Answer:
[tex]F_{net} = 4.22 N[/tex]
Explanation:
Since charge 1 and charge 2 are positive in nature so here we will have repulsion type of force between them
It is given as
[tex]F_{12} = \frac{kq_1q_2}{r^2}[/tex]
[tex]F_{12} = \frac{(9\times 10^9)(5 \mu C)(24 \mu C)}{0.23^2 + 0.69^2}\frac{-0.23\hat i + 0.69 \hat j}{\sqrt{0.23^2 + 0.69^2}}[/tex]
[tex]F_{12} = 2.81(-0.23\hat i + 0.69\hat j)[/tex]
Since charge three is a negative charge so the force between charge 1 and charge 3 is attraction type of force
[tex]F_{13} = \frac{(9\times 10^9)(5 \mu C)(5 \mu C)}{0.27^2 + 0^2} (-\hat i)[/tex]
[tex]F_{13} = 3.1(- \hat i)[/tex]
Now we will have net force on charge 1 as
[tex]F_{net} = F_{12} + F_{13}[/tex]
[tex]F_{net} = (-0.65 \hat i + 1.94 \hat j) + (-3.1 \hat i)[/tex]
[tex]F_{net} = (-3.75 \hat i + 1.94 \hat j)[/tex]
now magnitude of total force on the charge is given as
[tex]F_{net} = 4.22 N[/tex]
Consider a torque ~τ that is constant in both magnitude and direction, and acts on a rigid body of mass 10 kg at a point 1 m from the pivot. How much work does the torque do on the rigid body, if it turns through an angle of 180◦ while the torque is acting? Assume the acceleration due to gravity is 10 m/s2 .
Answer:
314 Joule
Explanation:
m = 10 kg, g = 10 m/s^2, d = 1 m, angle turn = 180 degree = π radian
work = torque x angle turn
torque = force x perpendicular distance
torque = m x g x d = 10 x 10 x 1 = 100 Nm
work = 100 x π
work = 100 x 3.14 = 314 Joule
Final answer:
The work done by a constant torque on a rigid body rotating through an angle of 180° can be found by multiplying the torque by the angle. The torque can be calculated using the equation τ = Iα, where I is the moment of inertia and α is the angular acceleration. Substituting the given values, we find that the work done by the torque on the rigid body is 18000 kg·m^2·rad/s^2.
Explanation:
The work done by a torque on a rigid body is given by the formula W = τθ, where τ is the torque and θ is the angle through which the body rotates. In this case, the torque is constant in both magnitude and direction, so we can use W = τθ. Given that the torque is constant and the body turns through an angle of 180°, we can calculate the work done as follows:
Since τ is constant, we can write W = τθ = τ (180° - 0°). The work done by the torque is equal to the torque multiplied by the change in angle. Substitute the given values into the formula: W = (τ) (180° - 0°) = (τ) (180°). The work done by the torque is equal to the torque multiplied by 180°.
To find the value of the torque, we need to use the equation τ = Iα, where I is the moment of inertia and α is the angular acceleration. In this case, the rigid body has a mass of 10 kg and a distance of 1 m from the pivot. The moment of inertia for a point mass rotating about a fixed axis is given by I = m(r^2), where m is the mass and r is the perpendicular distance from the axis. Substitute the given values into the formula: I = (10 kg)((1 m)^2) = 10 kg·m^2. Since α = a/r and the acceleration due to gravity is 10 m/s^2, we have α = (10 m/s^2)/(1 m) = 10 rad/s^2.
Substitute the values of τ and α into the equation τ = Iα: τ = (10 kg·m^2)(10 rad/s^2) = 100 kg·m^2·rad/s^2. Therefore, the torque is 100 kg·m^2·rad/s^2.
Finally, substitute the values of τ and θ into the equation W = τθ: W = (100 kg·m^2·rad/s^2 )(180°) = 18000 kg·m^2·rad/s^2.
Water is boiled in a pan on a stove at sea level. During 10 min of boiling, it is observed that 200 g of water has evaporated. Then the rate of heat transfer to the water is 225.7 kJ/min 45.1 kJ/min 53.5 kJ/min 0.84 kJ/min 41.8 kJ/min
Answer:
Rate of heat is 45.1 kJ/min
Explanation:
Heat required to evaporate the water is given by
Q = mL
here we know that
[tex]L = 2.25 \times 10^6 J/kg[/tex]
now we have
[tex]Q = (0.200)(2.25 \times 10^6 J/kg)[/tex]
[tex]Q = 452.1 kJ[/tex]
now the power is defined as rate of energy
[tex]P = \frac{Q}{t}[/tex]
[tex]P = \frac{452.1 kJ}{10}[/tex]
[tex]P = 45.1 kJ/min[/tex]
on a day when the speed of sound in air is 340 m/s a bat emits a shriek whose echo returns to it 0.0250 seconds later. How far away is the mosquito that reflects back the shriek?
Answer:
The distance of the mosquito from the bat is 4.25 m.
Explanation:
Given that,
Speed of sound in air v= 340 m/s
Time t = 0.0250 second
Let d be the distance of the mosquito from the bat.
The distance traveled by the sound when the echo heard is 2d.
We need to calculate the distance
Using formula of distance
[tex]v = \dfrac{2d}{t}[/tex]
Put the value into the formula
[tex]v = \dfrac{2d}{25\times10^{-3}}[/tex]
[tex]d =\dfrac{25\times10^{-3}\times340}{2}[/tex]
[tex]d=4.25\ m[/tex]
Hence, The distance of the mosquito from the bat is 4.25 m.
Utilizing the echo time of 0.0250 seconds and the sound speed of 340 m/s, the distance to the mosquito is calculated as 4.25 meters away from the bat.
Explanation:Calculating the Distance to the Mosquito using Echo Time
To determine the distance to the mosquito that reflects the bat's shriek, we need to use the speed of sound in air and the echo time. Since the shriek's echo returns in 0.0250 seconds and the speed of sound is given as 340 m/s, we can calculate the total distance traveled by the sound (to the mosquito and back to the bat) with the equation:
Distance = Speed × Time
Here, the time is the round-trip time for the sound, so the distance to the mosquito is half the total distance:
Total Distance = 340 m/s × 0.0250 s = 8.5 m
Distance to the Mosquito = Total Distance / 2 = 8.5 m / 2 = 4.25 m
Therefore, the mosquito is 4.25 meters away from the bat.
A heavy fuel oil has a specific gravity of 0.918. How much will 100 gallons(300 liters) of this oil weigh? INCLUDE UNITS!
Answer:
275.4 kg
Explanation:
specific gravity = 0.918
Density = 0.918 g/cm^3 = 918 kg/m^3
Volume = 100 gallons = 300 litres = 300 x 10^-3 m^3 = 0.3 m^3
Mass = volume x density = 0.3 x 918 = 275.4 kg
Which is not a simple harmonic motion (S.H.M.) (a) Simple Pendulum (b) Projectile motion (c) None (d) Spring motion
Answer:
b) Projectile MOTION
Explanation:
SHM is periodic motion or to and fro motion of a particle about its mean position in a straight line
In this type of motion particle must be in straight line motion
So here we can say
a) Simple Pendulum : it is a straight line to and fro motion about mean position so it is a SHM
b) Projectile motion : it is a parabolic path in which object do not move to and fro about its mean position So it is not SHM
d) Spring Motion : it is a straight line to and fro motion so it is also a SHM
So correct answer will be
b) Projectile MOTION
Final answer:
Projectile motion is not a simple harmonic motion because it does not meet the conditions for SHM.
Explanation:
Simple Harmonic Motion (SHM) is a special type of periodic motion where the restoring force is proportional to the displacement. The three conditions that must be met to produce SHM are: a linear restoring force, a constant force constant, and no external damping forces. Based on these conditions, the answer to the question is (b) Projectile motion, as it does not meet the conditions for SHM. A projectile follows a parabolic path and does not have a linear restoring force.
A uniformly charged sphere has a potential on its surface of 450 V. At a radial distance of 7.2 m from this surface, the potential is150 V What is the radius of the sphere?
Answer:
The radius of the sphere is 3.6 m.
Explanation:
Given that,
Potential of first sphere = 450 V
Radial distance = 7.2 m
If the potential of sphere =150 V
We need to calculate the radius
Using formula for potential
For 450 V
[tex]V=\dfrac{kQ}{r}[/tex]
[tex]450=\dfrac{kQ}{r}[/tex]....(I)
For 150 V
[tex]150=\dfrac{kQ}{r+7.2}[/tex]....(II)
Divided equation (I) by equation (II)
[tex]\dfrac{450}{150}=\dfrac{\dfrac{kQ}{r}}{\dfrac{kQ}{r+7.2}}[/tex]
[tex]3=\dfrac{(r+7.2)}{r}[/tex]
[tex]3r=r+7.2[/tex]
[tex]r=\dfrac{7.2}{2}[/tex]
[tex]r=3.6\ m[/tex]
Hence, The radius of the sphere is 3.6 m.
The radius of the sphere whose surface has a potential difference of 450 V is 3.6 m.
What is the radius of the sphere?We know that the potential difference can be written as,
[tex]V = k\dfrac{Q}{R}[/tex]
We know that at R= R, Potential difference= 450 V,
[tex]450 = k\dfrac{Q}{R}[/tex]
Also, at R = (R+7.2), Potential difference = 150 V,
[tex]150 = k\dfrac{Q}{(R+7.2)}[/tex]
Taking the ratio of the two,
[tex]\dfrac{450}{150} = \dfrac{kQ}{R} \times \dfrac{(R+7.2)}{kQ}\\\\\dfrac{450}{150} = \dfrac{(R+7.2)}{R}\\\\R = 3.6\ m[/tex]
Hence, the radius of the sphere whose surface has a potential difference of 450 V is 3.6 m.
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Absolute pressure in tank is P1 = 260 kPa and local ambient absolute pressure is P2 =100 kPa. If liquid density in pipe is 13600 kg/m3 , compute liquid height, h=..?.. m ? Use g =10 m/s2
Answer:
1.176m
Explanation:
Local ambient pressure(P1) = 100 kPa
Absolute pressure(P2)=260kPa
Net pressure=absolute pressure-local ambient absolute pressure
Net pressure=P1(absolute pressure)-P2(local ambient absolute pressure)
Net pressure=260-100=160kPa
Pressure= ρgh
160kPa=13600*10*h
h=[tex]\frac{160000}{136000}[/tex]
h=1.176m
A water trough is 8 m long and has a cross-section in the shape of an isosceles trapezoid that is 20 cm wide at the bottom, 80 cm wide at the top, and has height 60 cm. If the trough is being filled with water at the rate of 0.3 m3/min how fast is the water level rising when the water is 30 cm deep?
Answer:It is rising at a rate of [tex]7.5cm/min[/tex]
Explanation:
We have volume of trapezoid equals
[tex]V=Area\times Length\\\\V=\frac{1}{2}(a+b)h\times L[/tex]
Thus at any time 't' we have
[tex]V(t)=\frac{1}{2}(a(t)+b(t))h(t)\times L\\\\\therefore V(t)=\frac{1}{2}(20+b(t))\times h(t)\times L[/tex]
Differentiating both sides with respect to time we get
[tex]\frac{dV(t)}{dt}=\frac{1}{2}b'(t)h(t)L+\frac{1}{2}(20+b(t))\times h'(t)L[/tex]
Applying values we have
[tex]b(t)=20+h(t)\\b'(t)=h'(t)[/tex]
Thus we have
[tex]\frac{dV(t)}{dt}=\frac{1}{2}h'(t)h(t)L+\frac{1}{2}(20+20+h(t))\times h'(t)L\\\\2V'(t)=h'(t)L[h(t)+(40+h(t))]\\\\\therefore h'(t)=\frac{2V'(t)}{L(h(t)+(40+h(t)))}[/tex]
Applying values we get
[tex]h'(t)=0.075m/min=7.5cm/min[/tex]
The acceleration of a particle is given by a = −ks2 , where a is in meters per second squared, k is a constant, and s is in meters. Determine the velocity of the particle as a function of its position s. Evaluate your expression for s = 5 m if k = 0.1 m−1 s−2 and the initial conditions at time t = 0 are s0 = 3 m and v0 = 10 m /s
Answer:
[tex]v = \sqrt{v_0^2 - \frac{2k}{3}(s^3 - s_0^3)}[/tex]
v = 9.67 m/s
Explanation:
As we know that acceleration is rate of change in velocity
so it is defined as
[tex]a = \frac{dv}{dt}[/tex]
[tex]a = v\frac{dv}{ds}[/tex]
here we know that
[tex]a = - ks^2 = v\frac{dv}{ds}[/tex]
now we have
[tex]vdv = - ks^2ds[/tex]
integrate both sides we have
[tex]\int vdv = -k \int s^2ds[/tex]
[tex]\frac{v^2}{2} - \frac{v_0^2}{2} = -k(\frac{s^3}{3} - \frac{s_0^3}{3})[/tex]
[tex]v^2 = v_0^2 - \frac{2k}{3}(s^3 - s_0^3)[/tex]
here we know that
[tex]v_0 = 10 m/s[/tex]
[tex]s_0 = 3 m[/tex]
[tex]v^2 = 10^2 - \frac{2(0.10)}{3}(5^3 - 3^3)[/tex]
[tex]v = 9.67 m/s[/tex]
If the intensity of light that is incident on a piece of metal is increased, what else will be increased? Choose all that apply. number of electrons ejected stopping voltage cutoff frequency frequency KEmax work function wavelength
Answer:
explained
Explanation:
When the intensity of light is increased on a piece of metal only the number of electron ejected will increase because all other things independent of intensity of light.
Light below certain frequency will not cause any electron emission no matter how intense.
The intensity produces more electron but does not change the maximum kinetic energy of electrons.
Work function is independent of the intensity of light, because it is an intrinsic property of a material.
A proton experiences a force of 3.5x 10^-9 N when separated from a second charge by a distance of 1.6 mm. a) What is the size of the second charge? b) How many fundamental charges make up this charge in part a)?
Answer:
(a) 6.22 x 10^-6 C
(b) 3.8 x 10^13
Explanation:
Let the second charge is q2 = q
q1 = 1.6 x 10^-19 C
F = 3.5 x 10^9 N
d = 1.6 mm = 1.6 x 10^-3 m
(a) Use the formula of Coulomb's law
F = K q1 x q2 / d^2
3.5 x 10^-9 = 9 x 10^9 x 1.6 x 10^-19 x q / (1.6 x 10^-3)^2
q = 6.22 x 10^-6 C
(b)
Let the number of electrons be n
n = total charge / charge of one electron
n = 6.22 x 10^-6 / (1.6 x 10^-19) = 3.8 x 10^13
A typical adult ear has a surface area of 2.90 × 10-3 m2. The sound intensity during a normal conversation is about 2.19 × 10-6 W/m2 at the listener's ear. Assume that the sound strikes the surface of the ear perpendicularly. How much power is intercepted by the ear?
Answer:
[tex]6.35\cdot 10^{-9} W[/tex]
Explanation:
The relationship between power and intensity of a sound is given by:
[tex]I=\frac{P}{A}[/tex]
where
I is the intensity
P is the power
A is the area considered
In this problem, we know
[tex]A=2.90\cdot 10^{-3}m^2[/tex] is the surface area of the ear
[tex]I = 2.19\cdot 10^{-6} W/m^2[/tex] is the intensity of the sound
Re-arranging the equation, we can find the power intercepted by the ear:
[tex]P=IA=(2.19\cdot 10^{-6} W/m^2)(2.90\cdot 10^{-3} m^2)=6.35\cdot 10^{-9} W[/tex]
A 1500-kg car traveling east with a speed of 25.0 m/s collides at an intersection with a 2500-kg van traveling north at a speed of 20.0 m/s. Find the direction and magnitude of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (i.e. they stick together).
From conservation of linear momentum, the magnitude of the velocity of the wreckage after collision is 15.6 m/s while its direction is 53 degrees.
COLLISIONThere are four types of collision
Elastic collisionPerfectly elastic collisionInelastic collisionPerfectly inelastic collisionIn elastic collision, both momentum and energy are conserved. While in inelastic collision, only momentum is conserved.
From the given question, the following parameters are given.
[tex]m_{1}[/tex] = 1500kg[tex]v_{1}[/tex] = 25 m/s[tex]m_{2}[/tex] = 2500 kg[tex]v_{2}[/tex] = 20 m/sSince the collision is inelastic, they will both move with a common velocity after collision.
Horizontal component
[tex]m_{1}[/tex][tex]v_{1}[/tex] = ([tex]m_{1}[/tex] + [tex]m_{2}[/tex] ) V
1500 x 25 = (1500 + 2500) V
37500 = 4000V
V = 37500 / 4000
V = 9.375 m/s
Vertical component
[tex]m_{2}[/tex][tex]v_{2}[/tex] = ([tex]m_{1}[/tex] + [tex]m_{2}[/tex])V
2500 x 20 = (1500 + 2500)V
50000 = 4000V
V = 50000 / 4000
V = 12.5 m/s
The net velocity will be the magnitude of the velocity of the wreckage after collision
V = [tex]\sqrt{9.4^{2} + 12.5^{2} }[/tex]
V = [tex]\sqrt{244.61}[/tex]
V = 15.6 m/s
The direction will be
Tan Ф = 12.5 / 9.4
Ф = [tex]Tan^{-1}[/tex](1.329)
Ф = 53 degrees.
Therefore, the magnitude of the velocity of the wreckage after collision is 15.6 m/s while its direction is 53 degrees.
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The velocity of the wreckage after the collision is 32.015 m/s in the direction 38.66° north of east.
Explanation:This problem is related to the conservation of linear momentum which states the total momentum of an isolated system remains constant if no external forces act on it. Momentum is a vector quantity having both magnitude and direction. The total initial and final momentum in both horizontal (x) and vertical (y) directions must be equal.
Initial momentum of the car is (1500 kg)*(25.0 m/s) = 37500 kg*m/s in the east direction whereas initial momentum of the van is (2500 kg)*(20.0 m/s) = 50000 kg*m/s in the north direction.
Since the collision is perfectly inelastic, the two vehicles stick together after the collision and move as one. Therefore, the final momentum is the vector sum of the individual momenta. We therefore calculate the magnitude of the resultant velocity using Pythagoras theorem, √[(25.0 m/s)² + (20.0 m/s)²] = 32.015 m/s.
To find the direction of the final velocity, we use the tangent of the angle which is equal to the vertical component divided by the horizontal component, which gives us an angle of 38.66° north of east.
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A 2.7-kg cart is rolling along a frictionless, horizontal track towards a 1.1-kg cart that is held initially at rest. The carts are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is +3.7 m/s, and the second cart's velocity is -1.6 m/s. (a) What is the total momentum of the system of the two carts at this instant? (b) What was the velocity of the first cart when the second cart was still at rest?
Answer:
Part a)
P = 8.23 kg m/s
Part b)
v = 3.05 m/s
Explanation:
Part a)
momentum of cart 1 is given as
[tex]P_1 = m_1v_1[/tex]
[tex]P_1 = (2.7)(3.7) = 9.99 kg m/s[/tex]
Momentum of cart 2 is given as
[tex]P_2 = m_2v_2[/tex]
[tex]P_2 = (1.1)(-1.6) = -1.76 kg m/s[/tex]
Now total momentum of both carts is given as
[tex]P = P_1 + P_2[/tex]
[tex]P = 8.23 kg m/s[/tex]
Part b)
Since two carts are moving towards each other due to mutual attraction force and there is no external force on two carts so here momentum is always conserved
so here we will have
[tex]P_i = P_f[/tex]
[tex](2.7 kg)v = 8.23[/tex]
[tex]v = 3.05 m/s[/tex]
If a person is standing erect and flexes the trunk on the hip, the center of mass will move __________ and the line of gravity moves __________ within the base of support.
Answer:
anterior
anterior
Explanation:
In the given question is asked that
If a person is standing erect and flexes the trunk on the hip, the center of mass will move ___________ and the line of gravity moves___________ within the base of support.
The current answer to the blanks will be
anterior
anterior
hope this helps any further query can be asked in comment section.
A 16.0 kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 25.0 N. Starting from rest, the sled attains a speed of 1.00 m/s in 8.00 m. Find the coefficient of kinetic friction between the runners of the sled and the snow. You Answered
Answer:
[tex]\mu_k = 0.15[/tex]
Explanation:
according to the kinematic equation
[tex]v^{2} - u^{2} = 2aS[/tex]
Where
u is initial velocity = 0 m/s
a = acceleration
S is distance = 8.00 m
final velocity = 1.0 m/s
[tex]a = \frac {v^{2}}{2S}[/tex]
[tex]a = \frac {1{2}}{2*8.6}[/tex]
a = 0.058 m/s^2
from newton second law
Net force = ma
[tex]f_{net} = ma[/tex]
F - f = ma
2[tex]5 - \mu_kN = ma[/tex]
[tex]25 - \mu_kmg = ma[/tex]
[tex]\frac {25 - ma}{mg} =\mu_k[/tex]
[tex]\frac {25 - 16*0.058}{16*9.81} = 0.15[/tex]
[tex]\mu_k = 0.15[/tex]
A uniform rod of mass 2.55×10−2 kg and length 0.380 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.200 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.80×10−2 m on each side from the center of the rod, and the system is rotating at an angular velocity 35.0 rev/min . Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. What is the angular speed of the system at the instant when the rings reach the ends of the rod?
What is the angular speed of the rod after the rings leave it?
Answer:
[tex]\omega _{f}=0.3107rad/sec[/tex]
partb) [tex]\omega _{f}=14.93rad/sec[/tex]
Explanation:
Since the system is isolated it's angular momentum shall be conserved
[tex]L_{system}=I_{system}\omega \\\\I_{system}=I_{rod}+2\times m\times r^{2}\\\\I_{system}=\frac{1}{12}ml^{2}+2\times 0.2\times (4.8\times 10^{-2})^{2}\\\\I_{system}=1.22845\times 10^{-3}kgm^{2}\\\\L_{system}=1.22845\times 10^{-3}kgm^{2}\times 3.66rad/sec[/tex]
[tex]\L_{system}=4.496\times 10^{-3}kgm^{2}[/tex]
Now the final angular momentum of the system
[tex]L_{fianl}=I_{final}\omega \\\\I_{final}=I_{rod}+2\times m\times r_{f}^{2}\\\\I_{final}=\frac{1}{12}ml^{2}+2\times 0.2\times (0.19)^{2}\\\\I_{final}=0.01447kgm^{2}\\\\L_{final}=0.01447kgm^{2}\times \omega _{final}rad/sec[/tex]
Thus equating initial and final angular momentum we solve for final angular velocity of rod as
[tex]\omega _{f}=\frac{4.496\times 10^{-3}}{0.01447}\\\\\omega _{f}=0.3107rad/sec[/tex]
part b)
When the rings leave the system we again conserve the angular momentum just before the rings leave the system and the instant when they just leave
[tex]\therefore 0.01447=\frac{1}{12}ml^{2}w\\\\\therefore w=\frac{4.582\times 10^{-3}}{3.068\times10^{-4} }\\\\w_{final}=14.93rad/sec[/tex]
Answer:
A)[tex]\omega_{f}=2.92\frac{rev}{min}[/tex].
B)[tex]\omega_{f}=140\frac{rev}{min}[/tex].
Explanation:
A)
For this problem, we will use the conservation of angular momentum.
[tex]L_{0}=L_{f}\\[/tex].
In the beginning, we have that
[tex]L_{0}=I_{0}\omega_{0}\\[/tex]
where [tex]I_{0}[/tex] is the inertia moment of all the system at the starting position, this is the inertia moment of the rod plus the inertia moment of each ring ([tex]mr^{2}[/tex], with [tex]r[/tex] the distance from the ring to the fixed axis and, [tex]m[/tex] its mass) at the starting position and, [tex]\omega_{0}[/tex] is the initial angular velocity. So
[tex]L_{0}=(\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}[/tex].
When the rings are at the ends of the rod the angular momentum becomes
[tex]L_{f}=(\frac{1}{12}Ml^{2}+2mr_{f}^{2})\omega_{f}[/tex],
where [tex]r_{f}[/tex] is the distance from the fixed axis to the end of the rod (the final position of the rings).
Using conservation of angular momentum we get
[tex](\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}=(\frac{1}{12}Ml^{2}+2mr_{f}^{2})\omega_{f}[/tex].
thus
[tex]\omega_{f}=\frac{(\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}}{(\frac{1}{12}Ml^{2}+2mr_{f}^{2})}[/tex]
computing this last expresion we get
[tex]\omega_{f}=\frac{(\frac{1}{12}(2.55*10^{-2})(0.380)^{2}+2(0.200)(4.80*10^{-2})^{2})(35.0)}{(\frac{1}{12}(2.55*10^{-2})(0.380)^2+2(0.200)(0.19)^{2})}[/tex]
[tex]\omega_{f}=2.92\frac{rev}{min}[/tex].
B)
Again we use the conservation of angular momentum. The initial angular momentum if the same as before. The final angular momentum will be
[tex]L_{f}=(\frac{1}{12}Ml^{2})\omega_{f}[/tex],
this time we will not take into account the inertia moment of the rings because they are no longer part of the system (they leave the rod).
[tex](\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}=(\frac{1}{12}Ml^{2})\omega_{f}[/tex].
thus
[tex]\omega_{f}=\frac{(\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}}{(\frac{1}{12}Ml^{2})}[/tex]
computing this last expresion we get
[tex]\omega_{f}=\frac{(\frac{1}{12}(2.55*10^{-2})(0.380)^{2}+2(0.200)(4.80*10^{-2})^{2})(35.0)}{(\frac{1}{12}(2.55*10^{-2})(0.380)^2})[/tex]
[tex]\omega_{f}=140\frac{rev}{min}[/tex].
ml(d^2θ/dt^2) =-mgθ
1. From the linearized equation, justify Galileo’s observation that the period of a pendulum depends only on its length and not on the mass or on the initial displacement.
The equation of motion of a pendulum is:
[tex]\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\dfrac{g}{\ell}\sin\theta,[/tex]
where [tex]\ell[/tex] it its length and [tex]g[/tex] is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for small angles ([tex]\theta \ll 1[/tex]), we can use:
[tex]\sin\theta \simeq \theta.[/tex]
Additionally, let us define:
[tex]\omega^2\equiv\dfrac{g}{\ell}.[/tex]
We can now write:
[tex]\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.[/tex]
The solution to this differential equation is:
[tex]\theta(t) = A\sin(\omega t + \phi),[/tex]
where [tex]A[/tex] and [tex]\phi[/tex] are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:
[tex]T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.[/tex]
This justifies that the period depends only on the pendulum's length.
A 2.0 kg hanging mass stretches a coiled spring by 0.15 m. The spring constant, k, is: (A) 0.075 N/m, (B) 2.9 N/m (C) 131 N/m, (D) 1,742 N/m, (E) none of the above.
Answer:
C
Explanation:
Givens
m = 2 kg
F = 2 * 9.81
F = 19.62 N
x = 0.15 m
Formula
F = k*x
Solution
19.62 = k*0.15
k = 19.62/0.15
k = 130.8 which rounded to the nearest given answer is C
On takeoff, the combined action of the engines and the wings of an airplane exert a force of 8.00 × 103 N on the plane upward at an angle of 65.0" above the horizontal. The plane rises with constant velocity in the vertical direction while continuing to accelerate in the horizontal direction. (3 marks) a. What is the weight of the plane? b. What is the horizontal acceleration of the plane? .
Answer:
a) 7250.5 N
b) 4.6 m/s²
Explanation:
a)
F = applied force = 8000 N
θ = angle with the horizontal = 65 deg
Consider the motion along the vertical direction :
[tex]F_{y}[/tex] = Applied force in vertical direction in upward direction = F Sinθ = 8000 Sin65 = 7250.5 N
[tex]F_{g}[/tex] = weight of the plane in vertical direction in downward direction = ?
[tex]a_{y}[/tex] = Acceleration in vertical direction = 0 m/s²
Taking the force in upward direction as positive and in downward direction as negative, the force equation along the vertical direction can be written as
[tex]F_{y}-F_{g} = m a_{y}[/tex]
[tex]7250.5 -F_{g} = m (0)[/tex]
[tex]F_{g}[/tex] = 7250.5 N
b)
m = mass of the plane
force of gravity is given as
[tex]F_{g} = mg [/tex]
[tex]7250.5 = m(9.8) [/tex]
m = 739.85 kg
Consider the motion along the horizontal direction
[tex]F_{x}[/tex] = Applied force in horizontal direction = F Cosθ = 8000 Cos65 = 3381 N
[tex]a_{x}[/tex] = Acceleration in horizontal direction
Acceleration in horizontal direction is given as
[tex]a_{x}=\frac{F_{x}}{m}[/tex]
[tex]a_{x}=\frac{3381}{739.85}[/tex]
[tex]a_{x}[/tex] = 4.6 m/s²
A coil is made of 150 turns of copper wire wound on a cylindrical core. If the mean radius of the turns is 6.5 mm and the diameter of the wire is 0.4 mm, calculate the resistance of the coil!
Answer:
0.84 Ω
Explanation:
r = mean radius of the turn = 6.5 mm
n = number of turns of copper wire = 150
Total length of wire containing all the turns is given as
L = 2πnr
L = 2 (3.14)(150) (6.5)
L = 6123 mm
L = 6.123 m
d = diameter of the wire = 0.4 mm = 0.4 x 10⁻³ m
Area of cross-section of the wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (0.4 x 10⁻³)²
A = 1.256 x 10⁻⁷ m²
ρ = resistivity of copper = 1.72 x 10⁻⁸ Ω-m
Resistance of the coil is given as
[tex]R = \frac{\rho L}{A}[/tex]
[tex]R = \frac{(1.72\times 10^{-8}) (6.123))}{(1.256\times 10^{-7}))}[/tex]
R = 0.84 Ω
What is the force on your eardrum if its area is 1.00 cm^2, and you are swimming 3.0 m below water level?
Answer:
Force on eardrum
[tex]F=29400\times 1\times 10^{-4}=2.94N[/tex]
Explanation:
Force = Pressure x Area
Pressure = hρg
Height, h = 3 m
ρ = 1000 kg/m³
g = 9.8 m/s²
Pressure = hρg = 3 x 1000 x 9.8 = 29400 N/m²
Area = 1 cm²
Force on eardrum
[tex]F=29400\times 1\times 10^{-4}=2.94N[/tex]
Calculate the mass of a 0.9 m^3 block of a material having a density of 12500 kg/m^3.
Answer: The mass of the object will be 11250 kg.
Explanation:
Density is defined as the mass contained per unit volume.
[tex]Density=\frac{mass}{Volume}[/tex]
Given :
Density of the object= [tex]12500kg/m^3[/tex]
Mass of object = ?
Volume of the object = [tex]0.9m^3[/tex]
Putting in the values we get:
[tex]12500kg/m^3=\frac{mass}{0.9m^3}[/tex]
[tex]12500kg/m^3=\frac{mass}{0.9m^3}[/tex]
[tex]mass=11250kg[/tex]
Thus the mass of the object is 11250 kg.
How much heat transfer is required to raise the temperature of a 0.750-kg aluminum pot containing 2.50 kg of water from 30.0ºC to the boiling point and then boil away 0.750 kg of water? (b) How long does this take if the rate of heat transfer is 500 W
Answer:
Part a)
[tex]Q = 2.47 \times 10^6[/tex]
Part b)
t = 4950.3 s
Explanation:
As we know that heat required to raise the temperature of container and water in it is given as
[tex]Q = m_1s_2\Delta T_1 + m_2s_2\Delta T_2[/tex]
here we know that
[tex]m_1 = 0.750[/tex]
[tex]s_1 = 900[/tex]
[tex]m_2 = 2.50 kg[/tex]
[tex]s_2 = 4186[/tex]
[tex]\Delta T_1 = \Delta T_2 = 100 - 30 = 70^oC[/tex]
now we have
[tex]Q_1 = 0.750(900)(70) + (2.5)(4186)(70) = 779800 J[/tex]
now heat require to boil the water
[tex]Q = mL[/tex]
here
m = 0.750 kg
[tex]L = 2.25 \times 10^6 J/kg[/tex]
now we have
[tex]Q_2 = 0.750(2.25 \times 10^6) = 1.7 \times 10^6 J[/tex]
Now total heat required is given as
[tex]Q = Q_1 + Q_2[/tex]
[tex]Q = 779800 + 1.7 \times 10^6 = 2.47 \times 10^6 J[/tex]
Part b)
Time taken to heat the water is given as
[tex]t = \frac{Q}{P}[/tex]
here we know that
power = 500 W
now we have
[tex]t = \frac{2.47 \times 10^6}{500} = 4950.3 s[/tex]
The shortest air column inside a resonator vibrates with a frequency of 250 Hz, if the next harmonic is 750 Hz, and the speed of sound is 343 m/s.
a. Is this resonator closed at one end or open at both ends? Explain.
b. Find the length of the resonator.
Answer:
Part a)
the two frequencies are in ratio of odd numbers so it must be closed at one end
Part b)
L = 34.3 cm
Explanation:
Part a)
Since the shortest frequency is known as fundamental frequency
It is given as
[tex]f_o = 250 Hz[/tex]
next higher frequency is given as
[tex]f_1 = 750 Hz[/tex]
since the two frequencies here are in ratio of
[tex]\frac{f_1}{f_o} = \frac{750}{250} = 3 : 1[/tex]
since the two frequencies are in ratio of odd numbers so it must be closed at one end
Part b)
For the length of the pipe we can say that fundamental frequency is given as
[tex]f_o = \frac{v}{4L}[/tex]
here we have
[tex]250 = \frac{343}{4(L)}[/tex]
now we will have
[tex]L = \frac{343}{4\times 250}[/tex]
[tex]L = 34.3 cm[/tex]
Two forces are applied to a 5.0-kg crate; one is 3.0 N to the north and the other is 4.0 N to the east. The magnitude of the acceleration of the crate is: a. 1.0 m/s^2 b. 2.8 m/s^2 c.7.5 m/s^2 d. 10.0 m/s^2
Answer:
The acceleration of the crate is 1 m/s²
Explanation:
It is given that,
Mass of the crate, m = 5 kg
Two forces applied on the crate i.e. one is 3.0 N to the north and the other is 4.0 N to the east. So, there resultant force is :
[tex]F_{net}=\sqrt{3^2+4^2} =5\ N[/tex]
We need to find the acceleration of the crate. It is given by using the second law of motion as :
[tex]a=\dfrac{F_{net}}{m}[/tex]
[tex]a=\dfrac{5\ N}{5\ kg}[/tex]
a = 1 m/s²
So, the acceleration of the crate is 1 m/s². Hence, this is the required solution.
The magnitude of the acceleration of a crate with forces of 3.0 N north and 4.0 N east applied to it is 1.0 m/s². This is found using the Pythagorean theorem to calculate the resultant force and Newton's Second Law to calculate acceleration.
The forces are 3.0 N to the north and 4.0 N to the east on a 5.0-kg crate. Since the forces are perpendicular, we can use the Pythagorean theorem to find the resultant force. The resultant force (Fr) is √(3.02 + 4.02) N, which is 5.0 N. According to Newton's Second Law, F = ma, hence acceleration (a) is Fr divided by the mass (m). Calculating acceleration: a = 5.0 N / 5.0 kg = 1.0 m/s2. Therefore, the correct answer is a. 1.0 m/s2.
A sign is held in equilbrium by 7 vertically hanging ropes attached to the ceiling. If each rope has an equal tension of 53 Newtons, what is the mass of the sign in kg?
Answer:
37.86 kg
Explanation:
The weight of sign board is equally divided on each rope. It means the tension in all the ropes is equal to the weight of the sign board in equilibrium condition.
Tension in each rope = 53 N
Tension in 7 ropes = 7 x 53 N = 371 N
Thus, The weight of sign = 371 N
Now, weight = m g
where m is the mass of sign.
m = 371 / 9.8 = 37.86 kg
A 0.15 kg baseball is pushed with 100 N force. what will its acceleration be?
Answer:
The acceleration of the ball is 666.67 m/s²
Explanation:
It is given that,
Mass of the baseball, m = 0.15 kg
Applied force to it, F = 100 N
We need to find the acceleration of the ball. It can be calculated using Newton's second law of motion as :
F = ma
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{100\ N}{0.15\ kg}[/tex]
[tex]a=666.67\ m/s^2[/tex]
So, the acceleration of the ball is 666.67 m/s². Hence, this is the required solution.
What hanging mass will stretch a 3.0-m-long, 0.32 mm - diameter steel wire by 1.3 mm ? The Young's modulus of steel is 20×10^10 N/m^2.
Answer:
0.71 kg
Explanation:
L = length of the steel wire = 3.0 m
d = diameter of steel wire = 0.32 mm = 0.32 x 10⁻³ m
Area of cross-section of the steel wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (0.32 x 10⁻³)²
A = 8.04 x 10⁻⁸ m²
ΔL = change in length of the wire = 1.3 mm = 1.3 x 10⁻³ m
Y = Young's modulus of steel = 20 x 10¹⁰ Nm⁻²
m = mass hanging
F = weight of the mass hanging
Young's modulus of steel is given as
[tex]Y = \frac{FL}{A\Delta L}[/tex]
[tex]20\times 10^{10} = \frac{F(3)}{(8.04\times 10^{-8})(1.3\times 10^{-3})}[/tex]
F = 6.968 N
Weight of the hanging mass is given as
F = mg
6.968 = m (9.8)
m = 0.71 kg
Commander Shepard, an N7 spectre for Earth, weighs 799 N on the Earth's surface. When she lands on Noveria, a distant planet in our galaxy, she weighs 356 N. What is the acceleration of gravity on Noveria in m/s2? Round to one decimal place. (hint: first find her mass!)
Answer:
Acceleration of gravity on Noveria = 4.4 m/s²
Explanation:
Commander Shepard, an N7 spectre for Earth, weighs 799 N on the Earth's surface.
We have weight, W = mg
Acceleration due to gravity, g = 9.81m/s²
799 = m x 9.81
Mass of Shepard, m = 81.45 kg
She lands on Noveria, a distant planet in our galaxy, she weighs 356 N.
We have weight, W = mg'
356 = 81.45 xg'
Acceleration of gravity on Noveria, g' = 4.4 m/s²