To solve this problem it is necessary to apply the concept related to root mean square velocity, which can be expressed as
[tex]v_{rms} = \sqrt{\frac{3RT}{n}}[/tex]
Where,
T = Temperature
R = Gas ideal constant
n = Number of moles in grams.
Our values are given as
[tex]v_e =11.2km/s = 11200m/s[/tex]
The temperature is
[tex]T = 30\°C = 30+273 = 303K[/tex]
Therefore the root mean square velocity would be
[tex]v_{rms} = \sqrt{\frac{3(8.314)(303)}{0.002}}[/tex]
[tex]v_{rms} = 1943.9m/s[/tex]
The fraction of velocity then can be calculated between the escape velocity and the root mean square velocity
[tex]\alpha = \frac{v_{rms}}{v_e}[/tex]
[tex]\alpha = \frac{1943.9}{11200}[/tex]
[tex]\alpha = 0.1736[/tex]
Therefore the fraction of the scape velocity on the earth for molecula hydrogen is 0.1736
The earth has a radius of 6.38 × 106 m and turns on its axis once every 23.9 h.
(a) What is the tangential speed (in m/s) of a person living in Ecuador, a country that lies on the equator?
(b) At what latitude (i.e., the angle theta in the drawing) is the tangential speed one-third that of a person living in Ecuador?
Answer:
a) V = 465.9 m/s
b) θ = 70.529°
Explanation:
Let's first calculate angular velocity of earth:
[tex]\omega=\frac{2\pi}{23.9h}*1h/3600s[/tex]
Velocity of a person on Ecuador will be:
[tex]V_E = \omega*R[/tex]
[tex]V_E = 465.9 m/s[/tex]
For part b, since angular velocity is the same:
[tex]\frac{\omega*R}{3}=\omega*(R*cos\theta )[/tex]
Solving for θ:
[tex]\theta=acos(1/3)[/tex]
[tex]\theta=70.529\°[/tex]
(a) speed at the equator: 465.9 m/s
(b) Latitude for 1/3 speed: 50.6°
(a) The tangential speed of a person living in Ecuador is equal to the circumference of the Earth at the equator divided by the period of rotation. The circumference of the Earth at the equator is 2πr, where r is the radius of the Earth. The period of rotation is 23.9 hours, which is equal to 23.9 × 3600 seconds. Therefore, the tangential speed of a person living in Ecuador is:
v = 2πr / T = 2π(6.38 ×[tex]10^6[/tex] m) / (23.9 × 3600 s) ≈ 465.9 m/s
(b) The tangential speed of a person living at a latitude of θ is equal to vr * cos(θ), where vr is the tangential speed of a person living in Ecuador and r is the radius of the Earth. Therefore, we can solve for θ:
cos(θ) = vr / v = 465.9 m/s / v
Solving for θ, we get:
θ = arccos(vr / v) ≈ 50.6°
Therefore, the latitude at which the tangential speed is one-third that of a person living in Ecuador is approximately 50.6°.
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As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.19 mm apart and position your screen 3.53 m from the slits. While Young had to struggle to achieve a monochromatic light beam of sufficient intensity, you simply turn on a laser with a wavelength of 635 nm. How far on the screen are the first bright fringe and the second dark fringe from the central bright fringe? Express your answers in millimeters.
Answer:
First bright fringe is at 1.82 mm
First dark fringe is at 2.83 mm
Solution:
As per the question:
Slit width, d = 1.19 mm = [tex]1.19\times 10^{- 3}\ m[/tex]
Distance from the screen, x = 3.53 m
Wavelength of the light, [tex]\lambda = 635\ nm = 635\times 10^{- 9}\ m[/tex]
Now,
We know that the 1st bright fringe from the central fringe is given by:
[tex]y = \frac{n\lambda x}{d}[/tex]
where
n = 1
[tex]y = \frac{1\times 635\times 10^{- 9}\times 3.53}{1.19\times 10^{- 3}} = 1.88\ mm[/tex]
Also, we know that the 1st dark fringe from the central fringe is given by:
[tex]y = \frac{(n + \frac{1}{2})\lambda x}{d}[/tex]
[tex]y = \frac{(1 + \frac{1}{2})\times 635\times 10^{- 9}\times 3.53}{1.19\times 10^{- 3}} = 2.83\ mm[/tex]
One Newton is the force: 1. of gravity on a 1 g body. 2. of gravity on a 1 kg body. 3. that gives a 1 kg body an acceleration of 1 m/s2. 4. that gives a 1 kg body an acceleration of 9.8 m/s2. 5. that gives a 1 g body an acceleration of 1 cm/s.
Answer:
3. One Newton is the force that gives a 1 kg body an acceleration of 1 m/s².
Explanation:
The force is a vector magnitude that represents every cause capable of modifying the state of motion or rest of a body or of producing a deformation in it.
Its unit in the International System is the Newton (N). A Newton is the force that when applied on a mass of 1 Kg causes an acceleration of 1 m/s².
1 N = kg*(m/s²)
A car with tires of radius 25.0 cm accelerates from rest to 30.0 m/s in 10.0 s. When the car's speed is 2.00 m/s find the linear acceleration of the top of the wheel relative to (a) the center of the wheel, (b) the road. (Note that there will be a vertical and horizontal component to the acceleration.
Answer
given,
radius of tire = 25 cm = 0.25 m
initial velocity = 0 m/s
final velocity = 30 m/s
time = 10 s
a) acceleration with respect to center
v = u + a t
30 = 0 + 10 a
a = 3 m/s²
radial acceleration
[tex]a_r = \dfrac{v^2}{r}[/tex]
[tex]a_r = \dfrac{2^2}{0.25}[/tex]
[tex]a_r =16\ m/s^2[/tex]
[tex]a=\sqrt{a_r^2 + a_t^2}[/tex]
[tex]a=\sqrt{16^2 + 3^2}[/tex]
[tex]a=16.28\ m/s[/tex]
acceleration w.r.t center of tire = [tex]a=16.28\ m/s[/tex]
b) acceleration with respect to road will be equal to 3 m/s²
A solid brass (bulk modulus 6.7 x 1010 N/m2) sphere is subjected to a pressure of 1.0 x 105 Pa due to the earth's atmosphere. On Venus the pressure due to the atmosphere is 9.0 x 106 Pa. By what fraction r/r0 (including the algebraic sign) does the radius of the sphere change when it is exposed to the Venusian atmosphere? Assume that the change in radius is very small relative to the initial radius.
Answer:
[tex]\frac{r}{r_0}=-4.4x10^{-5}[/tex]
Explanation:
Using the equation to find the fraction does the radius of the sphere change when it is exposed to Venusian atmosphere.
[tex]\frac{r}{r_0}=\frac{1}{3}*\frac{V}{V_0}[/tex]
Replacing to find the relation between the radius
[tex]\frac{V}{V_0}=-\frac{P}{B}[/tex]
Replacing numeric to find the relation
[tex]\frac{r}{r_0}=\frac{1}{3}*\frac{8.9x10^6pa}{6.7x10^{10}pa}[/tex]
[tex]\frac{r}{r_0}=\frac{1}{3}*-1.33x10^{-4}[/tex]
So the relation is
[tex]\frac{r}{r_0}=-4.4x10^{-5}[/tex]
Vadim bought his wife a "gold" bracelet from a guy on the street for $100. If he drops it in a beaker of water and the volume raises 400 cm3 and the mass is determined to be 0.5 kg, then is the jewelry gold? If it is not gold then what do you think it is made of?
Answer: It is not gold. The density is 1.25 g/cm3, very little compared with gold.
Explanation:
According to Archimedes’ principle, any body submerged in a liquid, receives an upward force, called buoyant force, which magnitude is equal to the weight of the liquid that displaces.
This weight can be calcultated as follows:
Fb= δH2O . Vbody g
We are told that the volume of the liquid displaced (equal to the volume of the body as it is completely submerged in water) is 400 cm3.
If we know that the mass of the bracelet is 0.5 Kg, we can find out the density of the material, just applying the same relationship between mass, density and volume:
m = δbracelet. V
δbracelet = m / V = 500 g / 400 cm3= 1.25 g/cm3
As the gold density is approximately 19.3 g/cm3, it is clear that the bracelet is not made from gold, but from something less dense.
The density of Vadim's bracelet is calculated to be 1.25 g/cm³, which is significantly lower than the density of pure gold (19.3 g/cm³). Thus, the bracelet is not made of gold and could be made from a lighter metal or alloy.
To determine if Vadim's bracelet is gold, we must calculate its density and compare it to the density of pure gold. The density of a substance is the mass-to-volume ratio, and for gold, it is 19.3 g/cm³. Given that the bracelet has a mass of 0.5 kg (or 500 g) and a volume displacement of 400 cm³ when submerged in water, we can calculate its density using the formula:
Density = Mass / Volume
density = 500 g / 400 cm³ = 1.25 g/cm³
The calculated density of 1.25 g/cm³ is significantly lower than the density of pure gold, which suggests that the bracelet is not made of gold. Since the density is lower than that of gold, the bracelet could be made of a metal or alloy with a lower density, such as aluminium or a mixture that includes materials with lighter densities.
⦁ A speed skater increases her speed form 10 m/s to 12.5 m/s over a period of 3 seconds while coming out of a curve of 20 m radius. What are the magnitudes of her radial, tangential and total accelerations as she leaves the curve?
Answer:
7.85 m/s^2
Explanation:
linear or tangential acceleration= dv/dt
⇒[tex]a_t= \frac{12.5-10}{3}[/tex]
=0.83 m/s^2
radial acceleration is given by = [tex]\frac{v^2}{r}[/tex]
⇒[tex]a_r =\frac{12.5^2}{20}[/tex]
= 7.81 m/s^2
total acceleration
[tex]a_T= \sqrt{a_t^2+a_r^2}[/tex]
putting values we get
[tex]a_T= \sqrt{0.83^2+7.81^2}[/tex]
= 7.85 m/s^2
Answer:
Explanation:
Tangential acceleration = ( 12.5 - 10 )/ 3
a_t= .833 m /s²
radial acceleration
= v² / R
12.5² / 20 ( 12.5 m/s is the velocity when it leaves the curve )
a_r= 7.81 ms⁻²
Total acceleration √( .833² + 7.81²)
= √( 61.6939)
= 7.85 m/s⁻
A 25 - kg television set rests on four rubber pads , each having a height of 1 . 0 cm and a radius of 0 . 60 cm . A 200 - N horizontal force is applied to the television set . How far does it move sideways ? The shear modulus of rubber is 2 . 6 x 106 N / m2
The television set moves sideways approximately 0.15 µm.
Explanation:
To calculate the distance the television set moves sideways, we need to use the equation for shear deformation. The equation is Ax = (F * L0 * L2) / (S * A), where Ax is the displacement, F is the force applied, L0 is the original length, L2 is the height of the rubber pads, S is the shear modulus, and A is the cross-sectional area. Substituting the known values, we have Ax = (200 N * 1.0 cm) / (2.6 x 10^6 N/m^2 * (3.14 x (0.0060 m)^2)). Solving for Ax, we find Ax ≈ 1.50 x 10^-7 m, or 0.15 µm.
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A small rubber wheel drives the rotation of a larger pottery wheel by running along its edge. The small wheel radius is 1.2 cm, and it accelerates at 3 rad/s2. The pottery wheel has a radius of 36 cm. What is the angular acceleration of the pottery wheel? How long till the pottery wheel rotates at 60 rpm?
Answer:
α₂= 0.1 rad/s²
t= 62.8 s
Explanation:
Given that
For small wheel
r₁= 1.2 cm
α₁ = 3 rad/s²
For large wheel
r₂= 36 cm
Angular acceleration = α₂ rad/s²
The tangential acceleration for the both wheel will be same
a = α₁ r₁=α₂ r₂
Now by putting the values in the above equation
α₁ r₁=α₂ r₂
3 x 1.2 = 36 x α₂
α₂= 0.1 rad/s²
Given that
N = 60 rpm
Angular speed in rad/s ω
[tex]\omega = \dfrac{2\pi N}{60}[/tex]
[tex]\omega = \dfrac{2\pi \times 60}{60}[/tex]
ω = 6.28 rad/s
Time taken is t
ω = α₂ t
6.28 = 0.1 t
t= 62.8 s
The angular acceleration of the pottery wheel is 0.1 rad/s². It will rotate at 60 rpm after approximately 63 seconds.
Explanation:The concept involved in the question relates to the principle of angular momentum conservation. The angular acceleration of the pottery wheel can be calculated from the given angular acceleration of the small wheel and the ratio of their radii. The equation that connects linear acceleration (a), angular acceleration (alpha), and radius (r) is a = alpha*r. Given that the small wheel has an angular acceleration, alpha1=3 rad/s² and radius r1=1.2 cm, while the pottery wheel has radius r2=36 cm, the linear acceleration of both is the same. Hence, the angular acceleration of the pottery wheel, alpha2 = a/r2 = (alpha1*r1)/r2 = (3 * 1.2 cm)/(36 cm) = 0.1 rad/s².
Now, for how long until the pottery wheel rotates at 60 rpm, first convert 60 rpm to rad/s. Note that 1 rev = 2π rad, and 1 min = 60 s. So, 60 rev/min = (60 * 2π rad)/(60 s) = 2π rad/s = final angular velocity, omega.
Then, use the formula for time in angular motion having constant angular acceleration: t = (omega - omega_initial) / alpha, and since the pottery wheel starts from rest, omega_initial = 0. Hence, t = omega/alpha2 = (2π rad/s)/(0.1 rad/s²) = 20π s, or approximately 63 seconds.
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A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the poorly secured load falls off! Immediately after losing the load, the truck begins to accelerate at 1.5 m/s2. Part A What was the mass of the load? Ignore rolling friction.
Answer:
The load has a mass of 2636.8 kg
Explanation:
Step 1 : Data given
Mass of the truck = 7100 kg
Angle = 15°
velocity = 15m/s
Acceleration = 1.5 m/s²
Mass of truck = m1 kg
Mass of load = m2 kg
Thrust from engine = T
Step 2:
⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:
T = (m1+m2)*g*sinθ
⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes m1*gsinθ .
Resultant force on truck is F = T – m1*gsinθ
F causes the acceleration of the truck: F= m*a
This gives the equation:
T – m1*gsinθ = m1*a
T = m1(a + gsinθ)
Combining both equations gives:
(m1+m2)*g*sinθ = m1*(a + gsinθ)
m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ
m2*g*sinθ = m1*a
Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:
m2*g*sinθ = (7100 – m2)*a
m2*g*sinθ = 7100a – m2a
m2*gsinθ + m2*a = 7100a
m2* (gsinθ + a) = 7100a
m2 = 7100a/(gsinθ + a)
m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)
m2 = 2636.8 kg
The load has a mass of 2636.8 kg
The mass of the load is 2636.8 kg
given information:Mass of the truck m = 7100 kg
Angle θ = 15°
velocity v = 15m/s
Acceleration a = 1.5 m/s²
Now let us assume that the mass of the truck is m₁ and the mass of the load is m₂. The thrust from the engine be T.
resolving forces:Initially, the thrust (T) balances the total weight:
[tex]T = (m_1+m_2) g sin\theta[/tex]
When the load falls off, the thrust (T) remains the same
so the net force on the truck
[tex]F = T -m_1gsin\theta[/tex]
[tex]T-m_1gsin\theta = m_1a \\\\ T = m_1(a + gsin\theta)[/tex]
comparing both the equation for T we get:
[tex](m_1+m_2)gsin\theta = m_1(a + gsin\theta)\\\\m_1gsin\theta + m_2gsin\theta =m_1a + m_1gsin\theta\\\\m_2gsin\theta = m_1a[/tex]
Now, m₁+m₂ = 7100kg
m₁= 7100 – m₂
[tex]m_2gsin\theta = (7100-m_2)a\\\\m_2gsin\theta = 7100a-m_2a\\\\m_2gsin\theta + m_2a = 7100a\\\\m_2 (gsin\theta + a) = 7100a\\\\m_2 = 7100a/(gsin\theta + a)\\\\m_2 = (7100 \times1.5) / (9.8sin(15) + 1.5)\\\\m_2 = 2636.8 kg[/tex]
The load has a mass of 2636.8 kg
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Ethanol has a heat of vaporization of 38.56 kJ/mol and a vapor pressure of 760 torr at 78.4 oC. What is the vapor pressure of ethanol at 38.8 oC?
To develop this problem it is necessary to apply the concepts developed by Clausius - Claperyron.
This duet found the relationship between temperature and pressure expressed as,
[tex]ln P = constant - \frac{\Delta H}{RT}[/tex]
For the two states that we have then we could define the pressure and temperature in each of them as
[tex]ln(\frac{P_2}{P_1}) = \frac{-\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})[/tex]
Where,
[tex]P_{1,2}[/tex]= Pressure at state 1 and 2
[tex]T_{1,2}[/tex]= Temperature at state 1 and 2
[tex]\Delta H[/tex]= Enthalpy of Vaporization of a substance
R = Gas constant (8.134J/mol.K)
Our values are given by,
[tex]P_1 = 1atm \\\Delta H = 38.56*10^{-3} J/mol \\R = 8.134J/mol.K\\T_1 = 78.4\°C = 351.55K\\T_2 =38.8\°C = 311.95K[/tex]
Therefore replacing we have that,
[tex]ln(\frac{P_2}{P_1}) = \frac{-\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})[/tex]
[tex]ln(\frac{P_2}{1atm}) = \frac{-38.56*10^3}{8.314}(\frac{1}{311.95}-\frac{1}{351.55})[/tex]
[tex]ln(P_2) - Ln(1atm) = \frac{-38.56*10^3}{8.314}(\frac{1}{311.95}-\frac{1}{351.55})[/tex]
[tex]P_2 = e^{\frac{-38.56*10^3}{8.314}(\frac{1}{311.95}-\frac{1}{351.55})}[/tex]
[tex]P_2 = 0.187355atm[/tex]
Therefore the pressure of the Ethanol at 38.8°C is 0.187355atm
If the net work done on an object is positive, what can you conclude about the object's motion?
- The object is moving with a constant velocity.
- The object is speeding up.
- The object is slowing down.
- The object is at rest; its position is constant.
Since there is a positive work on an object, this directly indicates that there is an increase in Energy on the body.
Work can be expressed in terms of Force and distance as
W = Fd
Where,
F = Force
d = Distance
The force in turn is a description of how mass accelerates according to Newton's second law, that is
F = ma
The mass (m) remains constant, but the acceleration (a) is constant, which implies directly that there is a gradual change in the velocity of the object.
Therefore the correct answer is: The object is speeding up.
The work is positive so the energy of the object is increasing so the object is speeding up
What can you conclude about objects' motion?As we know that the work is the product of the force and the displacement
[tex]W=F\times D[/tex]
Where,
F = Force
D= Distance
And from newtons second law we can see that
[tex]F=m\times a[/tex]
Since here mass will be constant to there will be a change in the velocity that is acceleration in the body so the energy of the body will change
Thus work is positive so the energy of the object is increasing so the object is speeding up
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A hollow spherical shell with mass 1.65 kg rolls without slipping down a slope that makes an angle of 38.0 ∘ with the horizontal. Part A Find the magnitude of the acceleration acm of the center of mass of the spherical shell. Take the free-fall acceleration to be g = 9.80 m/s2 .
Answer:
The acceleration is 3.62 m/s²
Explanation:
Step 1: Data given
mass of the shell = 1.65 kg
angle = 38.0 °
Step 2: Calculate the acceleration
We have 2 forces working on the line of motion:
⇒ gravity down the slope = m*g*sinα
⇒ provides the linear acceleration
⇒ friction up the slope = F
⇒ provides the linear acceleration and also the torque about the CoM.
∑F = m*a = m*g*sin(α) - F
I*dω/dt = F*R
The spherical shell with mass m has moment of inertia I=2/3*m*R² Furthermore a pure rolling relates dω/dt and a through a = R dω/dt. So the two equations become
m*a = m*g sin(α) - F
2/3*m*a = F
IF we combine both:
m*a = m*g*sin(α) - 2/3*m*a
1.65a = 1.65*9.81 * sin(38.0) - 2/3 *1.65a
1.65a + 1.1a = 9.9654
2.75a = 9.9654
a = 3.62 m/s²
The acceleration is 3.62 m/s²
The magnitude of the acceleration of the center of mass of the spherical shell is 3.62 m/s².
What is Newton’s second law of motion?Newton’s second law of motion shows the relation between the force mass and acceleration of a body. It says, that the net force applied on the body is equal to the product of mass of the body and the acceleration of it.
It can be given as,
[tex]\sum F=ma[/tex]
Here, (m) is the mass of the body and (a) is the acceleration.
A hollow spherical shell with mass 1.65 kg rolls without slipping down a slope that makes an angle of 38.0 ∘ with the horizontal. The free-fall acceleration g is 9.80 m/s2 .
The total friction force is equal to the force of gravity acting downward of the slope.
[tex]\sum F=mg\sin(\theta)-F\\[/tex]
For the force acting on the rotating spherical shell is,
[tex]F=\dfrac{2}{3}ma[/tex]
Put this value in above equation,
[tex]\sum F=(1.65)(9.80)\sin(38)-\dfrac{2}{3}(1.65)a\\ma=(1.65)(9.80)\sin(38)-\dfrac{2}{3}(1.65)a\\(1.65)a=(1.65)(9.80)\sin(38)-\dfrac{2}{3}(1.65)a\\a=(9.80)\sin(38)-\dfrac{2}{3}a\\a=3.62\rm \; m/s^2[/tex]
Thus, the magnitude of the acceleration of the center of mass of the spherical shell is 3.62 m/s².
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A carnival game consists of a two masses on a curved frictionless track, as pictured below. The player pushes the larger object so that it strikes the stationary smaller object which then slides follows the curved track so that it rises vertically to a maximum height, h_{max}h max . The masses are equipped with elastic bumpers so that the impact between them is an elastic collision. If the larger object has a mass , M = 5.41 ~\text{kg}M=5.41 kg and the smaller object has a mass of m = 1.68~\text{kg}m=1.68 kg, then with what velocity, v_0v 0 should the player release the larger object so that the smaller object just reaches the target maximum height of h_{max} = 3.0 mh max =3.0m above the horizontal portion of the track?
The required velocity of the larger object for the smaller object to reach a maximum height of 3m in an elastic collision can be calculated using the equation v0 = sqrt((2*m*g*h_max)/M). This concept relies on the conservation of energy and momentum in an elastic collision.
Explanation:The central physics concept in this question is the conservation of energy and momentum in an elastic collision. In an elastic collision, both momentum and kinetic energy are conserved. According to the conservation of energy, the kinetic energy of the larger mass before the collision is converted into the potential energy of the smaller mass at its maximum height.
Expressing the conservation of energy mathematically, we have:
1/2*M*v0² = m*g*h_max
Solving for v0, we have:
v0 = sqrt((2*m*g*h_max)/M)
where v0 is the velocity with which the player should release the larger object, m is the mass of the smaller object (1.68 kg), g is the acceleration due to gravity (9.8 m/s²), h_max is the maximum height the smaller object will reach (3.0 m), and M is the mass of the larger object (5.41 kg).
Please do remember to plug in the correct values and units when solving the equation.
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The player must release the larger object with a velocity of 1.95 m/s so that the smaller object just reaches the target maximum height of 3.0 m.
Since the collision of the two masses is elastic, the total kinetic energy of the system before the collision is equal to the total kinetic energy of the system after the collision.
We can use the principle of conservation of mechanical energy to solve this problem. The mechanical energy of a system is the sum of its potential energy and kinetic energy. At the beginning of the collision, the larger object has only potential energy due to its height, while the smaller object is at rest on the ground and has no energy. After the collision, the larger object has some kinetic energy and the smaller object has both potential energy and kinetic energy.
Let's denote the velocity of the larger object before the collision as v_0 and the velocity of the smaller object after the collision as v.
The potential energy of the larger object before the collision is:
PE_initial = Mgh
where:
M is the mass of the larger object (kg)
g is the acceleration due to gravity (m/s^2)
h is the height of the larger object above the ground (m)
The kinetic energy of the larger object before the collision is:
KE_initial = (1/2)Mv_0^2
The total mechanical energy of the system before the collision is:
E_initial = PE_initial + KE_initial = Mgh + (1/2)Mv_0^2
After the collision, the potential energy of the smaller object is:
PE_final = mgh_{max}
where:
m is the mass of the smaller object (kg)
g is the acceleration due to gravity (m/s^2)
h_{max} is the maximum height reached by the smaller object (m)
The kinetic energy of the smaller object after the collision is:
KE_final = (1/2)mv^2
The total mechanical energy of the system after the collision is:
E_final = PE_final + KE_final = mgh_{max} + (1/2)mv^2
Since the collision is elastic, we know that E_initial = E_final. Therefore, we can set the two equations equal to each other and solve for v_0.
Mgh + (1/2)Mv_0^2 = mgh_{max} + (1/2)mv^2
Solving for v_0, we get:
v_0 = sqrt((2mg(h - h_{max}))/(M + m))
Plugging in the values, we get:
v_0 = sqrt((2(5.41 kg)(9.81 m/s^2)(0.3 m))/(5.41 kg + 1.68 kg)) = 1.95 m/s
Therefore, the player must release the larger object with a velocity of 1.95 m/s so that the smaller object just reaches the target maximum height of 3.0 m.
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A 86 g particle undergoes SHM with an amplitude of 1.3 mm, a maximum acceleration of magnitude 6.8 x 103 m/s2, and an unknown phase constant φ. What are (a) the period of the motion, (b) the maximum speed of the particle, and (c) the total mechanical energy of the oscillator? What is the magnitude of the force on the particle when the particle is at (d) its maximum displacement and (e) half its maximum displacement?
Answer:
Explanation:
mass, m = 86 g
Amplitude, A = 1.3 mm
Maximum acceleration, a = 6.8 x 10^3 m/s^2
(a) Let T be the period of motion and ω be the angular frequency.
maximum acceleration, a = ω²A
6.8 x 10^3 = ω² x 1.3 x 10^-3
ω = 2287.0875 rad/s
T = 2π/ω
T = ( 2 x 3.14) / 2287.0875
T = 2.75 x 10^-3 second
T = 2.75 milli second
(b) maximum speed = ωA
v = 2287.0875 x 1.3 x 10^-3
v = 2.97 m/s
(c) Total mechanical energy
T = 1/2 mω²A²
T = 0.5 x 86 x 10^-3 x 2287.0875 x 2287.0875 x 1.3 x 1.3 x 10^-6
T = 0.38 J
(d) At maximum displacement, the acceleration is maximum
F = m x a = 86 x 10^-3 x 6.8 x 1000
F = 541.8 N
(e) acceleration a = ω² y = ω² x A / 2
a = 2287.0875 x 2287.0875 x 1.3 x 10^-3 / 2
a = 3400 m/s^2
Force, F = 86 x 10^-3 x 3400
F = 292.4 N
A machine part has the shape of a solid uniform sphere of mass 220 g and diameter 4.50 cm . It is spinning about a frictionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of 0.0200 N at that point.
The sphere's angular acceleration is [tex]2.064 rad/s^2[/tex], and it will take approximately 10.17 seconds to decrease its rotational speed by 21.0 rad/s when experiencing a constant friction force.
Part A: Calculating Angular Acceleration
To find the angular acceleration, we must first calculate the torque. The torque ( au) caused by the friction force (F) is the product of the force and the radius (r) at which the force is applied, and torque is given by [tex]\tau = r \times F[/tex]. Using the diameter (d) to find the radius, we have r = d / 2 = 4.50 cm / 2 = 2.25 cm = 0.0225 m. The given friction force is 0.0200 N. Therefore, [tex]\tau = 0.0225 m \times 0.0200 N = 0.00045 Nm[/tex].
Next, we use the moment of inertia (I) for a sphere, [tex]I=\frac{2}{5} mr^2[/tex], to calculate the angular acceleration ([tex]\alpha[/tex]). The mass (m) is 220 g which is 0.220 kg. So, [tex]I = \frac{2}{5} \times 0.220 kg \times (0.0225 m)^2 = 0.000218025 kg m^2[/tex]. Angular acceleration, [tex]\alpha = \frac{\tau}{I} = \frac{0.00045 Nm}{0.000218025 kgm^2} \longrightarrow \alpha= 2.064 rad/s^2[/tex].
Part B: Time to Decrease Rotational Speed
To find the time (t) to decrease the rotational speed by 21.0 rad/s, we can use the equation [tex]\omega = \omega_0 + \alpha \times t[/tex], where [tex]\omega_0[/tex] is the initial angular velocity and [tex]\omega[/tex] is the final angular velocity. Since the sphere is slowing down due to friction, the angular acceleration will be negative, [tex]\alpha = -2.064 rad/s^2[/tex]. We are looking for the time it takes for the speed to decrease by 21.0 rad/s, so if the initial speed is [tex]\omega_0[/tex] and the final is [tex]\omega_0 - 21.0 rad/s[/tex], then [tex]0 = \omega_0 - 21.0 rad/s + \alpha \times t[/tex]. Solving for t gives [tex]t = \frac{21.0 rad/s}{2.064 rad/s^2} \approx t = 10.17 s[/tex].
** In Problem 4.26, the coefficient of static friction between the book and the vertical back of the wagon is μs. Determine an expression for the minimum acceleration of the wagon in terms of μs so that the book does not slide down. Does the mass of the book matter? Explain.
The minimum acceleration of the wagon required for the book not to slide down can be determined using the coefficient of static friction (μs) between the book and the vertical back of the wagon. The expression for the minimum acceleration is given by a = μs * g.
Explanation:The minimum acceleration of the wagon required for the book not to slide down can be determined using the coefficient of static friction (μs) between the book and the vertical back of the wagon. The expression for the minimum acceleration is given by a = μs * g, where g is the acceleration due to gravity.
The mass of the book does not matter in this scenario. The coefficient of static friction is a property that depends on the nature of the surfaces in contact, and it determines the maximum force of static friction that can act between the book and the wagon before the book starts to slide down. The acceleration required to prevent sliding is independent of the mass of the book.
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The femur is a bone in the leg whose minimum cross-sectional area is about 3.70 10-4 m2. A compressional force in excess of 6.60 104 N will fracture this bone. (a) Find the maximum stress that this bone can withstand. (b) What is the strain that exists under a maximum-stress condition
Answer:
178378378.37 Pa
0.01897
Explanation:
F = Force = [tex]6.6\times 10^4\ N[/tex]
A = Area = [tex]3.7\times 10^{-4}\ m^2[/tex]
Y = Young's modulus of bone under compression = [tex]9.4\times 10^{9}\ Pa[/tex]
[tex]\varepsilon[/tex] = Strain
Stress is given by
[tex]\sigma=\frac{F}{A}\\\Rightarrow \sigma=\frac{6.6\times 10^4}{3.7\times 10^{-4}}\\\Rightarrow \sigma=178378378.37\ Pa[/tex]
The maximum stress that this bone can withstand is 178378378.37 Pa
Compression force is given by
[tex]F=Y\varepsilon A\\\Rightarrow \varepsilon=\frac{F}{YA}\\\Rightarrow \varepsilon=\frac{6.6\times 10^4}{9.4\times 10^{9}\times 3.7\times 10^{-4}} \\\Rightarrow \varepsilon=0.01897[/tex]
The strain that exists under a maximum-stress condition is 0.01897
The maximum stress that the femur can withstand is 1.78 * 10^8 Pascals, and the strain under this maximum stress condition is approximately 0.012 or 1.2%.
Explanation:The subject of the question is the stress and strain on the femur bone, which is integral to the field of Physics. To address part (a) of the question, we must find the maximum stress that the femur can withstand. Stress is defined as the average force per unit area, or the force divided by the area. In this case, we are given a compressional force of 6.60 * 10^4 Newtons and a cross-sectional area of the femur as 3.70 * 10^-4 m2. We divide the force by the area to find the stress:
Stress = Force / Area = (6.60 * 10^4 N) / (3.70 * 10^-4 m2) = 1.78 * 10^8 Pa (Pascals).
For part (b) of the question, the strain under a maximum-stress condition can be calculated by dividing the stress by Young's modulus. For bone, the modulus can vary, and bones are brittle with the elastic region small. However, if we take an average value, say 1.5 * 10^10 Pa, then, the strain will be:
Strain = Stress / Young's modulus = (1.78 * 10^8 Pa) / (1.5 * 10^10 Pa) = 0.012.
So under maximum stress, the strain is 0.012 (dimensionless) or 1.2%. Remember, this is an approximation as the actual modulus can vary based on numerous factors such as age, diet and lifestyle of the individual.
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A proton and an alpha particle are momentarily at rest at adistance r from each other. They then begin to move apart.Find the speed of the proton by the time the distance between theproton and the alpha particle doubles. Both particles arepositively charged. The charge and the mass of the proton are,respectively, e and m. The e charge and the mass of the alphaparticle are, respectively, 2e and 4m.
Find the speed of the proton (vf)p by the time the distancebetween the particles doubles.
Express your answer in terms of some or all of the quantities,e, m, r, and ?0.
Which of the following quantities are unknown?
A initial separation of the particles
B final separation of the particles
C initial speed of the proton
D initial speed of the alpha particle
E final speed of the proton
F final speed of the alpha particle
G mass of the proton
H mass of the alpha particle
I charge of the proton
J charge of the alpha particle
Answer:
The unknown quantities are:
E and F
The final velocity of the proton is:
√(8/3) k e^2/(m*r)
Explanation:
Hello!
We can solve this problem using conservation of energy and momentum.
Since both particles are at rest at the beginning, the initial energy and momentum are:
Ei = k (q1q2)/r
pi = 0
where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)
and q1 = e and q2 = 2e
When the distance between the particles doubles, the energy and momentum are:
Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2
pf = m1v1 + m2v2
with m1 = m, m2 = 4m, v1=vf_p, v2 = vf_alpha
The conservation momentum states that:
pi = pf
Therefore:
m1v1 + m2v2 = 0
That is:
v2 = (1/4) v1
The conservation of energy states that:
Ei = Ef
Therefore:
k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2
Replacing
m1 = m, m2 = 4m, q1 = e, q2 = 2e
and v2 = (1/4)v1
We get:
(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 = k e^2/r + (1/8)mv1^2
(3/8) mv1^2 = k e^2/r
v1^2 = (8/3) k e^2/(m*r)
A wheel with a weight of 395 N comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at an angular velocity of 23.1 rad/s . The radius of the wheel is 0.652 m and its moment of inertia about its rotation axis is 0.800 MR2. Friction does work on the wheel as it rolls up the hill to a stop, at a height of h above the bottom of the hill; this work has a magnitude of 3470 J .
Answer:
h=12.04m
Explanation:
1) Notation and some important concepts
[tex]\omega_i[/tex] represent the initial angular velocity
[tex]v_i[/tex] represent the initial velocity
[tex]h[/tex] represent the final height reached by the mass
[tex]M[/tex] represent the mass of the object
[tex]W_f[/tex] represent the work due the friction force (variable of interest)
[tex]KE_{rot}[/tex] represent the rotational energy
[tex]KE_{tra}[/tex] represent the transitional kinetic energy
[tex]PE=mgh[/tex] represent the potential energy
[tex]I=0.8MR^2[/tex] represent the rotational inertia
W= 395 N is the weight of the object
For this problem we can use the principle of energy conservation, this principle states that "the total energy of an isolated system remains constant; it is said to be conserved over time".
At the begin the wheel had rotational energy defined as "The kinetic energy due to rotational motion. Is a scalar quantity and have units of energy usually Joules". And this energy is represented by the following formula: [tex]KE_{rot}=\frac{1}{2}I\omega^2_i[/tex]
At the starting point the wheel also had kinetic energy defined as "The energy of mass in motion" and is given by the formula : [tex]KE_{tran}=\frac{1}{2}mv_i^2_i[/tex]
At the end of the movement we have potential energy since the mass is at height h the potential energy is defined as "The energy stored within an object, due to the object's position, arrangement or state" and is given by the formula [tex]PE=mgh[/tex].
Since we have friction acting we have a work related to the force of friction and we need to subtract this from the formula of conservation of energy
2) Formulas to use
The figure attached is an schematic draw for the problem
Using the principle of energy conservation we have:
[tex]KE_{rot}+KE_{tran}-W_f =PE[/tex]
Replacing the formulas for each energy w ehave:
[tex]\frac{1}{2}I\omega^2_i+\frac{1}{2}mv_i^2_i-W_f =Mgh[/tex] (1)
We also know that [tex]v_i =\omega_i R[/tex] and [tex]I=0.8MR^2[/tex] so if we replace this into equation (1) we got:
[tex]0.8(\frac{1}{2})M(R\omega_i)^2 +\frac{1}{2}M(\omega_i R)^2-W_f=Mgh[/tex] (2)
We also know that the weight is defined as [tex]W=mg[/tex] so then [tex]M=\frac{W}{g}[/tex], so if we replace this into equation (2) we have:
[tex]0.8(\frac{1}{2})\frac{W}{g}(R\omega_i)^2 +\frac{1}{2}\frac{W}{g}(\omega_i R)^2-W_f=Wh[/tex] (3)
So then if we solve for h we got:
[tex]h=\frac{0.8(\frac{1}{2})\frac{W}{g}(R\omega_i)^2 +\frac{1}{2}\frac{W}{g}(\omega_i R)^2-W_f}{W}[/tex] (4)
3) Solution for the problem
Now we can replace the values given into equation (4):
[tex]h=\frac{0.8(\frac{1}{2})\frac{395N}{9.8\frac{m}{s^2}}(0.652m(23.1\frac{rad}{s}))^2 +\frac{1}{2}\frac{395}{9.8\frac{m}{s^2}}(23.1\frac{rad}{s}(0.652m))^2-3470J}{395N}=12.04m[/tex] (4)
So then our final answer would be h=12.04m
When 1.14 g of octane (molar mass = 114 g/mol) reacts with excess oxygen in a constant volume calorimeter, the temperature of the calorimeter increases by 10.0°C. The heat capacity of the calorimeter is 6.97 kJ/°C. Determine the energy flow, q (reaction).
Answer:
Q (reaction) = -69.7 kJ
Explanation:
Octane reacts with oxygen to give carbon dioxide and water.
C₈H₁₈ + 25 O₂ ---> 16 CO₂ +18 H₂O
This reaction is exothermic in nature. Therefore, the energy is released into the atmosphere. This reaction took place in a calorimeter, there the temperature (T) increases by 10 C. The heat capacity of the calorimeter is 6.97 kJ/C
The heat (q) of the reaction is calculated as follows:
Q= -cT, where c is the heat capacity of the calorimeter and T is the increase in temperature
q = -(6.97) x (10) = -69.7kJ
Since the heat capacity is given in kilo -joule per degree Celsius, therefore, the mass of octane is not required
A 70 kg person does a bungee jump from a bridge. The natural length of the bungee cord is 20m and it has a spring constant k of 80 N/m. When the bungee cord has a total length of 25m, what is the acceleration of the jumper (to 2 significant figures)? Hint: Draw a FBD showing the magnitudes and directions of all forces on the jumper at the instant described.
The bungee jumper experiences a downward acceleration of 4.09 m/s^2 at the point when the bungee cord has lengthened to 25m.
Explanation:The subject of this problem pertains to physics, specifically, the principles of dynamics and Hooke's Law. At the instant when the bungee cord has stretched to a total length of 25m, it has undergone an extension of 5m since its natural length is 20m. According to Hooke’s Law, this extension causes a restoring force F = kx, where k is the spring constant and x is the change in length.
In this case, F = 80 N/m × 5 m = 400 N which is the upward force exerted by the cord. The weight of the 70 kg person is mg, or 70 kg × 9.8 m/s² = 686 N downward. The net force acting on the jumper is the difference between the weight and the restoring force. This gives a net force of 686 N – 400 N = 286 N downward.
From Newton’s second law, force equals mass times acceleration, F = ma. Therefore, the acceleration of the jumper at the instant would be a = F/m, or 286 N ÷ 70 kg = 4.09 m/s² downward to 2 significant figures.
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A fatigue test was conducted in which the mean stress was 50 MPa (7250 psi) and the stress amplitude was 225 MPa (32,625 psi).
(a) Compute the maximum and minimum stress levels.
(b) Compute the stress ratio.
(c) Compute the magnitude of the stress range.
Answer:
275 MPa, -175 MPa
-0.63636
450 MPa
Explanation:
[tex]\sigma_{max}[/tex] = Maximum stress
[tex]\sigma_{min}[/tex] = Minimum stress
[tex]\sigma_m[/tex] = Mean stress = 50 MPa
[tex]\sigma_a[/tex] = Stress amplitude = 225 MPa
Mean stress is given by
[tex]\sigma_m=\frac{\sigma_{max}+\sigma_{min}}{2}\\\Rightarrow \sigma_{max}+\sigma_{min}=2\sigma_m\\\Rightarrow \sigma_{max}+\sigma_{min}=2\times 50\\\Rightarrow \sigma_{max}+\sigma_{min}=100\ MPa\\\Rightarrow \sigma_{max}=100-\sigma_{min}[/tex]
Stress amplitude is given by
[tex]\sigma_a=\frac{\sigma_{max}-\sigma_{min}}{2}\\\Rightarrow \sigma_{max}-\sigma_{min}=2\sigma_a\\\Rightarrow \sigma_{max}-\sigma_{min}=2\times 225\\\Rightarrow \sigma_{max}-\sigma_{min}=450\ MPa\\\Rightarrow 100-\sigma_{min}-\sigma_{min}=450\\\Rightarrow -2\sigma_{min}=350\\\Rightarrow \sigma_{min}=-175\ MPa[/tex]
[tex]\sigma_{max}=100-\sigma_{min}\\\Rightarrow \sigma_{max}=100-(-175)\\\Rightarrow \sigma_{max}=275\ MPa[/tex]
Maximum stress level is 275 MPa
Minimum stress level is -175 MPa
Stress ratio is given by
[tex]R=\frac{\sigma_{min}}{\sigma_{max}}\\\Rightarrow R=\frac{-175}{275}\\\Rightarrow R=-0.63636[/tex]
The stress ratio is -0.63636
Stress range is given by
[tex]\sigma_{max}-\sigma_{min}=450\ MPa[/tex]
Magnitude of the stress range is 450 MPa
The maximum and minimum stress levels could be calculated as 275 MPa and -175 MPa respectively. The stress ratio comes out to be -0.64, and the magnitude of the stress range is 450 MPa.
Explanation:In material fatigue studies, the stress levels in a material are often evaluated. In your case:
The mean stress (σm) is given as 50 MPa, and the stress amplitude (σa) is 225 MPa.(a) The maximum stress (σmax) can be calculated as σmax = σm + σa = 50 + 225 = 275 MPa, and the minimum stress (σmin) can be calculated as σmin = σm - σa = 50 - 225 = -175 MPa.(b) The stress ratio (R) is derived as the ratio of the minimum to the maximum stress. So R = σmin / σmax = -175 / 275 = -0.64(rounded to two decimal places).(c) The stress range (Δσ) is defined as the difference between the maximum and minimum stresses in a cycle. So Δσ = σmax - σmin = 275 - (-175) = 450 MPa.Learn more about Material Fatigue here:https://brainly.com/question/33293027
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Nitrogen (N2) undergoes an internally reversible process from 6 bar, 247°C during which pν1.2 = constant. The initial volume is 0.1 m3 and the work for the process is 50 kJ. Assuming ideal gas behavior, and neglecting kinetic and potential energy effects, determine heat transfer, in kJ, and the entropy change, in kJ/K.
The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.
Explanation:The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. In order to determine the heat transfer and entropy change, we need to use the first and second laws of thermodynamics. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.
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A conducting loop of radius 1.50 cm and resistance 8 × 10−6Ω is perpendicular to a uniform magnetic field of magnitude 23.0 × 10−6T. The field magnitude drops to zero in 7 ms. How much thermal energy is produced in the loop by the change in the magnetic field?
To solve this problem it is necessary to apply the concepts related to electromotive force or induced voltage.
By definition we know that the induced emf in the loop is equal to the negative of the change in the magnetic field, that is,
[tex]\epsilon = -A \times \frac{\Delta B}{\Delta t}[/tex]
[tex]\epsilon = -A \times (\frac{B_f-B_i}{t_f-t_i})[/tex]
Where A is the area of the loop, B the magnetic field and t the time.
Replacing with our values we have that
[tex]\epsilon = -(\pi (1.5*10^{-2})^2)(\frac{0-23*10^{-6}}{7*10^{-3}-0})[/tex]
[tex]\epsilon = 2.3225*10^{-6}V[/tex]
Therefore the thermal energy produced is given by
[tex]E = P*t = \frac{\epsilon^2}{R}t[/tex]
[tex]E = \frac{(2.3225*10^{-6})^2}{8*10^{-6}}*(7*10^{-3})[/tex]
[tex]E = 4.719*10^{-9}J[/tex]
The thermal energy produced in the loop is [tex]4.719*10^{-9}J[/tex]
A toy car having mass m = 1.50 kg collides inelastically with a toy train of mass M = 3.60 kg. Before the collision, the toy train is moving in the positive x-direction with a velocity of Vi = 2.45 m/s and the toy car is also moving in the positive x-direction with a velocity of vi = 4.75 m/s. Immediately after the collision, the toy car is observed moving in the positive x-direction with a velocity of 2.15 m/s.
(a) Determine Vf, the final velocity of the toy train.
Answer:
[tex]v_{f}[/tex] = 3,126 m / s
Explanation:
In a crash exercise the moment is conserved, for this a system formed by all the bodies before and after the crash is defined, so that the forces involved have been internalized.
the car has a mass of m = 1.50 kg a speed of v1 = 4.758 m / s and the mass of the train is M = 3.60 kg and its speed v2 = 2.45 m / s
Before the crash
p₀ = m v₁₀ + M v₂₀
After the inelastic shock
[tex]p_{f}[/tex]= m [tex]v_{1f}[/tex] + M [tex]v_{2f}[/tex]
p₀ = [tex]p_{f}[/tex]
m v₀ + M v₂₀ = m [tex]v_{1f}[/tex] + M [tex]v_{2f}[/tex]
We cleared the end of the train
M [tex]v_{2f}[/tex] = m (v₁₀ - v1f) + M v₂₀
Let's calculate
3.60 v2f = 1.50 (2.15-4.75) + 3.60 2.45
[tex]v_{2f}[/tex] = (-3.9 + 8.82) /3.60
[tex]v_{2f}[/tex] = 1.36 m / s
As we can see, this speed is lower than the speed of the car, so the two bodies are joined
set speed must be
m v₁₀ + M v₂₀ = (m + M) [tex]v_{f}[/tex]
[tex]v_{f}[/tex] = (m v₁₀ + M v₂₀) / (m + M)
[tex]v_{f}[/tex] = (1.50 4.75 + 3.60 2.45) /(1.50 + 3.60)
[tex]v_{f}[/tex] = 3,126 m / s
8. A solid wooden door, 90 cm wide by 2.0 m tall, has a mass of 35 kg. It is open and at rest. A small 500-g ball is thrown perpendicular to the door with a speed of 20 m/s and hits the door 60 cm from the hinged side, causing it to begin turning. The ball rebounds along the same line with a speed of 16.0 m/s relative to the ground. If the momentum of inertia of the door around the hinge is I=1/3 Ma 2 , where a is the width of the door, how much energy was lost during this collision?
Answer:
Explanation:
Kinetic energy of ball
= .5 x .5 x 20²
= 100 J
Original kinetic energy of door = 0
Total kinetic energy before ball hitting the door
= 100 J
We shall apply law of conservation of momentum to calculate angular velocity of the door after ball hitting it.
change in angular momentum of ball
= mvr - mur , u is initial velocity and v is final velocity of ball
= .5 ( 20 + 16 ) x .06
= 1.08
Change in angular momentum of door
= I x ω - 0
1/3 x 35 x .09² x ω
= .0945 x ω
so
.0945 x ω = 1.08
ω = 11.43
rotational K E of door after collision
= 1/2 I ω²
= .5 x .0945 x 11.43 ²
= 6.17 J
Kinetic energy of ball after collision
= 1/2 x .5 x 16²
= 64
Total KE of door and ball
= 64 + 6.17
= 70.17 J
LOSS OF ENERGY
= 100 - 70.17 J
= 29.83 J
Halogen lightbulbs allow their filaments to operate at a higher temperature than the filaments in standard incandescent bulbs. For comparison, the filament in a standard lightbulb operates at about 2900K, whereas the filament in a halogen bulb may operate at 3400K. Which bulb has the higher peak frequency? Calculate the ratio of the peak frequencies. The human eye is most sensitive to a frequency around 5.5x10^14 Hz. Which bulb produces a peak frequency close to this value?
Answer:
Halogen
0.85294
Explanation:
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
b = Wien's displacement constant = [tex]2.897\times 10^{-3}\ mK[/tex]
T = Temperature
From Wien's law we have
[tex]\lambda_m=\frac{b}{T}\\\Rightarrow \lambda_m=\frac{2.897\times 10^{-3}}{2900}\\\Rightarrow \lambda_m=9.98966\times 10^{-7}\ m[/tex]
Frequency is given by
[tex]\nu=\frac{c}{\lambda_m}\\\Rightarrow \nu=\frac{3\times 10^8}{9.98966\times 10^{-7}}\\\Rightarrow \nu=3.00311\times 10^{14}\ Hz[/tex]
For Halogen
[tex]\lambda_m=\frac{b}{T}\\\Rightarrow \lambda_m=\frac{2.897\times 10^{-3}}{3400}\\\Rightarrow \lambda_m=8.52059\times 10^{-7}\ m[/tex]
Frequency is given by
[tex]\nu=\frac{c}{\lambda_m}\\\Rightarrow \nu=\frac{3\times 10^8}{8.52059\times 10^{-7}}\\\Rightarrow \nu=3.52088\times 10^{14}\ Hz[/tex]
The maximum frequency is produced by Halogen bulbs which is closest to the value of [tex]5.5\times 10^{14}\ Hz[/tex]
Ratio
[tex]\frac{3.00311\times 10^{14}}{3.52088\times 10^{14}}=0.85294[/tex]
The ratio of Incandescent to halogen peak frequency is 0.85294
A horizontal uniform bar of mass 2.9 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 1.45 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:
given,
mass of uniform bar = 2.9 Kg
length of string = 3 m
mass of monkey = 1.45 Kg
monkey is sitting at the center
Weight of the beam will also consider to as the center
total downward force = 2.9 g + 1.45 g
now tension will balance the force
T₁ + T₂ = 2.9 g + 1.45 g
T₁ + T₂ = 4.35 g
taking moment about T₂ string
now,
T₁ x (3-0.74) - 4.35 g x (1.5 - 0.74) = 0
T₁ x 2.26 = 32.3988
T₁ = 14.34 N
wide tube that extends down from the bag of solution, which hangs from a pole so that the fluid level is 90.0 cm above the needle. The inner radius of the needle is 0.200 mm. The top of the fluid is exposed to the atmosphere, and the flow rate of the fluid (which has a density of 1025 kg/m^3 and a viscosity of 0.0010 Pass) through the needle is 0.200 L/h. What is the average gauge pressure inside the vein where the needle is? Use g = 9.8 m/s^2. _______ Pa
Answer:
The average gauge pressure inside the vein is 110270.58 Pa
Explanation:
This question can be solved using the Bernoulli's Equation. First, in order to determine the outlet pressure of the needle, we need to find the total pressure exerted by the atmosphere and the fluid.
[tex]P_f: fluid's\ pressure\\P_f= \rho g h=1025\frac{kg}{m^3} \times 9.8 \frac{m}{s^2} \times 0.9 m=9040.5 Pa \\P_T: total\ pressure\\P_T=P_{atm}+P_f\\P_T=101325 Pa + 9040.5 Pa=110275.5 Pa\\[/tex]
Then, we have to find the fluid's outlet velocity with the transversal area of the needle, as follows:
[tex]S: transversal\ area \\S= \pi r^2=\pi (0.200 \times 10^{-3})^2=5.65 \times 10^{-7} m^2\\v=\frac{F}{S}=\frac{5.55 \times 10^{-8} \frac{m^3}{s}}{5.65 \times 10^{-7} m^2}=0.98\times 10^{-1} \frac{m}{s}[/tex]
As we have all the information, we can complete the Bernoulli's expression and solve to find the outlet pressure as follows:
[tex]P_T-P_{out}=\frac{1}{2} \rho v^2\\P_{out}=P_T-\frac{1}{2} \rho v^2=110275.5 Pa-\frac{1}{2} 1025\frac{kg}{m^3} (0.98\times 10^{-1} \frac{m}{s})^2=110275.5 Pa-4.92 Pa =110270.58 Pa[/tex]