The measurement will be significantly affected.
Recall that the relationship between linear velocity and angular velocity is subject to the formula
[tex]v = \omega r[/tex],
Where r indicates the radius and [tex]\omega[/tex] the angular velocity.
As the radius increases, it is possible that the calibration is delayed and a higher linear velocity is indicated, that to the extent that the velocity is directly proportional to the radius of the tires.
Mounting larger-diameter tires on a truck leads to a systematic error in the speedometer reading, whereby the speedometer will underreport the truck's true linear speed due to an increase in linear distance covered per tire rotation at a constant angular velocity.
If larger-diameter tires are mounted on a truck, the speedometer reading will not be correct. This is due to the relationship between the angular velocity of the tires and the linear speed of the truck. Since the speedometer measures the angular speed of the tires to determine the truck's linear speed, changing to larger tires will cause a systematic error in the reading. For a fixed angular velocity, larger tires will cover a greater linear distance because of their larger circumference. Thus, the speedometer will display a speed lower than the truck's true linear speed. This affects the accuracy of the speedometer.
To further illustrate, consider a truck tire rotating with an angular velocity α. The linear (tangential) velocity v at the surface of the tire is given by v = rα, where r is the tire radius. With larger tires, for the same angular velocity α, r is bigger, hence v is larger. Therefore, when the tires are larger, the speedometer, which is calibrated for the standard tire size, underestimates the truck's speed because it is based on an incorrect assumption about the tire circumference.
If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon. Assume that the beam leaves the flashlight through a 7.0-cm aperture, that its white light has an average wavelength of 550nm, and that the beam spreads due to diffraction only. The distance from the Earth to the Moon is 384x10^3km.
To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution. By definition is defined as:
[tex]\theta = 1.22\frac{\lambda}{d}[/tex]
Where,
[tex]\lambda[/tex]= Wavelength
d = Width of the slit
[tex]\theta[/tex]= Angular resolution
Through the arc length we can find the radius, which would be given according to the length and angle previously described.
The radius of the beam on the moon is
[tex]r = l\theta[/tex]
Relacing [tex]\theta[/tex]
[tex]r = l(\frac{1.22\lambda}{d})[/tex]
[tex]r = 1.22\frac{l\lambda}{d}[/tex]
Replacing with our values we have that,
[tex]r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})[/tex]
[tex]r = 3680.91m[/tex]
Therefore the diameter of the beam on the moon is
[tex]d = 2r[/tex]
[tex]d = 2 * (3690.91)[/tex]
[tex]d = 7361.8285m[/tex]
Hence, the diameter of the beam when it reaches the moon is 7361.82m
Final answer:
Using diffraction formulas, it is estimated that a flashlight beam with an aperture of 7.0cm and an average wavelength of 550nm would have a diameter of approximately 3.7 km when it reaches the Moon.
Explanation:
To estimate the diameter of a flashlight beam when it reaches the Moon, we'll use the concept of diffraction spreading, as the beam is assumed to spread due to diffraction only. According to the diffraction formula for a circular aperture, θ = 1.22 λ/D. The angle θ is the angle of the spread, λ is the average wavelength of the light, and D is the diameter of the aperture through which the light passes.
Using the provided average wavelength of λ = 550nm and an aperture diameter of D = 7.0cm for the flashlight, we can calculate the minimum angular spread of the beam, θ.
θ = 1.22 × (550 x 10^-9 m) / (7.0 x 10^-2 m) = 0.0000095714 radians.
Then, to find out how broad the beam would be when it reaches the Moon, we use the angular spread to calculate the diameter of the beam at the distance to the Moon, which is 384,000 km (or 384 x 10^6 m):
Beam Diameter on the Moon = θ × Distance to the Moon = 0.0000095714 radians × 384 x 10^6 m = 3677.388 m.
Therefore, the flashlight's beam would have an estimated diameter of approximately 3.7 km when it reaches the Moon, considering only the diffraction spreading.
Young's double slit experiment is one of the classic tests for the wave nature of light. In an experiment using red light (λ = 641 nm) the second dark fringe on either side of the central maximum is θ = 4.4 degrees relative to the central bright fringe.
(a) Write an expression for the separation distance between the slits.
Answer:
The separation distance between the slits is 16710.32 nm.
Explanation:
Given that,
Wavelength = 641 nm
Angle =4.4°
(a). We need to calculate the separation distance between the slits
Using formula of young's double slit
[tex]d\sin\theta=m\lambda[/tex]
[tex]d=\dfrac{m\lambda}{\sin\theta}[/tex]
Where, d = the separation distance between the slits
m = number of order
[tex]\lambda[/tex] =wavelength
Put the value into the formula
[tex]d=\dfrac{2\times641\times10^{-9}}{\sin4.4}[/tex]
[tex]d=0.00001671032\ m[/tex]
[tex]d=16710.32\ nm[/tex]
Hence, The separation distance between the slits is 16710.32 nm.
The separation distance between the slits in Young's double slit experiment can be calculated using the formula d = λx / sin(θ), where d is the separation distance, λ is the wavelength of light, θ is the angle of the fringe, and x is the distance between the fringe and the screen.
Explanation:In Young's double slit experiment, the separation distance between the slits can be determined using the formula:
d = λx / sin(θ)
Where d is the separation distance between the slits, λ is the wavelength of light, θ is the angle of the fringe, and x is the distance between the fringe and the screen. In this example, the second dark fringe on either side of the central maximum is given as θ = 4.4 degrees. To calculate the separation distance between the slits, we also need to know the wavelength of the red light, which is given as λ = 641 nm.
Using the formula, we have:
d = (641 nm) * x / sin(4.4 degrees)
To find the value of d, we need to know the value of x, which is the distance between the fringe and the screen.
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Which of the following is the reason why the magnetic flux through the bracelet is changing?
a. The magnitude of the magnetic field is changing.
b. The magnetic field is changing direction with respect to the bracelet.
c. The bracelet is moving in the magnetic field.
The magnetic flux through an object like a bracelet can change a) because the bracelet is moving in the magnetic field, field direction, or if the bracelet moves through the field. Examples include loops moving into or rotating within a magnetic field.
The reason why the magnetic flux through the bracelet is changing can be due to various factors. These include a change in the magnitude of the magnetic field, a) a change in the field's direction relative to the bracelet, or the bracelet moving through the magnetic field. Magnetic flux changes when a loop moves into a magnetic field or rotates within it. Specifically, in part (a) as the loop moves into the field, and in part (b) as the loop rotates, thus changing the orientation.
It is also indicated that the change in magnetic flux through a loop or coil can result from the spinning of a magnet nearby, causing the magnetic field within the coil to change rapidly, as described with a rotating magnet affecting coil current due to the variation in the number and direction of magnetic field lines passing through it.
The correct option is c. The bracelet is moving in the magnetic field.
Magnetic flux through a surface, such as a bracelet, is defined as the product of the magnetic field (B), the area of the surface (A), and the cosine of the angle (θ) between the magnetic field lines and the normal to the surface. Mathematically, this is expressed as:
[tex]\[ \Phi_B = B \cdot A \cdot \cos(\theta) \][/tex]
For the magnetic flux to change, one or more of the following must occur:
1. The magnitude of the magnetic field (B) changes.
2. The area of the surface (A) changes.
3. The orientation of the surface with respect to the magnetic field changes, i.e., the angle (θ) changes.
4. The surface moves within the magnetic field, changing the amount of field lines passing through it.
Given the options:
a. The magnitude of the magnetic field is changing. - This would indeed change the magnetic flux, but it is not one of the given scenarios.
b. The magnetic field is changing direction with respect to the bracelet. - This would also change the magnetic flux, as it would change the angle θ between the magnetic field lines and the normal to the surface of the bracelet. However, this option does not explicitly state that the direction of the magnetic field is changing; it only mentions the direction with respect to the bracelet, which could imply a change in orientation of the bracelet itself.
c. The bracelet is moving in the magnetic field. - This option indicates that the bracelet is changing its position within the magnetic field. As the bracelet moves, the amount of magnetic field lines passing through it can change, even if the magnetic field itself is uniform and constant. This movement can alter the effective area through which the field lines pass (A) and/or the angle (θ), thus changing the magnetic flux.
Since the question asks for the reason why the magnetic flux through the bracelet is changing, and option c directly describes a scenario where the bracelet's movement within the field would cause a change in flux, it is the correct answer. The other options either do not apply to the given scenario or are not explicitly described as occurring.
A voltmeter and an ammeter are used to respectively monitor the voltage across and the current through a resistor connected to an AC source. If the resistor has a value of 10.0 Ω and the ammeter reads 8.02 A, determine the following.
(a) the rms voltage (in V) across the resistor(b) the peak voltage (in V) of the source(c) the maximum current (in A) in the resistor(d) the average power (in W) delivered to the resistor
Final answer:
To determine the values requested - (a) the rms voltage across the resistor is 80.2 V, (b) the peak voltage of the source is 113.4 V, (c) the maximum current in the resistor is 8.02 A, and (d) the average power delivered to the resistor is 641 W.
Explanation:
In order to determine the values requested, we can use Ohm's law which states that V = IR, where V is the voltage, I is the current, and R is the resistance. Let's go through each part:
(a) To find the rms voltage across the resistor, we use the formula V = IR, where I is the ammeter reading and R is the resistance. Plugging in the values, we get V = (8.02 A) x (10.0 Ω) = 80.2 V.
(b) The peak voltage of the source can be found by multiplying the rms voltage by √2. So the peak voltage is (80.2 V) x √2 = 113.4 V.
(c) The maximum current in the resistor is equal to the ammeter reading, which is 8.02 A.
(d) The average power delivered to the resistor can be calculated using the formula P = IV, where I is the rms current and V is the rms voltage. Since the ammeter reading is the rms current, we can plug in the values to get P = (8.02 A) x (80.2 V) = 641 W.
A 47.0 kg uniform rod 4.25 m long is attached to a wall with a hinge at one end. The rod is held in a horizontal position by a wire attached to its other end. The wire makes an angle of 30.0 ∘ with the horizontal, and is bolted to the wall directly above the hinge.
If the wire can withstand a maximum tension of 1350 N before breaking, how far from the wall can a 69.0 kg person sit without breaking the wire?
Answer:
2.79 m
Explanation:
Use the static equilibrium condition, net torque actin on the system is zero.
[tex]\sum \tau= 0[/tex]
[tex]T(Lsin\theta)- Mg\frac{L}{2}-mgx=0[/tex]
solve for the distance of the person from the wall x
[tex]x= \frac{TLsin\theta-Mg(L/2)}{Mg}[/tex]
now putting the values we get
[tex]x= \frac{1350\times4.25sin30-47\times9.81\times(4.25/2)}{69\times9.80}[/tex]
= 2.79 m
A 100 kg object hangs from two steel cables, both with radius 1.2 mm. The first cable is 2.5 m long and 2 mm shorter than the second, but the object is horizontal (so they are now the same length). What is the force on the first cable? Young's Modulus for steel is 2.0 x 1011 N/m2.A :470 NB :500 NC :850 ND :1000 N
Answer:
850N
Explanation:
The step by step is in the attachment.
A glider of mass 5.0 kg hits the end of a horizontal rail and bounces off with the same speed, in the opposite direction. The collision is elastic and takes place in a time interval of 0.2s, with an average force of 100N. What was the speed, in m/s, of the glider
To solve the exercise it is necessary to apply the concepts given in Newton's second law and the equations of motion description.
Let's start by defining acceleration based on speed and time, that is
[tex]a = \frac{v}{t}[/tex]
On the other hand according to Newton's second law we have to
F=ma
where
m= Mass
a = Acceleration
Replacing the value of acceleration in this equation we have
[tex]F=m(\frac{v}{t})[/tex]
Substituting with our values we have
[tex]100N=(5Kg)\frac{v}{0.2}[/tex]
Re-arrange to find v
[tex]v=\frac{100*0.2}{5}[/tex]
[tex]v = 2m/s[/tex]
Therefore the speed of the glider is 2m/s
What is the linear size of the smallest box in which you can confine an electron if you want to know for certain that the electron's speed is no more than 13 m/s ? Express your answer to two significant figures and include the appropriate units.
Using the Heisenberg Uncertainty Principle in quantum mechanics, which states that position and momentum of a particle cannot both be precisely measured at the same time, we can calculate the smallest box in which an electron can be confined with a specified maximum speed. The mass of the electron and the given speed are used to compute the uncertainty in momentum, which is then used in the Heisenberg inequality formula to find the size of the box.
Explanation:This question can be addressed using the Heisenberg Uncertainty Principle, which in quantum mechanics, states that the position and the momentum of a particle cannot both be precisely measured at the same time. The more precisely we know the position (Δx), the less precisely we can know the velocity (Δv), and vice versa.
Considering an electron, which has a mass (m) of 9.11×10-31 kg, and wanting the speed to be no more than 13 m/s, we can use the Heisenberg inequality formula:
Δx* Δp ≥ ℏ/2,
where ℏ is the reduced Planck constant, and Δp is the uncertainty in momentum. Since momentum (p) is equal to the mass (m) times the velocity (v), we find the uncertainty in momentum (Δp) to be m* Δv which is (9.11×10-31 kg * 13 m/s). Now, we can solve for the smallest uncertainty in position (Δx), also known as the linear size of the box.
Using these calculations, we must first find Δp, then substitute into the above inequality and solve for Δx. This will give you the minimum size of the box in which the electron can be confined.
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The Tevatron acceleator at the Fermi National Accelerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 GeV) in five stages (the numbers given in parentheses represent the total kinetic energy at the end of each stage): Cockcroft-Walton (750 keV), Linac (400 MeV), Booster (8 GeV), Main ring or injector (150 Gev) and finally the Tevatron itself (1 TeV). What is the speed of the proton at the end of each stage?
Answer:
a) [tex] v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s[/tex]
b) [tex] v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s[/tex]
c) [tex] v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s[/tex]
d) [tex] v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s[/tex]
e) [tex] v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s[/tex]
Explanation:
At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.
[tex] KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)[/tex] (1)
Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.
Let's solve (1) for β.
[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}} [/tex]
We can write the mass of a proton in MeV/c².
[tex] m_{p}=938.28 MeV/c^{2} [/tex]
Now we can calculate the speed in each stage.
a) Cockcroft-Walton (750 keV)
[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]
[tex] \beta = 0.04 [/tex]
[tex] v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s[/tex]
b) Linac (400 MeV)
[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]
[tex] \beta = 0.71 [/tex]
[tex] v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s[/tex]
c) Booster (8 GeV)
[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]
[tex] \beta = 0.994 [/tex]
[tex] v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s[/tex]
d) Main ring or injector (150 Gev)
[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]
[tex] \beta = 0.999 [/tex]
[tex] v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s[/tex]
e) Tevatron (1 TeV)
[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]
[tex] \beta = 0.9999 [/tex]
[tex] v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s[/tex]
Have a nice day!
The heating element of a water heater in an apartment building has a maximum power output of 30 kW. Four residents of the building take showers at the same time, and each receives heated water at a volume flow rate of 1.49 10-4 m3/s. If the water going into the heater has a temperature of 10° C, what is the maximum possible temperature of the hot water that each showering resident receives
Answer:
22.02°C
Explanation:
P = Power output = 30 kW
Volume flow rate of water = [tex]1.49\times 10^{-4}\ m^3/s[/tex]
S = Specific heat of water = 4.186 kJ/mol°C
[tex]T_i[/tex] = Initial temperature = 10°C
[tex]T_f[/tex] = Final temperature
Total volume of water is
[tex]v=4\times 1.49\times 10^{-4}\\\Rightarrow v=5.96\times 10^{-4}\ m^3/s[/tex]
Mass flow rate is
[tex]m=\rho\times v\\\Rightarrow m=1000\times 5.96\times 10^{-4}\\\Rightarrow m=0.596\ kg/s[/tex]
Heat is given by
[tex]\frac{q}{t}=\frac{m}{t}S\Delta T\\\Rightarrow 30=0.596\times 4.186(T_f-10)\\\Rightarrow T_f=\frac{30}{0.596\times 4.186}+10\\\Rightarrow T_f=22.02\ ^{\circ}C[/tex]
The maximum possible temperature of the hot water that each showering resident receives is 22.02°C
Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edge of the tires. If the car takes 6 seconds to come to a stop, calculate the average angular acceleration of a point on the outer edge of the tires.
To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through
[tex]\omega = \frac{v}{R}[/tex]
Where,
[tex]\omega =[/tex]Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is
[tex]\alpha = \frac{\omega}{t}[/tex]
Where
[tex]\alpha =[/tex]Angular acceleration
[tex]\omega =[/tex] Angular velocity
t = Time
Our values are
[tex]v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})[/tex]
[tex]v = 16.67m/s[/tex]
[tex]r = 0.25m[/tex]
[tex]t=6s[/tex]
Replacing at the previous equation we have that the angular velocity is
[tex]\omega = \frac{v}{R}[/tex]
[tex]\omega = \frac{ 16.67}{0.25}[/tex]
[tex]\omega = 66.67rad/s[/tex]
Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be
[tex]\alpha = \frac{\omega}{t}[/tex]
[tex]\alpha = \frac{66.67}{6}[/tex]
[tex]\alpha = 11.11rad/s^2[/tex]
Therefore the angular acceleration of a point on the outer edge of the tires is [tex]11.11rad/s^2[/tex]
The average human has a density of 945 kg/m3 after inhaling and 1020 kg/m3 after exhaling. (a) Without making any swimming movements, what percentage of the human body would be above the surface in the Dead Sea (a body of water with a density of about 1230 kg/m3) in each of these cases?
To determine the percentage of the human body above the surface in the Dead Sea, compare the body's density with the water's density. Use the formula to calculate the percentage.
Explanation:To determine the percentage of the human body that would be above the surface in the Dead Sea, we need to compare the density of the human body with the density of the water. The average density of the human body is 945 kg/m3 after inhaling and 1020 kg/m3 after exhaling, whereas the density of the Dead Sea water is about 1230 kg/m3. Since the density of the human body is lower than the density of the Dead Sea water, a certain percentage of the body would be above the surface.
In the case of inhaling, the density of the human body is lower, so a larger percentage of the body would be above the surface compared to when exhaling. The exact percentage can be calculated using the formula:
Percentage above surface = (density of human body - density of water) / density of human body × 100
Once you plug in the values, you will get the percentage of the body above the Dead Sea surface.
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Plane microwaves are incident on a thin metal sheet that has a long, narrow slit of width 4.8 cm in it. The microwave radiation strikes the sheet at normal incidence. The first diffraction minimum is observed at θ = 42°. What is the wavelength of the microwaves?
Answer:
[tex] \lambda = 3.21 \ cm[/tex]
Explanation:
given,
width of narrow slit = 4.8 cm
minimum angle of diffraction = θ = 42°
wavelength of the microwave = ?
condition for the diffraction for single slit diffraction
[tex]d sin \theta = m \lambda[/tex]
for the first minima m = 1
[tex]d sin \theta = \lambda[/tex]
wavelength of microwave radiation is equal to
[tex] \lambda = d sin \theta[/tex]
[tex] \lambda = 4.8\times sin 42^0[/tex]
[tex] \lambda = 3.21 \ cm[/tex]
the wavelength of microwaves is equal to [tex] \lambda = 3.21 \ cm[/tex]
The wavelength of the microwaves is approximately 0.032 m or 3.2 cm.
To find the wavelength of the microwaves given the slit width and the angle of the first diffraction minimum, you can use the diffraction formula for a single slit:
[tex]\[a \sin \theta = m \lambda\][/tex]where:
[tex]\(a\)[/tex] is the slit width,[tex]\(\theta\)[/tex] is the angle of the first diffraction minimum,[tex]\(m\)[/tex] is the order of the minimum for the first minimum, [tex]\(m = 1\)[/tex],[tex]\(\lambda\)[/tex] is the wavelength of the microwaves.Given:
[tex]\(a = 4.8 \text{ cm} = 0.048 \text{ m}\)[/tex][tex]\(\theta = 42^\circ\)[/tex][tex]\(m = 1\)[/tex]Rearrange the formula to solve for the wavelength [tex]\(\lambda\)[/tex]:
[tex]\[\lambda = \frac{a \sin \theta}{m}\][/tex]Plug in the values:
[tex]\[\lambda = \frac{0.048 \text{ m} \times \sin 42^\circ}{1}\][/tex]First, calculate [tex]\(\sin 42^\circ\)[/tex]:
[tex]\[\sin 42^\circ \approx 0.6691\][/tex]Now compute the wavelength:
[tex]\[\lambda = 0.048 \text{ m} \times 0.6691 \approx 0.032 \text{ m}\][/tex]So, the wavelength of the microwaves is approximately [tex]\(0.032 \text{ m}\) or \(3.2 \text{ cm}\)[/tex].
A 20-kg crate is sitting on a frictionless ice rink. A child standing at the side of the ice rink uses a slingshot to launch a 1.5-kg beanbag at the crate. Assume that the beanbag strikes the crate horizontally in the +x direction with a speed of 10 m/s. The beanbag bounces straight back in the -x direction with a speed of 6 m/s. Using conservation of momentum, calculate the speed of the crate after the beanbag hits it.
Answer:
The speed of the crate after the beanbag hits it is 1.2 m/s.
Explanation:
Hi there!
The momentum of the system beanbag-crate remains the same after the collision, i.e., the momentum of the system is conserved. The momentum of the system is calculated by adding the momenta of each object. So, the initial momentum (before the collision) is calculated as follows:
initial momentum = momentum of the crate + momentum of the beanbag
initial momentum = mc · vc + mb · vb
Where:
mc = mass of the crate.
vc = initial velocity of the crate.
mb = mass of the beanbag
vb = initial mass of the beanbag
With the data we have, we can calculate the initial momentum:
initial momentum = 20 kg · 0 m/s + 1.5 kg · 10 m/s = 15 kg · m/s
Now, let´s write the equation of the momentum of the system after the collision:
final momentum = mc · vc´ + mb · vb´
Where vc´ and vb´ are the final velocity of the crate and the beanbag respectively. Let´s replace with the data we have:
final momentum = 20 kg · vc´ + 1.5 · (-6 m/s)
Since
initial momentum = final momentum
Then:
15 kg · m/s = 20 kg · vc´ + 1.5 kg · (-6 m/s)
Solving for vc´:
(15 kg · m/s + 9 kg · m/s) / 20 kg = vc´
vc´ = 1.2 m/s
The speed of the crate after the beanbag hits it is 1.2 m/s.
Using the law of conservation of momentum, the speed of the crate after the beanbag hits it is found to be 1.2 m/s in the +x direction.
Explanation:The situation described can be solved using the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision. The momentum (p) of an object is its mass (m) times its velocity (v), or p = mv.
Before the collision, the 1.5-kg beanbag has a momentum of 1.5 kg * 10 m/s = 15 kg*m/s in the +x direction, and the 20-kg crate has a momentum of 0, as it is at rest. Therefore, the total momentum before the collision is 15 kg*m/s.
After the collision, the beanbag has a momentum of 1.5 kg * -6 m/s = -9 kg*m/s in the -x direction. Since the total momentum must be conserved, the crate must have a momentum of 15 kg*m/s (total initial momentum) + 9 kg*m/s (final momentum of beanbag) = 24 kg*m/s in the +x direction. Therefore, the speed (v) of the crate after the collision is its momentum divided by its mass, or v = p/m = 24 kg*m/s / 20 kg = 1.2 m/s.
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A small solid sphere and a small thin hoop are rolling along a horizontal surface with the same translational speed when they encounter a 20° rising slope. If these two objects roll up the slope without slipping,which will rise farther up the slope?
a. The sphere.
b. More information about the objects' mass and diameter is needed.
c. The hoop.
d. Both the same.
Answer:C.
Explanation: the hoop is the right answer
calculate the speed of a car that covers a distance of 108km in 30 mins
The speed of the car is 60 m/s
Explanation:
The speed of an object is a scalar quantity telling "how fast" the object is moving, regardless of its direction. It can be calculated as follows:
[tex]speed=\frac{d}{t}[/tex]
where
d is the distance covered
t is the time taken
For the car in this problem, we have
[tex]d=108 km =1.08\cdot 10^5 m[/tex] is the distance covered
[tex]t=30 min \cdot 60 =1800 s[/tex] is the time taken
Subsituting into the equation, we find the speed:
[tex]speed=\frac{1.08\cdot 10^5}{1800}=60 m/s[/tex]
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When a single source of light shines through an extremely thin rectangular slit and projects on a far away viewing screen, a single rectangular region of the viewing screen is illuminated (matching the shape of the thin rectangular slit).
O True O False
True
If it helps you plz brainlest me
Oil at 150°C flows slowly through a long, thin-walled pipe of 25-mm inner diameter. The pipe is suspended in a room for which the air temperature is 20°C and the convection coefficient at the outer tube surface is 11 W/m2·K. Estimate the heat loss per unit length of tube, in W/m.
Final answer:
The heat loss per unit length of the tube due to convection is approximately 141.7 W/m, calculated using the formula for heat transfer by convection with the provided temperatures and convection coefficient.
Explanation:
To estimate the heat loss per unit length of tube in W/m due to convection, we use the formula for heat transfer by convection which is Q = hA(Tsur - Tair). The convection coefficient 'h' is given as 11 W/m²·K, the temperature of the oil 'Tsur' is 150°C, and the air temperature 'Tair' is 20°C.
We need to find the surface area 'A' which for a pipe is calculated using A = π x d x L, where 'd' is the diameter and 'L' is the length of the pipe. Since we need the heat loss per unit length, L will be 1 meter.
Firstly, calculate the surface area per meter of the pipe: A = π x 0.025 m x 1 m = 0.0785 m². Then, plug these values into the convection formula to find heat loss per meter: Q = 11 W/m²·K x 0.0785 m² x (150°C - 20°C) = 141.7 W/m. Therefore, the heat loss per unit length of the tube is approximately 141.7 W/m.
A 45.8-kg girl is standing on a 151-kg plank. Both originally at rest on a frozen lake that constitutes a frictionless, flat surface. The girl begins to walk along the plank at a constant velocity of 1.46î m/s relative to the plank.
(a) What is her velocity relative to the surface of ice?1 m/s(b) What is the velocity of the plank relative to the surface of ice?2 m/s
Answer:
a. 1.12m/s
b. 0.34m/s
Explanation:
Let Vgi = the velocity of the girl relative to the ice
Let Vgp = the velocity of the girl relative to the plank
Let Vpi = the velocity of the plank relative to the ice
Vgi = Vgp + Vpi
From the question, Vgp = 1.46/s
So, Vgi = 1.46 + Vpi
Conservation of Momentum (Relative to the ice), we have
Initial Momentum = Final Momentum
Because the girl and the plank are at rest initially, the initial Momentum = 0
So, 0 = Mg * Vgi + Mp * Vpi
Where Mg = Mass of the girl = 45.8kg
Mo = Mass of the Plank = 151kg
Make Vpi the subject of the formula, we then have
-Mg*Vgi = Mp*Vpi ---------- Divide through by Mp
Vpi = -Mg * Vgi/Mp
Vpi =( -45.8 * Vgi)/151
Vpi = -45.8Vgi/151
Remember that, we have (Vgi = 1.46 + Vpi)
We then substitute the expression of Vpi in the above equation
That is;
Vgi = 1.46 + (-45.8Vgi/151) ------- Open the bracket
Vgi = 1.46 - 45.8Vgi/151 ----------- Multiply through by 151
151 * Vgi = 151 * 1.46 - 45.8Vgj
151Vgi = 220.46 - 45.8Vgi --------- Collect like terms
151Vgi + 45.8Vgi = 220.46
196.8Vgi = 220.46 --------------- Divide through by 196.8
Vgi = 220.46/196.6
Vgi = 1.1213631739572736
Vgi = 1.12 m/s (Approximated)
So, the velocity of the girl relative to the ice is 1.12m/s
b. Velocity of the plank, relative to the ice
We can solve this using (Vgi = 1.46 + Vpi)
All we need to do is substitute 1.12 for Vgi in the equation
So, we have
Vgi = 1.46 + Vpi becomes
1.12 = 1.46 + Vpi -------- Collect like terms
1.12 - 1.46 = Vpi
-0.34 = Vpi
So, the Velocity of the plank is 0.34m/s to the left
Answer: (a) 2.095 m/s (b) 0.635 m/s
Explanation: The initial momentum is = 0. Also, the final momentum is = 0. At the initial momentum she is at rest hence it is = 0. Whereas, the final momentum is = 0 because it has to equal the initial momentum.
This helps us to understand that when final momentum is = 0 then the girl's direction would cancel the affect of the plank's momentum in the opposite direction.
Therefore,
Vg = Velocity of girl
Vp= Velocity of plank
(45.8)(Vg) + (151)(Vp) = 0
Girl's velocity relative to the plank is 1.46 m/s
Therefore, Vg - Vp = 1.46
Vg = 1.46 + Vp
(45.8)(1.46 + Vp) = 151Vp
66.868 + 45.8Vp = 151Vp
66.868 = 105.2Vp
Vp = 0.635 m/s
Vg = 1.46 + 0.635
Vg = 2.095
A bowler throws a bowling ball of radius R = 11.0 cm down the lane with initial speed = 8.50 m/s. The ball is thrown in such a way that it skids for a certain distance before it starts to roll. It is not rotating at all when it first hits the lane, its motion being pure translation. The coefficient of kinetic friction between the ball and the lane is 0.210.
(a) For what length of time does the ball skid? (Hint: As the ball skids, its speed v decreases and its angular speed ω increases; skidding ceases when v = Rω.)
(b) How far down the lane does it skid?
(c) How many revolutions does it make before it starts to roll?
(d) How fast is it moving when it starts to roll?
Answer:
a) 1.18 seconds
b) 8.6 m
c) 5.19 revolutions
d) 6.07 m/s
Explanation:
Step 1: Data given
radius of the ball = 11.0 cm
Initial speed of the ball = 8.50 m/s
The coefficient of kinetic friction between the ball and the lane is 0.210.
(a) For what length of time does the ball skid?
The velocity at time t can be written as v(t) = v0 + at
⇒ with v(t) = the velocity at time t
⇒ with v0 : the initial velocity = 8.50 m/s
⇒ with a = the acceleration (in m/s²)
⇒The acceleration (negative) due to friction: a = -µg
⇒ with µ = 0.210
⇒ with g = 9.81 m/s²
v(t) =8.5m/s - 0.21*9.81m/s² * t = 8.5 - 2.06t
Torque τ = Iα = (2m(0.11m)²/5)α = 0.00484m*α
τ = F * r = µm*g*R = 0.21 * M * 9.81m/s² * 0.11m = 0.227m
so α = 0.227m / 0.00484m = 46.9 rad/s²
angular velocity ω(t) = ωo + αt = 0 + 46.9 rad/s² * t
The ball stops sliding when v(t) = ω(t) * r
8.5 - 2.06t = 46.9*0.11*t = 5.159t
7.219t = 8.5
t = 1.18 seconds
b) How far down the lane does it skid?
s = Vo*t + ½at² = 8.5m/s * 1.18s - ½* 2.06 m/s² * (1.18s)² = 8.6 m
c) How many revolutions does it make before it starts to roll?
The angular acceleration of the ball is:
α = τ/I
⇒ with τ = the torque experienced by the ball due the frictional force
⇒ τ = fk*R
α = fk*R /I
⇒ I = 2/5 m*R²
⇒ fk = µk*m*g
α = (µk*m*g*R)/(2/5mR²)
α = 5µk*g /2R
The angular displacement of the ball is:
∅ = 1/2αt²
⇒ The ball does not have an initial angular velocity
∅ =1/2*(5µk*g/2)*t²
∅ = 5µkgt²/4R
∅ = (5*0.21*9.81*1.18²)/(4*11.0 *10^-2)
∅ = 32.6 rad
Number of revolutions = 32.6 rad /2π
Number of revolutions = 5.19
(d) How fast is it moving when it starts to roll?
v = Vo + at = 8.5m/s - 2.06m/s² * 1.18s = 6.07 m/s
To find the time, distance, number of revolutions, and speed at which the bowling ball starts to roll, we can use equations of motion and the relationship between linear and angular speed.
Explanation:To find the answers to the given questions, we need to use the equations of linear and rotational motion as well as the relationship between linear and angular speed for the bowling ball.
(a) To find the time for the ball to skid, we can use the equation v = u + at, where v = 0 (since the ball stops skidding), u = 8.50 m/s (initial speed), and a = μg, where μ = 0.210 (coefficient of kinetic friction) and g = 9.8 m/s2 (acceleration due to gravity). Solving for t, we find t = u / (μg).
(b) To find the distance the ball skids, we can use the equation s = ut + (1/2)at2, where s is the distance skidded. Substituting the values we know, we find s = u2 / (2μg).
(c) To find the number of revolutions before the ball starts to roll, we can use the relationship between linear speed v and angular speed ω: v = Rω, where R is the radius of the ball. Solving for ω, we find ω = v / R. Since the ball is not rotating initially, the angular speed is 0. Therefore, the number of revolutions is 0.
(d) To find the speed of the ball when it starts to roll, we can use the equation v = Rω. Substituting the values we know, we find v = Rω = R(u / (Rμg)) = u / (μg).
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You are using a rotary encoder to determine the distance a robot has traveled. The rotary encoder is directly connected to one of the wheels that has a diameter of 10cm. Assume the encoder has 16 lines/revolution and that both the rising and falling edges (2x mode) are detected. What is the smallest detectable change of the rover traveling distance? If you want the rover to travel 2 meters before making a turn, how many encoder counts (rise and fall) are needed before sending a turn command? If you want to upgrade the encoder so that the smallest detectable distance change is ±0.20cm, what would be the required lines/revolution?
The smallest detectable distance with the current setup is 0.98125cm. To make the robot travel 2m, the encoder needs to count roughly 204 times. If the smallest detectable change should be ±0.20cm, the encoder requires 80 lines per revolution.
Explanation:The smallest detectable change in travel distance for the robot can be calculated based on the circumference of the wheel and the resolution of the rotary encoder. Given the diameter of the wheel is 10cm, the circumference can be calculated using the formula 2πr (r= radius of wheel), which results in 31.4cm. With 16 lines per revolution, this yields a distance per count (rise and fall) of 31.4cm/32 = 0.98125cm.
For the robot to travel 2m before making a turn, we will need to calculate the number of encoder counts. 2m equals 200cm, and if each count is representing a distance of 0.98125cm, then the number of encoder counts would be 200/0.98125, roughly equal to 204 counts.
Upgrading the EncoderIf the goal is to reduce the smallest detectable distance to ±0.20cm, then we need to increase the lines per revolution of the encoder. If one line equates to 0.98125cm right now, that means we need roughly 5 times more lines on the encoder to achieve a resolution of 0.20cm. Therefore, the required lines per revolution would be 5*16 = 80 lines.
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A person who weighs 625 N is riding a 98-N mountain bike. Suppose that the entire weight of the rider and bike is supported equally by the two tires. If the pressure in each tire is 7.60 x 10^5 Pa, what is the area of contact between each tire and the ground?
Answer:
A = 4.76 x 10⁻⁴ m²
Explanation:
given,
weight of the person = 625 N
weight of the bike = 98 N
Pressure on each Tyre = 7.60 x 10⁵ Pa
Area of contact on each Tyre = ?
total weight of the system = 625 + 98
= 723 N
Let F be the force on both the Tyre
F + F = W
2 F = 723
F = 361.5 N
F = P A
[tex]A = \dfrac{F}{P}[/tex]
[tex]A = \dfrac{361.5}{7.60 \times 10^5}[/tex]
A = 4.76 x 10⁻⁴ m²
The area of contact between each tire and the ground is 4.76×10⁻⁴ m².
To calculate the area of contact between each tire and the ground, we use the formula of pressure
What is pressure?Pressure is the force acting perpendicular per unit area to the surface of an object.
Formula:
P = F/A.............. Equation 1Where:
P = Pressure on each tireF = Weight acting on each tireA = Area of contact between each tire and the ground.Make A the subject of the equation
A = F/P.......... Equation 2From the question,
Given:
F = (625+98)/2 (supported equally by the two tires) = 361.5 NP = 7.6×10⁵ PaSubstitute these values into equation 2
A = 361.5/(7.6×10⁵)A = 4.76×10⁻⁴ m²Hence, The area of contact between each tire and the ground is 4.76×10⁻⁴ m².
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Find a recurrence relation for the number of ways to go n miles by foot walking at 2 miles per hour or jogging at 4 miles per hour or running at 8 miles per hour; at the end of each hour a choice is made of how to go the next hour. (b) How many ways are there to go 12 miles?
Answer:
(a) [tex]x_{n} = x_{n - 2} + x_{n - 4} + x_{n - 8}[/tex]
(b) 18 ways
Solution:
As per the question:
We need to find the no.of ways to go 'n' miles by foot:
From the question it is clear that, at each hours even miles can be traveled by a person, If the distance to be covered is odd, then there in no way 3 miles can be covered.
Thus defining recurrence as:
[tex]x_{o} = 1[/tex]
[tex]x_{n} = 0[/tex]
where
n: negative or odd
Then
[tex]x_{n} = x_{n - 2} + x_{n - 4} + x_{n - 8}[/tex]
(b) No. of ways to travel 12 miles:
[tex]x_{o} = 1[/tex]
[tex]x_{2} = 1 + 0 = 1[/tex]
[tex]x_{4} = x_{2} + x_{o} + 0 = 1 + 1 = 2[/tex]
[tex]x_{6} = x_{4} + x_{2} + 0 = 2 + 1 + 0 = 3[/tex]
[tex]x_{8} = x_{6} + x_{4} + 0 = 3 + 2 = 6[/tex]
[tex]x_{10} = x_{8} + x_{6} + x_{2} = 6 + 3 + 1 = 10[/tex]
[tex]x_{12} = x_{10} + x_{8} + x_{4} = 10 + 6 + 2 = 18[/tex]
Thus clearly there are 18 ways to travel 12 miles
The recurrence relation for the number of ways to go n miles by foot at different speeds is An = An-2 + An-4 + An-8. To find the number of ways to go 12 miles, we can compute the values of An using the recurrence relation.
Explanation:To find a recurrence relation for the number of ways to go n miles by foot walking at 2 miles per hour or jogging at 4 miles per hour or running at 8 miles per hour, we need to consider the choices made at each hour. Let's denote the number of ways to go n miles as An. At each hour, there are three choices: walk, jog, or run. If we choose to walk, we have An-2 ways to go n miles. If we choose to jog, we have An-4 ways. If we choose to run, we have An-8 ways. Therefore, the recurrence relation is:
An = An-2 + An-4 + An-8
To find the number of ways to go 12 miles, we need to find A12. Starting from the base cases, A0 = 1, A2 = 1, A4 = 2, and A8 = 4. Using the recurrence relation, we can compute the values of A12 as follows:
A12 = A10 + A8 + A4= A8 + A6 + A2 + A6 + A4 + A0 + A4 + A2 + A0 + A2 + A0= 4 + 1 + 1 + 1 + 2 + 1 + 2 + 1 + 1 + 1 + 1= 16Learn more about Recurrence relation here:https://brainly.com/question/36494057
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On the other side of the gorge, at the highest point of his swing, the vine makes an angle of \theta=40^\circθ=40 ∘ from the vertical. At this point, Tarzan's speed is instantaneously zero (as he transitions from swinging up to swinging back again). What is the tension in the vine at the highest point of his swing?
Answer:
[tex]T = m*g*cos\theta[/tex]
Explanation:
Since tarzan moves in a circular trajectory we can sum all forces on the centripetal-axis:
[tex]T - m*g*cos\theta = m*a_c[/tex]
[tex]T - m*g*cos\theta = m*V^2/R[/tex] Since the speed is zero:
[tex]T - m*g*cos\theta = 0[/tex]
[tex]T = m*g*cos\theta[/tex] Having Tarzan's mass we could calculate the module of the tension in the vine.
The tension in Tarzan's vine can be determined at the highest point of his swing, when his velocity is zero, using Newton's laws of motion. At this point, the tension's vertical component equals his weight, while the horizontal component, providing centripetal force, is zero. The equation T=mg/cosθ can be used to calculate the tension.
Explanation:The subject of your question involves physics, particularly mechanics. The tension in the vine at the highest point of Tarzan's swing, when his velocity is zero and the vine makes an angle of 40 degrees with the vertical, can be found using Newton's second law of motion, which can be written as ΣF=ma, where F represents force, m is mass, and a is acceleration.
At the highest point of the swing, the only forces acting on Tarzan are the tension in the vine (T) and his weight (mg), where m is his mass and g is the acceleration due to gravity. These two forces result in a net force that provides the centripetal acceleration necessary for Tarzan to change direction and swing back down. We have:
The vertical component of tension (Tcosθ) equals mg.The horizontal component of tension (Tsinθ) provides the centripetal force, which is zero at the highest point.So we can get the tension in vine (T) from the equation Tcosθ=mg. Rearranging, we have T=mg/cosθ.
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A rock of mass m = 0.0450 kg is attached to one end of a string and is whirled around in a horizontal circle. If the radius of the circle is 0.580 m and the angular speed is 2.34 rad/s. What is the tension in the string?
Answer:
Tension, T = 0.1429 N
Explanation:
Given that,
Mass of the rock, m = 0.0450 kg
Radius of the circle, r = 0.580 m
Angular speed, [tex]\omega=2.34\ rad/s[/tex]
The tension in the string is balanced by the centripetal force acting on it. It is given by :
[tex]T=\dfrac{mv^2}{r}[/tex]
Since, [tex]v=r\omega[/tex]
[tex]T=\dfrac{m(r\omega)^2}{r}[/tex]
[tex]T=\dfrac{0.0450\times (0.580\times 2.34)^2}{0.580}[/tex]
T = 0.1429 N
So, the tension in the string is 0.1429 N. Hence, this is the required solution.
A satellite in circular orbit around the Earth moves at constant speed. This orbit is maintained by the force of gravity between the Earth and the satellite, yet no work is done on the satellite. How is this possible?
a. No work is done if there is no contact between objects.
b. No work is done because there is no gravity in space.
c. No work is done if the direction of motion is perpendicular to the force.
d. No work is done if objects move in a circle.
Answer:
C. No work is done if the direction of motion is perpendicular to the force.
Explanation:
We know that work is the dot product of force and displacement.
Let's take the angle between the force and the displacement = θ
W= F . d cosθ
F=Force , d=Displacement
If θ = 0° then W= F.d
If θ = 90° then W= 0
So we can say that when force is perpendicular to the displacement then the work done by force will be zero.
Therefore the answer is C.
Water at 298 K discharges from a nozzle and travels horizontally hitting a flat, vertical wall. The nozzle diameter is 12 mm and the water exits the nozzle with a flat velocity profile at a velocity of 6.0 m/s. Use macroscopic balance for momentum to compute the force [N] on the wall neglecting gravitational and frictional effects
The force on the wall can be calculated using the macroscopic balance for momentum. We need to calculate the mass flow rate of water and use the equation: mass flow rate = density × velocity × cross-sectional area of the nozzle. With the given values, we can find the force on the wall.
Explanation:The force on the wall can be computed using the macroscopic balance for momentum. Since water discharges from a nozzle and travels horizontally, the initial momentum of the water is zero. The change in momentum of the water is equal to the force exerted on the wall. To calculate the force, we need to find the mass flow rate of the water. Assuming the water is an ideal fluid with a flat velocity profile, we can use the equation: mass flow rate = density × velocity × cross-sectional area of the nozzle. Plugging in the given values, we can then calculate the force on the wall.
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Final answer:
To calculate the force exerted on the wall by the water discharging from the nozzle, we use the mass flow rate and velocity of the water to find the momentum change. This calculation applies the macroscopic momentum balance principle and yields the force in Newtons.
Explanation:
The student is asking to calculate the force on a wall when water from a nozzle hits it horizontally. To find this force, we can use the macroscopic momentum balance principle which states that the rate of change of momentum of the fluid is equal to the sum of the external forces applied to the fluid.
As gravity and frictional effects are negligible, the only external force is the force on the wall exerted by the water. Key information given includes the diameter of the nozzle (12 mm) and the velocity of the water (6.0 m/s).
To calculate the force, we need to find the mass flow rate of the water, which can be determined by multiplying the density of water (ρ = 1000 kg/m³ at standard conditions) by the cross-sectional area of the nozzle (A) and the velocity (v) of the water.
The cross-sectional area, A, of the nozzle is A = π * (d/2)², where d is the diameter of the nozzle. Substituting the given values, the area A is π * (0.012/2)² m². The mass flow rate (ṭ) then is ρ * A * v.
With the mass flow rate known, the force (F) exerted on the wall can be calculated by taking the rate of change of momentum as ṭ * v, because the water's velocity is reduced to zero upon hitting the wall, thus the change in momentum is just the initial momentum. Therefore, F = ṭ * v.
After calculating the mass flow rate and applying the equation for force, the student will get the desired force in Newtons (N) on the wall.
A piece of taffy slams into and sticks to an identical piece of taffy at rest. The momentum of the combined pieces after the collision is the same as before the collision, but this is not true of the kinetic energy, which partly degrades into heat. What percentage of the kinetic energy becomes heat?
A) 25%
B) 0%
C) 75%
D) 50%
E) need more information.
Answer:
D) 50%
Explanation:
According to conservation of momentum:
[tex]m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}[/tex]
Assuming that the second taffy started at rest, both pieces have the same mass and that they combined after the collision, their final velocity is:
[tex]m v_{1i} = (m+m) v_{f}\\v_{f} = 0.5 v_{1i}[/tex]
The initial kinetic energy of the system is:
[tex]E_{ki} = \frac{m*v_{1i}^2}{2}[/tex]
Since the second taffy was not moving, it had no kinetic energy at first.
The initial kinetic energy of the system is:
[tex]E_{kf} = \frac{2m*v_{f}^2}{2}\\E_{kf} = \frac{2m*(0.5v_{1i})^2}{2}\\E_{kf} = \frac{0.5m*v_{1i}^2}{2}[/tex]
The percentage of kinetic energy that becomes heat is given by:
[tex]H=1 - \frac{E_{kf}}{E_{ki}}\\H=1 - \frac{\frac{0.5m*v_{1i}^2}{2}}{\frac{m*v_{1i}^2}{2}}\\\\H=1- 0.5 = 0.5[/tex]
THerefore, 50% of the kinetic energy becomes heat
In inelastic collisions, such as the described scenario where two objects stick together, 75% of the initial kinetic energy is lost to other forms, including heat.
The question involves a type of collision known as an inelastic collision, where two objects stick together post-collision. In these scenarios, while momentum is conserved, kinetic energy is not. The initial kinetic energy before the collision gets partly converted into other forms, such as heat.
Considering a scenario where a moving object collides with a stationary one of identical mass and they stick together, the resulting velocity of the combined mass is half of the original moving object's velocity. This result emerges from the conservation of momentum. Since kinetic energy is proportional to the square of the velocity, the kinetic energy of the combined object will be one-fourth of the initial kinetic energy. Consequently, 75% of the initial kinetic energy is converted into other forms of energy like heat, which answers the original question.
Calculate the net filtration pressure if the glomerular hydrostatic pressure measures 46 mmHg, the colloid osmotic pressure 34 mmHg, and the capsular hydrostatic pressure 10 mmHg. Does this differ from the normal value? If so, how? What effect, if any, would this have on the GFR?
Answer:
Net filtration pressure is 2 mmHg
Solution:
As per the question:
Glomerular Pressure, [tex]P_{gh} = 46\ mmHg[/tex]
Colloidal Osmotic Pressure, [tex]P_{co} = 34\ mmHg[/tex]
Capsular Hydrostatic Pressure, [tex]P_{ch} = 10\ mmHg[/tex]
Now,
The net filtration Pressure is given by:
[tex]P_{net} = P_{gh} - (P_{co} + P_{ch})[/tex]
[tex]P_{net} = 46 - (34 + 10) = 2\ mmHg[/tex]
The normal value of the net filtration pressure is 20 mmHg and the calculated value is very low as compared to the normal value.GFR is low for increased hydrostatic pressure.The lower rate of the glomerular filtration can result in lower value of the GFRThe net filtration pressure is 2 mmHg, which is lower than the normal value. This decrease would result in a decrease in the GFR.
Explanation:The net filtration pressure is calculated by subtracting the colloid osmotic pressure and the capsular hydrostatic pressure from the glomerular hydrostatic pressure. In this case, the net filtration pressure would be 46 mmHg - 34 mmHg - 10 mmHg = 2 mmHg. This differs from the normal net filtration pressure of around 10 mmHg. A decrease in net filtration pressure would lead to a decrease in the glomerular filtration rate (GFR).
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A toy train of m=0.60 kg moves at 20m/s along a straight track. It bumps into another train of M=1.5kg moving in the same direction. They stick together and continue on the track at a speed 12 m/s. What was the speed in m/s of the second train just before the collision?
Answer:8.8 m/s
Explanation:
Given
mass of train [tex]m_1=0.6 kg[/tex]
velocity of Train [tex]v_1=20 m/s[/tex]
mass of another train [tex]m_2=1.5 kg[/tex]
Final velocity of both train [tex]v=12 m/s[/tex]
Let [tex]v_2[/tex] be the velocity of [tex]m_2[/tex] before collision
As external Force is zero therefore change in momentum is zero
conserving Momentum
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]
[tex]0.6\times 20+1.5\times v_2=(0.6+1.5)\cdot 12[/tex]
[tex]1.5\times v_2=25.2-12[/tex]
[tex]v=\frac{13.2}{1.5}=8.8 m/s[/tex]