Answer:
height solid / height hollow = 3/4
Explanation:
the height reached is directly proportional to the original KE of the object.
Total KE solid
=[tex]1/2mv^2+1/2\times1/2m r^2(v^2 / r^2 )=3/4mv^2[/tex]
Total KE hollow = [tex]1/2mv^2+1/2mr^2(v^2 / r^2 )=mv^2[/tex]
height solid / height hollow = 3/4
A sled plus passenger with total mass m = 52.3 kg is pulled a distance d = 21.8 m across a horizontal, snow-packed surface for which the coefficient of kinetic friction with the sled is μk = 0.13. The pulling force is constant and makes an angle of φ = 36.7 degrees above horizontal. The sled moves at constant velocity.a) Write an expression for the work done by the pulling force in terms of m, g (acceleration due to gravity), φ, μk, and d.b) What is the work done by the pulling force, in joules?c) Write an expression for the work done on the sled by friction in terms of m, g (acceleration due to gravity), φ, μk, and d.d) What is the work done on the sled by friction, in joules?
Answer:
Explanation:
Since the sled plus passenger moves with constant velocity , force applied will be equal to frictional force. Let the force applied be F
a ) Frictional force = μ R = F cosφ
R = mg - F sinφ
μ(mg - F sinφ) = F cosφ
μmg = F (μsinφ+cosφ)
F = μmg / (μsinφ+cosφ)
Work done
= F cosφ x d
= μmg x cosφ x d / (μsinφ+cosφ)
b )Work done
= 0.13 x 52.3 x 9.8 cos36.7 x 21.8 / ( 0.13 sin36.7 +cos36.7)
= 1164.61 / .87946
1324.23 J
c ) work done on the sled by friction
= - (work done by force)
= - μmg x cosφ x d / (μsinφ+cosφ)
d ) work done on the sled by friction
= - 1324.23 J
The work done by the pulling force is given by the formula W = F * d * cos(φ), and in this case, the force of friction can be calculated as Ff = μk * m * g * cos(φ).
Explanation:The work done by the pulling force is given by the formula W = F * d * cos(φ), where F is the magnitude of the pulling force, d is the distance, and φ is the angle above the horizontal. In this case, the force of friction can be calculated as Ff = μk * m * g * cos(φ). Therefore, the work done by the pulling force can be written as W = Ff * d.
Using the given values, we can substitute in the values to calculate the work done by the pulling force, which is W = μk * m * g * cos(φ) * d.
The work done on the sled by friction can be calculated by multiplying the force of friction by the distance, since the force is acting in the opposite direction. Therefore, the work done on the sled by friction is given by Wfr = -Ff * d. Substituting the given values, we can calculate the work done by friction, which is Wfr = -μk * m * g * cos(φ) * d.
Inserting given values in this formula, we can calculate the work done by friction as Wf = - (0.13) * (52.3 kg) * (9.8 m/s²) * (21.8 m) = -1475.24 Joules.
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A hoop, a disk, and a solid sphere each have mass 1.4 kg and diameter 16 cm. Starting from rest, all three objects roll down a 7° slope. If the slope is 3 m long and all bodies roll without slipping, find the speed of each at the bottom.
I know I have to use rotational kinetic energy and translational kinetic energy to get the answer but im not sure how.
The answers are Hoop=1.89 m/s disk=2.18 m/s and sphere=2.26 m/s
Answer:
The velocities of hoop , disk and sphere are 1.89 m/s , 2.18 m/s , 2.26 m/s.
Explanation:
Lets find the speed of any general body of mass 'm' , moment of inertia 'I' , radius 'r'.
Let 'v' be the speed and 'ω' be the angular velocity of the body at bottom of the slope.
Since there is no external force acting on the system (Eventhough friction is acting at the point of contact of the body and slope , it does no work as the point of contact is always at rest and not moving) , we can conserve energy for this system.
Initially the body is at rest and at a vertical height 'h' from the ground.
Here , h=3sin(7°)
Initial energy = mgh.
Finally on reaching bottom h=0 but the body has both rotational and translational kinetic energy.
∴ Final energy = [tex]\dfrac{1 }{2}[/tex]I[tex]ω^{2}[/tex] + [tex]\dfrac{1 }{2}[/tex]m[tex]v^{2}[/tex].
Since the body is rolling without slipping.
v=rω
and
Initial Energy = Final Energy
mgh = [tex]\dfrac{1 }{2}[/tex]I[tex]ω^{2}[/tex] + [tex]\dfrac{1 }{2}[/tex]m[tex]v^{2}[/tex]
∴ mgh = [tex]\dfrac{1 }{2}[/tex]I[tex]\dfrac{v^{2} }{r^{2} }[/tex] + [tex]\dfrac{1 }{2}[/tex]m[tex]v^{2}[/tex]
∴ v = [tex]\sqrt{\frac{2mgh}{\frac{I}{r^{2} }+m } }[/tex]
For a hoop ,
I = m[tex]r^{2}[/tex]
Substituting above value of I in the expression of v.
We get,
v = [tex]\sqrt{gh}[/tex] = [tex]\sqrt{9.81×3sin(7°) }[/tex] = 1.89 m/s
Similarly for disk,
I = [tex]\dfrac{1}{2}[/tex]m[tex]r^{2}[/tex]
We get,
v = [tex]\sqrt{\frac{4gh}{3} }[/tex] = 2.18 m/s
For solid sphere ,
I = [tex]\dfrac{2}{5}[/tex]m[tex]r^{2}[/tex]
v = [tex]\sqrt{\frac{10gh}{7} }[/tex] = 2.26 m/s.
On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on the moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 14 m/s at an angle 20 ∘ above the horizontal. For how much more time was the ball in flight?
Answer:
4.86 seconds
Explanation:
Velocity of projection, u = 14 m/s
angle of projection, θ = 20°
Formula for the time of flight
[tex]T=\frac{2uSin\theta }{g}[/tex]
For earth
Te = (2 x 14 x Sin 20) / 9.8
Te = 0.98 s
For moon
g' = g/6 = 1.64 m/s^2
Tm = ( 2 x 14 x Sin 20) / 1.64
Tm = 5.84 seconds
Tm - Te = 5.84 - 0.98 = 4.86 s
So, it takes 4.86 s more time of flight on moon than the earth.
The golf ball hit by Alan Shepard was in flight approximately 4.87 seconds longer on the moon than on Earth.
To analyze the flight duration of a projectile under different gravitational conditions.
1. Initial Parameters:
Initial velocity of the golf ball, [tex]v_0 = 14 \, \text{m/s}[/tex]
Launch angle, [tex]\theta = 20^{\circ}[/tex]
Acceleration due to gravity on Earth, [tex]g = 9.81 \, \text{m/s}^2[/tex]
Acceleration due to gravity on the moon, [tex]g_{moon} = \frac{g}{6} = 1.635 \, \text{m/s}^2[/tex]
2. Determine the Vertical Component of Initial Velocity:
The vertical component of the initial velocity can be calculated using the formula:
[tex]v_{0y} = v_0 \sin(\theta)[/tex]
Substituting the known values:
[tex]v_{0y} = 14 \, \text{m/s} \cdot \sin(20^{\circ}) \approx 4.78 \, \text{m/s}[/tex]
3. Calculate Time of Flight on Earth:
The time of flight [tex]T[/tex] for a projectile launched and landing on a flat surface can be calculated using the formula:
[tex]T = \frac{2 v_{0y}}{g}[/tex]
For Earth:
[tex]T_{Earth} = \frac{2 \cdot 4.78 \, \text{m/s}}{9.81 \, \text{m/s}^2} \approx 0.975 \, \text{s}[/tex]
4. Calculate Time of Flight on the Moon:
Using the same formula but with the moon's gravity:
[tex]T_{Moon} = \frac{2 v_{0y}}{g_{moon}}[/tex]
Substituting the values:
[tex]T_{Moon} = \frac{2 \cdot 4.78 \, \text{m/s}}{1.635 \, \text{m/s}^2} \approx 5.84 \, \text{s}[/tex]
5. Calculate the Difference in Flight Time:
Finally, to find how much more time the ball was in flight on the moon compared to Earth:
[tex]\Delta T = T_{Moon} - T_{Earth}[/tex]
[tex]\Delta T = 5.84 \, \text{s} - 0.975 \, \text{s} \approx 4.87 \, \text{s}[/tex]
A power plant uses 365 GJ/hr of energy from a boiler and extracts 18 GJ/hr as work from a turbine and rejects the remaining energy to a large cold resevoir of air (the atmosphere). What is the rate of heat rejection to the atmosphere for this plant? Give your answer in GJ/hr
Answer:
The rate of heat rejection will be 347 GJ/hr
Explanation:
We have given that power plant uses energy of 365 GJ/hr
So energy uses [tex]Q_B=365GJ/hr[/tex]
Work done from turbine [tex]W_T=18GJ/hr[/tex]
We have to find the rate of heat rejection to the atmosphere [tex]Q_C[/tex]
We know that [tex]Q_B=W_T+Q_C[/tex]
[tex]365=18+Q_C[/tex]
[tex]Q_C=347GJ/hr[/tex]
So the rate of heat rejection will be 347 GJ/hr
A coil 3.95 cm radius, containing 520 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×10−2 T/s )t+( 3.45×10−5 T/s4 )t4. The coil is connected to a 560-Ω resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil.
a-)Find the magnitude of the induced emf in the coil as a function of time.
Find the magnitude of the induced emf in the coil as a function of time.
E= 1.07×10−2 V +( 1.05×10−4 V/s3 )t3
E= 3.36×10−2 V +( 8.26×10−5 V/s3 )t3
E= 3.36×10−2 V +( 3.30×10−4 V/s3 )t3
E= 1.07×10−2 V +( 3.30×10−4 V/s3 )t3
b-)What is the current in the resistor at time t0 = 5.25 s ?
Final answer:
The magnitude of the induced emf in the coil as a function of time is given by the equation E = 3.36×10^(-2) V + (3.30×10^(-4) V/s^3)t^3. The current in the resistor at time t0 = 5.25 s can be calculated using Ohm's law by dividing the induced emf in the coil by the resistance.
Explanation:
In order to find the magnitude of the induced emf in the coil as a function of time, we can use Faraday's law of electromagnetic induction. Faraday's law states that the induced emf in a coil is equal to the rate of change of magnetic flux through the coil. The magnetic flux through the coil can be determined by multiplying the magnetic field by the area of the coil. Therefore, the induced emf in the coil is given by the equation E = -N * dΦ/dt, where E is the induced emf, N is the number of turns in the coil, and dΦ/dt is the rate of change of magnetic flux.
In this case, the magnetic field varies with time according to B = (1.20×10^(-2) T/s)t + (3.45×10^(-5) T/s^4)t^4. The magnetic field is perpendicular to the plane of the coil, so the area of the coil through which the magnetic flux passes is given by A = πr^2, where r is the radius of the coil.
Substituting these values into the equation for the induced emf, we get:
E = -N * (A * dB/dt)
E = -N * (πr^2 * dB/dt)
E = -520 * (π(0.0395)^2 * ((1.20×10^(-2) T/s) + (3.45×10^(-5) T/s^4)(4t^3)))
Simplifying this equation gives the magnitude of the induced emf in the coil as a function of time: E = 3.36×10^(-2) V + (3.30×10^(-4) V/s^3)t^3.
To find the current in the resistor at time t0 = 5.25 s, we can use Ohm's law, which states that the current through a resistor is equal to the voltage across the resistor divided by the resistance. The voltage across the resistor is equal to the induced emf in the coil. Therefore, the current in the resistor is given by the equation I = E/R, where I is the current, E is the induced emf, and R is the resistance.
Substituting the values given in the question into this equation, we get:
I = (3.36×10^(-2) V + (3.30×10^(-4) V/s^3)(5.25 s)^3) / 560 Ω
Calculating this gives the current in the resistor at time t0 = 5.25 s.
A 350 kg mass, constrained to move only vertically, is supported by two springs, each having a spring constant of 250 kN/m. A periodic force with a maximum value of 100 N is applied to the mass with a frequency of 2.5 rad/s. Given a damping factor of 0.125, the amplitude of the vibration is
Answer:
Explanation:
Expression for amplitude of forced damped oscillation is as follows
A = [tex]\frac{F_0}{\sqrt{m(\omega^2-\omega_0^2)^2+b^2\omega^2} }[/tex]
where
ω₀ =[tex]\sqrt{\frac{k}{m} }[/tex]
ω₀ = [tex]\sqrt{\frac{500000}{350} }[/tex]
=37.8
b = .125 ,
ω = 2.5
m = 350
A = [tex]\frac{100}{\sqrt{350(37.8^2-2.5^2)^2+.125^2\times2.5^2} }[/tex]
A = 3.75 mm . Ans
In the year 1178, five monks at Canterbury Cathedral in England observed what appeared to be an asteroid colliding with the moon, causing a red glow in and around it. It is hypothesized that this event created the crater Giordano Bruno, which is right on the edge of the area we can usually see from Earth. How long after the asteroid hit the Moon, which is 3.84 x 105 km away, would the light first arrive on Earth in seconds?
To solve this problem it is necessary to apply the kinematic equations of description of the movement in which it is understood that the velocity is the travel of a particle in a fraction of time, that is to say
[tex]v = \frac{x}{t}[/tex]
Where,
x = Displacement
t = time
In our case the speed is equivalent to that of the Light, and the distance is necessary to reach the moon by the asteroid.
[tex]v = 3.8*10^8m/s[/tex]
[tex]x = 3.84*10^5km = 3.84*10^8m[/tex]
Re-arrange to find t,
[tex]t = \frac{x}{V}[/tex]
[tex]t = \frac{3.84*10^8}{3.8*10^8}[/tex]
[tex]t = 1.28s[/tex]
Therefore will take 1.28 s for the light arrive on Earth.
The light from the asteroid collision on the Moon would take approximately 1.28 seconds to reach Earth.
Explanation:The speed of light in a vacuum is approximately 3 x 10^8 meters per second. Since the Moon is 3.84 x 10^5 km away from Earth, we can calculate the time it takes for light to travel from the Moon to Earth using the formula:
Time = Distance / Speed
Converting the distance from km to meters, we get 3.84 x 10^8 meters. Plugging in the values:
Time = (3.84 x 10^8 meters) / (3 x 10^8 meters per second) = 1.28 seconds
So, it would take approximately 1.28 seconds for the light from the asteroid collision on the Moon to reach Earth.
) Titan’s Evolving Atmosphere: Titan’s exopshere lies nearly 1400 km above its surface. What is the escape velocity from this altitude? What is the thermal speed of a hydrogen atom at the exospheric temperature of about 200 K?
Answer:
Explanation:
first we calculate the escape velocity using [tex]V_{esc} =\sqrt{\frac{2GM}{R} }[/tex]
Thermal velocity of an atom [tex]V_{therm} =\sqrt{\frac{2KT}{M} }[/tex]
Given [tex]G= 6.67*10^{-11} m^{3}/kgS^{2}\\\\Mx_{titan} = 1.35*10^{23}kg\\[/tex]
Resistance from the centr of the titan is:
R= R(titan) + height of exosphere = 2575 + 1400=3.975*10^6m
escape vel = [tex]V_{esc} =\sqrt{\frac{2GM}{R} }\\\\V_{esc} =\sqrt{\frac{2*6.67*10^{-11}*1.35*10^{23}}{3.975*10^{6}} }=2.129*10^{3}m/s[/tex]
Thermal velocity of an atom [tex]V_{therm} =\sqrt{\frac{2KT}{M} }\\\\V_{therm} =\sqrt{\frac{2*1.38*10^{-23}*200}{1.67*10^{-27}} }=1.8*10^{3}m/s[/tex]
The escape velocity from Titan's exosphere and the thermal speed of a hydrogen atom at the exospheric temperature of about 200 K can be calculated using specific formulas.
Explanation:The escape velocity from Titan's exosphere can be calculated using the formula:
V = sqrt((2GM)/r)
Where V is the escape velocity, G is the gravitational constant, M is the mass of Titan, and r is the distance from the center of Titan to the exosphere. The thermal speed of a hydrogen atom at the exospheric temperature can be calculated using the formula:
v = sqrt((3kT)/m)
Where v is the thermal speed, k is the Boltzmann constant, T is the temperature, and m is the mass of the hydrogen atom.
Given the information provided, the escape velocity from Titan's exosphere and the thermal speed of a hydrogen atom at the exospheric temperature of about 200 K can be determined.
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An Atwood machine consists of a mass of 3.5 kg connected by a light string to a mass of 6.0 kg over a frictionless pulley with a moment of inertia of 0.0352 kg m2 and a radius of 12.5 cm. If the system is released from rest, what is the speed of the masses after they have moved through 1.25 m if the string does not slip on the pulley?
Please note: the professor has told us that the correct answer is 2.3 m/s. how does one arrive at this answer?
Answer:
[tex]v=2.28m/s[/tex]
Explanation:
For the first mass we have [tex]m_1=3.5kg[/tex], and for the second [tex]m_2=6.0kg[/tex]. The pulley has a moment of intertia [tex]I_p=0.0352kgm^2[/tex] and a radius [tex]r_p=0.125m[/tex].
We solve this with conservation of energy. The initial and final states in this case, where no mechanical energy is lost, must comply that:
[tex]K_i+U_i=K_f+U_f[/tex]
Where K is the kinetic energy and U the gravitational potential energy.
We can write this as:
[tex]K_f+U_f-(K_i+U_i)=(K_f-K_i)+(U_f-U_i)=0J[/tex]
Initially we depart from rest so [tex]K_i=0J[/tex], while in the final state we will have both masses moving at velocity v and the tangential velocity of the pully will be also v since it's all connected by the string, so we have:
[tex]K_f=\frac{m_1v^2}{2}+\frac{m_2v^2}{2}+\frac{I_p\omega_p^2}{2}=(m_1+m_2+\frac{I_p}{r_p^2})\frac{v^2}{2}[/tex]
where we have used the rotational kinetic energy formula and that [tex]v=r\omega[/tex]
For the gravitational potential energy part we will have:
[tex]U_f-U_i=m_1gh_{1f}+m_2gh_{2f}-(m_1gh_{1i}+m_2gh_{2i})=m_1g(h_{1f}-h_{1i})+m_2g(h_{2f}-h_{2i})[/tex]
We don't know the final and initial heights of the masses, but since the heavier, [tex]m_2[/tex], will go down and the lighter, [tex]m_1[/tex], up, both by the same magnitude h=1.25m (since they are connected) we know that [tex]h_{1f}-h_{1i}=h[/tex] and [tex]h_{2f}-h_{2i}=-h[/tex], so we can write:
[tex]U_f-U_i=m_1gh-m_2gh=gh(m_1-m_2)[/tex]
Putting all together we have:
[tex](K_f-K_i)+(U_f-U_i)=(m_1+m_2+\frac{I_p}{r_p^2})\frac{v^2}{2}+gh(m_1-m_2)=0J[/tex]
Which means:
[tex]v=\sqrt{\frac{2gh(m_2-m_1)}{m_1+m_2+\frac{I_p}{r_p^2}}}=\sqrt{\frac{2(9.8m/s^2)(1.25m)(6.0kg-3.5kg)}{3.5kg+6.0kg+\frac{0.0352kgm^2}{(0.125m)^2}}}=2.28m/s[/tex]
Final answer:
To solve this problem, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system (kinetic energy + potential energy) remains constant if no external forces are acting on the system.
Explanation:
Let's denote the initial position of the masses as position 1 and the final position as position 2. At position 1, the 3.5 kg mass is at a height h above the ground, and the 6.0 kg mass is at a height 1.25 m - h above the ground. At position 2, both masses are at a height 1.25 m above the ground.
The potential energy (PE) of the system at position 1 is:
PE1 = m1 * g * h + m2 * g * (1.25 m - h)
The potential energy (PE) of the system at position 2 is:
PE2 = (m1 + m2) * g * 1.25 m
Since the system is released from rest, the initial kinetic energy (KE) of the system is zero. The final kinetic energy (KE) of the system is:
KE2 = (1/2) * (m1 + m2) * v^2
where v is the speed of the masses after they have moved through 1.25 m.
According to the principle of conservation of mechanical energy, the total mechanical energy at position 1 is equal to the total mechanical energy at position 2:
PE1 + KE1 = PE2 + KE2
Substituting the expressions for PE1, PE2, and KE2, we get:
m1 * g * h + m2 * g * (1.25 m - h) = (m1 + m2) * g * 1.25 m + (1/2) * (m1 + m2) * v^2
Solving for v^2, we get:
v^2 = 2 * g * (m1 * h + m2 * (1.25 m - h) - (m1 + m2) * 1.25 m)
Substituting the given values for g, m1, m2, and h, we get:
v^2 = 2 * 9.8 m/s^2 * (3.5 kg * 1.25 m + 6.0 kg * (1.25 m - 1.25 m) - (3.5 kg + 6.0 kg) * 1.25 m)
Solving for v, we get:
v ≈ 2.3 m/s
So, the speed of the masses after they have moved through 1.25 m is approximately 2.3 m/s.
A long, thin solenoid has 700 turns per meter and radius 2.50 cm. The current in the solenoid is increasing at a uniform rate of 36.0 A/s . a. What is the magnitude of the induced electric field at a point near the center of the solenoid? b. What is the magnitude of the induced electric field at a point 0.500 cm from the axis of the solenoid?
To solve this problem it is necessary to apply the concepts related to the magnetic field in a solenoid.
By definition we know that the magnetic field is,
[tex]B = \mu_0 n I[/tex]
[tex]\frac{dB}{dt} = \frac{d}{dt} \mu_0 NI[/tex]
[tex]\frac{dB}{dt} = \mu_0 n\frac{dI}{dt}[/tex]
At the same tome we know that the induced voltage is defined as
[tex]\epsilon = \frac{\Phi}{dt}[/tex]
[tex]\epsilon = A \frac{dB}{dt}[/tex]
Replacing
[tex]\epsilon = A \mu_0 n\frac{dI}{dt}[/tex]
[tex]\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}[/tex]
PART A) Substituting with our values we have that
[tex]\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}\\\epsilon = \frac{(4\pi*10^{-7})700(0)(36)}{2}\\\epsilon = 0V/m\\[/tex]
Therefore there is not induced electric field at the center of solenoid.
PART B) Replacing the radius for 0.5cm
[tex]\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}\\\epsilon = \frac{(4\pi*10^{-7})700(0.5*10^{-2})(36)}{2}\\\epsilon = 7.9168*10^{-5}V/m[/tex]
Therefore the magnitude of the induced electric field at a point 0.5cm is [tex]7.9168*10^{-5}V/m[/tex]
The magnitude of the induced emf near the center of the solenoid is 0 V/m.
The magnitude of the induced electric field at a point 0.500 cm from the axis of the solenoid is 8 x 10⁻⁵ V/m.
The given parameters;
number of turns of the solenoid, N = 700 turns/mradius of the wire, r = 2.5 cmcurrent in the solenoid, I = 36 A/sThe magnitude of the induced emf near the center of the solenoid is calculated as follows;
[tex]B = \mu_0 nI\\\\\frac{dB}{dt} = \mu_0 n \frac{dI}{dt} \\\\\frac{dB}{dt} = (4\pi \times 10^{-7} \times 700 \times 36) = 0.032 \ T/s[/tex]
[tex]E = \frac{r}{2} (\frac{dB}{dt} )\\\\\ E = \frac{0}{2} (0.032)\\\\ E = 0 \ V/m[/tex]
The magnitude of the induced electric field at a point 0.500 cm from the axis of the solenoid is calculated as follows;
[tex]E = \frac{r}{2} (\frac{dB}{dr} )\\\\ E= \frac{0.5\times 10^{-2} }{2}( 0.032)\\\\E = 8\times 10^{-5} \ V/m[/tex]
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A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheric pressure is Pa = 1.01 × 105 Pa. The volume of the container is V0 = 4.4 × 10-4 m3. The maximum difference between the pressure inside and outside that this particular container can withstand before bursting or imploding is ΔPmax = 2.26 × 105 Pa. For this problem, assume that the density of air maintains a constant value of rhoa = 1.20 kg / m3 and that the density of seawater maintains a constant value of rhos = 1025 kg / m3.What is the maximum height h in meters above the ground that the container can be lifted before bursting?
To determine the maximum height h that the container can be lifted before bursting, we can use the formula ΔP = ρgh, where ΔP is the pressure difference, ρ is the density of the fluid inside the container, g is the acceleration due to gravity, and h is the height difference.
Explanation:To determine the maximum height h that the container can be lifted before bursting, we need to consider the difference in pressure inside and outside the container.
The maximum pressure difference that the container can withstand before bursting is given as ΔPmax = 2.26 × 105 Pa.
To calculate the maximum height h, we can use the formula: ΔP = ρgh, where ΔP is the pressure difference, ρ is the density of the fluid inside the container (assuming it is the same as seawater density, rhos = 1025 kg/m3), g is the acceleration due to gravity (approximately 9.81 m/s2), and h is the height difference.
Plugging in the values, we get ΔPmax = 1025 * 9.81 * h. Solving for h, we find h ≈ 231 meters.
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Four point charges are placed at the corners of a square. Each charge has the identical value +Q. The length of the diagonal of the square is 2a.
What is the electric potential at the center of the square?
Answer:[tex]V_{net}=4\frac{kQ}{a}[/tex]
Explanation:
Given
charge on each Particle is Q
Length of diagonal of the square is 2a
therefore distance between center and each charge is [tex]\frac{2a}{2}=a[/tex]
Electric Potential of charged Particle is given by
For First Charge
[tex]V_1=\frac{kQ}{a}[/tex]
[tex]V_2=\frac{kQ}{a} [/tex]
[tex]V_3=\frac{kQ}{a}[/tex]
[tex]V_4=\frac{kQ}{a}[/tex]
total Electric Potential At center is given by
[tex]V_{net}=V_1+V_2+V_3+V_4[/tex]
[tex]V_{net}=4\times \frac{kQ}{a}[/tex]
[tex]V_{net}=4\frac{kQ}{a}[/tex]
Two bumper cars in an amusement park ride collide elastically as one approaches the other directly fromthe rear. Car A has a mass of 435 kg and car B a mass of 495 kg, owing to differences in passenger mass. Ifcar A approaches at 4.50 m/s and car B is moving at 3.70 m/s calculatea) their velocities after the collisionb) the change in momentum of each.
Answer:
a) The velocity of car B after the collision is 4.45 m/s.
The velocity of car A after the collision is 3.65 m/s.
b) The change of momentum of car A is - 370.45 kg · m/s
The change of momentum of car B is 370.45 kg · m/s
Explanation:
Hi there!
Since the cars collide elastically, the momentum and kinetic energy of the system do not change after the collision.
The momentum of the system is calculated adding the momenta of each car:
initial momentum = final momentum
mA · vA + mB · vB = mA · vA´ + mB · vB´
Where:
mA = mass of car A
vA = initial velocity of car A
mB = mass of car B
vB = initial velocity of car B
vA´= final velocity of car A
vB´ = final velocity of car B
Let´s replace with the data we have and solve the equation for vA´:
mA · vA + mB · vB = mA · vA´ + mB · vB´
435 kg · 4.50 m/s + 495 kg · 3.70 m/s = 435 kg · vA´ + 495 kg · vB´
3789 kg · m/s = 435 kg · vA´ + 495 kg · vB´
3789 kg · m/s - 495 kg · vB´ = 435 kg · vA´
(3789 kg · m/s - 495 kg · vB´)/435 kg = vA´
Let´s write this expression without units for a bit more clarity:
vA´= (3789 - 495 vB´)/435
The kinetic energy of the system is also conserved, then, the initial kinetic energy is equal to the final kinetic energy:
initial kinetic energy of the system = final kinetic energy of the system
1/2 · mA · vA² + 1/2 · mB · vB² = 1/2 · mA · (vA´)² + 1/2 · mB · (vB´)²
Replacing with the data:
initial kinetic energy = 1/2 · 435 kg · (4.50 m/s)² + 1/2 · 495 kg · (3.70)²
initial kinetic energy = 7792.65 kg · m²/s²
7792.65 kg · m²/s² = 1/2 · 435 kg · (vA´)² + 1/2 · 495 kg · (vB´)²
multiply by 2 both sides of the equation:
15585.3 kg · m²/s² = 435 kg · (vA´)² + 495 kg · (vB´)²
Let´s replace vA´ = (3789 - 495 vB´)/435
I will omit units for clarity in the calculation:
15585.3 = 435 · (vA´)² + 495 · (vB´)²
15585.3 = 435 · (3789 - 495 vB´)²/ 435² + 495 (vB´)²
15585.3 = (3789² - 3751110 vB´ + 245025 vB²) / 435 + 495 (vB´)²
multiply both sides of the equation by 435:
6779605.5 = 3789² - 3751110 vB´ + 245025 vB² + 215325 vB´²
0 = -6779605.5 + 3789² - 3751110 vB´ + 460350 vB´²
0 = 7576915.5 - 3751110 vB´ + 460350 vB´²
Solving the quadratic equation:
vB´ = 4.45 m/s
vB´ = 3.70 m/s (the initial velocity)
a) The velocity of car B after the collision is 4.45 m/s
The velocity of car A will be teh following:
vA´= (3789 - 495 vB´)/435
vA´= (3789 - 495 (4.45 m/s))/435
vA´ = 3.65 m/s
The velocity of car A after the collision is 3.65 m/s
b) The change of momentum of each car is calculated as the difference between its final momentum and its initial momentum:
ΔpA = final momentum of car A - initial momentum of car A
ΔpA = mA · vA´ - mA · vA
ΔpA = mA (vA´ - vA)
ΔpA = 435 kg (3.648387097 m/s - 4.50 m/s) (I have used the value of vA´ without rounding).
ΔpA = - 370.45 kg · m/s
The change of momentum of car A is - 370.45 kg · m/s
ΔpB = mB (vB´ - vB)
ΔpB = 495 kg (4.448387097 m/s - 3.70 m/s) (I have used the value of vB´ without rounding).
ΔpB = 370.45 kg · m/s
The change of momentum of car B is 370.45 kg · m/s
I have used the values of the final velocities without rounding so we can notice that the change of momentum of both cars is equal but of opposite sign.
In Physics, when discussing collisions, the total momentum of a closed and isolated system is conserved. The change in momentum during a collision can be calculated using the principles of conservation of momentum and energy, with specifics depending on whether the collision is elastic or inelastic.
Explanation:The subject of this question is Physics, specifically dealing with the concepts of elastic and inelastic collisions, momentum, and the conservation of momentum principle. When two objects collide, momentum is exchanged between them, but the total momentum of the system remains constant if the system is closed and isolated. The change in momentum depends on whether the collision is elastic, where kinetic energy is conserved, or inelastic, where the objects stick together and kinetic energy is not conserved.
The example given with cars of equal mass indicates that after an elastic collision, the cars will trade velocities, and after a completely inelastic collision, they will move together with a combined momentum equal to their total initial momentum.
However, for the specific question asked by the student, which involves two bumper cars with different masses and velocities, the velocities after the collision and the change in momentum for each car would be determined by applying the elastic collision equations. These include conservation of momentum and conservation of kinetic energy. The precise calculations were not provided as part of the task.
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Our two intrepid relacar drivers are named Pam and Ned. We use these names to make it easy to remember: measurements made by Pam are primed (x', t') and those made by Ned are not primed (x, t). v is the velocity of Pam (the other frame of reference) as measured by Ned. What is the interpretation of v'?
This is a problem based on the logic and interpretation of the variables. From the measured data taken
what is collected by the two individuals is expressed as,
- NED reference system: (x, t)
- PAM reference system: (x ', t')
From the reference system we know that ν is the speed of PAM (the other reference system) as a measurement by NED.
Then ν' is the speed of NED (from the other system of the reference) as a measurement by PAM.
In the context of Special Relativity, v' represents the relative velocity of Ned as observed from Pam's reference frame. This velocity takes into account the motion of both Ned and Pam and can be computed using the principles of Conservation of Momentum.
Explanation:The symbol v' can be interpreted as the velocity of Ned as measured from Pam's frame of reference. This concept forms the basis of Special Relativity as proposed by Einstein, where observations are taken from different frames of reference, in this case, Pam and Ned. Relative velocity, like v', is the velocity of an object (in this case, Ned) as observed from a particular frame of reference (in this case, Pam's), and it can change depending on who is making the observation. It takes into account the motion of both Ned and Pam and describes how fast Ned is moving and in what direction from Pam's perspective.
For example, if Pam and Ned were moving towards each other, the relative velocity v' would be the sum of their individual speeds as observed from Pam's perspective. If they were moving in the same direction, v' would be the difference in their individual speeds. Also, Pam's velocity in Ned's frame can be defined as -v'. This negative sign indicates they are moving in opposite directions.
To compute v', you would need to understand the principle of Conservation of Momentum, represented by equations defining momentum conservation in the x and y directions. Using this principle, you can solve for V.
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Determine the ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy (1/2mv2) when a particle has a speed of (a) 2.71 x 10-3c. and (b) 0.855c.
Final answer:
The ratio of relativistic kinetic energy to nonrelativistic kinetic energy depends on the speed of the particle. For speeds much less than the speed of light, the ratio is close to 1, but for speeds approaching the speed of light, the relativistic effect becomes significant and the ratio increases.
Explanation:
The ratio of relativistic kinetic energy to nonrelativistic kinetic energy can be calculated using the formula for relativistic kinetic energy KErel = mc2((1/(1 - v2/c2)1/2) - 1), where m is the particle mass, v is its velocity, and c is the speed of light. The nonrelativistic kinetic energy is given by KEnr = (1/2)mv2. Firstly, to determine the ratios at given speeds, we calculate the relativistic factor γ, which is (1/(1 - v2/c2)1/2), for each speed
For part (a) with v = 2.71 x 10-3c, the relativistic factor is close to 1, indicating that the speeds are nonrelativistic, and the ratio KErel/KEnr would be close to 1. For part (b), at v = 0.855c, the relativistic factor is significantly greater than 1, and the relativistic kinetic energy will be much larger than the classical value, yielding a ratio much greater than 1.
Find the radius R of the orbit of a geosynchronous satellite that circles the Earth. (Note that R is measured from the center of the Earth, not the surface of the Earth.) Use the following values if needed in this problem: The universal gravitational constant G is 6.67
The radius R of a geosynchronous satellite's orbit is 42.2 x 10⁶ meters, or approximately 6.63 times the radius of the Earth.
The radius R of the orbit of a geosynchronous satellite that circles the Earth can be computed using the principles of circular motion and the law of universal gravitation, considering a geostationary satellite takes 23 hours 56 minutes and 4 seconds to complete one orbit. The gravitational force provides the necessary centripetal force to keep the satellite in orbit. Given the mass of the Earth (me = 5.98 x 10²⁴kg), the mean radius of the earth (R2 = 6.37 x 10⁶ m), and the universal gravitational constant (G = 6.67 x 10⁻¹¹ N m²kg⁻²), the radius R is already provided as 42.2 x 10⁶ m. This is the distance from the center of the Earth to the satellite. When expressed in terms of Earth's radii, this is approximately 6.63 Earth radii, since the radius of the Earth is 6.37 x 10⁶ m.
The speed of sound in air is around 345 m/s. A tuning fork vibrates at 610 Hz above the open end of the sound resonance tube. What is the wavelength (in cm) of the sound waves in the tube? Never include units with a numerical answer.
Answer:
[tex]\lambda=56.5cm[/tex]
Explanation:
Wavelength is calculated as:
[tex]\lambda=\frac{V}{f}[/tex]
Replacing the given values:
[tex]\lambda=\frac{345}{610}[/tex]
[tex]\lambda=0.565m[/tex]
Converting the result into cm:
[tex]\lambda=56.5cm[/tex]
A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is 6.68 × 10-3 Wb. What is the flux that passes through the circular loop?
Answer:
0.0085 Wb
Explanation:
a = Side of square
r = Radius of circle
[tex]\phi_s[/tex] = Magnetic flux through square loop = [tex]6.68\times 10^{-3}\ Wb[/tex]
Magnetic flux is given by
[tex]\phi=BA[/tex]
For square
[tex]\phi_{s}=Ba^2[/tex]
The length of the square will be equal to the circumference of the circle
[tex]4a=2\pi r\\\Rightarrow r=\frac{2a}{\pi}[/tex]
For circle
[tex]\phi_{c}=B\pi r^2\\\Rightarrow \phi_{c}=B\pi \left(\frac{2a}{\pi}\right)^2\\\Rightarrow \phi_c=Ba^2\frac{4}{\pi}\\\Rightarrow \phi_c=\phi_s\frac{4}{\pi}\\\Rightarrow \phi_c=6.68\times 10^{-3}\frac{4}{\pi}\\\Rightarrow \phi_c=0.0085\ Wb[/tex]
The flux that passes through the circular loop is 0.0085 Wb
The magnetic flux that passes through the circular loop is 0.00848 Wb.
What is the magnetic flux?The magnetic flux is defined as the measurement of the total magnetic field which passes through a given area.
Given that the magnetic flux that passes through the square loop is 6.68 × 10-3 Wb.
Let's consider that r is the radius of the circle and a is the side of the square.
The magnetic flux through the square loop is given below.
[tex]\phi_s = Ba^2[/tex]
Where B is the magnetic field.
The length of the square will be equal to the circumference of the circle
[tex]4a = 2\pi r[/tex]
[tex]r = \dfrac {2a}{\pi}[/tex]
The magnetic flux through the circular loop is given below.
[tex]\phi = BA[/tex]
Where A is is the cross-sectional area of the circular loop.
[tex]\phi = B \pi r^2[/tex]
[tex]\phi = B \pi (\dfrac {2a}{\pi})^2[/tex]
[tex]\phi = B\pi \times \dfrac {4a^2}{\pi^2}[/tex]
[tex]\phi = Ba^2 \dfrac {4}{\pi}[/tex]
[tex]\phi = \phi_s \dfrac {4}{3.14}[/tex]
[tex]\phi = 6.68 \times 10^{-3} \times 1.27[/tex]
[tex]\phi = 0.00848 \;\rm Wb[/tex]
Hence we can conclude that the magnetic flux that passes through the circular loop is 0.00848 Wb.
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You have two springs. One has a greater spring constant than theother. You also have two objects, one with a greater mass than theother. Which object should be attached to which spring, so that theresulting spring-object system has the greatest possible period ofoscillation?
The object with the smaller mass should be attached to the springwith the greater spring constant.
The object with the smaller mass should be attached to the springwith the smaller spring constant.
The object with the greater mass should be attached to the springwith the greater spring constant.
The object with the greater mass should be attached to the springwith the smaller spring constant.
The object with the greater mass should be attached to the spring with the smaller spring constant, so that the resulting spring-object system has the greatest possible period of oscillation.
Answer: Option D
Explanation:
According to the simple harmonic motions, from physics, it gives a relation between deformation force and the deflection. The more deflection results in more time period of oscillation.
F = - k x
where ‘k’ is the spring constant, and ‘F’ is the deformation force.
So, deflection is directly proportionate to forces, and inversely proportionate to its spring constant. Hence, we can derive that the force must be maximum, and hence weight must be maximum, with the spring constant lesser. Then, the deflection will be high. So, time period increases.
An airplane is flying at Mach 1.48 at an altitude of 7,800.00 meters, where the speed of sound is v = 311.83 m/s. How far away from a stationary observer will the plane be when the observer hears the sonic boom? (Enter the total distance in m.)
Answer:
The plane will be 11545.46 m far when the observer hears the sonic boom
Explanation:
Step 1: Data given
Altitude of the plane = 7800 meters
speed of sound = 311.83 m/s
Step 2:
The mach number M = vs/v
This means v/vs = 1/M
Half- angle = ∅
sin∅= v/vs
∅ = sin^-1 (v/vs)
∅ = sin^-1 (1/M)
∅ = sin^-1(1/1.48)
∅= 42.5 °
tan ∅ = h/x
⇒ with h= the altitude of the plane = 7800 meter
⇒ with x = the horizontal distance moved by the plane
x = h/tan ∅
x = 7800 / tan 42.5
x = 8512.2 meters
d = the distance between the observer and the plane when the observer hears the sonic boom is:
d = √(8512.2² + 7800²)
d = 11545.46 m
The plane will be 11545.46 m far when the observer hears the sonic boom
(a) How far is the center of mass of the Earth-Moon system from the center of the Earth? (Appendix C gives the masses of the Earth and the Moon and the distance between the two.) m
(b) Express the answer to (a) as a fraction of the Earth's radius.'
Answer:
a) [tex]r_{cm} = 4672 km[/tex]
b)r_cm = 0.733 R_{earth}
Explanation:
Location of center of mass of Earth- moon system is
[tex]r_{cm}= \frac{M_{earth}r_{earth}+M_{moon}r_{moon}}{M_{moon}+M{earth}}[/tex]
[tex]r_{cm}= \frac{6\times10^{24}(0)+7.34\times10^{22}384000\times1000}{6\times10^{24}+7.34\times10^{22}}[/tex]
=4.672×10^(6) m
a) [tex]r_{cm} = 4672 km[/tex]
b) [tex]\frac{r_{cm}}{R{earth}} = \frac{4672}{6371} = 0.733[/tex]
therefore
r_cm = 0.733 R_{earth}
The center of mass of the Earth-Moon system is the point around which the two bodies orbit and it is located within Earth but not at the center. The distance from the Earth's center to this point can be calculated using the masses of the Moon and Earth, and the distance between them. This distance expressed as a fraction of the Earth's radius is approximately 0.73.
Explanation:The center of mass of the Earth-Moon system is the point about which the two bodies orbit, and it's located inside the Earth, not at the Earth's center. The distance from the Earth's center to the system's center of mass can be calculated using the formula: d = D*(m/(M+m)), where D is the distance between the Earth and the Moon, m is the Moon's mass, M is the Earth's mass.
To express the previous value as a fraction of the Earth's radius, you must divide it by the radius of the Earth. For instance, let's assume the distance turned out to be 4671 km and the Earth's radius is 6371 km. The resulting fraction would be 4671/6371 which simplifies to approximately 0.73.
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A quantity of an ideal gas is kept in a rigid container of constant volume. If the gas is originally at a temperature of 28 °C, at what temperature (in °C) will the pressure of the gas triple from its base value?
Answer:
[tex]T_2=630^{\circ}C[/tex]'
Explanation:
Original temperature of the gas, [tex]T_1=28^{\circ}C=301\ K[/tex]
From the ideal gas equation,
[tex]P_1V_1=nRT_1[/tex]
Since,
[tex]P_2=3P_1[/tex]
[tex]nRT_2=3(nRT_1)[/tex]
[tex]T_2=3T_1[/tex]
[tex]T_2=3\times 301[/tex]
[tex]T_2=903\ K[/tex]
or
[tex]T_2=630^{\circ}C[/tex]
So, the new temperature of the gas is 630 degree Celsius. Hence, this is the required solution.
A heat engine receives an amount of energy Qh= 790 kJ by heat transfer from a high temperature thermal reservoir at Th=950 K. Energy is rejected by heat transfer to a lower temperature thermal reservoir at T1=590 K. If waste heat in the amount of Q1=160 kJ is rejected to the low temperature thermal reservoir during each cycle.
a) Solve for the maximum theoretical efficiency that an engine in this situation could operate with. ANSWER: 0.379
b) Solve for actual efficiency that the engine is operating with.
c) Which of the following best describes the manner in which the cycle is operating...
-Reversibly or Impossibly?
Answer:
(a) [tex]\eta_{max}=37.895\%[/tex]
(b) [tex]\eta=79.75\%[/tex]
(c) Impossibly
Explanation:
Given:
temperature of source reservoir, [tex]T_H=950\ K[/tex]temperature of sink reservoir, [tex]T_L=590\ K[/tex]heat absorbed by the engine, [tex]Q_H=790\ kJ[/tex]heat rejected by the engine, [tex]Q_L=160\ kJ[/tex](a)
Now the maximum theoretical efficiency of the engine:
[tex]\eta_{max}=\frac{T_H-T_L}{T_H}\times 100\% [/tex]
[tex]\eta_{max}=\frac{950-590}{950}\times 100\% [/tex]
[tex]\eta_{max}=37.895\%[/tex]
(b)
Actual efficiency of the heat engine:
[tex]\eta=\frac{Q_H-Q_L}{Q_H}\times 100\% [/tex]
[tex]\eta=\frac{790-160}{790}\times 100\% [/tex]
[tex]\eta=79.75\%[/tex]
(c)
This is impossible because the actual efficiency can never be greater than the ideal (Carnot) efficiency of a heat engine.
The answers are:
(a) the maximum theoretical efficiency is 37.9%
(b) the actual efficiency is 79.7%
(c) The cycle is operating impossibly
The following data is provided to us:
The temperature of the source, [tex]T_1=950K[/tex]
The temperature of the sink, [tex]T_2=590K[/tex]
Heat absorbed by the engine, [tex]Q_1=790kJ[/tex]
Heat rejected by the engine, [tex]Q_2=160kJ[/tex]
(a) Now, the maximum efficiency of the engine is:
[tex]\eta _{max}=1-\frac{T_2}{T1}\\ \\\eta_{max}=1-\frac{590}{950}\\ \\\eta_{max}=0.379[/tex]
maximum efficiency ( [tex]\eta_{max}[/tex] ) = 37.9%
(b) actual efficiency:
[tex]\eta=1-\frac{Q_2}{Q_1}\\ \\\eta=1-\frac{160}{790}\\ \\\eta=0.797[/tex]
actual efficiency ([tex]\eta[/tex]) = 79.7%
(c) it is not possible for a Carnot engine to have more efficiency than the maximum possible efficiency.
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The space shuttle, in circular orbit around the Earth, collides with a small asteroid which ends up in the shuttle's storage bay. For this collision,
a. neither momentum nor kinetic energy is conserved.
b. only momentum is conserved.
c. both momentum and kinetic energy are conserved.
d. only kinetic energy is conserved.
Answer:
b
Explanation:
The space shuttle, in circular orbit around the Earth, collides with a small asteroid which ends up in the shuttle's storage bay.
This form of collision is called inelastic collision. And inelastic collision momentum is conserved but the kinetic energy is not conserved. Hence the correct option is b. only momentum is conserved.
Answer:
b
Explanation:
This form of collision is called inelastic collision.
Two balls have the same mass of 5.00 kg. Suppose that these two balls are attached to a rigid massless rod of length 2L, where L = 0.550 m.
One is attached at one end of the rod and the other at the middle of the rod.
If the rod is held by the open end and rotates in a circular motion with angular speed of 45.6 revolutions per second,
(a) What is the tension for the first half of the rod, i.e., between 0 and L if the pivot point is chosen as origin?
(b) What is the tension for the second half of the rod, i.e., between L and 2L if the pivot point is chosen as origin?
Answer:
Explanation:
Given
mass of balls [tex]m= 5 kg[/tex]
[tex]N=45.6 rev/s[/tex]
angular velocity [tex]\omega =2\pi N=286.55 rad/s[/tex]
Length of Rod [tex]2L=1.1 m[/tex]
Tension in the Second half of rod
[tex]T_2=m\omega ^2(2L)=2m\omega ^2L[/tex]
[tex]T_2=5\times (286.55)^2\times 1.1[/tex]
[tex]T_2=451.609 kN[/tex]
For First Part
[tex]T_1-T_2=m\omega ^2L[/tex]
[tex]T_1=T_2+m\omega ^2L[/tex]
[tex]T_1=3 m\omega ^2L[/tex]
[tex]T_1=3\times 5\times (286.55)^2\times 0.55[/tex]
[tex]T_1=677.41 kN[/tex]
Final answer:
The question deals with the tension in a rotating rod with attached masses. Tension in the first half of the rod is dictated by the centripetal force for one mass, while the second half must account for two masses. It's a problem in the field of Physics, specifically rotational dynamics at the college level.
Explanation:
The student's question revolves around the concepts of rotational motion and tension in a system consisting of a massless rod with balls of equal mass attached at different points. Given the angular speed and the positions of the masses, we are to find the tension in two parts of the rod during rotation.
The tension for the first half of the rod can be calculated by considering the centripetal force required to keep the ball rotating in a circle of radius L, the middle of the rod. Since the ball at L is the only mass in the first segment, we only need to consider its centripetal force requirement.
For the second segment of the rod, from L to 2L, the tension must accommodate the centripetal force for both masses, one at the middle and the other at the end. The ball at the end experiences more tension because it is further from the pivot point and thus has a larger radius of rotation.
Note that actual equations and calculations are not provided here, as the question seems to request conceptual explanations rather than specific numerical solutions.
Neutron activation analysis for a sample of a rock revealed the presence of 131 53I, which has a half-life of 8.08 days . Assuming the isotope was freshly separated from its decay products, what is the mass of 131 53I in a sample emitting 1.00 mCi of radiation?
Answer:
The mass of [tex]_{53}^{131}I[/tex] is [tex]8.09\times10^{-9}\ g[/tex].
Explanation:
Given that,
Half life [tex]t_{\frac{1}{2}}=8.08\ days[/tex]
Sample emitting radiation = 1.00 mCi = [tex]3.7\times10^{7}\ dps[/tex]
We need to calculate the rate constant
Using formula of rate constant
[tex]\lambda=\dfrac{0.693}{t_{\frac{1}{2}}}[/tex]
[tex]\lambda=\dfrac{0.693}{8.08\times24\times60\times60}[/tex]
[tex]\lambda=9.92\times10^{-7}\ s^{-1}[/tex]
We need to calculate the numbers of atoms
Using formula of numbers of atoms
[tex]N_{0}=\dfrac{N}{\lambda}[/tex]
[tex]N_{0} =\dfrac{3.7\times10^{7}}{9.92\times10^{-7}}[/tex]
[tex]N_{0}=3.72\times10^{13}\ atoms[/tex]
We need to calculate the mass of [tex]_{53}^{131}I[/tex]
Using formula for mass
[tex]m=\dfrac{131\times3.72\times10^{13}}{6.023\times10^{23}}[/tex]
[tex]m=8.09\times10^{-9}\ g[/tex]
Hence, The mass of [tex]_{53}^{131}I[/tex] is [tex]8.09\times10^{-9}\ g[/tex].
The problem is related to the concept of radioactive decay, specifically of the isotope Iodine-131. The activity of the sample is calculated to be 37,000,000 decays/second. Through a series of calculations using formulas for decay constant, the number of atoms and finally the sample mass, we find the mass to be 9.3x10^-16 g/day.
Explanation:The problem given relates to nuclear physics and utilizes the concept of radioactive decay. It specifically involves the isotope Iodine-131 (131 53I), which decays with a half-life of 8.08 days.
Assuming the activity of the sample is 1 mCi, which is equivalent to 37,000,000 decays/second (since 1 Ci=3.7x10^10 decays/sec, 1 mCi = 1/1000 Ci), we can use the activity formula which is given by A = λN, where A is activity, λ is the decay constant, and N is the number of atoms in the sample. We first need to calculate λ, which we can find using the formula λ = Ln(2)/T(1/2), where T(1/2) is the half-life.
λ = Ln(2)/8.08 days = 0.086/day. Therefore, N = A/ λ = 37,000,000 / 0.086 = 4.3x10^8 atoms/day.
The remaining calculation is the mass of the sample. The atomic mass of Iodine-131 is approximately 131 g/mol. Using Avogadro's number (6.02x10^23 atoms/mol), we can calculate the mass: M = N * (131 g/mol) / (6.02x10^23 atoms/mol) = 9.3x10^-16 g/day.
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A spacecraft is placed in a circular orbit around a planet with mass 6.4 x 1023 kg. The spacecraft orbits at a height of 4.5 x 107 m above the planet’s surface. What additional information is needed to calculate the speed of the spacecraft in the orbit?
Answer:
It is necessary to know the radius of the planet.
Explanation:
The speed of the spacecraft can be found by means of the equation of the Universal law of gravity:
[tex]F = G \frac{M.m}{r^{2}}[/tex] (1)
Where F is the gravitational force, G is gravitational constant, M is the mass of the planet, m is the mass of the spacecraft and r is the orbital radius of the spacecraft.
Equation 1 can be express in terms of the speed by using Newton's second law and the equation for centripetal acceleration:
[tex]F = ma[/tex] (2)
Replacing equation 2 in equation 1 it is gotten:
[tex]ma = G \frac{M.m}{r^{2}}[/tex] (3)
the centripetal acceleration is defined as:
[tex]a = \frac{v^{2}}{r}[/tex] (4)
Replacing equation 4 in equation 3 it is gotten:
[tex]m\frac{v^{2}}{r} = G \frac{M.m}{r^{2}}[/tex] (5)
Then, v can be isolated from equation 5:
[tex]mv^{2} = G \frac{M.m}{r}[/tex]
[tex]v^{2} = G \frac{M.m}{rm}[/tex]
[tex]v^{2} = G \frac{M}{r}[/tex]
[tex]v = \sqrt{\frac{GM}{r}}[/tex]
However, the orbital radius of the spacecraft is obtained by the sum of the radius of the planet and the height of the spacecraft above the surface of the planet (r = R+h)
[tex]v = \sqrt{\frac{GM}{R+h}}[/tex] (6)
Hence, by equation 6 it can be concluded that it is necessary to know the radius of the planet in order to calculate the speed of the spacecraft.
Which one of the following statements is false?
(a) The orbits in the Bohr model have precise sizes, whereas in the quantum mechanical picture of the hydrogen atom they do not.
(b) In the absence of external magnetic fields, both the Bohr model and quantum mechanics predict the same total energy for the electron in the hydrogen atom.
(c) The spin angular momentum of the electron plays a role in both the Bohr model and the quantum mechanical picture of the hydrogen atom.
(d) The magnitude of the orbital angular momentum cannot be zero in the Bohr model, but it can be zero in the quantum mechanical picture of the hydrogen atom.
Answer:
d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero
Explanation:
Affirmations
a) true. The orbits are accurate in the Bohr model and probabilistic in quantum mechanics
b) True. If both give the same results and use the same quantum number (n)
c) True. If in angular momentum it is quantized, in the Bohr model too but it does not justify it
d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero
A historian claims that a cannonball fired at a castle wall would melt on impact with the wall. Let's examine this claim. Assume that the kinetic energy of the cannonball is completely transformed into the internal energy of the cannonball on impact with no energy transferred to the wall. The cannonball is made of iron, which has a specific heat capacity of 450 J/kg-K and a melting point of 1811 K.
1) If the cannonball has an initial temperature of 298 K, how fast would the cannonball need to travel in order to reach its melting point on impact?
_________m/s
Answer:
v = 1166.9 m / s
Explanation:
The equation for caloric energy is
Q = m [tex]c_{e}[/tex] ΔT = m ce ([tex]T_{f}[/tex]-T₀)
Where m is the mass, ce is the specific heat and DT is the temperature variation
In this case the cannonball has kinetic energy
Em = K = ½ m v²
They indicate that mechanical energy is completely transformed into heat
Q = Em
m [tex]c_{e}[/tex] ([tex]T_{f}[/tex] - To) = ½ m v²
v = √ 2 [tex]c_{e}[/tex] ([tex]T_{f}[/tex]-To)
Let's calculate
v = √ (2 450 (1811-298))
v = 1166.9 m / s
A satellite explodes in outer space, far from any other body, sending thousands of pieces in all directions. Is the linear momentum of the satellite before the explosion less than, equal to, or greater than the total linear momentum of all the pieces after the explosion?
The answer to this problem can be given through energy conservation as well as Newton's first law.
Newton's first law states that as long as there is no force exerted on a body, its movement will be constant or at rest. In this way the amount of linear movement of the satellite before the explosion will be equal to the sum of the movement generated in the pieces after the explosion. In the absence of external forces but the preservation of the internal force as the only one to act, these will not change the total momentum of the system.