex]N=\dfrac{0.188}{\pi\times(4\times10^{-6})^2}[/tex]
[tex]N=3.74\times10^{9}[/tex]
Hence, (I). The effective cross sectional area of the capillaries is 0.188 m².
(II). The approximate number of capillaries is [tex]3.74\times10^{9}[/tex]
At one particular moment, a 15.0 kg toboggan is moving over a horizontal surface of snow at 4.80 m/s. After 7.00 s have elapsed, the toboggan stops. Use a momentum approach to find the magnitude of the average friction force (in N) acting on the toboggan while it was moving
Answer:
10.28571 N
Explanation:
m = Mass of toboggan = 15 kg
u = Initial velocity = 4.8 m/s
v = Final velocity = 0
t = Time taken = 7 seconds
Friction force is given by the change in momentum over time
[tex]F=\frac{m(v-u)}{t}\\\Rightarrow F=\frac{15(0-4.8)}{7}\\\Rightarrow F=-10.28571\ N[/tex]
The magnitude of the average friction force acting on the toboggan while it was moving is 10.28571 N
Using the principle of conservation of momentum, the magnitude of the average friction force acting on the toboggan can be found. The initial momentum of the toboggan is equal to the change in momentum, which is equal to the mass of the toboggan multiplied by the change in velocity. Dividing the change in momentum by the time interval gives the magnitude of the average friction force as 10.29 N.
Explanation:To find the magnitude of the average friction force acting on the toboggan, we can use the principle of conservation of momentum. The initial momentum of the toboggan is given by P = m * v, where m is the mass (15.0 kg) and v is the velocity (4.80 m/s). The final momentum is zero, as the toboggan comes to a stop. Therefore, the change in momentum is equal to the initial momentum.
The change in momentum is given by δP = m * δv, where δv is the change in velocity. Since the velocity changes from 4.80 m/s to 0 m/s, the change in velocity is -4.80 m/s. Therefore, the change in momentum is -15.0 kg * 4.80 m/s = -72.0 kg*m/s.
The average friction force is equal to the change in momentum divided by the time interval. The time interval is given as 7.00 s. Therefore, the magnitude of the average friction force is |-72.0 kg*m/s / 7.00 s| = 10.29 N.
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Mantles for gas lanterns contain thorium, because it forms an oxide that can survive being heated to incandescence for long periods of time. Natural thorium is almost 100% 232Th, with a half-life of 1.405 ✕ 1010 y. If an average lantern mantle contains 200 mg of thorium, what is its activity (in Bq)?
Answer:
The activity is 811.77 Bq
Solution:
As per the question:
Half life of Thorium, [tex]t_{\frac{1}{2}} = 1.405\times 10^{10}\ yrs[/tex]
Mass of Thorium, m = 200 mg = 0.2 g
M = 232 g/mol
Now,
No. of nuclei of Thorium in 200 mg of Thorium:
[tex]N = \frac{N_{o}m}{M}[/tex]
where
[tex]N_{o }[/tex] = Avagadro's number
Thus
[tex]N = \frac{6.02\times 10^{23}\times 0.2}{232} = 5.19\times 10^{20}[/tex]
Also,
Activity is given by:
[tex]\frac{0.693}{t_{\frac{1}{2}}}\times N[/tex]
[tex]A= \frac{0.693}{1.405\times 10^{10}}\times 5.19\times 10^{20} = 2.56\times 10^{10}\ \yr[/tex]
[tex]A = \frac{2.56\times 10^{10}}{365\times 24\times 60\times 60} = 811.77\ Bq[/tex]
You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 76.2 kg hop on board for a ride through the woods and the springs (one for each wheel) each compress by 6.17 cm. When you pull the trailer over a tree root in the trail, it oscillates with a period of 2.09 s. Determine the following. (a) force constant of the springs? N/m
(b) mass of the trailer? kg
(c) frequency of the oscillation? Hz
(d) time it takes for the trailer to bounce up and down 10 times? s
a) The spring constant is 12,103 N/m
b) The mass of the trailer 2,678 kg
c) The frequency of oscillation is 0.478 Hz
d) The time taken for 10 oscillations is 20.9 s
Explanation:
a)
When the two children jumps on board of the trailer, the two springs compresses by a certain amount
[tex]\Delta x = 6.17 cm = 0.0617 m[/tex]
Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:
[tex]2mg = k'\Delta x[/tex] (1)
where
m = 76.2 kg is the mass of each children
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]k'[/tex] is the equivalent spring constant of the 2-spring system
For two springs in parallel each with constant k,
[tex]k'=k+k=2k[/tex]
Substituting into (1) and solving for k, we find:
[tex]2mg=2k\Delta x\\k=\frac{mg}{\Delta x}=\frac{(76.2)(9.8)}{0.0617}=12,103 N/m[/tex]
b)
The period of the oscillating system is given by
[tex]T=2\pi \sqrt{\frac{m}{k'}}[/tex]
where
And for the system in the problem, we know that
T = 2.09 s is the period of oscillation
m is the mass of the trailer
[tex]k'=2k=2(12,103)=24,206 N/m[/tex] is the equivalent spring constant of the system
Solving the equation for m, we find the mass of the trailer:
[tex]m=(\frac{T}{2\pi})^2 k'=(\frac{2.09}{2\pi})^2 (24,206)=2,678 kg[/tex]
c)
The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:
[tex]f=\frac{1}{T}[/tex]
where
f is the frequency
T is the period
In this problem, we have
T = 2.09 s is the period
Therefore, the frequency of oscillation is
[tex]f=\frac{1}{2.09}=0.478 Hz[/tex]
d)
The period of the system is
T = 2.09 s
And this time is the time it takes for the trailer to complete one oscillation.
In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.
Therefore, the time needed for 10 oscillations is:
[tex]t=10T=10(2.09)=20.9 s[/tex]
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Suppose the Earth's magnetic field at the equator has magnitude 0.00005 T and a northerly direction at all points. How fast must a singly ionized uranium atom (m=238u, q=e) move so as to circle the Earth 1.44 km above the equator? Give your answer in meters/second.
Answer:
Velocity will be [tex]v=1.291\times 10^8m/sec[/tex]
Explanation:
We have given magnetic field B = 0.00005 T
Mass m = 238 U
We know that [tex]1u=1.66\times 10^{-27}kg[/tex]
So 238 U [tex]=238\times 1.66\times 10^{-27}=395.08\times 10^{-27}kg[/tex]
Radius [tex]=R+1.44=6378+1.44=6379.44KM[/tex]
We know that magnetic force is given by
[tex]F=qvB[/tex] which is equal to the centripetal force
So [tex]qvB=\frac{mv^2}{r}[/tex]
[tex]1.6\times 10^{-19}\times v\times 0.00005=\frac{395.08\times 10^{-27}v^2}{6379.44}[/tex]
[tex]v=1.291\times 10^8m/sec[/tex]
A 2.8-carat diamond is grown under a high pressure of 58 × 10 9 N / m 2 .
(a) By how much does the volume of a spherical 2.8-carat diamond expand once it is removed from the chamber and exposed to atmospheric pressure?
(b) What is the increase in the diamond’s radius? One carat is 0.200 g, and you can use 3.52 g/cm3 for the density of diamond, and 4.43 × 10 11 N / m 2 for the bulk modulus of diamond.
Final answer:
To calculate the change in volume of a diamond exposed to atmospheric pressure, we can use the formula for bulk modulus. The increase in the diamond's radius can be found using the formula for increase in volume.
Explanation:
(a) To calculate the change in volume, we can use the formula for bulk modulus:
ΔV = V * (Pf - Pi) / B
Where ΔV is the change in volume, V is the initial volume, Pf is the final pressure, Pi is the initial pressure, and B is the bulk modulus.
Substituting the given values, we get:
ΔV = (4/3) * π * r^3 * (Pf - Pi) / B
Since the sphere is symmetrical, the change in radius (Δr) is the same in all directions. So, we can calculate it as:
Δr = ΔV / ((4/3) * π * (r^2) * ΔP)
(b) To find the increase in the diamond's radius, we can use the formula for increase in volume:
ΔV = (4/3) * π * (Rf^3 - Ri^3)
Substituting the given values, we get:
Rf = (3 * ΔV + 4 * π * Ri^3) / (4 * π * Ri^2)
The diamond's volume expands by 2.08 x 10⁻⁹ m³ and its radius increases by 10 micrometers upon being exposed to atmospheric pressure. These calculations utilize the mass, density, and bulk modulus of the diamond.
First, we convert the mass of the diamond into kilograms:
2.8 carats * 0.200 g/carat = 0.56 g = 0.00056 kg
Next, calculate the initial volume ([tex]V_i[/tex]) of the diamond using the density formula:
Density = Mass / Volume
So, the initial volume:
[tex]V_i[/tex] = Mass / Density
[tex]V_i[/tex] = 0.56 g / 3.52 g/cm³
[tex]V_i[/tex] = 0.1591 cm³ = 1.591 × 10⁻⁷ m³
The bulk modulus (B) of diamond is given as 4.43 × 10¹¹ N/m² and the pressure change (ΔP) is:
ΔP = 58 × 10⁹ N/m²
Using the formula for volume change (ΔV), where:
ΔV / V = -ΔP / B
we get:
ΔV = -[tex]V_i[/tex] * ΔP / B
ΔV = -1.591 × 10⁻⁷ m³ * 58 × 109 N/m² / 4.43 × 10¹¹ N/m²
ΔV ≈ -2.08 × 10⁻⁹ m³
Therefore, the volume expansion when exposed to atmospheric pressure is:
ΔV = 2.08 × 10⁻⁹ m³
(b) For the increase in radius (Δr), use the formula for the volume of a sphere:
V = (4/3)πr³
The new volume ([tex]V_f[/tex]) is:
[tex]V_f[/tex] = Vi + ΔV
[tex]V_f[/tex] ≈ 1.591 × 10⁻⁷ m³ + 2.08 × 10⁻⁹ m³
[tex]V_f[/tex] ≈ 1.611 × 10⁻⁷ m³
Setting the initial and final volumes equal to the sphere volume formula, we solve for the radii:
[tex]r_{i[/tex] = (3[tex]V_i[/tex] / 4π)^(1/3)
[tex]r_{i[/tex] ≈ (3 * 1.591 × 10⁻⁷ m³ / 4π)^(1/3)
[tex]r_{i[/tex] ≈ 3.37 × 10-3 m
[tex]r_f[/tex] = (3[tex]V_f[/tex] / 4π)^(1/3)
[tex]r_f[/tex] ≈ (3 * 1.611 × 10⁻⁷ m³ / 4π)^(1/3)
[tex]r_f[/tex] ≈ 3.38 × 10⁻³ m
Thus the increase in radius Δr is:
Δr ≈ [tex]r_f[/tex] - [tex]r_{i[/tex]
Δr ≈ 3.38 × 10⁻³ m - 3.37 × 10⁻³ m
Δr = 0.01 × 10⁻³ m
Δr = 10 × 10⁻⁶ m = 10 μm
A 2.0-kg block travels around a 0.40-m radius circle with an angular speed of 16 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.60 m from the origin. What is the magnitude of the component in the xy plane of the angular momentum around the origin?
The magnitude of the component in the xy plane of the angular momentum around the origin is 7.68 kg・m²/s. This is calculated using the formula for angular momentum, with the velocity determined from the product of the radius and angular speed.
Explanation:The magnitude of the component in the xy plane of the angular momentum around the origin can be calculated using the formula for angular momentum, L = mvr, where m is the mass, v is the velocity (which can be obtained from v = rw where r is the radius, and w is the angular speed), and r is the perpendicular distance from the center of the circular path to the origin. In this case, m = 2.0 kg, r = 0.60 m (the distance from the z-axis, not the radius of the circle), and the velocity v = (0.40 m)(16 rad/s) = 6.4 m/s.
Plugging these values into the angular momentum formula gives us, L = mvr = (2.0 kg)(6.4 m/s)(0.60 m) = 7.68 kg・m²/s as the magnitude of the component in the xy plane of the angular momentum around the origin.
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A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy train is moving in the positive x-direction with a velocity of Vi = 2.20 m/s and the toy car is also moving in the positive x-direction with a velocity of vi = 4.95 m/s. Immediately after the collision, the toy car is observed moving in the positive x-direction with a velocity of 1.80 m/s.
(a) Determine the final velocity of the toy train. cm/s
(b) Determine the change ake in the total kinetic energy.
Answer:
[tex]V_{ft}= 317 cm/s[/tex]
ΔK = 2.45 J
Explanation:
a) Using the law of the conservation of the linear momentum:
[tex]P_i = P_f[/tex]
Where:
[tex]P_i=M_cV_{ic} + M_tV_{it}[/tex]
[tex]P_f = M_cV_{fc} + M_tV_{ft}[/tex]
Now:
[tex]M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}[/tex]
Where [tex]M_c[/tex] is the mass of the car, [tex]V_{ic}[/tex] is the initial velocity of the car, [tex]M_t[/tex] is the mass of train, [tex]V_{fc}[/tex] is the final velocity of the car and [tex]V_{ft}[/tex] is the final velocity of the train.
Replacing data:
[tex](1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}[/tex]
Solving for [tex]V_{ft}[/tex]:
[tex]V_{ft}= 3.17 m/s[/tex]
Changed to cm/s, we get:
[tex]V_{ft}= 3.17*100 = 317 cm/s[/tex]
b) The kinetic energy K is calculated as:
K = [tex]\frac{1}{2}MV^2[/tex]
where M is the mass and V is the velocity.
So, the initial K is:
[tex]K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2[/tex]
[tex]K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2[/tex]
[tex]K_i = 22.06 J[/tex]
And the final K is:
[tex]K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2[/tex]
[tex]K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2[/tex]
[tex]K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2[/tex]
[tex]K_f = 19.61 J[/tex]
Finally, the change in the total kinetic energy is:
ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J
You are given two carts A and B. They look identical and you are told they are made of the same material. You place A at rest at an air track and give B a constant velocity directed to the right so that it collides elastically with A. After the collision cart B moves to the left. What do you conclude?
(A) Cart A is hollow.
(B) The two carts are identical.
(C) Cart B is hollow.
Based on the motion after the collision, cart A is concluded to be hollow.
Let's analyze the situation step by step.
Initial State: Collision: The collision is elastic, meaning both kinetic energy and momentum are conserved.After the Collision:Now, let's consider the possible scenarios:
If the two carts were completely identical (same mass, same structure), and there were no external forces, they would move together after the collision (due to conservation of momentum). The fact that Cart B moves to the left suggests that there might be a difference between the two carts.If Cart A is hollow, it would have less mass than Cart B. After the collision, the two carts would move in the direction of the heavier cart (Cart B) due to conservation of momentum. The fact that Cart B moves to the left supports the idea that there is a mass difference.If Cart B is hollow, it would have less mass than Cart A. After the collision, the two carts would move in the direction of the heavier cart (Cart A) due to the conservation of momentum. However, this contradicts the observed motion of Cart B moving to the left.Therefore, based on the information provided, the conclusion is: (A) Cart A is hollow.
Air at 207 kPa and 200◦C enters a 2.5-cm-ID tube at 6 m/s. The tube is constructed ofcopper with a thickness of 0.8 mm and a length of 3 m. Atmospheric air at 1 atm and 20◦Cflows normal to the outside of the tube with a free-stream velocity of 12 m/s. Calculate theair temperature at exit from the tube. What would be the effect of reducing the hot-air flowing half?
Answer:
Temperature of air at exit = 24.32 C, After reducing hot air the temperature of the exit air becomes = 20.11 C
Explanation:
ρ = P/R(Ti) where ρ is the density of air at the entry, P is pressure of air at entrance, R is the gas constant, Ti is the temperature at entry
ρ = (2.07 x 10⁵)/(287)(473) = 1.525 kg/m³
Calculate the mass flow rate given by
m (flow rate) = (ρ x u(i) x A(i)) where u(i) is the speed of air, A(i) is the area of the tube (πr²) of the tube
m (flow rate) = 1.525 x (π x 0.0125²) x 6 = 4.491 x 10⁻³ kg/s
The Reynold's Number for the air inside the tube is given by
R(i) = (ρ x u(i) x d)/μ where d is the inner diameter of the tube and μ is the dynamic viscosity of air (found from the table at Temp = 473 K)
R(i) = (1.525) x (6) x 0.025/2.58 x 10⁻⁵ = 8866
Calculate the convection heat transfer Coefficient as
h(i) = (k/d)(R(i)^0.8)(Pr^0.3) where k is the thermal conductivity constant known from table and Pr is the Prandtl's Number which can also be found from the table at Temperature = 473 K
h(i) = (0.0383/0.025) x (8866^0.8) x (0.681^0.3) = 1965.1 W/m². C
The fluid temperature is given by T(f) = (T(i) + T(o))/2 where T(i) is the temperature of entry and T(o) is the temperature of air at exit
T(f) = (200 + 20)/2 = 110 C = 383 K
Now calculate the Reynold's Number and the Convection heat transfer Coefficient for the outside
R(o) = (μ∞ x do)/V(f) where μ∞ is the speed of the air outside, do is the outer diameter of the tube and V(f) is the kinematic viscosity which can be known from the table at temperature = 383 K
R(o) = (12 x 0.0266)/(25.15 x 10⁻⁶) = 12692
h(o) = K(f)/d(o)(0.193 x Ro^0.618)(∛Pr) where K(f) is the Thermal conductivity of air on the outside known from the table along with the Prandtl's Number (Pr) from the table at temperature = 383 K
h(o) = (0.0324/0.0266) x (0.193 x 12692^0.618) x (0.69^1/3) = 71.36 W/m². C
Calculate the overall heat transfer coefficient given by
U = 1/{(1/h(i)) + A(i)/(A(o) x h(o))} simplifying the equation we get
U = 1/{(1/h(i) + (πd(i)L)/(πd(o)L) x h(o)} = 1/{(1/h(i) + di/(d(o) x h(o))}
U = 1/{(1/1965.1) + 0.025/(0.0266 x 71.36)} = 73.1 W/m². C
Find out the minimum capacity rate by
C(min) = m (flow rate) x C(a) where C(a) is the specific heat of air known from the table at temperature = 473 K
C(min) = (4.491 x 10⁻³) x (1030) = 4.626 W/ C
hence the Number of Units Transferred may be calculated by
NTU = U x A(i)/C(min) = (73.1 x π x 0.025 x 3)/4.626 = 3.723
Calculate the effectiveness of heat ex-changer using
∈ = 1 - е^(-NTU) = 1 - e^(-3.723) = 0.976
Use the following equation to find the exit temperature of the air
(Ti - Te) = ∈(Ti - To) where Te is the exit temperature
(200 - Te) = (0.976) x (200 - 20)
Te = 24.32 C
The effect of reducing the hot air flow by half, we need to calculate a new value of Number of Units transferred followed by the new Effectiveness of heat ex-changer and finally the exit temperature under these new conditions.
Since the new NTU is half of the previous NTU we can say that
NTU (new) = 2 x NTU = 2 x 3.723 = 7.446
∈(new) = 1 - e^(-7.446) = 0.999
(200 - Te (new)) = (0.999) x (200 - 20)
Te (new) = 20.11 C
A toy balloon, which has a mass of 2.90 g before it is inflated, is filled with helium (with a density of 0.180 kg/m^3) to a volume of 8400 cm^3. What is the minimum mass that should be hung from the balloon to prevent it from rising up into the air? Assume the air has a density of 1.29 kg/m^3.
Answer:
[tex]M=6.4243\ g[/tex]
Explanation:
Given:
mass of deflated balloon, [tex]m_b=2.9\ g=0.0029\ kg[/tex]density of helium, [tex]\rho_h=0.180\ kg.m^{-3}[/tex]volume of inflation, [tex]V=8400\ cm^3=0.0084\ m^3[/tex]density of air, [tex]\rho_a=1.29\ kg.m^{-3}[/tex]To stop this balloon from rising up we need to counter the buoyant force.
mass of balloon after inflation:
[tex]m=m_h+m_b[/tex]
[tex]m=0.0084\times 0.180+0.0029[/tex]
[tex]m=0.004412\ kg[/tex]
Now the density of inflated balloon:
[tex]\rho_b=\frac{m}{V}[/tex]
[tex]\rho_b=\frac{0.004412}{0.0084}[/tex]
[tex]\rho_b=0.5252\ kg.m^{-3}[/tex]
Now the buoyant force on balloon
[tex]F_B=V(\rho_a-\rho_b).g[/tex]
[tex]F_B=0.0084(1.29-0.5252)\times 9.8[/tex]
[tex]F_B=0.063\ N[/tex]
∴Mass to be hung:
[tex]M=\frac{F_B}{g}[/tex]
[tex]M=0.00642432\ kg[/tex]
[tex]M=6.4243\ g[/tex]
A uniform thin circular ring rolls without slipping down an incline making an angle θ with the horizontal. What is its acceleration? (Enter the magnitude. Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)
Answer:[tex]a=\frac{g\sin \theta }{2}[/tex]
Explanation:
Given
inclination is [tex]\theta [/tex]
let M be the mass and r be the radius of uniform circular ring
Moment of Inertia of ring [tex]I=mr^2[/tex]
Friction will Provide the Torque to ring
[tex]f_r\times r=I\times \alpha [/tex]
[tex]f_r\times r=mr^2\times \alpha [/tex]
in pure Rolling [tex]a=\alpha r[/tex]
[tex]\alpha =\frac{a}{r}[/tex]
[tex]f_r=ma[/tex]
Form FBD [tex]mg\sin \theta -f_r=ma[/tex]
[tex]mg\sin \theta =ma+ma[/tex]
[tex]2ma=mg\sin \theta [/tex]
[tex]a=\frac{g\sin \theta }{2}[/tex]
Two UFPD are patrolling the campus on foot. To cover more ground, they split up and begin walking in different directions. Office A is walking at 5 mph directly south while Office B is walking at 6 mph directly west. How long would they need to walk before they are 2 miles away from each other?
Answer:
0.256 hours
Explanation:
Vectors in the plane
We know Office A is walking at 5 mph directly south. Let [tex]X_A[/tex] be its distance. In t hours he has walked
[tex]X_A=5t\ \text{miles}[/tex]
Office B is walking at 6 mph directly west. In t hours his distance is
[tex]X_B=6t\ \text{miles}[/tex]
Since both directions are 90 degrees apart, the distance between them is the hypotenuse of a triangle which sides are the distances of each office
[tex]D=\sqrt{X_A^2+X_B^2}[/tex]
[tex]D=\sqrt{(5t)^2+(6t)^2}[/tex]
[tex]D=\sqrt{61}t[/tex]
This distance is known to be 2 miles, so
[tex]\sqrt{61}t=2[/tex]
[tex]t =\frac{2}{\sqrt{61}}=0.256\ hours[/tex]
t is approximately 15 minutes
Astronomers discover an exoplanet, a planet obriting a star other than the Sun, that has an orbital period of 3.27 Earth years in a circular orbit around its star, which has a measured mass of 3.03×1030 kg . Find the radius of the exoplanet's orbit.
Answer:
r = 3.787 10¹¹ m
Explanation:
We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration
F = ma
G m M / r² = m a
The centripetal acceleration is given by
a = v² / r
For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship
v = d / t
The distance traveled Esla orbits, in a circle the distance is
d = 2 π r
Time in time to complete the orbit, called period
v = 2π r / T
Let's replace
G m M / r² = m a
G M / r² = (2π r / T)² / r
G M / r² = 4π² r / T²
G M T² = 4π² r3
r = ∛ (G M T² / 4π²)
Let's reduce the magnitudes to the SI system
T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)
T = 1.03 10⁸ s
Let's calculate
r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]
r = ∛ (21.44 10³⁵ / 39.478)
r = ∛(0.0543087 10 36)
r = 0.3787 10¹² m
r = 3.787 10¹¹ m
A rectangular loop of wire with width w = 5 cm, length L = 10cm, mass m = 40 g, and resistance R = 20 mΩ has an initial velocity v0 = 1 m/s to the right. It crosses from a region with zero magnetic field to a region with B = 2T pointing out of the page. How far does the loop penetrate into the magnetic field?
Answer:
The loop penetrate 4 cm into the magnetic field.
Explanation:
Given that,
Width w= 5 cm
Length L= 10 cm
mass m = 40 g
Resistance R = 20 mΩ
Initial velocity = 1 m/s
Magnetic field = 2 T
We need to calculate the induced emf
Using formula of emf
[tex]\epsilon=v_{0}Bw[/tex]
Put the value into the formula
[tex]\epsilon =1\times2\times5\times10^{-2}[/tex]
[tex]\epsilon =10\times10^{-2}\ volt[/tex]
We need to calculate the current
Using Lenz's formula
[tex]i=\dfrac{\epsilon}{R}[/tex]
[tex]i=\dfrac{10\times10^{-2}}{20\times10^{-3}}[/tex]
[tex]i=5\ A[/tex]
We need to calculate the force
Using formula of force
[tex]F=i(\vec{w}\times\vec{B})[/tex]
[tex]F=iwB[/tex]
Put the value into the formula
[tex]F=5\times5\times10^{-2}\times2[/tex]
[tex]F=0.5\ N[/tex]
We need to calculate the acceleration
Using formula of acceleration
[tex]a=\dfrac{F}{m}[/tex]
Put the value in to the formula
[tex]a=\dfrac{0.5}{40\times10^{-3}}[/tex]
[tex]a=12.5\ m/s^2[/tex]
We need to calculate the distance
Using equation of motion
[tex]v^2=u^2+2as[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{0-1^2}{2\times(-12.5)}[/tex]
[tex]s=0.04\ m[/tex]
[tex]s=4\ cm[/tex]
Hence, The loop penetrate 4 cm into the magnetic field.
Sound level B in decibels is defined as
B= 10 log (i/i)
where i = 1 × 10-12 W/m2 . The decibel
scale intensity for busy traffic is 80 dB. Two
people having a loud conversation have a deci-
bel intensity of 70 dB.
What is the approximate combined sound
intensity?
Answer in units of W/m2
Answer:
The approximate combined sound intensity is [tex]I_{T}=1.1\times10^{-4}W/m^{2}[/tex]
Explanation:
The decibel scale intensity for busy traffic is 80 dB. so intensity will be
[tex]10log(\frac{I_{1}}{I_{0}} )=80[/tex], therefore [tex]I_{1}=1\times10^{8}I_{0}=1\times10^{8} * 1\times10^{-12}W/m^{2}=1\times10^{-4}W/m^{2}[/tex]
In the same way for the loud conversation having a decibel intensity of 70 dB.
[tex]10log(\frac{I_{2}}{I_{0}} )=70[/tex], therefore [tex]I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}[/tex]
Finally we add both of them [tex]I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}[/tex], is the approximate combined sound intensity.
Identify the procedure to determine a formula for self-inductance, or inductance for short. Using the formula derived in the text, find the inductance in henries for a solenoid with 900 loops of wire wound on a rod 9 cm long with radius 3 cm.
Answer:
L = 0.0319 H
Explanation:
Given that,
Number of loops in the solenoid, N = 900
Radius of the wire, r = 3 cm = 0.03 m
Length of the rod, l = 9 cm = 0.09 m
To find,
Self inductance in the solenoid
Solution,
The expression for the self inductance of the solenoid is given by :
[tex]L=\dfrac{\mu_o N^2 A}{l}[/tex]
[tex]L=\dfrac{4\pi\times10^{-7}\times(900)^{2}\times\pi(0.03)^{2}}{0.09}[/tex]
L = 0.0319 H
So, the self inductance of the solenoid is 0.0319 henries.
A thin-walled, hollow sphere of mass M rolls without slipping down a ramp that is inclined at an angle β to the horizontal. Find the magnitude of the acceleration of the sphere along the ramp. Express your answer in terms of β and acceleration due to gravity g.
Answer:
Explanation:
Given
inclination [tex]=\beta [/tex]
Assuming radius of sphere is r
Now from Free Body Diagram
[tex]Mg\sin \theta -f_r=Ma[/tex]
where [tex]f_r=friction\ force[/tex]
[tex]a=acceleration\ of\ system[/tex]
Now friction will Provide the Torque
[tex]f_r\times r=I\cdot \alpha [/tex]
where [tex]I=moment\ of\ inertia[/tex]
[tex]\alpha =angular\ acceleration [/tex]
[tex]f_r\times r=\frac{2}{3}Mr^2\times \frac{a}{r}[/tex]
in pure rolling [tex]a=\alpha r[/tex]
[tex]f_r=\frac{2}{3}Ma[/tex]
[tex]mg\sin \beta -\frac{2}{3}Ma=Ma[/tex]
[tex]Mg\sin \beta =\frac{5}{3}Ma[/tex]
[tex]a=\frac{3g\sin \beta }{5}[/tex]