Answer:
[tex]r_{P}=2.58 m[/tex]
Explanation:
If we want to find how far up the ladder can a person climb before the ladder begins to slip, the total force in the y-direction and the total torque must be zero, it is an equilibrium condition.
So, let's start whit y-direction forces.
[tex]\Sigma F_{y}=0[/tex]
[tex]N_{L}-W_{L}-W_{P}=0[/tex]
NL is the normal forceWL is the weight of the ladderWP is the weight of the personSo, we can find the normal force:
[tex]N_{L}=m_{L}g+m_{P}g=g(m_{L}+m_{P})=9.8(267+1068)=13083 N[/tex]
We know that the friction force is the product of the coefficient of static friction times the normal force. Let's find it.
[tex]F_{f}=\mu N_{L}=0.53*13083=6934 N[/tex]
Referent to the x-direction forces, we can say that the total sum must be equal to zero too.
[tex]\Sigma F_{x}=0[/tex]
[tex] -N_{wall}+F_{f}=0[/tex]
[tex] N_{wall}=F_{f}=6934 N[/tex]
Where N(wall) is the normal force due to the wall. (negative direction)Now, as we said before, the total torque must be zero, so we have:
[tex]\Sum \tau=\tau_{wall}+\tau_{L}+\tau_{P}=0[/tex] (1)
We will choose the pivot point at the bottom of the ladder, then we have:
Each force acting on the ladder must be perpendicular to the ladder, so we need to use the components.
[tex]\tau_{wall}=r_{wall}*N_{wall}sin(32.7)=8.6*6934*sin(32.7)=32215.8Nm[/tex] (positive due to the counterclockwise rotation)
[tex]\tau_{L}=r_{L}*W_{L}sin(57.3)=(8.6/2)*2616.6*sin(57.3)=-9468.2 Nm[/tex] (negative due to the clockwise rotation)
[tex]\tau_{P}=r_{P}*W_{P}sin(57.3)=r_{P}*10466.4*sin(57.3)=-r_{P}8807.6 Nm[/tex] (negative due to the clockwise rotation)
We use 57.3° because is the angle between the ladder weight vector and the ladder surface, we use the same angle to the person.
Now we can use (1) to find the distance
[tex]\Sum \tau=\tau_{wall}+\tau_{L}+\tau_{P}=32215.8-9468.2-r_{P}8807.6=0[/tex]
Therefore [tex]r_{P}=2.58 m[/tex]
I hope it helps you!
A car is traveling at 33.0 m/s when the driver uses the brakes to slow down to 29.0 m/s in 2.50 seconds. How many meters will it travel during that time?
Explanation:
We have equation of motion v = u + at
Initial velocity, u = 33 m/s
Final velocity, v = 29 m/s
Time, t = 2.50 s
Substituting
v = u + at
29 = 33 + a x 2.50
a = -1.6 m/s²
We have equation of motion v² = u² + 2as
Initial velocity, u = 33 m/s
Acceleration, a = -1.6 m/s²
Final velocity, v = 29 m/s
Substituting
v² = u² + 2as
29² = 33² + 2 x -1.6 x s
s = 77.5 m
Distance traveled during that time is 77.5 m
Final answer:
The car will travel 77.5 meters while decelerating from 33.0 m/s to 29.0 m/s over a period of 2.50 seconds by using the average speed during the deceleration period.
Explanation:
The question deals with calculating the distance covered by a car while decelerating from 33.0 m/s to 29.0 m/s over 2.50 seconds. We can solve this by finding the car's average speed during this period and multiplying it by the time duration. Here's how we can solve it:
Step-by-Step Calculation
Find the average speed: (Initial speed + Final speed) / 2 = (33.0 m/s + 29.0 m/s) / 2 = 31.0 m/s.
Multiply the average speed by the time taken to decelerate to find the distance covered: Distance = Average speed * Time = 31.0 m/s * 2.50 s = 77.5 meters.
Therefore, the car will travel 77.5 meters during that time.
Vector A is 3.00 units in length and points along the positive x-axis. Vector B is 4.00 units in length and points along the negative y-axis. Use graphical methods to find the magnitude and direction of the following vectors:
To find the magnitude and direction of the resultant vector R, we can use graphical methods. First, find the components of vectors A and B along the x and y axes. Then, use the Pythagorean theorem to find the magnitude of R and the inverse tangent function to find the direction of R.
Explanation:To find the magnitude and direction of the resultant Vector B, with a magnitude of 4 units, points along the negative y-axis, so its x-component is 0 and its y-component is -4.
To find the components of R, we can simply add the corresponding components of A and B: Rx = Ax + Bx
= 3 + 0 = 3, Ry = Ay + By = 0 + (-4) = -4.
Using the Pythagorean theorem, we can find the magnitude of R: R = sqrt(Rx^2 + Ry^2)
= sqrt(3^2 + (-4)^2) = sqrt(9 + 16) = sqrt(25) = 5 units.
To find the direction of R, we can use the inverse tangent function: Rtheta = atan(Ry/Rx)
= atan((-4)/3)
= atan(-4/3) = -53.13 degrees.
However, since vector B points along the negative y-axis, the direction of R is 90 degrees minus the calculated angle: Rtheta = 90 - 53.13 = 36.87 degrees.
Therefore, the magnitude of R is 5 units and it points at an angle of 36.87 degrees north of the x-axis.
If a substance can be separated by physical means and it is not the same throughout, what is it?
A: a homogeneous solution
B: a heterogeneous mixture
C: a pure substance
D: an element
E: a compound
Answer:
Option (B)
Explanation:
A heterogeneous mixture is usually defined as a combination of two or more chemical substances. It can be also elements as well as compounds. These contrasting components can be easily separated from one another by means of physical process. These are comprised of substances that are not even everywhere. There is variation in it.
Thus, the correct answer is option (B).
Answer:
B: a heterogeneous mixture
Explanation:
If the components of a substance can be separated by the physical means then the substance is a mixture and its component have not undergone any kind of chemical change with their original molecular structure.
The mixtures are of two types homogeneous and heterogeneous.
Homogeneous mixtures have a uniform composition of its components throughout the mixture whereas heterogeneous mixtures have a non-uniform composition of its constituents in the mixture.A rectangular block floats in pure water with 0.400 in. above the surface and 1.60 in. below the surface. When placed in an aqueous solution, the block of material floats with 0.800 in. below the surface. Estimate the specific gravities of the block and the solution.
Answer:
specific gravity = 0.8
specific gravity of solution = 2
Explanation:
given data
rectangular block above water = 0.400 in
rectangular block below water = 1.60 in
material floats below water = 0.800 in
solution
first we get here specific gravity of block that is
specific gravity = block vol below ÷ total block vol × specific gravity water ..............1
put here value we get
specific gravity = [tex]\frac{1.60}{1.60+0.400}[/tex] × 1
specific gravity = 0.8
and now we get here specific gravity of solution that is express as
specific gravity of solution = total block vol ÷ block vol below × specific gravity block ........................2
put here value we get
specific gravity of solution = [tex]\frac{1.60+0.400}{0.800}[/tex] × 0.8
specific gravity of solution = 2
Describe a situation that includes no less than four charges of any magnitude, but they combine so that another location, p, has no net electric field at that point. Describe where those charges could be or what magnitudes they could be
Answer:
Four charges of equal magnitude sitting at the vertices of a square
Explanation:
We can arrive at such a situation by thinking of a simple example first, a configuration of two charges. The force acting on the middle point of a straight line joining the two points(charges) will be zero. That is, the net Electric field will be zero as they cancel out being equal in magnitude and opposite in direction.
Now, we can extend this idea to a square having charge q at each vertex. If we put 'p' at the geometric center, we can see that the Electric fields along the diagonals cancel out due to the charges at the diagonally opposite vertices(refer to the figure attached). Actually, the only requirement is that the diagonally opposite charges are equal.
We can further take this to 3 dimensions. Consider a cube having charges of equal magnitude at each vertex. In this case, the point 'p' will yet again be the geometric center as the Electric field due to the diagonally opposite charges will cancel out.
An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is twice the magnitude of the tangential acceleration. What is the angle?
Answer:
Explanation:
Given
magnitude of centripetal acceleration is twice the magnitude of tangential acceleration
Suppose [tex]\theta [/tex] is theta angle rotated by electric drill
it is given that it starts from rest i.e. [tex]\omega _0=0[/tex]
suppose [tex]\omega [/tex] and [tex]\alpha [/tex] is the final angular velocity and angular acceleration
using rotational motion equation
[tex]\omega ^2-\omega _0^2=2\times \alpha \times \theta [/tex]
where [tex]\theta [/tex]=angle turned by drill
[tex]\omega _0[/tex]=initial angular velocity
[tex]\omega [/tex]=final angular velocity
[tex]\alpha [/tex]=angular acceleration
[tex]\omega ^2-0=2\times \alpha \times \theta [/tex]
[tex]\omega ^2=2\alpha \theta ---1[/tex]
It is also given that centripetal acceleration is twice the magnitude of tangential i.e.
[tex]\omega ^2r=\alpha \times r[/tex]
where r=radial distance of any point from axis of drill
i.e. [tex]\omega ^2=\alpha [/tex]
substitute this value to equation 1
we get
[tex]\theta =\frac{\omega ^2}{2\alpha }[/tex]
[tex]\theta =1\ rad[/tex]
A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s. What is the average friction force opposing its motion?
Answer: [tex]f_{r}[/tex] = 16.49N
Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,
[tex]W_{x}[/tex] = horizontal component of the weight = mgsinФ
[tex]W_{y}[/tex] = vertical component of weight = mgcosФ
Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.
by using newton's law of motion, we have that
mgsinФ - [tex]f_{r}[/tex] = ma
where m = mass of object=5kg
a = acceleration= unknown
Ф = angle of inclination = 37°
g = acceleration due to gravity = 9.8[tex]m/s^{2}[/tex]
[tex]f_{r}[/tex] = frictional force = unknown
we need to first get the acceleration before the frictional force which is gotten by using the equation below
[tex]v^{2} = u^{2} + 2aS[/tex]
where v = final velocity = 2m/s
u = initial velocity = 0m/s (because the object started from rest)
a= unknown
S= distance covered = length of plane = 5m
[tex]2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}[/tex]
we slot in a into the equation below to get frictional force
mgsinФ - [tex]f_{r}[/tex] = ma
3 * 9.8 * sin 37 - [tex]f_{r}[/tex] = 3* 0.4
17.9633 - [tex]f_{r}[/tex] = 1.2
[tex]f_{r}[/tex] = 17.9633 - 1.2
[tex]f_{r}[/tex] = 16.49N
We should stress again that the Carnot engine does not exist in real life: It is a purely theoretical device, useful for understanding the limitations of heat engines. Real engines never operate on the Carnot cycle; their efficiency is hence lower than that of the Carnot engine. However, no attempts to build a Carnot engine are being made. Why is that?
A) A Carnot engine would generate too much thermal pollution.
B) Building the Carnot engine is possible but is too expensive.
C) The Carnot engine has zero power.
D) The Carnot engine has too low an efficiency.
Answer:c
Explanation:
Although the Carnot engine is the most efficient it is not feasible in real life. Carnot engine contains processes that are reversible which makes it difficult to build in real life.
For a system to be in equilibrium a system must be in equilibrium with its surroundings in each step and every step which allows it to be infinitely slow.
Due to this slow rate Power of the engine is zero as the work is being done at an infinitely slow rate.
Answer:
good
Explanation:
What is the unit of measurement that defines the space available in a rack? How tall are standard racks?
Answer:
The unit of measurement that defines the space available in a rack are the inches for the width and for the height the units of Rack (U), this unit is equivalent to 44.45 mm. Standard racks are 19 inches wide and 42U high. A rack is a type of shelf where servers can be stacked on top of each other.
The unit of measurement that defines rack space is 'rack unit' or 'U', with each unit equal to 1.75 inches or 44.45 millimeters. Standard racks can have different heights, with 42U being the most common.
Explanation:The unit of measurement that defines the space available in a rack is called 'rack unit' or 'U'. Each rack unit is equal to 1.75 inches or 44.45 millimeters. So, when we talk about the height of a standard rack, we refer to the number of rack units it can accommodate.
The height of standard racks can vary, but the most common height for a rack is 42U, which means it can accommodate equipment up to 73.5 inches or 1866.9 millimeters tall. Other common rack heights include 48U and 45U.
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u are holding the axle of a bicycle wheel with radius30 cm and mass 1.05 kg. You get the wheel spinning at arate of 77 rpm and then stop it by pressing the tire againstthe pavement. You notice that it takes 1.37 s for the wheelto come to a complete stop. What is the angular accelerationof the wheel
Answer:
-5.9 rad/s^{2}
Explanation:
radius (r) = 30 cm = 0.3 m
mass (m) = 1.05 kg
initial speed (u) = 77 rpm
final speed (v) = 0 rpm
time (t) = 1.37 s
angular acceleration =[tex]\frac{(final speed-initial speed)rad/s}{time}[/tex]
therefore
initial speed (U) = 77 rpm = 77 x (2π/60) = 8.06 rad/s
final speed (v) = 0 rpm = 0 rad/s
angular acceleration = [tex]\frac{0-8.06}{1.37}[/tex] = -5.9 rad/s^{2}
Answer:
-5.886 rad/s^2.
Explanation:
radius, r = 30 cm
= 0.3 m
mass, m
= 1.05 kg
initial speed, wo = 77 rpm
Converting from rpm to rad/s,
= 77 rpm * 2pi rad * 1 min/60 s
= 8.063 rad/s
final speed, wi = 0 rad/s
time, t = 1.37 s
angular acceleration = Δw/Δt
= (wi - wo)/t
= 8.063/1.37
= -5.886 rad/s^2.
A motorcycle with two riders weaves dangerously between parked cars in a crowded shopping center parking lot. As the motorcyclists dart between cars, they confront a moving car. Both the car and motorcycle veer to avoid a head-on collision. The motorcycle strikes the side of the oncoming car, throwing riders to the ground. The car stops abruptly, throwing the driver into the windshield. Nearby, Lisa and Paul (two college students) hear the sound of crunching metal and blaring horns and decide to join the small group that has gathered?
Answer:
a
Explanation:
Answer:
A
Explanation:
Suppose that the voltage of the battery in the circuit is 3.3 V, the magnitude of the magnetic field (directed perpendicularly into the plane of the screen) is 0.66 T, and the length of the rod between the rails is 0.19 m. Assuming that the rails are very long and have negligible resistance, find the maximum speed attained by the rod after the switch is closed.
Answer:
Maximum velocity of rod will be equal to 26.315 m/sec
Explanation:
We have given voltage of the battery in the circuit V = 3.3 volt
Magnetic field B = 0.66 Tesla
Length of the rod l = 0.19 m
We know that emf is given by e = BVl
We have to find the maximum velocity of the rod
Here velocity is maximum
So [tex]e=v_{max}Bl[/tex]
So [tex]3.3=v_{max}\times 0.66\times 0.19[/tex]
[tex]v_{max}=26.315m/sec[/tex]
So maximum velocity of rod will be equal to 26.315 m/sec
A certain spacecraft is x AU (Astronomical Units) from Earth. How long in seconds does it take for a signal to reach the Earth after it is transmitted from the spacecraft? Hint: An AU is about 149.9 Million Km, and light moves at 299,800 Km/s. Indicate your answer to the nearest whole second.
Answer:
The time taken for a signal to reach the Earth after it is transmitted from the spacecraft is (500x) seconds
Explanation:
Distance of spacecraft from Earth = x AU = x × 149.9×10^6 Km = (149.9×10^6x) Km
Speed of light = 299,800Km/s
Time taken for a signal to reach the Earth after it is transmitted from the spacecraft = distance of spacecraft from Earth ÷ speed of light = (149.9×10^6x)Km ÷ 299,800Km/s = (500x) seconds
answers If visible light and radio waves are both examples of electromagnetic waves, why can’t we see radio waves?
Answer:
Radio Waves:
Radio waves being the lowest-energy form of light and are known to be produced by electrons spiraling around magnetic fields. These Magnetic fields are generated by stars, including our sun, and many weird celestial objects like black holes and neutron stars.
Explanation:
All electromagnetic radiation is light, but we can only see a small portion of this radiation that is the portion we call visible light. Cone-shaped cells in our eyes act as receivers tuned to the wavelengths in this narrow band of the spectrum
Visible light and radio waves are both examples of electromagnetic waves, but visible light has shorter wavelengths that our eyes can detect, while radio waves have longer wavelengths that are outside the range of what our eyes can see.
Explanation:Visible light and radio waves are both examples of electromagnetic waves, but they have different wavelengths. Visible light has a shorter wavelength than radio waves, which is why we can see it. Our eyes are sensitive to the wavelengths of visible light, but not to the longer wavelengths of radio waves.
The human eye can detect wavelengths within a certain range, known as the visible spectrum, which includes the colors of the rainbow. Radio waves have much longer wavelengths, ranging from meters to kilometers. These longer wavelengths are outside the range of what our eyes can detect, so we cannot see radio waves.
However, even though we can't see radio waves, we can still use them for various purposes, such as wireless communication, radar, and satellite transmission.
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A train travels due south at 25 m/s (relative to the ground) in a rain that is blown toward the south by the wind. The path of each raindrop makes an angle of 66° with the vertical, as measured by an observer stationary on the ground. An observer on the train, however, sees the drops fall perfectly vertically. Determine the speed of the raindrops relative to the ground.
Answer:
Explanation:
Given
Train travels towards south with a velocity if [tex]v_t=25\ m/s[/tex]
Rain makes an angle of [tex]\theta =66^{o}[/tex] with vertical
If an observer sees the drop fall perfectly vertical i.e. horizontal component of rain velocity is equal to train velocity
suppose [tex]v_r[/tex] is the velocity of rain with respect to ground then
[tex]v_r\sin\theta =v_t[/tex]
[tex]v_r\times \sin (66)=25[/tex]
[tex]v_r=27.36\ m/s[/tex]
Therefore velocity of rain drops is 27.36 m/s
A water break at the entrance to a harbor consists of a rock barrier with a 50.0-m-wide opening. Ocean waves of 20.0-m wavelength approach the opening straight on. At what angles to the incident direction are the boats inside the harbor most protected against wave action?
The angles to the incident direction are the boats inside the harbor most protected against wave action will be 23.57 °
What is diffraction ?Diffraction is the phenomenon that occur when a wave of light encounter an obstacle or a slit generally.
considering wide opening of harbor as thickness d
the ocean wave as light source (coherent )
boats inside the harbor as screen where diffraction pattern is going to happen
so , destructive interference should happen (to minimize the amplitude of wave ) in order to save the boats from its effect
n * lambda = d sin (theta )
n=1 ( first order minima )
sin(theta ) = lambda / d
sin( theta ) = 20 /50
sin(theta) = 2/5
theta = sin inverse (2/5)
theta = 23.57 °
The angles to the incident direction are the boats inside the harbor most protected against wave action will be 23.57 °
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Three resistors are connected into the section of the circuit described by the diagram. Then the wire is cut at point x, and the two cut ends of the wire are separated. Through which of the three resistors, if any, does current still flow?
A. None of the resistors
B. All three resistors
C. R1 only
D. R1 and R2 only
Answer:
C. R1 only
Explanation:
As the wire is cut at x, there will be no current through the resistors R2 and R3. Then the current will only go from a to b through the R1 resistor.
Another way to think about this is that once the wire is cut at x, there is now infinite resistance at the point of cutting; therefore, the current can no longer flow through R2 and R3 resistors, but now it only flows through the R1 resistor.
Therefore, only choice C is correct.
To have a negative ion, you must have: A. added a positron to the outer electron shell. B. taken away a proton from the nucleus. C. added an electron to the outer electron shell. D. added a positron to the nucleus. E. None of these; only positive ions can exist in nature.
C. added an electron to the outer electron shell.
Explanation:
Atoms consist of three particles:
- Protons: they are located in the nucleus, they have positive charge of [tex]+e[/tex], and mass of [tex]1.67\cdot 10^{-27}kg[/tex]
- Neutrons: they are also located in the nucleus, they have no electric charge, and mass similar to that of the proton
- Electrons: they orbit around the nucleus, they have negative charge of [tex]-e[/tex], and mass around 1800 smaller than the proton
Normally, atoms are neutral (no electric charge), because they have an equal number of protons and electrons.
However, sometimes atoms can give off or take electrons from other atoms. We have two cases:
If an atom gives off an electron, it remains with an excess of positive charge, so it becomes a positive ionIf an atom takes an electron from another atom, it remains with an excess of negative charge, so it becomes a negative ionTherefore, in order to form a negative ion, the atom must have
C. added an electron to the outer electron shell
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Which of the following are molecules?
Answer:
b c and e
Explanation:
Answer: C, D, and E are all molecules!
In The Funeral of St. Bonaventure, Francisco de Zurbarán used the principle of __________ to create emphasis and focal point.
Answer:
In The Funeral of St. Bonaventure, Francisco de Zurbarán used the principle of _contrast_ to create emphasis and focal point
Explanation:
Contrast relates to the combination of opposing components and effects as an art theory. Lighter and darker colours, polished and rough materials, large and small shapes. Contrast can be used for the production of diversity visual interest and excitement.
A train, traveling at a constant speed of 25.8 m/s, comes to an incline with a constant slope. While going up the incline, the train slows down with a constant acceleration of magnitude 1.66 m/s². What is the speed of the train after 8.30 s on the incline?
The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Answer:
Explanation:
So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.
So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then
v = 25.8 + (-1.66×8.3)
v =12.022 m/s.
So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Water at 1 atm pressure is compressed to 430 atm pressure isothermally. Determine the increase in the density of water. Take the isothermal compressibility of water to be 4.80 × 10−5 atm−1. The density of water at 20°C and 1 atm pressure is rho1 = 998 kg/m3.
Answer:
Explanation:
compressibility = 1 / bulk modulus of elasticity ( B )
B = 1 / 4.8 x 10⁻⁵ = Δp / Δv /v ( Δp is change in pressure , Δv is change in volume )
1 / 4.8 x 10⁻⁵ = 429 / Δv /v
Δv /v = 429 x 4.8 x 10⁻⁵
= 2059.2 x 10⁻⁵
= .021
v = m / d ( d is density and m is mass of the water taken )
taking log and then differentiating
Δv /v = - Δd / d
- .021 = - Δd / d
Δd =.021 x d
= .021 x 998
= 20.9
new density
= 998 + 20.9
1018.9 kg/m3
The increase in the density of water when it is compressed from 1 atm to 430 atm pressure isothermally is 20.6 kg/m³.
Explanation:In this problem, we are asked to find the increase in the density of water when it is compressed from 1 atm to 430 atm pressure isothermally. The isothermal compressibility of water is given as 4.80 × 10−5 atm−1. The density of water at 1 atm pressure and 20°C is given as rho1 = 998 kg/m^3.
The formula that connects these quantities is Δρ = - ρ * β * ΔP, where Δρ is the change in density, β is the isothermal compressibility (in atm^−1), ΔP is the change in pressure (in atm), and ρ is the initial density (in kg/m^3).
Substituting the given values into the formula we get Δρ = - (998 kg/m³) * (4.80 × 10−5 atm−1) * (430 atm - 1 atm) = - ((998 kg/m³) * (4.80 × 10−5 atm−1) * (429 atm)) = -20.6 kg/m^3. The negative sign indicates an increase in density due to compression. So the increase in density of the water is 20.6 kg/m³.
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A model airplane with mass 0.750-kg is tethered by a wire so that it flies in a circle of radius 30.0-m. The airplane engine provides a force of 0.800-N perpendicular to the tethering wire. (Consider the airplane to be a point mass) (a) Find the torque that the net thrust produces about the center of the circle. (b) Find the angular acceleration of the airplane. (c) Find the linear acceleration of the airplane tangent to its flight path.
Answer:
a) 24.0 N.m b) 3.6*10⁻² rad/s² c) 1.07 m/s²
Explanation:
a) If the force that produces the torque is perpendicular to the tethering wire, we can determine its magnitude just as follows:
τ = F*r = 0.800 N * 30.0 m = 24.0 N*m (1)
b) We can express the torque we found above, using the rotational form of Newton´s 2nd Law, as follows:
τ = I* α (2)
where I is the rotational inertia regarding an axis passing through the center of the circle and α is the angular acceleration of the airplane.
If we consider the airplane as a point mass, the rotational inertia I can be calculated as follows:
I = m*r² = 0.750 Kg * (30.0)² m² = 675 Kg*m²
From (1) and (2), we can solve for α, as follows:
[tex]\alpha = \frac{T}{I} = \frac{24.0 N*m}{675.0 kg*m2} = (3.6e-2) rad/s2[/tex]
⇒α = 3.6*10⁻² rad/s²
c) Applying the definition of the angular velocity, and the definition of an angle, we can find the following realtionship between the linear and angular velocity:
v = ω*r
Dividing both sides by Δt, we can extend this relationship to the linear and angular acceleration, as follows:
a = α*r
a = 3.6*10⁻² rad/s²* 30.0 m = 1.07 m/s²
a. The torque that the net thrust produces about the center of the circle is 24 Newton.
b. The angular acceleration of the airplane is equal to [tex]0.036 \;rad/s^2[/tex]
c. The linear acceleration of the airplane tangent to its flight path is[tex]1.08 \;m/s^2[/tex]
Given the following data:
Mass of airplane = 0.750 kgRadius = 30.0 mForce = 0.800 Newtona. To find the torque that the net thrust produces about the center of the circle:
Mathematically, the torque produced by a perpendicular force is given by:
[tex]T = Fr[/tex]
Where:
T is the torque.F is the perpendicular force.r is the radius.Substituting the given parameters into the formula, we have;
[tex]T = 0.8 \times 30[/tex]
Torque, T = 24 Newton
b. To find the angular acceleration of the airplane, we would use Newton's Second Law of rotational motion:
[tex]T = I\alpha[/tex]
But, [tex]I = mr^2[/tex]
[tex]I = 0.750 \times 30^2\\\\I = 0.750 \times 900[/tex]
Moment of inertia, I = [tex]675\;Kgm^2[/tex]
[tex]\alpha = \frac{T}{I} \\\\\alpha = \frac{24}{675}\\\\\alpha = 0.036 \;rad/s^2[/tex]
c. To find the linear acceleration of the airplane tangent to its flight path:
[tex]a = r\alpha \\\\a = 30 \times 0.036[/tex]
Linear acceleration, a = [tex]1.08 \;m/s^2[/tex]
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A flea jumps straight up to a maximum height of 0.490 m . How long is the flea in the air from the time it jumps to the time it hits the ground?
Answer:
0.62 s
Explanation:
given,
maximum height of the flea = 0.49 m
velocity at maximum height = 0 m/s
now, calculating initial velocity
using equation of motion
[tex]v^2 = u^2 + 2 g h[/tex]
[tex]0^2 = u^2 - 2\times 9.8 \times 0.49[/tex]
[tex]u^2 = 9.604[/tex]
[tex]u = 3\ m/s[/tex]
now, calculating time he take to reach at the highest point
v = u + g t
0 = 3 - 9.8 x t
9.8 t = 3
t = 0.31 s
Time it will be in air will be twice the time it took to reach to the maximum height.
Time for which it was in air = 2 x 0.31 s = 0.62 s
A traffic expert wants to estimate the maximum number of cars that can safely travel on a particular road at a given speed. He assumes that each car is 15 feet long, travels at speed s, and follows the car in front of it at a safe distance for that speed. He finds that the number N of cars that can pass a given spot per minute is modeled by the function N(s)=88s/16+16(s19)2
At what speed can the greatest number of cars travel safely on that road?
Answer:
Speed s= 19
Explanation:
Take note of the following parameters:
Speed= s,
Number of cars= N,
Number N of cars that can pass a given spot per minute=
N(s)=88s/16+16(s19)2
The principle of differentiation is here:
We let N(s) = N
N = 86s / (17 + 17((s/19)^2))
17N = 86s /(1 + s²/19²)
17N = 361* 86s /(361 + s²)
17N = 31046s /(361 + s²)
Next step;
Differentiate with respect to s
17N = 31046s /(361 + s²)
Remember the quotient rule [u/v]’ = (vu’ - uv’) / v²
Therefore,
u = 31046s =====> du/ds = u’ = 31046
v = (361 + s²) ====>dv/ds = v’ = 2s
17N = 31046s /(361 + s²)
17 dN/ds = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²
The maximum when dN/ds = 0
17 dN/ds = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²
17 * 0 = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²
( 31046(361 + s²) - 31046s(2s) ) = 0
31046 [ (361 + s²) - 2s² ] = 0
31046 (361 - s²) = 0
(361 - s²) = 0
(19 - s)(19 + s) = 0
either s = -19 or s = 19, but s > 0
s = 19
The speed at which the greatest number of cars travel safely on that road is; s = 19
What is the speed required?We are given the function to represent Number N of cars that can pass a given spot per minute as;
N(s) = 88s/(16 + 16(s/19)²)
where;
s is Speed
N is number of cars
Differentiating the function gives;
N' = -3971(s² - 361)/(2(s² + 361)²
Now, the speed at the greatest number of cars would be gotten when N' = 0. Thus;
-3971(s² - 361)/(2(s² + 361)² = 0
Cross multiply to get;
-3971(s² - 361) = 0
divide both sides by -3971 to get;
s² - 361 = 0
s = √361
s = 19
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What is the physical meaning of the physics equation of a spring force vs displacement graph?
Explanation:
The force of a spring is described by Hooke's law:
F = kx
where k is the spring stiffness in N/m, and x is the displacement in m.
A spring force vs displacement graph is a line passing through the origin with a slope of k.
Final answer:
A spring force vs. displacement graph represents Hooke's law, showing the restorative force exerted by a spring as it is deformed. The slope of the straight line on the graph is the spring's force constant 'k'. The area under the line represents the work done on the spring.
Explanation:
Physical Meaning of Spring Force vs. Displacement
The physical meaning of a spring force vs. displacement graph is that it shows how the force exerted by a spring changes as the spring is stretched or compressed. This relationship is governed by Hooke's law, which states that the force (F) exerted by a spring is directly proportional to the displacement (x) from its equilibrium position, given by the equation F = -kx. Here, 'k' is the spring's force constant, representing its stiffness, with units of newtons per meter (N/m). The negative sign indicates that the force is restorative, meaning it acts in the opposite direction of displacement.
The graph typically portrays a straight line passing through the origin, with the slope equivalent to the spring's force constant 'k'. The steeper the slope, the greater the force constant, and vice versa. Additionally, the work done on the spring, which is equal to the energy stored in it, can be calculated from the area under the force versus displacement graph, given by W = (1/2)kx², where W is the work done and x is the displacement.
A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 130-g pebble during the same amount of time. Which object has the greatest value for each of these quantities?
Answer: pebble has the greatest value of acceleration due to its low mass.
Explanation:
According to newton's second law of motion,
Force = mass × acceleration
For boulder,
Acceleration = Force/mass
Force acting on boulder is 200N
Its mass is 100kg
Acceleration = 200/100
Acceleration = 2m/s²
Similarly for pebble,
Force on pebble = 200N (the same as boulder)
Its mass is 130g = 0.13kg(has to be converted to the standard unit which is kg)
Its acceleration = 200/0.13
Acceleration of pebble = 1538.5m/s²
Since the question doesn't specify what to compare, we will compare their accelerations.
Therefore, pebble has the greatest value of acceleration due to its low mass.
If you fire a projectile from the ground, it hits the ground some distance R away (called "the range"). If you keep the launch angle fixed, but double the initial launch speed, what happens to the range?
Answer:
range becomes 4 times
Explanation:
We know that the range of a projectile is given as:
[tex]R=\frac{u^2.\sin(2\theta)}{g}[/tex]
where:
[tex]R=[/tex] range of the projectile
[tex]u=[/tex] initial velocity of projectile
[tex]\theta=[/tex] initial angle of projection form the horizontal
g = acceleration due to gravity
When the initial velocity of launch is doubled:
[tex]R'=\frac{(2u)^2.\sin(2\theta)}{g}[/tex]
[tex]R'=\frac{4u^2.\sin(2\theta)}{g}[/tex]
[tex]R'=4R[/tex]
range becomes 4 times
Doubling the initial launch speed of a projectile, while keeping the angle of launch fixed, results in the range being quadrupled.
Explanation:The range R of a projectile motion is given by the formula R = ((v^2)*sin(2*theta))/g, where v is the initial launch speed, theta is the launch angle, and g is the acceleration due to gravity. If the initial launch speed v is doubled, the new range R' would be R' = ((2v)^2)*sin(2*theta))/g, which simplifies to R' = 4R.
Thus, if you double the initial launch speed, the range of the projectile is quadrupled, assuming the launch angle is fixed.
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A uniform plank of length 6.1 m and mass 33 kg rests horizontally on a scaffold, with 1.6 m of the plank hanging over one end of the scaffold. L l x How far can a painter of mass 60 kg walk on the overhanging part of the plank x before it tips? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.
Answer:
Explanation:
consider the principle of moment
when a system is in equilibrium, the clockwise moment (torque) about the pivot is equal to the counterclockwise moment ( torque). Since the plank is uniform the weight of the plank act at the middle which = 6.1 m / 2 = 3.05 m
the distance that can support the weight of the man = d
mass of the man = 70
70 × d = 33 × ( 3.05 - 1.6)
d = 47.85 / 60 = 0.798 m, if the man work beyond this point he will fall.
A record is spinning at the rate of 25rpm. If a ladybug is sitting 10cm from the center of the record.
A-What is the rotational speed of the ladybug? (in rev/sec)
B-What is the frequency of the ladybug's revolutions? (in Hz)
C-What is the tangential speed of the ladybug? (in cm/sec)
D-After 20 seconds. how far has the ladybug traveled? (in cm)
A) Angular speed: 0.42 rev/s
B) Frequency: 0.42 Hz
C) Tangential speed: 26.4 cm/s
D) Distance travelled: 528 cm
Explanation:
A)
In this problem, the ladybug is rotating together with the record.
The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is
[tex]\omega = 25 rpm[/tex]
where here it is measured in revolutions per minute.
Keeping in mind that
1 minute = 60 seconds
We can rewrite the angular speed in revolutions per second:
[tex]\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s[/tex]
B)
The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation
[tex]\omega = 2 \pi f[/tex] (1)
where
[tex]\omega[/tex] is the angular speed
f is the frequency of revolution
For the ladybug in this problem,
[tex]\omega=0.42 rev/s[/tex]
Keeping in mind that [tex]1 rev = 2\pi rad[/tex], the angular speed can be rewritten as
[tex]\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42[/tex]
And re-arranginf eq.(1), we can find the frequency:
[tex]f=\frac{\omega}{2\pi}=\frac{(2\pi)0.42}{2\pi}=0.42 Hz[/tex]
And the frequency is the number of complete revolutions made per second.
C)
For an object in circular motion, the tangential speed is related to the angular speed by the equation
[tex]v=\omega r[/tex]
where
[tex]\omega[/tex] is the angular speed
v is the tangential speed
r is the distance of the object from the axis of rotation
For the ladybug here,
[tex]\omega = 2\pi \cdot 0.42 rad/s[/tex] is the angular speed
r = 10 cm = 0.10 m is the distance from the center of the record
So, its tangential speed is
[tex]v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s[/tex]
D)
The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:
[tex]d=vt[/tex]
where
v is the tangential speed
t is the time elapsed
Here we have:
v = 26.4 cm/s (tangential speed)
t = 20 s
Therefoe, the distance covered by the ladybug is
[tex]d=(26.4)(20)=528 cm[/tex]
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