Answer:
14.43° or 0.25184 rad
Explanation:
v = Speed of sound in air = 343 m/s
f = Frequency = 1240 Hz
d = Width in doorway = 1.11 m
Wavelength is given by
[tex]\lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{343}{1240}\\\Rightarrow \lambda=0.2766\ m[/tex]
In the case of Fraunhofer diffraction we have the relation
[tex]dsin\theta=\lambda\\\Rightarrow \theta=sin^{-1}\frac{\lambda}{d}\\\Rightarrow \theta=sin^{-1}\frac{0.2766}{1.11}\\\Rightarrow \theta=14.43^{\circ}\ or\ 0.25184\ rad[/tex]
The minimum angle relative to the center line perpendicular to the doorway will someone outside the room hear no sound is 14.43° or 0.25184 rad
You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,500 N across the carpet to a spot 5 m away on the opposite wall. Hoping to just slide your dresser easily across the floor, you do not empty your clothes out of the drawers before trying to move it. You push with all your might but cannot move the dresser before becoming completely exhausted. How much work do you do on the dresser?
Answer:
0 Joules
Explanation:
The work done is given by
[tex]W=F\times s\times cos\theta[/tex]
where,
F = Force applied
s = Displacement of the object = 0 m
[tex]\theta[/tex] = Angle between the force applied and the horizontal = 0
[tex]W=F\times 0\times cos0\\\Rightarrow W=0\ J[/tex]
Work is only observed when there is a displacement.
The work done by me is 0 Joules as I was unable to move it.
A 1400 kg car moving at 6.6 m/s is initially traveling north in the positive y direction.
After completing a 90° right-hand turn to the positive x direction in 4.9 s, the inattentive operator drives into a tree, which stops the car in 430 ms.
What is the magnitude of the impulse on the car :
(a) due to the turn and (b) due to the collision?
What is the magnitude of the average force that acts on the car (c) during the turn and (d) during the collision?
(e) What is the angle between the average force in (c) and the positive x direction?
Answer:
a) J = 13067 kg*m/s
b) J = 9240 kg*m/s
c) F = 2666.73 N
d) F = 21488.37 N
e) 135°
Explanation:
We know that the impulse could be calculated by:
J = ΔP
where ΔP is the change in the linear momentum:
ΔP = [tex]P_f -P_i[/tex]
Also:
P = MV
Where M is the mass and V is the velocity.
so:
J= [tex]MV_f-MV_i[/tex]
where [tex]V_f[/tex] is the final velocity and [tex]V_i[/tex] is the inicial velocity
a)The impluse from turn is:
[tex]J_x=MV_{fx}-MV_{ix}[/tex]
[tex]J_y=MV_{fy}-MV_{iy}[/tex]
On the turn, [tex]V_{ix}=0[/tex] and [tex]V_{fy}=0[/tex], the magnitude of the impulse on direction x and y are:
[tex]J_x=9240 kg*m/s[/tex]
[tex]J_y=-9240 kg*m/s[/tex]
So, using pythagoras theorem the magnitude of the impulse is:
J = [tex]\sqrt{9240^2+9240^2}[/tex]
J = 13067 kg*m/s
b) The impluse from the collision is:
[tex]J=MV_{f}-MV_{i}[/tex]
J = 0 - (1400)(6.6)
J = 9240 kg*m/s
c) Using the next equation:
FΔt = J
where F is the force, Δt is the time and J is the impulse.
Replacing J by the impulse due to the turn, Δt by 4.9s and solving for F we have that:
F = J / Δt
F = 13067 / 4.9 s
F = 2666.73 N
d) At the same way, replacing J by the impulse during the collision, Δt by 0.43s and solving for F we have that:
F = J / Δt
F = 9240 / 0.43
F = 21488.37 N
e) The force have the same direction than the impulse due to the turn, Then, if the impulse have a direction of -45°, the force have -45° or 135°
Two spheres have the same radius and equal mass. One sphere is solid, and the other is hollow and made of a denser material. Which one has the bigger moment of inertia about an axis through its center?
a. Both the same.
b. The solid one.
c. The hollow one.
Answer:
Explanation:
Two sphere have same mass and radius but one is solid and another is hollow
Moment of inertia is the distribution of mass about the axis of rotation
For Hollow sphere more mass is at outside therefore its moment of inertia is more than solid sphere
[tex]I_{hollow}=\frac{2}{3}mr^2[/tex]
[tex]I_{solid}=\frac{2}{5}mr^2[/tex]
The force exerted by an electric charge at the origin on a charged particle at a point (x, y, z) with position vector r = x, y, z is F(r) = Kr/|r|3 where K is a constant. Find the work done as the particle moves along a straight line from (2, 0, 0) to (2, 4, 5).
Answer:
Explanation:
F(r) = Kr/|r|3
= k / r²
Work done
= ∫F(r) dr
= ∫ k / r² dr
limit from
r₁ = 2 to r₂ = √ ( 2² + 4² + 5² ) = 6.7
Work done
∫ k / r² dr
[ - k / r ]
Taking limit from 2 to 6.7
Work done = K ( - 1 / 6.7 + 1 / 2 )
= 0.35 K
The work done as a particle moves along a straight line from one point to another in the presence of an electric field is 0.
To find the work done as a particle moves along a straight line from one point to another in the presence of an electric field, you can use the following formula for the work done:
[tex]\[W = \int_{\mathbf{r_1}}^{\mathbf{r_2}} \mathbf{F} \cdot d\mathbf{r}\][/tex]
Where:
- [tex]\(W\)[/tex] is the work done.
- [tex]\(\mathbf{F}\)[/tex] is the force acting on the particle.
- [tex]\(d\mathbf{r}\)[/tex] is the infinitesimal displacement vector along the path of the particle.
- [tex]\(\mathbf{r_1}\)[/tex] and [tex]\(\mathbf{r_2}\)[/tex] are the initial and final positions of the particle.
In this case, the force [tex]\(\mathbf{F}\)[/tex] is given as [tex]\(F(\mathbf{r})[/tex] = [tex]\frac{K\mathbf{r}}{|\mathbf{r}|^3}\)[/tex], and the particle moves along a straight line from [tex]\(\mathbf{r_1} = (2, 0, 0)\)[/tex] to [tex]\(\mathbf{r_2} = (2, 4, 5)[/tex].
We can rewrite [tex]\(\mathbf{F}[/tex] as a vector:
[tex]\[\mathbf{F} = \frac{K}{|\mathbf{r}|^3}\mathbf{r}\][/tex]
Now, let's calculate the differential displacement [tex]\(d\mathbf{r}\)[/tex] along the path from [tex]\(\mathbf{r_1}\)[/tex] to [tex]r_2[/tex]:
[tex]\[d\mathbf{r} = (dx, dy, dz)\][/tex]
Now, let's calculate the dot product [tex]\(\mathbf{F} \cdot d\mathbf{r}\)[/tex]:
[tex]\[\mathbf{F} \cdot d\mathbf{r} = \frac{K}{|\mathbf{r}|^3}\mathbf{r} \cdot (dx, dy, dz)\][/tex]
Since the path is a straight line, [tex]\(\mathbf{r}\) and \(d\mathbf{r}\)\\[/tex] are parallel, so[tex]\(\mathbf{r} \cdot d\mathbf{r} = |\mathbf{r}||d\mathbf{r}|\cos(0^\circ) = |\mathbf{r}||d\mathbf{r}|\)[/tex].
Therefore, we have:
[tex]\[\mathbf{F} \cdot d\mathbf{r} = \frac{K}{|\mathbf{r}|^3}|\mathbf{r}||d\mathbf{r}| = \frac{K}{|\mathbf{r}|^2}|d\mathbf{r}|\][/tex]
Now, we can integrate [tex]\(\mathbf{F} \cdot d\mathbf{r}\)[/tex] from [tex]\(\mathbf{r_1}\)[/tex] to [tex]\(\mathbf{r_2}\)[/tex] along the straight line path:
[tex]\[W = \int_{\mathbf{r_1}}^{\mathbf{r_2}} \frac{K}{|\mathbf{r}|^2}|d\mathbf{r}|\][/tex]
The integration limits and the path are straight, so this simplifies to:
[tex]\[W = K \int_{2}^{2} \frac{1}{(2^2+4^2+5^2)^{3/2}}|d\mathbf{r}|\][/tex]
Since the particle moves from [tex]\(\mathbf{r_1}\) to \(\mathbf{r_2}\)[/tex], the displacement [tex]\(|d\mathbf{r}| = |\mathbf{r_2} - \mathbf{r_1}| = |(0, 4, 5)| = \sqrt{0^2 + 4^2 + 5^2} = \sqrt{41}\)[/tex].
So, the work done is:
[tex]\[W = K \int_{2}^{2} \frac{1}{(2^2+4^2+5^2)^{3/2}}\sqrt{41} dr\][/tex]
Since the limits of integration are the same, the integral evaluates to zero. Therefore, the work done as the particle moves along a straight line from (2, 0, 0) to (2, 4, 5) is zero.
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g An object with mass m=2 kg is completely submerged, and tethered, to the bottom of a large body of water. If the density of the water is rhow = 1000 kg/m3and the density of the object is rhoob j=500 kg/m3, find the tension in the rope. Take g=10 m/s2and assume the object has a uniform mass density
Answer:
The tension in the rope is 20 N
Solution:
As per the question:
Mass of the object, M = 2 kg
Density of water, [tex]\rho_{w} = 1000\ kg/m^{3}[/tex]
Density of the object, [tex]\rho_{ob} = 500\kg/m^{3}[/tex]
Acceleration due to gravity, g = [tex]10\ m/s^{2}[/tex]
Now,
From the fig.1:
'N' represents the Bouyant force and T represents tension in the rope.
Suppose, the volume of the block be V:
V = [tex]\frac{M}{\rho_{ob}}[/tex] (1)
Also, we know that Bouyant force is given by:
[tex]N = \rho_{w}Vg[/tex]
Using eqn (1):
[tex]N = \rho_{w}\frac{M}{\rho_{ob}}g[/tex]
[tex]N = 1000\frac{2}{500}\times 10 = 40\ N[/tex]
From the fig.1:
N = Mg + T
40 = 2(10) + T
T = 40 - 20 = 20 N
[tex]N = \rho_{w}Vg[/tex]
A square plate has mass 0.600 kg and sides of length 0.150 m. It is free to rotate without friction around an axis through its center and perpendicular to the plane of the plate. How much work must you do on the plate to change its angular speed from 0 to 40.0 rad/s?
Answer:
W = 1.8 J
Explanation:
given,
mass of square plate = 0.6 Kg
length of side of square plate = 0.15 m
speed is change from 0 to 40.0 rad/s
work done = ?
moment of inertia through the plane perpendicular to plane
[tex]I = \dfrac{Mr^2}{6}[/tex]
r will be equal to 0.15 m
[tex]I = \dfrac{0.6 \times 0.15^2}{6}[/tex]
[tex]I =0.00225\ kgm^2[/tex]
work done is equal to change in kinetic energy
[tex]W = \dfrac{1}{2}I(\omega_f^2-\omega_i^2)[/tex]
[tex]W = \dfrac{1}{2}\times 0.00225(40^2-0^2)[/tex]
W = 1.8 J
work done when speed changes from 0 to 40 rad/s is equal to W = 1.8 J
. A satellite S travels in a circular orbit around the earth. A rocket is located at the apogee of its elliptical orbit for which = 0.58. (a) Determine the sudden change in speed that must occur at A so that the rocket can enter the satellite’s orbit while in free flight along the blue elliptical trajectory. (b) When it arrives at B, determine the sudden adjustment in speed that must be given to the rocket in order to maintain the circular orbit.
so it is not bad not good. pefectly balanced just as all things should be
The radius of the orbits and the rocket's mass, The changes would involve transitioning from the elliptical to the circular orbit, taking into account Earth's gravitational pull.
Explanation:This question delves into detail regarding the physics of celestial movements, in particular, satellites and their orbits. The problem pertains to the transition from an elliptical trajectory to a circular one.
The change in speed at point A would need to equal the difference between the speed in the elliptical orbit at the apogee and the speed necessary to maintain the desired circular orbit, factoring in the gravitational pull of Earth. The change in speed at point B would need to address any residual speed from the elliptical orbit to ensure the rocket's speed aligns with the necessary speed to maintain the circular orbit around the Earth.
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An electric motor has an eccentric mass of 10 kg (10% of the total mass) and is set on two identical springs (k=3200 N/m). The motor runs at 1750 rpm, and the mass eccentricity is 100 mm from the center. The springs are mounted 250 mm apart with the motor shaft in the center. Neglect damping and determine the amplitude of vertical vibration.
Answer:
The amplitude is 0.010 m
Solution:
As per the question:
Eccentric Mass, [tex]m_{e} = 10\ kg[/tex]
[tex]m_{e}[/tex] = 10%M kg
0.1M = 10
Total mass, M = 100 kg
Spring constant, k = 3200 N/m
Eccentric center, e = 100 mm = 0.1 m
Speed of motor, N = 1750 rpm
Distance between two springs, d = 250 mm = 0.25 m
Now,
Angular velocity, [tex]\omega = \frac{2\pi N}{60} = \frac{2\pi \times 1750}{60} = 183.26\ rad/s[/tex]
For the vertical vibrations:
[tex]\omega_{n} = \sqrt{\frac{2k}{M}} = \sqrt{\frac{6400}{100}} = 8\ rad/s[/tex]
Now, the frequency ratio is given by:
r = [tex]\frac{\omega}{\omega_{n}} = \frac{183.26}{8} = 22.91[/tex]
Thus the amplitude is given by:
A = [tex]e^{\frac{m_{e}}{M}}.\frac{r^{2}}{|1 - r^{2}|}[/tex]
A = [tex]e^{\frac{10}{100}}.\frac{22.91^{2}}{|1 - 22.91^{2}|} = 0.010 m[/tex]
Final answer:
The amplitude of vertical vibration for the electric motor mounted on springs is 100 mm or 0.1 meters, considering the eccentric mass and neglecting damping.
Explanation:
The subject of this question is Physics, and it pertains to a college-level study of mechanics in the context of oscillations and waves, specifically relating to a system experiencing simple harmonic motion (SHM). The motor with an eccentric mass which causes vertical vibrations can be modeled as a mass-spring system, characteristic of simple harmonic oscillators.
Given the mass and the spring constant, we can calculate the natural angular frequency (not represented in this problem due to lack of damping) of this system. The driving frequency can also be converted to an angular frequency to check against the natural frequency. However, the question directly asks for the amplitude of vertical vibration, which in the undamped case of SHM, depends on the initial displacement of the mass from its equilibrium position. Since the question neglects damping, the amplitude can be considered equal to the eccentricity of the mass in the static case. Therefore, the amplitude of vertical vibration is 100 mm, or 0.1 meters.
An astronaut is being tested in a centrifuge. The centrifuge has a radius of 9 m and, in starting, rotates according to ? = 0.30t2, where t is in seconds and ? is in radians.
(a) When t = 9.0 s, what is the magnitude of the astronaut's angular velocity?
(b) When t = 9.0 s, what is the magnitude of the astronaut's linear speed?
(c) When t = 9.0 s, what is the magnitude of the astronaut's tangential acceleration?
(d) When t = 9.0 s, what is the magnitude of the astronaut's radial acceleration?
Answer:
5.4 rad/s
48.6 m/s
5.4 m/s²
262.44 m/s²
Explanation:
r = Radius of centrifuge = 9 m
[tex]\theta=0.3t^2[/tex]
Differentiating with respect to time
[tex]\frac{d\theta}{dt}=0.6t=\omega[/tex]
At t = 9 s
[tex]\omega=0.6\times 9=5.4\ rad/s[/tex]
Angular velocity is 5.4 rad/s
Linear speed
[tex]v=r\omega\\\Rightarrow v=9\times 5.4=48.6\ m/s[/tex]
The linear speed of the astronaut is 48.6 m/s
Differentiating [tex]\omega[/tex] with respect to time
[tex]\frac{d\omega}{dt}=0.6=\alpha[/tex]
Tangential acceleration is given by
[tex]a_t=\alpha r\\\Rightarrow a_t=0.6\times 9=5.4\ m/s^2[/tex]
Tangential acceleration of the astronaut is 5.4 m/s²
Radial acceleration is given by
[tex]a_r=\frac{v^2}{r}\\\Rightarrow a_r=\frac{48.6^2}{9}\\\Rightarrow a_r=262.44\ m/s^2[/tex]
The radial acceleration of the astronaut is 262.44 m/s²
A tightened string vibrates with a standing wave. Which of the following statements is correct? Group of answer choices Points on the string undergo the same displacement. Points on the string vibrates with different amplitude. Points on the string vibrates with different frequencies. Points on the string undergo the same speed. Points on the string vibrate with the same energy..
Nodes and antinodes on a vibrating string standing wave has different amplitudes.
Option B
Explanation:
For a standing wave produced from a tightened string vibration, points of maximum and minimum displacement are formed termed as nodes and antinodes. Hence (A) is false.The amplitudes at nodes are zero and increases gradually to the maximum when it approaches the antinode. Statement (B) is true. The string vibrates with a single natural frequency and a number of resonant frequencies. The nodes and antinodes have same frequencies. (C) is false.Velocity at the nodes is zero and it increases gradually at the antinodes. Statement (D) is false. Since nodes have a displacement equal to zero, no energy is present. All energy is confined in between two nodes. Energy at any point on the string is always constant and is not transferred.(E) is false.Final answer:
Points on a tightened string vibrating with a standing wave have different amplitudes, with nodes having no displacement and antinodes having the maximum. The frequency and speed of the wave are constant along the string, but energy varies with amplitude.
Explanation:
When a tightened string vibrates with a standing wave, points on the string vibrate with different amplitudes. This variation in amplitude can be seen by observing the points of maximum displacement (antinodes) and the points of no displacement (nodes). While points at the antinodes experience the greatest amplitude, the points at the nodes do not move at all. The frequency of oscillation is the same for all points on the string because a standing wave is established by the interference of waves traveling in opposite directions with the same frequency. Consequently, the speed of the wave along the string is also the same because it is determined by the tension and the linear mass density of the string which we assume are constant. However, the energy of the points is not the same due to the varying amplitudes; points at the antinodes have more energy as compared to those at the nodes.
A cylinder is closed by a piston connected to a spring of constant 2.40 103 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C.
(a) If the piston has a cross-sectional area of 0.0110 m2 and negligible mass, how high will it rise when the temperature is raised to 250°C? m
(b) What is the pressure of the gas at 250°C?
Answer:
a) The correct solution would be x=0.1544m.
b) [tex]P_f = P_i +\frac{Kx}{A}=101325Pa +\frac{2400N/m (0.1544m)}{0.011m^2}=135012.27 Pa[/tex]
Explanation:
Initial states
[tex]P_i =1atm=101325 Pa[/tex] represent the initial pressure
[tex]V_i =0.005m^3 = 5L[/tex] represent the initial volume
N represent the number of moles
[tex]T_i = 20+273.15=293.15K[/tex] represent the initial temperature
Final states
[tex]P_f[/tex] represent the final pressure
[tex]V_f[/tex] represent the final volume
[tex]T_f = 250+273.15=523.15K[/tex]
Part a
From the initial states and with the law of ideal gases we can find the moles like this
[tex]n=\frac{P_iV_i}{RT_i}=\frac{101325Pa(0.005m^3)}{8.314\frac{J}{mol K} 293.15K}=0.2079mol[/tex]
In order to find the final pressure we need to take in count the atmospheric pressure and the pressure related to the force that the springs applies to the piston, like this:
[tex]P_f =P_i +\frac{F}{A}= P_i +\frac{Kx}{A}[/tex]
Where x is the distance that the piston is displaced upward, since increase the temperature and the gas inside tends to expand.
And we can find the final volume on this way [tex]V_f = V_i + Ax[/tex]
For the final state we have the following equation:
[tex]P_f V_f=nR T_f [/tex]
And if we replace [tex]V_f[/tex] we got:
[tex](P_i +\frac{Kx}{A})(V_i +Ax)=nRT_f[/tex]
[tex]P_i V_i +P_i A x +\frac{KxV_i}{A}+kx^2=nRT_f[/tex]
And x can be solved with a quadratic equation:
[tex]kx^2+(P_i A +\frac{V_i K}{A})x+(P_i V_i-nRT_f)=0[/tex]
[tex]2400 k^2 +(101325Pa)(0.0110m^2)+\frac{0.005m^3 x2400N/m}{0.0110m^2})x+(101325Pa(0.005m^3)-0.2079mol(8.314J/molK)(523.15K))=0[/tex]
[tex]2400 k^2 +(2205.484)x-397.63=0[/tex]
And solving for x we got x=0.1544 or x=-1.07331 m. The correct solution would be x=0.1544m.
Part b
And the final pressure would be given by:
[tex]P_f = P_i +\frac{Kx}{A}=101325Pa +\frac{2400N/m (0.1544m)}{0.011m^2}=135012.27 Pa[/tex]
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 m/s. If the faucet and the nozzle 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?
I keep coming up with 4 cm. Please someone confirm my answer. If incorrect please explain why.
Answer:
D₂ = 2.738 cm
Explanation:
Continuity equation
The continuity equation is nothing more than a particular case of the principle of conservation of mass. It is based on the flow rate (Q) of the fluid must remain constant throughout the entire pipeline.
Since the flow rate is the product of the surface of a section of the duct because of the speed with which the fluid flows, we will have to comply with two points of the same pipeline:
Q = v*A : Flow Equation
where:
Q = Flow in (m³/s)
A is the surface of the cross sections of points 1 and 2 of the duct.
v is the flow velocity at points 1 and 2 of the pipe.
It can be concluded that since the flow rate must be kept constant throughout the entire duct, when the section decreases, the flow rate increases in the same proportion and vice versa.
Data
D₁= 6.0 cm : faucet diameter
v₁ = 5 m/s : speed of fluid in the faucet
v₂ = 20 m/s : speed of fluid in the nozzle
Area calculation
A = (π*D²)/4
A₁ = (π*D₁²)/4
A₂ = (π*D₂²)/4
Continuity equation
Q₁ = Q₂
v₁A₁ = v₂A₂
v₁(π*D₁²)/4 = v₂(π*D₂²)/4 : We divide by (π/4) both sides of the equation
v₁ (D₁)² = v₂(D₂)²
We replace data
6 *(5)² = 20*(D₂)²
150 = 20*(D₂)²
(150 /20) = (D₂)²
7.5 = (D₂)²
[tex]D_{2} = \sqrt{7.5}[/tex]
D₂ = 2.738 cm : nozzle diameter
A geothermal plant uses geothermal water (yes a liquid) extracted from beneath the Earth using drilling apparatus. The geothermal water (a condensed liquid) comes up at a temperature of 200 C at a rate of 200 kg/s. It is used to drive a turbine that extracts 6000 kw of power. The water leaves the turbine at 80 C and enters a lake at 25 C. What is the actual thermal efficiency of the turbine in percent to three significant digits
Answer:
4.1%
Explanation:
Given that water specific heat c = 4186 J/kgC. If water is flowing with mass rate of 200kg/s and temperature is dropping from 200C to 25C. Then the thermal energy rate from the water should be
[tex]P_i = mc\Delta t = 200*4186*(200 - 25) = 146510000 J/s[/tex] or 146510kW
Since the turbine is only able to extract 6000kW of power, then the thermal efficiency is:
[tex]Eff = \frac{P_o}{P_i} = \frac{6000}{146510} = 0.041[/tex] or 4.1%
A coil has N_a turns enclosing an area of A. In a physics laboratory experiment, the coilis rotated during the time interval Deltat from a position in which theplane of each turn is perpendicular to Earth's magnetic field toone in which the plane of each turn is parallel to the field. Themagnitude of Earth's magnetic field at the lab location isB.
a.)What is the total magnitude Phi_initial of the magneticflux through the coil before it is rotated?
b.)What is the magnitude Phi_final of the total magneticflux through the coil after it is rotated?
c.)What is the magnitude of the average emf induced in the coil?
Answer:
Answered
Explanation:
a) The magnitude of the magnetic flux through the coil before it is rotated is,
[tex]\phi_{initial}= BAsin\theta[/tex]
=BAsin90° = BA
therefore, the magnitude of the magnetic flux= BA
b) The magnitude of the magnetic flux through the coil after it is rotated is, Ф_final = BAsinθ
= BA sin0= 0
therefore, the magnitude of magnetic flux through the coil after it is rotated is, zero.
c) The magnitude of the average emf induced in the coil is,
[tex]\epsilon = N\frac{d\phi}{dt}[/tex]
[tex]\epsilon = N\frac{NBA}{\Delta t}[/tex]
because Ф = BA
A small ball of mass 62 g is suspended from a string of length 72 cm and whirled in a circle lying in the horizontal plane. r v 9.8 m/s 2 72 cm 62 g 32◦ If the string makes an angle of 32◦ with the vertical, find the centripetal force experienced by the ball. The acceleration of gravity is ____.
Answer:
Centripetal force, [tex]F_x=0.379\ N[/tex]
Explanation:
It is given that,
Mass of the ball, m = 62 g = 0.062 kg
Length of the string, l = 72 cm = 0.72 m
Angle, [tex]\theta=32^{\circ}[/tex]
To find,
The centripetal force experienced by the ball.
Solution,
According to the attached free body diagram,
[tex]T\ cos\theta-mg=0[/tex]
[tex]T=\dfrac{mg}{cos\theta}[/tex]
[tex]T=\dfrac{0.062\times 9.8}{cos(32)}[/tex]
T = 0.716 N
The centripetal force will be experienced by the ball in horizontal direction. It is equal to :
[tex]F_x=T\ sin\theta[/tex]
[tex]F_x=0.716\ N\ sin(32)[/tex]
[tex]F_x=0.379\ N[/tex]
So, the centripetal force experienced by the ball is 0.379 N.
The centripetal force experienced by the small ball if mass 62 g and attached to the string is 0.379 N.
What is centripetal force?Centripetal force is the force which is required to keep rotate a body in a circular path. The direction of the centripetal force is inward of the circle towards the center of rotational path.
Given information-
The mass of the small bag is 62 grams.
The length of the string is 72 cm.
Suppose T is the tension in the string, m is the mass of ball and g is the gravitational force.
The horizontal force applied on the body is,
[tex]\sum F_x=T\cos \theta-mg=0\\T=\dfrac{mg}{\cos \theta}[/tex]
The centripetal force applied on the body is,
[tex]F_c=T\sin \theta\\T=\dfrac{F_c}{\sin \theta}[/tex]
Compare both the values of the T, we get,
[tex]\dfrac{F_c}{\sin\theta}=\dfrac{mg}{\cos \theta}\\{F_c}=\dfrac{mg\times{\sin\theta}}{\cos \theta}\\{F_c}={mg\times{\tan\theta}}[/tex]
Put the values as,
[tex]F_c=0.062\times9.8\times\tan(32)\\F_c=0.379\rm N[/tex]
Thus, the centripetal force experienced by the small ball if mass 62 g and attached to the string is 0.379 N.
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A skateboarder is attempting to make a circular arc of radius r = 11 m in a parking lot. The total mass of the skateboard and skateboarder is m = 87 kg. The coefficient of static friction between the surface of the parking lot and the wheels of the skateboard is μs = 0.62 .
(a) What is the maximum speed, in meters per second, he can travel through the arc without slipping?
Answer:
V = 8.26 m/s
Explanation:
The sum of forces on the centripetal-axis is:
[tex]Ff = m*a_c[/tex]
Maximum speed happens when friction force is maximum, so:
[tex]\mu*N = m*V^2/R[/tex]
[tex]\mu*m*g = m*V^2/R[/tex]
[tex]V=\sqrt{R*\mu*g}[/tex]
V = 8.26 m/s
The maximum speed that he can travel through the arc without slipping is mathematically given as
v = 9.93m/s
The maximum speedQuestion Parameters:
Generally the equation for the Maximum velocity is mathematically given as
[tex]\frac{mv^2}{r} = \mu_s mg[/tex]
Therefore
[tex]v = \sqrt{(0.63)(9.8)(16)}[/tex]
v = 9.93m/s
Therefore, the maximum speed that he can travel through the arc without slipping is
v = 9.93m/s
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A bowling ball encounters a 0.760-m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 8.57 m/s at the bottom of the rise. Find the translational speed at the top.
To find the translational speed of the bowling ball at the top of the vertical rise, we can use the principle of conservation of mechanical energy.
Explanation:To find the translational speed of the bowling ball at the top of the vertical rise, we can use the principle of conservation of mechanical energy. Initially, the ball has kinetic energy due to its translational speed. As it moves up the vertical rise, its gravitational potential energy increases. Since there is no friction, we can assume there is no loss of energy.
Using the equation for conservation of mechanical energy:
Initial kinetic energy = final potential energy
0.5 * m * v^2 = m * g * h
Where m is the mass of the ball, v is the translational speed at the bottom, g is the acceleration due to gravity, and h is the height of the vertical rise.
Plugging in the given values:
0.5 * m * (8.57 m/s)^2 = m * 9.8 m/s^2 * 0.760 m
Simplifying and solving for the mass:
m = (9.8 m/s^2 * 0.760 m) / (0.5 * (8.57 m/s)^2)
Finally, we can plug the mass back into the initial equation to find the translational speed at the top of the rise.
7. A stick of length L and mass M is hanging at rest from its top edge from a ceiling hinged at that point so that it is free to rotate. Find the distance from the top of the stick where an impulse, FΔt, is applied such that there is no horizontal component to the force of the hinge on the stick. This point is called the center of percussion.
Answer:
The distance from the top of the stick would be 2l/3
Explanation:
Let the impulse 'FΔt' acts as a distance 'x' from the hinge 'H'. Assume no impulsive reaction is generated at 'H'. Let the angular velocity of the rod about 'H' just after the applied impulse be 'W'. Also consider that the center of percussion is the point on a bean attached to a pivot where a perpendicular impact will produce no reactive shock at the pivot.
Applying impulse momentum theorem for linear momentum.
FΔt = m(Wl/2), since velocity of center of mass of rod = Wl/2
Similarly applying impulse momentum theorem per angular momentum about H
FΔt * x = I * W
Where FΔt * x represents the impulsive torque and I is the moment of inertia
F Δt.x = (ml² . W)/3
Substituting FΔt
M(Wl/2) * x = (ml². W)/3
1/x = 3/2l
x = 2l/3
A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subject to some friction, so it gradually slows down. In the 10-s period following the inital spin, the bike wheel undergoes 65.0 complete rotations. Assuming the frictional torque remains constant, how much more time delta ts will it take the bike wheel to come to a complete stop? The bike wheel has a mass of 0.725 kg and a radius of 0.315 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque τf that was acting on the spinning wheel.
After 10 seconds, the bike wheel will take roughly 15.98 seconds to come to a complete halt. The frictional torque operating on the spinning bike wheel is -0.112 N*m.
Explanation:The process to solve this problem involves several steps. The first one is to determine the initial and final angular velocities of the wheel. Angular velocities are usually measured in radians per second, but we have them in rotations per second. So, firstly these values need to be converted by multiplying the number of rotations by 2π radians. The initial angular velocity, ωi, hence becomes
9 * 2π = 56.55 rad/s,
and after 10 seconds, the final angular velocity, ωf, is
65 / 10 * 2π = 40.84 rad/s.
Now, we already know 10 seconds.
Using these values, the angular acceleration (α) can be calculated as
(ωf - ωi) / Δt = (40.84 rad/s - 56.55 rad/s) / 10 s = -1.571 rad/s².
Note that this is negative because it's a deceleration. We're also given the wheel's mass (m) and radius (r), so the moment of inertia (I), which can be calculated as
I = m*r², will be 0.725 kg * (0.315 m)² = 0.071 kg*m².
From here, frictional torque (τf), which is the product of the moment of inertia and angular acceleration, can be found as
I * α = 0.071 kg*m² * -1.571 rad/s² = -0.112 N*m.
The negative sign just indicates the torque acts in the direction to slow down the wheel.
Finally, to find the time Δts it takes to come to rest, we can use the equation for angular velocity:
ω = ωi + α * t.
Setting ω to 0 (as the wheel stops) and solving for t gives us:
t = (0 - ωi) / α = - (40.84 rad/s) / -1.571 rad/s² = 25.98 s.
But our Δts is the time after the first 10 seconds, so Δts = t - 10 s = 15.98 s.
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You place three nonflammable objects in a fire. They are identical in shape and size, but one object is black, the second is white, and the third is shiny silver. After a few minutes, all three objects are at the same temperature: 1800° C. They remain solid and are now glowing with thermal radiation. Which one is glowing most brightly? The first object (black). They are all glowing with equal brightness. The third object (silver). The second object (white).
Answer:
The first object (black)
Explanation:
The thermal radiation of a body is governed by the Stefan Boltzmann Law, which is mathematically given as:
[tex]E=\epsilon. \sigma.T^4[/tex]
where:
[tex]\sigma =5.67\times 10^{-8}\ W.m^{-2}.K^{-4}[/tex] (Stefan Boltzmann constant)
Also the absorptivity and emissivity of a perfectly black body is maximum ( =1 ) therefore it absorbs all the heat energy incident on it and glows the brightest due to the perfect emission.So here also the black object will glow brighter than the others.
Answer:
The guy above me is definitely correct just read it.
Explanation:
it is what it is.
Blood takes about 1.65 s to pass through a 2.00 mm long capillary. If the diameter of the capillary is 5.00 μm and the pressure drop is 2.85 kPa , calculate the viscosity of blood. Assume laminar flow.
Answer:
Viscosity of blood will be [tex]918.54\times 10^{-6}Pa-sec[/tex]
Explanation:
We have given time to pass the capillary = 1.65 sec
Length of capillary L = 2 mm = 0.002 m
Pressure drop [tex]\Delta P=2.85kPa=2.85\times 10^3Pa[/tex]
Diameter [tex]d=5\mu m=5\times 10^{-6}m[/tex]
So radius [tex]r=\frac{5\times 10^{-6}}{2}=2.5\times 10^{-6}m[/tex]
Velocity will be [tex]v=\frac{L}{t}=\frac{0.002}{1.65}=1.212\times 10^{-3}m/sec[/tex]
We know that viscosity is given by [tex]\eta =\frac{r^2\Delta P}{8Lv}=\frac{(2.5\times 10^{-6})^2\times 2.85\times 10^3}{8\times 0.002\times 1.212\times 10^{-3}}=918.54\times 10^{-6}Pa-sec[/tex]
Viscosity of blood will be [tex]918.54\times 10^{-6}Pa-sec[/tex]
The viscosity of the blood flowing in the given capillary is [tex]9.27 \times 10^{-4} \ Pa.s[/tex]
The given parameters;
time taken for the blood to pass, t = 1.65 sdiameter of the capillary, d = 5 μmlength of the capillary, l = 2.00 mmpressure drop, ΔP = 2.85 kPaThe velocity of the blood is calculated as follows;
[tex]v = \frac{L}{t} = \frac{2\times 10^{-3} \ m}{1.65 \ s} = 0.0012 \ m/s[/tex]
The radius of the capillary is calculated as follows;
[tex]r = \frac{d}{2} \\\\r = \frac{5\times 10^{-6}}{2} = 2.5 \times 10^{-6} \ m[/tex]
Assuming laminar flow, the viscosity of the blood is calculated as;
[tex]\mu = \frac{r^2 \times \Delta P}{8lv} \\\\\mu = \frac{(2.5\times 10^{-6})^2 \times 2850}{8 \times 2\times 10^{-3} \times 0.0012} \\\\\mu= 9.27 \times 10^{-4} \ Pa.s[/tex]
Thus, the viscosity of the blood flowing in the given capillary is [tex]9.27 \times 10^{-4} \ Pa.s[/tex]
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A ball is thrown straight up. At what point does the ball have the most energy? Ignore air resistance.
a. At the highest point of its path.
b. When it is first thrown.
c. Just before it hits the ground.
d. Everywhere; the energy of the ball is the same at all of these points.
e. When the ball is halfway to the highest point of its path.
Answer:
Everywhere; the energy of the ball is the same at all of these points.
Explanation:
A ball is thrown straight up. The total energy of an object remains constant as per the law of conservation of energy. At the highest point the object have only potential energy and at lowest point it have only kinetic energy. As a result, the total energy remains constant.
So, at every point the energy of the ball is the same at all of these points. Hence, the correct option is (d).
The ball has the most kinetic energy when it is first thrown. As it rises, its kinetic energy decreases while its potential energy increases, but the total energy is conserved and remains constant unless air resistance is factored in. So the correct option is b.
Explanation:The answer is b. When it is first thrown. The total energy of the ball in this context is its kinetic energy (energy of motion) and potential energy (energy of position). At the moment the ball is thrown, it has the maximum kinetic energy and zero potential energy. As it rises, its kinetic energy decreases while its potential energy increases, but the total energy (kinetic plus potential) is conserved and remains constant if air resistance is ignored. So, the ball doesn't really have 'the most energy' at any one point, but it has the maximum kinetic energy when it is first thrown, which is probably what the question is asking.
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In order to rendezvous with an asteroid passing close to the earth, a spacecraft must be moving at 8.50×103m/s relative to the earth at a distance 2.50×108m from the center of the earth. At what speed must this spacecraft be launched from the earth's surface? Neglect air resistance and the gravitational pull of the moon.
Answer:
v₀ = 13.9 10³ m / s
Explanation:
Let's analyze this exercise we can use the basic kinematics relationships to love the initial velocity and the acceleration we can look for from Newton's second law where force is gravitational attraction.
F = m a
G m M / x² = m dv / dt = m dv/dx dx/dt
G M / x² = dv/dx v
GM dx / x² = v dv
We integrate
v² / 2 = GM (-1 / x)
We evaluate between the lower limits where x = Re = 6.37 10⁶m and the velocity v = vo and the upper limit x = 2.50 10⁸m with a velocity of v = 8.50 10³ m/s
½ ((8.5 10³)² - v₀²) = GM (-1 /(2.50 10⁸) + 1 / (6.37 10⁶))
72.25 10⁶ - v₀² = 2 G M (+0.4 10⁻⁸ - 1.57 10⁻⁷)
72.25 10⁶ - v₀² = 2 6.63 10⁻¹¹ 5.98 10²⁴ (-15.3 10⁻⁸)
72.25 10⁶ - v₀² = -1.213 10⁸
v₀² = 72.25 10⁶ + 1,213 10⁸
v₀² = 193.6 10⁶
v₀ = 13.9 10³ m / s
A flat, rectangular coil consisting of 60 turns measures 23.0 cm by 31.0 cm . It is in a uniform, 1.20-T, magnetic field, with the plane of the coil parallel to the field. In 0.210 s , it is rotated so that the plane of the coil is perpendicular to the field.Part A What is the change in the magnetic flux through the coil due to this rotation?Part B Find the magnitude of the average emf induced in the coil during this rotation.
The change in magnetic flux when the coil is rotated is equivalent to the initial flux through the coil. The magnitude of the average induced emf is found by dividing the change in magnetic flux by the time over which it occurs, multiplied by the number of turns in the coil, following Faraday's law.
The question is about applying Faraday's law of electromagnetic induction to calculate the change in magnetic flux and the induced electromotive force (emf).
Part A: Change in Magnetic Flux
The magnetic flux \\(\Phi\\) through a coil is given by \\(\Phi = B \cdot A \cdot \cos(\theta)\\), where \\(B\\) is the magnetic field strength, \\(A\\) is the area of the coil, and \\(\theta\\) is the angle between the field and the normal to the coil's plane. Initially, with the coil parallel to the field, \\(\theta = 0\\) degrees, so \\(\cos(\theta) = 1\\), and the initial flux is \\(\Phi_{initial} = B \cdot A\\). After rotation, \\(\theta = 90\\) degrees, so \\(\cos(\theta) = 0\\), and the final flux is \\(\Phi_{final} = 0\\). The change in flux is the final minus the initial. Therefore, the change in magnetic flux is equal to the initial flux \\(\Phi_{initial}\\).
Part B: Magnitude of the Average Induced Emf
To find the magnitude of the average induced emf, Faraday's law is used, which states that the induced emf is proportional to the rate of change of magnetic flux. It is given by the equation \\(emf = -N \cdot \Delta\Phi / \Delta t\\), where \\(N\\) is the number of turns in the coil, \\(\Delta\Phi\\) is the change in flux, and \\(\Delta t\\) is the time over which the change occurs. Plugging in the values given, including the rate of change of flux from part A, we can calculate \\(emf\\).
At a height of fourteen meters above the surface of a freshwater lake, a sound pulse is generated. The echo from the bottom of the lake returns to the point of origin 0.107 s later. The air and water temperature is 20 °C. How deep is the lake?
To solve this problem it is necessary to resort to concepts related to the kinematic equations of movement description which define speed as
[tex]v = \frac{x}{t}[/tex]
Where,
x = Distance
t=Time
Re-arrange to find the time we have,
[tex]t = \frac{x}{v}[/tex]
At 20°C the speed of sound in air is 343.2m/s and the speed of sound in water is 1484.3m/s
The time in which the sound lasts in traveling in the air would be given by the shipping and return (even when it touches the water) which is 28m (14 way and 14 back), in other words
[tex]t_a = \frac{28}{343.2} = 0.08158s[/tex]
In the case of the lake, the same logic is followed, only that there is no net distance here, therefore
[tex]t_w = \frac{2d}{1484.3}[/tex]
The total time would be given by
[tex]t_T = 0.08158 + \frac{2d}{1484.3}[/tex]
The total time is given in the statement so we clear to find d
[tex]0.108-0.08158= \frac{2d}{1484.3}[/tex]
[tex]d = \frac{1484.3(0.108-0.08158)}{2}[/tex]
[tex]d = 18.86m[/tex]
Therefore the lake has a depth of 18.86 m
Because of its metabolic processes, your body continually emits thermal energy. Suppose that the air in your bedroom absorbs all of this thermal energy during the time you sleep at night (8.0 hours). Assume your metabolic rate to be P = 120 J/s. Estimate the temperature change you expect in this air. Let the volume of your room be 36 m2 and Heat capacity of air is 1000 J/kg C.
Answer:
77.83783 °C
Explanation:
V = Volume of room = 36 m²
c = Heat capacity = 1000 J/kg C
P = Power = 120 J/s = 120 W
t = Time = 8 hours
[tex]\rho[/tex] = Density of air = 1.225 kg/m³
[tex]\Delta T[/tex] = Change in temperature
Density is given by
[tex]\rho=\frac{m}{V}\\\Rightarrow m=\rhoV\\\Rightarrow m=1.225\times 36\\\Rightarrow m=44.1\ kg[/tex]
Heat is given by
[tex]Q=mc\Delta T\\\Rightarrow Pt=mc\Delta T\\\Rightarrow \Delta T=\frac{Pt}{mc}\\\Rightarrow \Delta T=\frac{120\times 8\times 3600}{44.4\times 1000}\\\Rightarrow \Delta T=77.83783^{\circ}C[/tex]
The temperature change you expected in this air 77.83783 °C
Final answer:
The estimated temperature change in the air of your bedroom while you sleep is approximately 78.3 °C.
Explanation:
In order to estimate the temperature change in the air of your bedroom while you sleep, we need to calculate the amount of thermal energy emitted by your body and the amount of energy absorbed by the air. Your metabolic rate is given as P = 120 J/s. Assuming you sleep for 8.0 hours, the total energy emitted is 120 J/s * 8 hours * 3600 seconds/hour = 3456000 J.
The specific heat capacity of air is given as 1000 J/kg C and the volume of your room is 36 m³. Assuming the density of air is 1.225 kg/m³, the mass of the air in your room is 1.225 kg/m³ * 36 m³ = 44.1 kg.
Using the equation Q = mcΔθ, where Q is the thermal energy absorbed/lost, m is the mass, c is the specific heat capacity, and Δθ is the change in temperature, we can calculate the temperature change as follows:
Q = mcΔθ
Δθ = Q / mc
Δθ = 3456000 J / (44.1 kg * 1000 J/kg C)
Δθ ≈ 78.3 °C
Therefore, we can estimate that the temperature change in the air of your bedroom is approximately 78.3 °C.
The brass pendulum of a particular grandfather clock has a mass of 2 kilograms and completes its swing once every 2 seconds.
If you were to replace the brass pendulum with an identically shaped gold pendulum having a mass of 4 kilograms, it would complete its swing once every
2 seconds.
4 seconds.
8 seconds.
16 seconds.
To solve this problem it is necessary to apply the concept of period in a pendulum. Its definition is framed in the equation
[tex]T= 2\pi \sqrt{\frac{l}{g}}[/tex]
Where,
l = Length
g = Gravity
As you can see, the Period is dependent on both the gravitational acceleration and the length of the string. The period does not depend on mass. Therefore the period must remain the same at 2 seconds.
You come across an open container that is filled with two liquids. Since the two liquids have different densities, there is a distinct separation between them. Water, which has a density of rho w = 1.00 × 10 3 kg/m 3 , fills the lower portion of the container to a depth of 0.221 m . The fluid that is floating on top of the water is 0.335 m deep. If the absolute pressure on the bottom of the container is 1.049 × 10 5 Pa , what is the density, rho l , of the unknown fluid? The acceleration due to gravity is g = 9.81 m/s 2 and atmospheric pressure is P 0 = 1.013 × 10 5 Pa .
Answer:
496.57492 kg/m³
Explanation:
[tex]P_a[/tex] = Atmospheric pressure = 101300 Pa
[tex]\rho_w[/tex] = Density of water = [tex]1000 kg/m^3[/tex]
[tex]h_w[/tex] = Height of water = 0.221 m
[tex]h_l[/tex] = Height of fluid = 0.335 m
g = Acceleration due to gravity = 9.81 m/s²
[tex]\rho_l[/tex] = Density of the unknown fluid
Absolute pressure at the bottom
[tex]P_{abs}=P_a+\rho_wgh_w+\rho_lgh_l\\\Rightarrow \rho_l=\frac{P_{abs}-P_a-\rho_wgh_w}{gh_l}\\\Rightarrow \rho_l=\frac{104900-101300-1000\times 9.81\times 0.221}{9.81\times 0.335}\\\Rightarrow \rho_l=435.73873\ kg/m^3[/tex]
The density of the unknown fluid is 496.57492 kg/m³
To find the density of the unknown fluid in the container, we subtract the atmospheric pressure and the pressure due to water from the absolute pressure at the bottom of the container and divide the result by the product of gravity and the height of the unknown fluid.
Explanation:The question is asking about the determination of the density of an unknown fluid in an open container where you have two separated fluids due to their different densities. In this setup, water fills the lower portion of the container and the unknown fluid fills the upper part.
The pressure at the bottom of this container (P) can be expressed as the sum of the atmospheric pressure (P0), pressure due to water (ρw * g * h1), and pressure due to the unknown fluid (ρl * g * h2) where 'g' is the acceleration due to gravity and 'h' represents the depth of each fluid. Given in the problem are P, P0, ρw, g, h1, and h2 and to find the density of the unknown fluid (ρl), we rearrange the equation into ρl = (P - P0 - ρw * g * h1) / (g * h2). By substituting the given values into this equation, we can find the density of the unknown fluid.
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A large uniform chain is hanging from the ceiling, supporting a block of mass 46 kg. The mass of the chain itself is 19 kg, and the length of the chain is 1.9 m. The acceleration of gravity is 9.81 m/s 2 . Find the tension in the chain at the point where the chain is supporting the block. Answer in units of N.
Answer:
T = 451.26 N
Explanation:
It is given that,
The mass of block, m = 46 kg
Mass of the chain, m' = 19 kg
Length of the chain, l = 1.9 m
Let T is the the tension in the chain at the point where the chain is supporting the block. It is clearly equal to the product of mass and acceleration.
[tex]T=mg[/tex]
[tex]T=46\ kg\times 9.81\ m/s^2[/tex]
T = 451.26 N
So, the tension in the chain at the point where the chain is supporting the block is 451.26 N. Hence, this is the required solution.
A combination lock has a 1.3-cm-diameter knob that is part of the dial you turn to unlock the lock. To turn that knob, you twist your wrist as you grip it between your thumb and forefinger with a force of 5.0 N . Suppose the coefficient of static friction between the knob and your fingers is 0.65.
What is the most torque that you can exert on the knob without having it slip between your fingers?
Answer:
0.04225 Nm
Explanation:
N = Force applied = 5 N
[tex]\mu[/tex] = Coefficient of static friction = 0.65
d = Diameter of knob = 1.3 cm
r = Radius of knob = [tex]\frac{d}{2}=\frac{1.3}{2}=0.65\ cm[/tex]
Force is given by
[tex]F=N\mu\\\Rightarrow F=5\times 0.65\\\Rightarrow F=3.25\ N[/tex]
When we multiply force and radius we get torque
Torque on thumb
[tex]\tau_t=F\times r\\\Rightarrow \tau_t=3.25\times 0.0065\\\Rightarrow \tau_t=0.021125\ Nm[/tex]
Torque on forefinger
[tex]\tau_f=F\times r\\\Rightarrow \tau_f=3.25\times 0.0065\\\Rightarrow \tau_f=0.021125\ Nm[/tex]
The total torque is given by
[tex]\tau=\tau_t+\tau_f\\\Rightarrow \tau=0.021125+0.021125\\\Rightarrow \tau=0.04225\ Nm[/tex]
The most torque that exerted on the knob is 0.04225 Nm
The torque is dependent on the force applied to the system. The torque exerted on the knob is 0.04225 Nm.
What is torque?
The torque is defined as the measure of the force that can cause an object to rotate about its axis.
Given that the applied force F is 5 N and the coefficient of static friction [tex]\mu[/tex] between the knob and your fingers is 0.65. The diameter d of the knob is 1.3 cm.
The radius r of the knob is given as,
[tex]r = \dfrac {d}{2}[/tex]
[tex]r = \dfrac {1.3}{2}[/tex]
[tex]r = 0.65 \;\rm cm[/tex]
The force F' applied to the system is calculated as given below.
[tex]F' = F\mu[/tex]
[tex]F' = 5\times 0.65[/tex]
[tex]F'=3.25 \;\rm N[/tex]
Now the torque on the thumb is given as,
[tex]\tau_t = F'\times r[/tex]
[tex]\tau_t = 3.25\times 0.65\times 10^{-2}[/tex]
[tex]\tau_t = 0.021125 \;\rm Nm[/tex]
The torque on the forefinger is,
[tex]\tau_f = F'\times r[/tex]
[tex]\tau_f= 3.25\times 0.65\times 10^{-2}[/tex]
[tex]\tau_f = 0.021125 \;\rm Nm[/tex]
The total torque of the system is calculated as,
[tex]\tau = \tau_t + \tau_f[/tex]
[tex]\tau= 0.021125 + 0.021125[/tex]
[tex]\tau = 0.04225 \;\rm Nm[/tex]
Hence we can conclude that the torque exerted on the knob is 0.04225 Nm.
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