Answer:
The third ball emerges from the right side.
Explanation:
This is a conservation of Momentum problem
In an explosion or collision, the momentum is always conserved.
Momentum before explosion = Momentum after explosion
Since the rigid pipe was initially at rest,
Momentum before explosion = 0 kgm/s
- Taking the right end as the positive direction for the velocity of the balls
- And calling the speed of the 6 g ball after explosion v
- This means the velocity of the 4 g ball has to be -2v
- Mass of the third ball = m
- Let the velocuty of the third ball be V
Momentum after collision = (6)(v) + (4)(-2v) + (m)(V)
Momentum before explosion = Momentum after explosion
0 = (6)(v) + (4)(-2v) + (m)(V)
6v - 8v + mV = 0
mV - 2v = 0
mV = 2v
V = (2/m) v
Note that since we have established that the sign on m and v at both positive, the sign of the velocity of the third ball is also positive.
Hence, the velocity of the third ball according to our convention is to the right.
Hope this Helps!!!
Answer:
the third ball emerge from right end .
Explanation:
initially the three balls are at rest so u = 0
Given m1 = 6g
v1 = v(i) (Right end indicates positive X axis so it is represented as (i))
m2 = 4g
v2 = 2(v)(-i) (left end indicates negative X axis so it is represented as (-i))
m3 = m
v3 = v'
Applying conservation of momentum
m1(u1) + m2(u2) + m3(u3) = m1(v1) + m2(v2) + m3(v3)
u1 = u2 = u3 = u = 0 ( at rest)
0 = (6×10⁻³)(v)(i) + (4×10⁻³)(2v)(-i) + m(v')
m(v') = (2×10⁻³)(v)(i)
v' = (2×10⁻³)(v)(i)/(m)
So positive (i) indicates right end
So the third ball emerge from right end .
2. On January 21 in 1918, Granville, North Dakota, had a surprising change in temperature. Within 12 hours, the temperature changed from 237 K to 283 K. What is this change in temperature in the Celsius and Fahrenheit scales?
Answer:
The temperature change in Celsius is 46°C.
The temperature change in Fahrenheit is 82.8°F.
Explanation:
A degree of Celsius scale is equal to that of kelvin scale; therefore,
[tex]\Delta C = \Delta K = 283K-237K\\\\ \boxed{\Delta C = 46^oC}[/tex]
A degree in Fahrenheit is 1.8 times the Celsius degree; therefore
[tex]\Delta F = 1.8(46^o)[/tex]
[tex]\boxed{\Delta F = 82.8^oF}[/tex]
Hence, the temperature change in Celsius is 46°C, and the temperature change in Fahrenheit is 82.8°F.
Final answer:
The temperature change of 46 degrees from Kelvin to Celsius in Granville, North Dakota, is equivalent to a change of 82.8 degrees Fahrenheit.
Explanation:
On January 21, 1918, the temperature in Granville, North Dakota, changed from 237 K to 283 K within 12 hours. To convert the change in temperature to the Celsius scale, we first recognize that 0 degrees Celsius is equivalent to 273.15 K. Therefore, the initial temperature in Celsius would have been 237 K - 273.15 K = -36.15 extdegree C, and the final temperature would have been 283 K - 273.15 K = 9.85 extdegree C. The change in the Celsius scale is then 9.85 extdegree C - (-36.15 extdegree C) = 46 extdegree C
To convert the change in temperature to the Fahrenheit scale, we start with the change in Celsius and use the conversion formula (F = C imes \frac{9}{5} + 32). Since we are looking for the change, we only need to convert the difference in Celsius to Fahrenheit. Therefore, the change in Fahrenheit is 46 extdegree C imes \frac{9}{5} = 82.8 extdegree F.
4. Analyze: What can you say about the acceleration of dividers when the pressure increases
from left to right?
Velocity changes with time and this is called acceleration.
What is acceleration?The question is incomplete but I will try to explain the meaning of acceleration to you. The term acceleration refers to the change of velocity with time.
a = Δv/t
a = accelerationΔv = change in velocityt = time intervalHence, velocity changes with time and this is called acceleration.
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The increase of the pressure of the system would lead to an increase of acceleration.
What happens when the pressure equates to the force of action?As pressure equals to the force upon action. Here F is the force and the A the area. As Newton's second law F = ma We substitute P A = m a
Hence the pressure is directly proportional to acceleration.
Find out more information about acceleration.
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In a collision between two unequal masses, how does the impulse imparted to the smaller mass by the larger mass compare with the impulse imparted to the larger mass by the smaller one?
Answer:
They are equal
Explanation:
Newton 3rd Law of motion states that for every force applied or action there is usually an equal and opposite force.
Impulse = Force * time
It is measured in Newton seconds.
The force and time of collision is the same which translates to an equal impulse by both scenarios.
The impulse imparted to the smaller mass by the larger mass is equal in magnitude to the impulse imparted to the larger mass by the smaller one, due to the conservation of momentum and Newton's third law. Both experience equal and opposite momentum transfers, ensuring the total momentum of the system remains constant.
Explanation:In the context of collisions, impulse is defined as the change in momentum of an object when it is subjected to a force over a period of time. According to Newton's third law, 'For every action, there is an equal and opposite reaction,' meaning that the impulse imparted to the smaller mass by the larger mass is exactly equal in magnitude to the impulse imparted to the larger mass by the smaller one, although the direction of the impulses will be opposite. When considering the conservation of momentum, the total momentum before the collision must equal the total momentum after the collision if no external forces are acting on the system (assuming a closed system). Therefore, if two cars collide, such as described in the provided text, regardless of their masses, the momentum transfer will be the same for both, thus the total momentum of the system remains constant.
How much current does a 100-W lamp draw when connected to 120 V?
P/V = I
Answer:
0.83 amps is what I believe is the answer.
a 500 pound metal star is hanging on two cables which are attached to the ceiling. the left hand cable makes a 13 degree angle with the ceiling while the rigth hand cables makes a 77 degree angle with the ceiling. What is the tension on each of the cables
Answer:
T1 = 112.07[lb]
T2 = 487.3 [lb]
Explanation:
To solve this problem we must perform a static balance analysis, for this we perform a free body diagram. In this free body diagram we use the angles mentioned in the description of the problem.
Performing a sum of forces on the X-axis equal to zero, we can find an equation that relates the tension of the T1 & T2 cables.
Then we perform a summation of forces on the Y-axis, in which we can find another equation. In this new equation, we replace the previous one and we can find the tension T2.
T1 = 112.07[lb]
T2 = 487.3 [lb]
A mass spectrometer applies a voltage of 2.00 kV to accelerate a singly charged positive ion. A magnetic field of B = 0.400 T then bends the ion into a circular path of radius 0.305 m. What is the mass of the ion?
The mass of the ion is 5.96 X 10⁻²⁵ kg
Explanation:
The electrical energy given to the ion Vq will be changed into kinetic energy [tex]\frac{1}{2}mv^2[/tex]
As the ion moves with velocity v in a magnetic field B then the magnetic Lorentz force Bqv will be balanced by centrifugal force [tex]\frac{mv^2}{r}[/tex].
So,
[tex]Vq = \frac{1}{2}mv^2[/tex]
and
[tex]Bqv = \frac{mv^2}{r}[/tex]
Right from these eliminating v, we can derive
[tex]m = \frac{B^2r^2q}{2V}[/tex]
On substituting the value, we get:
[tex]m = \frac{(0.4)^2X (0.305)^2 X1.602X 10^-^1^9}{2X 2000}\\\\[/tex]
m = 5.96 X 10⁻²⁵ kg.
The mass of the ion is found using the principles of kinetic and magnetic forces, is approximately 9.65 x 10⁻²⁷ kg.
To find the mass of the ion, we use the principles of energy and magnetic force. When a singly charged positive ion is accelerated through a voltage (V), it gains kinetic energy equal to the electrical potential energy provided by the voltage:
Kinetic Energy (KE) = qV
Since the ion starts from rest, this kinetic energy is also given by:
Kinetic Energy (KE) = (1/2)mv²
Equating these two expressions for kinetic energy gives:
qV = (1/2)mv²
Solving for the velocity (v), we get:
v =√(2qV/m)
When the ion enters the magnetic field (B) and bends into a circular path, the centripetal force required is provided by the magnetic force:
Magnetic Force = qvB
The centripetal force is also given by:
Centripetal Force = mv²/r
Equating these two expressions for the forces, we get:
qvB = mv²/r
Solving for the mass (m) of the ion:
m = qBr/v
We already have v = √(2qV/m), substituting this into the equation above to eliminate v, we get:
m = (qBr) / √(2qV/m)
Square both sides to simplify and solve for m:
m² = (q²B²r²) / (2qV)
m = (qB²r²) / (2V)
Using the given values (q = 1.602 x 10⁻¹⁹ C as charge of a singly charged positive ion, V = 2000 V, B = 0.400 T, and r = 0.305 m), we get:
m = (1.602 x 10⁻¹⁹ C x (0.400 T)² x (0.305 m)²) / (2 x 2000 V)
Calculating this gives the mass of the ion approximately equal to:
m ≈ 9.65 x 10⁻²⁷ kg
A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.827 m and 2.89 kg, respectively. When the propeller rotates at 507 rpm (revolutions per minute), what is its rotational kinetic energy K
Answer:
3241.35J
Explanation:
No. Of rods = 5
Mass = 2.89kg
Length (L) = 0.827m
W = 507rpm
Kinetic energy of rotation = ½I*ω²
For each rod, the moment of inertia (I) = ML² / 3
I = ML² / 3
I = [2.89*(0.827)²] / 3
I = 1.367 / 3 = 0.46kgm²
ω = 507 rev/min. Convert rev/min to rev/sec.
507 * 2Πrads/60s = 53.09rad/s
ω = 53.09rad/s
k.e = ½ I * ω²
K.E = ½ * 0.46 * (53.09)²
K.E = 648.27.
But there five (5) rods, so kinetic energy is equal to
K.E = 5 * 648.27 = 3241.35J
The rotational kinetic energy of a propeller with five blades, each modeled as a uniform rod, can be determined using formulas for moment of inertia and rotational kinetic energy.
Explanation:The kinetic energy of rotating objects depends on their moment of inertia and angular velocity. Given that each propeller blade is modeled as a uniform rod rotating about its end, we can calculate the moment of inertia (I) of all five blades using the formula I=5*(1/3*m*L^2), where m is the mass and L is the length of each rod. After we find the moment of inertia, we can determine the angular velocity (ω) in radian per second, given that the propeller rotates at 507 rpm, by using the conversion factor ω=(507*2*pi)/60. Finally, we calculate the rotational kinetic energy (K) using the formula K=1/2*I*ω^2.
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The ultracentrifuge is an important tool for separating and analyzing proteins. Because of the enormous centripetal accelerations, the centrifuge must be carefully balanced, with each sample matched by a sample of identical mass on the opposite side. Any difference in the masses of opposing samples creates a net force on the shaft of the rotor, potentially leading to a catastrophic failure of the apparatus. Suppose a scientist makes a slight error in sample preparation and one sample has a mass 10 mg larger than the opposing sample.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The net force is [tex]F_{net}= 6.44 *10^{-4} N[/tex]
Explanation:
Generally the net force is a force that come up due to the unequal centripetal force(A difference in centripetal force ) and it is mathematically represented as
[tex]F_{net} = \Delta F_{cen}[/tex]
and the difference in centripetal force [tex]\Delta F_{cen}[/tex] is mathematically represented as
[tex]\Delta F_{cen} = \Delta m* rw^2[/tex]
Which the difference in mass multiplied by the centripetal acceleration
Substituting 10 mg = [tex]10 *10^{-3}g[/tex] for [tex]\Delta m[/tex] , 12 cm = [tex]\frac{12}{100} = 0.12m[/tex] for radius
and 70,000 rpm = [tex]70,000 *[\frac{2 \pi rad}{1 rev}][\frac{1 min}{60s} ] = 7326.7 rad/s[/tex]
[tex]F_{net} = \Delta F_{cen} = 10*10^{-3} * 0.12 * 7326.7[/tex]
[tex]F_{net}= 6.44 *10^{-4} N[/tex]
The speed of light in a vacuum is 300,000.0 kmls. How long does it take light to travel
150,000,000 km from the Sun to Earth?
Answer:
8 minutes
Explanation:
The time it takes for light to travel from the Sun to Earth can be found using the formula time = distance/speed. Given the speed of light is known to be 300,000 km/s and the distance is 150,000,000 km, it will take light about 500 seconds to travel from the Sun to the Earth.
Explanation:To calculate the time it takes for light to travel from the Sun to Earth, you need to use the formula for time which is distance/speed. The speed of light is 300,000 km/s and the distance from the Sun to Earth is 150,000,000 km.
So, Time = Distance / Speed = 150,000,000 km / 300,000 km/s = 500 seconds
Therefore, it takes approximately 500 seconds for light to travel from the Sun to Earth.
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The magnetic dipole moment of Earth has magnitude 8.00 1022 J/T.Assume that this is produced by charges flowing in Earth’s molten outer core. If the radius of their circular path is 3500 km, calculate the current they produce.
Answer:
[tex]2.08\cdot 10^9 A[/tex]
Explanation:
The magnetic dipole moment of a circular coil with a current is given by
[tex]\mu = IA[/tex]
where
I is the current in the coil
[tex]A=\pi r^2[/tex] is the area enclosed by the coil, where
[tex]r[/tex] is the radius of the coil
So the magnetic dipole moment can be rewritten as
[tex]\mu = I\pi r^2[/tex] (1)
Here we can assume that the magnetic dipole moment of Earth is produced by charges flowing in Earth’s molten outer core, so by a current flowing in a circular path of radius
[tex]r=3500 km = 3.5\cdot 10^6 m[/tex]
Here we also know that the Earth's magnetic dipole moment is
[tex]\mu = 8.0\cdot 10^{22} J/T[/tex]
Therefore, we can re-arrange eq (1) to find the current that the charges produced:
[tex]I=\frac{\mu}{\pi r^2}=\frac{8.00\cdot 10^{22}}{\pi (3.5\cdot 10^6)^2}=2.08\cdot 10^9 A[/tex]
A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 6.3 m/s2. For what value of the coefficient of static friction between the truck bed and a cabinet will the cabinet slip along the bed surface?
Answer:
Coefficient of static friction will be equal to 0.642
Explanation:
We have given acceleration [tex]a=6.3m/sec^2[/tex]
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
We have to find the coefficient of static friction between truck and a cabinet will
We know that acceleration is equal to [tex]a=\mu g[/tex], here [tex]\mu[/tex] is coefficient of static friction and g is acceleration due to gravity
So [tex]\mu =\frac{a}{g}=\frac{6.3}{9.8}=0.642[/tex]
So coefficient of static friction will be equal to 0.642
Consider a completely elastic head-on collision between two particles that have the same mass and the same speed. What are the velocities of the particles after the collision?
Answer:
The magnitude of their velocities will be the same but their direction will be reversed.
Final answer:
In a completely elastic collision between two particles of identical mass and equal but opposite velocities, both particles simply swap their velocities as a result of the collision, conserving both momentum and kinetic energy.
Explanation:
In a completely elastic head-on collision between two particles with identical mass and equal but opposite speeds, the outcome is quite straightforward. Since the collision is elastic, both conservation of momentum and conservation of kinetic energy apply.
Before the collision, let's assume particle 1 with mass m has velocity v, and particle 2 also with mass m has velocity -v. The total momentum before collision would be m*v + m*(-v) = 0, and the total kinetic energy would be (1/2)*m*v² + (1/2)*m*(-v)².
For identical masses and equal and opposite velocities, the particles simply swap velocities after collision. Therefore, the first particle will have a velocity of -v and the second a velocity of v post-collision, as momentum and kinetic energy are conserved. They essentially 'bounce' off each other and move in the opposite directions with the same speed they had before the collision.
A 0.450-kg hockey puck, moving east with a speed of 5.25 m/s , has a head-on collision with a 0.850-kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed (magnitude of the velocity) of each object after the collision
Explanation:
Given that,
Mass of the hockey puck, m₁ = 0.45 kg
Initial peed of the hockey puck, u₁ = 5.25 m/s (east)
Mass of other puck, m₁ = 0.85 kg
Initial speed of other puck, u₂ = 0 (at rest)
Let v₁ and v₂ are the final speeds of both pucks after the collision respectively. Using the conservation of momentum as :
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\m_1v_1+m_2v_2=0.45\times 5.25+0.85\times 0\\\\m_1v_1+m_2v_2=2.36\\\\0.45v_1+0.85v_2=2.36.........(1)[/tex]
The coefficient of restitution for elastic collision is equal to 1.
[tex]C=\dfrac{v_2-v_1}{u_1-u_2}\\\\1=\dfrac{v_2-v_1}{u_1-u_2}\\\\1=\dfrac{v_2-v_1}{5.25-0}\\\\v_2-v_1=5.25.......(2)[/tex]
On solving equation (1) and (2) we get :
[tex]v_1=-1.611\ m/s\\\\v_2=3.63 m/s[/tex]
Hence, this is the required solution.
Ender and Shen are flying at each other during a battle in space. Ender moves with a velocity v1 of 12 m/s and Shen, with a mass of 45 kg, moves at a velocity v2 of 9.0 m/s. When the two collide, they hold on to each other and move with a velocity v3 of 6.4 m/s. What is Ender’s mass, in kg?
Final answer:
Using the conservation of momentum for an inelastic collision, Ender's mass is calculated to be approximately 51.43 kg when the given velocities and Shen's mass are considered.
Explanation:
The subject of this question is physics, specifically dealing with the concept of conservation of momentum during collisions. We are asked to find Ender's mass given the velocities and mass involved in a collision in space. Since the collision here is inelastic (Ender and Shen hold on to each other), the total momentum before the collision must be equal to the total momentum after the collision.
According to the conservation of momentum, m1 × v1 + m2 × v2 = (m1 + m2) × v3, where m1 and m2 are the masses of Ender and Shen respectively, and v1, v2, and v3 are their velocities.
We know Shen's mass (m2) is 45 kg, Ender's velocity (v1) is 12 m/s, Shen's velocity (v2) is 9 m/s, and the combined velocity after the collision (v3) is 6.4 m/s. Using the momentum conservation formula, we can solve for Ender's mass (m1) as follows:
m1 × 12 m/s + 45 kg × 9.0 m/s = (m1 + 45 kg) × 6.4 m/s
Expanding this and rearranging the terms, we get:
12m1 = 6.4m1 + 6.4 × 45
12m1 - 6.4m1 = 6.4 × 45
m1(12 - 6.4) = 6.4 × 45
m1 = (6.4 × 45) / (12 - 6.4)
m1 = 288 / 5.6
m1 = 51.43 kg
Therefore, Ender's mass is approximately 51.43 kg.
Angular oscillation of the slotted link is achieved by the crank OA, which rotates clockwise at the steady speed N = 118 rev/min. Find the angular velocity of the slotted link when θ = 61°
Answer:
The angular velocity is 4.939rad/sec
Explanation:
θ = 61°
N = 118 rev/min
The formula for angular velocity is given as;
w = θ/t
where;
w = angular velocity
θ = angle
t = time in rad/sec
1 rev/ min = 0.1047 radian/second.
Therefore 118 rev/min = 118*0.1047rad/sec
=12.35rad/sec
Substituting into the formula, we have
w = 61/12.35
w = 4.939rad/sec
A helicopter lifts a 81 kg astronaut 19 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/15. How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her
Answer:
a) The work done on the astronaut by the force from the helicopter is [tex]W_{h}=16087.68\ J[/tex].
b) The work done on the astronaut by the gravitational force is [tex]W_{g}=-15082.2\ J[/tex] .
Explanation:
We are told that the mass of the astronaut is [tex]m=81\ kg[/tex], the displacement is [tex]\Delta x=19\ m[/tex], the acceleration of the astronaut is [tex]|\vec{a}|=\frac{g}{15}[/tex] and the acceleration of gravity is [tex]g=9.8\ \frac{m}{s^{2}}[/tex] .
We suppose that in the vertical direction the force from the helicopter [tex]F_{h}[/tex] is upwards and the gravitational force [tex]F_{g}[/tex] is downwards. From the sum of forces we can get the value of [tex]F_{h}[/tex]:
[tex]F_{h}-F_{g}=m.a[/tex]
[tex]F_{h}-mg=m.\frac{g}{15}[/tex]
[tex]F_{h}=mg(1+\frac{1}{15})[/tex]
[tex]F_{h}=(\frac{16}{15}).81\ kg.\ 9.8\ \frac{m}{s^{2}}\ \Longrightarrow\ F_{h}=846.72\ N[/tex]
We define work as the product of the force, the displacement of the body and the cosine of the angle [tex]\theta[/tex] between the direction of the force and the displacement of the body:
[tex]W=F.\Delta x.\ cos(\theta)[/tex]
a) The work done on the astronaut by the force from the helicopter
[tex]W_{h}=F_{h}.\Delta x[/tex]
[tex]W_{h}=846.72\ N.\ 19\ m[/tex]
[tex]W_{h}=16087.68\ J[/tex]
b) The work done on the astronaut by the gravitational force
[tex]W_{g}=-F_{g}.\Delta x[/tex]
[tex]W_{g}=-mg\Delta x[/tex]
[tex]W_{g}=-81\ kg.\ 9.8\ \frac{m}{s^{2}}.\ 19\ m[/tex]
[tex]W_{g}=-15082.2\ J[/tex]
Answer:
a) Work done on the astronaut by the force from the helicopter = 16.104 kJ
b) Work done on the astronaut by the gravitational force = -15.082 kJ
Explanation:
mass of the astronaut, m = 81 kg
height, h = 19 m
acceleration of the astronaut, a = g/15
Since the astronaut is lifted up, using the third law of motion:
T - mg = ma
T = mg + ma
T = (81*9.81) + 81*(9.81/15)
T = 847.584 N
Work done on the astronaut by the helicopter
Work done = Tension * height
W = T* h
W = 847.584 * 19
Work done, W = 16104.096 Joules
W = 16.104 kJ
b) Work done on the astronaut by the gravitational force on her
[tex]W = -f_{g} h[/tex]
[tex]f_{g} = mg = 81 * 9.8\\f_{g} = 793.8 N[/tex]
[tex]W = -793.8 * 19\\W =- 15082.2 J[/tex]
W = -15.082 kJ
Charge is distributed uniformly along a long straight wire. The electric field 2.00 cm from the wire is 20.0 NC/ directed radially inward towards the axis of symmetry. The linear charge density on the wire is
The linear charge density on the wire is approximately 0.002 N/Cm.
Explanation:The linear charge density on the wire can be calculated using the formula:
λ = E × r / 2πk
where λ is the linear charge density, E is the electric field, r is the distance from the wire, and k is the Coulomb's constant. Plugging in the known values of E = 20.0 N/C and r = 2.00 cm = 0.02 m, we can solve for λ:
λ = (20.0 N/C) × (0.02 m) / (2π × 8.99 × 10^9 Nm^2/C^2) ≈ 0.002 N/Cm
Therefore, the linear charge density on the wire is approximately 0.002 N/Cm.
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The uniform slender bar AB has a mass of 7 kg and swings in a vertical plane about the pivot at A. If angular velocity of the bar is 3 rad/s when θ = 35o , compute the force supported by the pin at A at that instant
Answer:
49 N
Explanation:
The diagram of the bar is obtained online and attached to this solution.
The free body diagram is also attached.
Since the weight of the bar acts at the middle of the bar, the torque due to the weight of the bar is given by
τ = mgx
where x = (L/2) cos 35° = 0.45 × cos 35° = 0.3686 m
τ = (7)(9.8)(0.3686) = 25.29 Nm
The force acting on pin A = torque ÷ (length × sin 35°) = 25.29 ÷ (0.9 × sin 35°)
= 25.29 ÷ 0.5162 = 48.99 N = 49 N
Hope this Helps!!!
Answer:
The force supported by the pin at A is 69.081 N
Explanation:
The diagram is in the figure attach. The angular acceleration using the moment expression is:
[tex]-mg(\frac{Lcos\theta }{2} )=I\alpha \\\alpha =\frac{-3g}{2L} cos\theta[/tex]
Where
L = length of the bar = 900 mm = 0.9 m
[tex]\alpha =\frac{-3*9.8cos35}{2*0.9} =-13.38rad/s^{2}[/tex]
The acceleration in point G is equal to:
[tex]a_{G} =a_{A} +\alpha kr_{G/A} -w^{2} r_{G/A}[/tex]
Where
aA = acceleration at A = 0
w = angular velocity of the bar = 3 rad/s
rG/A = position vector of G respect to A = [tex]\frac{L}{2} cos\theta i-\frac{L}{2} cos\theta j[/tex]
[tex]a_{G} =(\frac{L}{2}\alpha sin\theta -\frac{w^{2}Lcos\theta }{2} )i+(\frac{L}{2}\alpha cos\theta +\frac{w^{2}Lsin\theta }{2} )j=(\frac{0.9*(-13.38)*sin35}{2} -(\frac{3^{2}*0.9*cos35 }{2} )i+(\frac{0.9*(-13.38)*cos35}{2} +\frac{3^{2} *0.9*sin35)}{2} )j=-6.76i-2.61jm/s^{2}[/tex]
The force at A in x is equal to:
[tex]A_{x} =ma_{G} =7*(-6.76)=-47.32N[/tex]
The force at A in y is:
[tex]A_{y} =ma_{G} +mg=(7*(-2.61))+(7*9.8)=50.33N[/tex]
The magnitude of force A is equal to:
[tex]A=\sqrt{A_{x}^{2}+A_{y}^{2} } =\sqrt{(-47.32^{2})+50.33^{2} } =69.081N[/tex]
A standing wave on a string that is fixed at both ends has frequency 80.0 Hz. The distance between adjacent antinodes of the standing wave is 20.0 cm. What is the speed of the waves on the string, in m/s?
The speed of waves on the string is 32 m/s.
Explanation:The speed of waves on a string can be calculated using the formula:
velocity = frequency × wavelength
In this case, the frequency is given as 80.0 Hz and the distance between adjacent antinodes is 20.0 cm. Since the wavelength is twice the distance between adjacent antinodes, we have:
wavelength = 2 × 20.0 cm = 40.0 cm = 0.4 m
Plugging these values into the formula, we get:
velocity = 80.0 Hz × 0.4 m = 32 m/s
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The compressive strength of our bones is important in everyday life. Young’s modulus for bone is about 1.4 * 1010 Pa. Bone can take only about a 1.0% change in its length before fracturing. (a) What is the maximum force that can be applied to a bone whose minimum cross-sectional area is 3.0 cm2 ? (This is approximately the cross-sectional area of a tibia, or shin bone, at its narrowest point.) (b) Estimate the maximum height from which a 70 kg man could jump and not fracture his tibia. Take the time between when he first touches the floor and when he has stopped to be 0.030 s, and assume that the stress on his two legs is distributed equally.
Answer:
Explanation:
Young modulus ε = 1.4 × 10¹⁰ Pa
ΔL = 1% of the original length = 0.01 x where x is the original length
cross sectional area = 3.0 cm² =( 3 .0 / 10000) m²= 0.0003 m²
ε = Stress / strain
stress = ε × strain
stress = F /A
F force = ε × A × ( ΔL / L) = 1.4 × 10¹⁰ Pa × 0.0003 m² × 0.01 = 4.2 × 10⁴ N
b) F net = F max - mg ( weight) = 84000 - ( 70 × 9.8 m/s² ) ( F is double since the stress on the two leg is equally distributed)
f net = ma = 84000 - ( 70 × 9.8 m/s² )
a = (84000 - ( 70 × 9.8 m/s² )) = 1190.2 m/s²
v = u + at
where final velocity equal zero
- u = -at since it coming downwards
u = at = 1190.2 m/s² × 0.03s = 35.706 m/s
using conservation of energy
1/2 mv² = mgh
1/2v²/ g = h
h = 0.5 × (35.706 m/s )² / 9.8 = 65.04 m
A magnetic dipole with a dipole moment of magnitude 0.0243 J/T is released from rest in a uniform magnetic field of magnitude 57.5 mT. The rotation of the dipole due to the magnetic force on it is unimpeded. When the dipole rotates through the orientation where its dipole moment is aligned with the magnetic field, its kinetic energy is 0.458 mJ. (a) What is the initial angle between the dipole moment and the magnetic field
Answer:
47.76°
Explanation:
Magnitude of dipole moment = 0.0243J/T
Magnetic Field = 57.5mT
kinetic energy = 0.458mJ
∇U = -∇K
Uf - Ui = -0.458mJ
Ui - Uf = 0.458mJ
(-μBcosθi) - (-μBcosθf) = 0.458mJ
rearranging the equation,
(μBcosθf) - (μBcosθi) = 0.458mJ
μB * (cosθf - cosθi) = 0.458mJ
θf is at 0° because the dipole moment is aligned with the magnetic field.
μB * (cos 0 - cos θi) = 0.458mJ
but cos 0 = 1
(0.0243 * 0.0575) (1 - cos θi) = 0.458*10⁻³
1 - cos θi = 0.458*10⁻³ / 1.397*10⁻³
1 - cos θi = 0.3278
collect like terms
cosθi = 0.6722
θ = cos⁻ 0.6722
θ = 47.76°
Answer:
The initial angle between the dipole moment and the magnetic field is 47.76⁰
Explanation:
Given;
magnitude of dipole moment, μ = 0.0243 J/T
magnitude of magnetic field, B = 57.5 mT
change in kinetic energy, ΔKE = 0.458 mJ
ΔKE = - ΔU
ΔKE = - (U₂ -U₁)
ΔKE = U₁ - U₂
U₁ -U₂ = 0.458 mJ
[tex](-\mu Bcos \theta_i )- (-\mu Bcos \theta_f) = 0.458 mJ\\\\-\mu Bcos \theta_i + \mu Bcos \theta_f = 0.458 mJ\\\\\mu Bcos \theta_f -\mu Bcos \theta_i = 0.458 mJ\\\\\mu B(cos \theta_f - cos \theta_i ) = 0.458 mJ[/tex]
where;
θ₁ is the initial angle between the dipole moment and the magnetic field
[tex]\theta_f[/tex] is the final angle which is zero (0) since the dipole moment is aligned with the magnetic field
μB(cos0 - cosθ₁) = 0.458 mJ
Substitute the given values of μ and B
0.0243 x 0.0575 (1 - cosθ₁) = 0.000458
0.00139725 (1 - cosθ₁) = 0.000458
(1 - cosθ₁) = 0.000458 / 0.00139725
(1 - cosθ₁) = 0.327787
cosθ₁ = 1 - 0.327787
cosθ₁ = 0.672213
θ₁ = cos⁻¹ (0.672213)
θ₁ = 47.76⁰
Thus, the initial angle between the dipole moment and the magnetic field is 47.76⁰
2. A 3.5 m long string is fixed at both ends and vibrates in its 7th harmonic with an amplitude (at an antinode) of 2.4 cm. If the speed of waves on the string is 150 m/s, what is the maximum speed for a point on the string at an antinode?
Answer:
Maximum speed for a point on the string at anti node will be 22.6 m/sec
Explanation:
We have given length of string L = 3.5 m
For 7th harmonic length of the string [tex]L=\frac{7\lambda }{2}[/tex]
So [tex]\lambda =\frac{2L}{7}[/tex]
Speed of the wave in the string is 150 m/sec
Frequency corresponding to this wavelength [tex]f=\frac{v}{\lambda }=\frac{7v}{2L}[/tex]
So angular frequency will be equal to [tex]\omega =2\pi f=2\pi \times \frac{7v}{2L}=2\times 3.14\times \frac{7\times 150}{2\times 3.5}=942rad/sec[/tex]
Maximum speed is equal to [tex]v_m=A\omega =0.024\times 942=22.60m/sec[/tex]
So maximum speed for a point on the string at anti node will be 22.6 m/sec
Assuming that 10.0% of a 100-W light bulb's energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, and that the photons spread out uniformly and are not absorbed by the atmosphere, how far away would you be if 500 photons per second enter the 3.00-mm diameter pupil of your eye? (This number easily stimulates the retina.) (answer in km)
Answer:
R = 1.81 10² km
Explanation:
Let's start by looking for the power in the visible range emitted this is 10W, the energy of that power is one second is
P = E₁ / t
E₁ = P t
E₁ = 10 J
Let's find the energy of a photon with Planck's equation
E = h f
c = λ f
we substitute
E = h c /λ
E = 6.63 10⁻³⁴ 3 10⁸/580 10⁻⁹
E = 3.42 10⁻¹⁹ J
we can use a direct proportions rule to find the number of photons in the energy E₁
#_photon = E₁ / E
#_photon = 10 / 3.42 10⁻¹⁹
#_photon = 2.92 10¹⁹ photons
This number of photons is distributed on the surface of a sphere. Let's find what the distance is so that there are 500 photons in 3 mm = 0.003 m.
the area of the sphere is
A = 4π R²
area of the circle is
A´ = π r²
as the intensity is constant over the entire sphere
P = #_photon / A = 500 / A´
# _photon / 4π R² = 500 / π r²
R² = #_photon r² / 4 500
r = d / 2 = 0.003 / 2 = 0.0015 m
R² = 2.92 10¹⁹ 0.0015 2/2000
R = √ (3,285 10¹⁰)
R = 1.81 10⁵ m
R = 1.81 10² km
A cyclist coasts up a 9.00° slope, traveling 12.0 m along the road to the top of the hill. If the cyclist’s initial speed is 9.00 m/s, what is the final speed? Ignore friction and air resistance.
Answer:
[tex]v_{f} \approx 6.647\,\frac{m}{s}[/tex]
Explanation:
The final speed of the cyclist is determined by applying the Principle of Energy Conservation:
[tex]\frac{1}{2}\cdot m\cdot v_{o}^{2} + m\cdot g\cdot h_{o} = \frac{1}{2}\cdot m\cdot v_{f}^{2} + m\cdot g\cdot h_{f}[/tex]
[tex]\frac{1}{2} \cdot v_{o}^{2} + g\cdot (h_{o}-h_{f}) = \frac{1}{2}\cdot v_{f}^{2}[/tex]
[tex]v_{f}^{2}=v_{o}^{2} + 2\cdot g \cdot (h_{o}-h_{f})[/tex]
[tex]v_{f} = \sqrt{v_{o}^{2}+2\cdot g \cdot (h_{o}-h_{f})}[/tex]
[tex]v_{f} = \sqrt{v_{o}^{2}-2\cdot g \cdot \Delta s \cdot \sin \theta}[/tex]
[tex]v_{f} = \sqrt{(9\,\frac{m}{s} )^{2}-2\cdot (9.807\,\frac{m}{s} )\cdot (12\,m)\cdot \sin 9^{\textdegree}}[/tex]
[tex]v_{f} \approx 6.647\,\frac{m}{s}[/tex]
Answer:
the final speed of the 10.85 m/s.
Explanation:
Given that,
Slope with respect to horizontal, [tex]\theta=9^{\circ}[/tex]
Distance travelled, d = 12 m
Initial speed of the cyclist, u = 9 m/s
We need to find the final speed of the cyclist. Let h is the height of the sloping surface such that,
[tex]h=d\times sin\thetah\\\\=12\times sin(9) \\\\h = 1.877 m[/tex]
v is the final speed of the cyclist. It can be calculated using work energy theorem as
[tex]\dfrac{1}{2}m(v^2-u^2)\\\\=mgh\dfrac{1}{2}(v^2-u^2)\\\\=gh\dfrac{1}{2}\times (v^2-(9.0)^2)\\\\=9.8\times 1.87\\\\v = 10.85 m/s[/tex]
Thus,the final speed of the 10.85 m/s.
What are potential impacts of pollution on a watershed? Check all that apply.
excess runoff
loss of farmland
loss of areas for tourism
contaminated drinking water
deaths of sea creatures that are used as a food source
limits to potential economic activities such as a fishing
Answer:
contaminated drinking water
deaths of sea creatures that are used as a food source
limits to potential economic activities such as a fishing
Explanation:
A watershed is a large area that comprises of drainage area of all the surrounding water bodies meeting at a common affluence point before draining into sea or ocean or any other large water body. Pollution in this area can pollute the small water streams flowing through it, thereby polluting the larger water body into which it drains.
Thus, the water extracted for drinking from such area will be contaminated. Pollution in larger water body can cause death of water creature and hence pose a threat to fishing.
Answer:
loss of areas for tourism
contaminated drinking water
deaths of sea creatures that are used as a food source
limits to potential economic activities such as fishing
For the questions below, answer if the statement is true or false and explain how you reach to your conclusion. a. The Clausius statement denies the possibility of heat transfer from a cooler to a hotter body. b. The COP of a reversible refrigeration cycles is equal or greater than the COP of an irreversible refrigeration cycle if both cycles operate between the same thermal reservoirs. c. Mass, energy, and temperature are the examples of intensive properties. d. For reversible refrigeration and heat pump cycles operating between the same hot and cold reservoirs, the relation between their coefficients of performance is COPHP = COPR + 1. e. For a reversible heat pump that operates between cold and hot thermal reservoirs at 350°C and 550°C, respectively, the COP is equal to 4.11. f. In the absence of any friction and other irreversibilities, a heat engine can achieve an efficiency of 100%. g. A refrigerator, with a COP of 1.2, rejects 60 kJ/min from a refrigerated space when the electric power consumed by the refrigerator is 50 kJ/min. This refrigerator violates the first law of thermodynamics.
Answer:
a) Seems true at first glance, but on further inspection, the statement is false.
b) True
c) False
d) True
e) True
f) False
g) False
Explanation:
Taking the statements one by one
a.) The Clausius statement denies the possibility of heat transfer from a cooler to a hotter body.
The Clausius Statement denies the possibility of heat transfer from a cooler to a hotter body without extra work. It does not outrightly state that there is no possibility of heat transfer from a cooler to a hotter body.
For example, an Air conditioner or refrigerator rejects heat from a cold reservoir to a hot reservoir.
So, this statement is false.
b) The COP of a reversible refrigeration cycles is equal or greater than the COP of an irreversible refrigeration cycle if both cycles operate between the same thermal reservoirs.
The Coefficient of Performance of a reversible cycle is the maximum efficiency possible. It is the efficiency of a Carnot Engine.
Hence, it is greater than or equal to the Coefficient of Performance of an irreversible cycle.
COP(reversible) ≥ COP(irreversible)
This statement is true.
c. Mass, energy, and temperature are the examples of intensive properties.
Intensive properties are properties of thermodynamic systems that do not depend on the extent of the system. They are the same for a particular size of substance and stay the same if the size of the substance is doubled or halved. Examples include temperature, specific capacity, specific volume, every specific property basically, etc.
Extensive properties depend on the extent of the system. They double or half when the size of the extent doubles or halves respectively.
Mass and Energy are Extensive properties.
Temperature is the only intensive property among these options.
This statement is false.
d. For reversible refrigeration and heat pump cycles operating between the same hot and cold reservoirs, the relation between their coefficients of performance is
COPHP = COPR + 1.
The coefficients of performance for reversible refrigeration and heat pump cycles operating between the same hot and cold reservoirs are indeed related through
COP(HP) = COP(R) + 1
COP of a heat pump = COP(HP) = (Qh/W)
COP of a refrigerator = COP(R) = (Qc/W)
But, Qh = Qc + W
Divide through by W
(Qh/W) = (Qc/W) + (W/W)
COP(HP) = COP(R) + 1 (Proved!)
This Statement is true.
e. For a reversible heat pump that operates between cold and hot thermal reservoirs at 350°C and 550°C, respectively, the COP is equal to 4.11.
The COP is given as (1/efficiency).
Efficiency = 1 - (Tc/Th)
Tc = temperature of cold thermal reservoir in Kelvin = 350°C = 623.15 K
Th = temperature of hot thermal reservoir in Kelvin = 550°C = 823.15 K
Efficiency = 1 - (623.15/823.15)
= 1 - 0.757 = 0.243
COP = (1/Efficiency) = (1/0.243) = 4.11
This statement is true.
f. In the absence of any friction and other irreversibilities, a heat engine can achieve an efficiency of 100%.
In the absence of friction and other irreversibilities and for a heat engine to have 100% efficiency, the temperature of its cold reservoir has to be 0 K or the tempersture of its hot reservoir has to be infinity.
Efficiency = 1 - (Tc/Th)
For efficiency to be 1,
(Tc/Th) = 0; that is, Tc = 0 or Th = infinity
These two aren't physically possible and for 100% efficiency to happen, the heat engine will have to violate the Kelvin-Planck's statement of the second law of thermodynamics.
According to Kelvin-Planck's statement of the second law of thermodynamics, net amount of work cannot be produced by exchanging heat with single reservoir i.e. there will be another reservoir to reject heat.
Hence, a heat engine cannot have an efficiency of 100%.
This statement is false.
g. A refrigerator, with a COP of 1.2, rejects 60 kJ/min from a refrigerated space when the electric power consumed by the refrigerator is 50 kJ/min. This refrigerator violates the first law of thermodynamics.
The COP of a refrigerator is given as
COP = (Qcold)/W
Qcold = Heat rejected from the cold reservoir = Heat rejected from refrigerated space = 60 KJ/min
W = work done on the system = electrical power consumed = 50 KJ/min
COP = (60/50) = 1.2
This system does not violate the first law of thermodynamics.
This statement is false.
Hope this Helps!!!
An excited hydrogen atom releases an electromagnetic wave to return to its normal state. You use your futuristic dual electric/magnetic field tester on the electromagnetic wave to find the directions of the electric field and magnetic field. Your device tells you that the electric field is pointing in the negative z direction and the magnetic field is pointing in the positive y direction. In which direction does the released electromagnetic wave travel
Answer:
The answer is -x direction.
Explanation:
According to the right hand rule for electromagnetism, when the thumb points up, index finger points forward and the middle finger is perpendicular to the index finger, the thumb points in the direction of the magnetic force, the index finger in the direction of the charge movement and the middle finger in the direction of the electromagnetic lines. If the electric field is pointing in the -z direction which is into the screen and the magnetic field is in the +y direction which is upwards, then the magnetic lines are in the -x direction.
I hope this answer helps.
Starting at t = 0 s , a horizontal net force F⃗ =( 0.285 N/s )ti^+(-0.460 N/s2 )t2j^ is applied to a box that has an initial momentum p⃗ = ( -3.10 kg⋅m/s )i^+( 3.90 kg⋅m/s )j^ .
What is the momentum of the box at t = 1.90 s ?
Enter the x and y components of the momentum separated by a comma.
The momentum of the box at t = 1.90 s is - 2.586 i + 2.85j
And, the x and y component is - 2.586 and 2.85
How to calculate momentum?Since we know that
impulse = force * time
Here impulse means the change in momentum
Now we can write as in different way like
change in momentum = force * time
Also, Force = F = .285 t -.46t²
Due to variable force
change in momentum = ∫ F dt
So,
= ∫ .285ti - .46t²j dt
= .285 t² / 2i - .46 t³ / 3 j
Sine t = 1.9
So,
change in momentum = .285 * 1.9² /2 i - .46 * 1.9³ / 3 j
= .514i - 1.05 j
And,
final momentum
= - 3.1 i + 3.9j +.514i - 1.05j
= - 2.586 i + 2.85j
So, finally
x component = - 2.586
y component = 2.85
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the momentum of the box at t = 1.90 s is approximately (-2.8435 kg·m/s)i^ + (1.4007 kg·m/s)j^.
To find the momentum of the box at t = 1.90 s, we can integrate the net force over time to get the change in momentum.
The net force is given as:
F⃗ = (0.285 N/s)ti^ + (-0.460 N/s^2)t^2j^
To find the change in momentum, we integrate this force with respect to time from t = 0 to t = 1.90 s:
Δp⃗ = ∫(0 to 1.90) F⃗ dt
Δp⃗ = ∫(0 to 1.90) (0.285ti^ - 0.460t^2j^) dt
Now, we integrate each component separately:
Δpx = ∫(0 to 1.90) (0.285t) dt
Δpx = 0.285 * (t^2/2)|[0, 1.90]
Δpx = 0.285 * (1.90^2/2 - 0)
Δpx = 0.2565 N·s
Δpy = ∫(0 to 1.90) (-0.460t^2) dt
Δpy = -0.460 * (t^3/3)|[0, 1.90]
Δpy = -0.460 * (1.90^3/3 - 0)
Δpy = -2.4993 N·s
Now, we need to add the change in momentum to the initial momentum to find the momentum at t = 1.90 s:
p⃗(t = 1.90 s) = p⃗(t = 0 s) + Δp⃗
p⃗(t = 1.90 s) = (-3.10 kg·m/s)i^ + (3.90 kg·m/s)j^ + (0.2565 N·s)i^ + (-2.4993 N·s)j^
Now, add the x and y components separately:
p⃗(t = 1.90 s) = (-3.10 + 0.2565)i^ + (3.90 - 2.4993)j^
p⃗(t = 1.90 s) = (-2.8435 kg·m/s)i^ + (1.4007 kg·m/s)j^
So, the momentum of the box at t = 1.90 s is approximately (-2.8435 kg·m/s)i^ + (1.4007 kg·m/s)j^.
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A horizontal uniform bar of mass 2.5 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the end of the bar, and string 2 is attached a distance 0.74 m from the other end. A monkey of mass 1.25 kg walks from one end of the bar to the other. Find the tension T1 in string 1 at the moment that the monkey is halfway between the ends of the bar.
Answer:
The tension T₁ in string 1 at the moment that the monkey is halfway between the ends of the bar is 34.68 N
Explanation:
Given;
mass of the uniform horizontal bar, m₁ = 2.5 kg
length of the bar, L = 3.0 m
mass of the monkey, m₂ = 1.25 kg
distance from the right end of the second string, d = 0.74 m
For a body to remain in rotational equilibrium, the net external torque acting on it due to applied external forces must be equal.
∑τ = 0
For vertical equilibrium of bar-string system, in which T₁ and T₂ are the tension on both ends of the string;
T₁ + T₂ = (m₁ + m₂)g
T₁ + T₂ = (2.5 + 1.25) 9.8
T₁ + T₂ = 36.75 N
For rotational equilibrium when the monkey is halfway between the ends of the bar, take moment about the left end of the string.
(m₁ + m₂) L /2 = T₂(L - d)
(m₁ + m₂)0.5L = T₂( L - d)
(2.5 + 1.25)0.5 x 3 = T₂ ( 3 - 0.74)
4.6875 = T₂ (2.26)
T₂ = (4.6875) / (2.26)
T₂ = 2.074 N
Thus, T₁ = 36.75 N - T₂
T₁ = 36.75 N - 2.074 N
T₁ = 34.68 N
Sketch the solid whose volume is given by the integral and evaluate the integral. $ \int_{0}^{{\color{red}6}}\int_{0}^{2\pi }\int_{{\color{red}4} r}^{{\color{red}24}}\,r\,dz\,d\theta \,dr
Answer:
Explanation:
Attached is the evaluation
The given integral represents the volume of a solid bounded by specific surfaces and coordinates. To sketch this solid, imagine a cylindrical shape centered around the z-axis with varying radius and height. To evaluate the integral, integrate with respect to z, then θ, and finally r.
Explanation:First, let's consider the given integral: ∫_0^6 ∫_0^(2π) ∫_4^24 r dz dθ dr. This triple integral represents the volume of a solid. The limits of integration indicate that the solid is bounded by the surfaces r = 4 and r = 24, the angle θ ranges from 0 to 2π, and the z-coordinate extends from 0 to 6.
To sketch this solid, imagine a cylindrical shape centered around the z-axis, with a varying radius (r) extending from 4 to 24, and a height (z) ranging from 0 to 6. The angle θ represents rotation around the z-axis.
To evaluate the integral, we can first integrate with respect to z, then θ, and finally r. This involves applying the limits of integration and performing the calculations step by step. The final result will give us the volume of the solid.
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