In Lab 127, the objectives are to experimentally determine the rotational inertia of a rotating body and practice computing rotational inertias for objects with different shapes.
Step 1: Set up the experiment by choosing a rotating body with a known mass and shape. Attach it to an appropriate apparatus that allows you to measure its angular acceleration.
Step 2: Apply a known torque to the rotating body, either by applying a force at a specific distance from the axis of rotation or using a torque measurement device.
Step 3: Measure the angular acceleration of the rotating body as the torque is applied. This can be done using tools like a motion sensor, a protractor, or a photogate.
Step 4: Use the measured angular acceleration and the applied torque to calculate the rotational inertia of the rotating body using the relation:
τ = Iα
where τ is the torque, I is the rotational inertia, and α is the angular acceleration.
Step 5: Repeat steps 1-4 for different rotating bodies with various shapes and mass distributions.
Step 6: Practice computing the rotational inertias for these different shapes using standard formulas. Compare your experimental results with the theoretical values to check for consistency and improve your understanding of rotational inertia.
By following these steps, you will be able to experimentally determine the rotational inertia of a rotating body and practice computing rotational inertias for objects with different shapes.
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How does head to tail rule helps us to find the resultant of forces ?? Explain with the help of a diagram.
Head to tail rule helps us to find the resultant of forces because in this rule we join many vectors with each other forming resultant vector.
What is head to tail rule?
The joining of tail of the first vector with the head of the second vector represents the resultant vector whereas the direction of the resultant vector is from the tail of the first vector towards the head of the second.
So we can conclude that Head to tail rule helps us to find the resultant of forces because in this rule we join many vectors with each other forming resultant vector.
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An air conditioner runs 15 minutes each hour on a hot summer day. It is on a 240 volt circuit and uses 21 amps. Rate is $.10/kWh
How much does it cost to run for one day? Round to the nearest cent. $Answer
Answer:
Approximately [tex]\$ 3.02[/tex].
Explanation:
Note that the electric rate in this question is in the unit dollar-per-[tex]{\rm kWh}[/tex], where [tex]1\; {\rm kWh}[/tex] is the energy to run an appliance of power [tex]1\; {\rm kW}[/tex] for an hour.
Number of minutes for which the air conditioner is running in that day: [tex]15 \times 24 = 360\; \text{minute}[/tex]. Apply unit conversion and ensure that this time is measured in hours (same as the unit of the electric rate.)
[tex]\begin{aligned} \text{time} &= 360\; \text{minute} \times \frac{1\; \text{hour}}{60\; \text{minute}} = 6\; \text{hour} \end{aligned}[/tex].
The power of this air conditioner is:
[tex]\begin{aligned} \text{power} &= \text{voltage} \times \text{current} \\ &= 240\; {\rm V} \times 21\; {\rm A} \\ &= 5040\; {\rm W} \\ &= 5.04\; {\rm kW} \end{aligned}[/tex].
Thus, the energy that this air conditioner would consume would be:
[tex]\begin{aligned}\text{energy} &= \text{power} \times \text{time} \\ &= 5.04\; {\rm kW} \times 6\; \text{hour} \\ &= 30.24\; {\rm kWh} \end{aligned}[/tex].
At a rate of [tex]0.1[/tex] dollar-per-[tex]{\rm kWh}[/tex], the cost of that much energy would be approximately [tex]3.02[/tex] dollars (rounded to the nearest cent.)
A carnival ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the acceleration of a passenger at his lowest point during the ride?.
The acceleration of the person is 4.108 m/s^2.
What is acceleration?
The acceleration is the rate of change in the velocity in a unit of time.
Angular velocity: The change in angular displacement in a unit of time is called angular velocity.
Tangential velocity: The tangential velocity can be defined as the velocity of an object which is perpendicular to the radius in the rotational motion.
The relation between angular velocity ω, tangential velocity v, and radius r from the axis when the radius is perpendicular to the tangential velocity is,
v=ω*r
Given r=15 m, and ω=5 turns/minutes, substitute these values in the above formula.
Note: 1 turn/minute = 2π/60 rad/s.
v=(5 turns/minutes)*15 m
v=(5*2π/60 rad/s)* 15 m
v=7.85 m/s
Since the motion is circular, and the person is at the lowest point of the wheel, so the acceleration due to gravity will have no effect as it is perpendicular to tangential velocity here. The acceleration a for rotational motion is given by,
a=v^2/r
Substitute v=7.85 m/s, and r=15 m in this equation and solve it.
a=(2.094)^2/(15)
a=4.108 m/s^2
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State Newton's second law of motion and establish the relation F = m x a
F = ma is the formula of Newton's Second Law of Motion. Newton's Second Law of Motion is described as Force is equal to the rate of change of momentum. For a constant mass, force equals mass duration acceleration.
What is the significance of Newton's second law of motion explain?
Newton's Second Law of Motion F=ma is very significant because it indicates the relationship between forces and motion. It permits you to calculate the acceleration (and therefore velocity and position) of an object with known forces. This is extremely helpful for scientists, engineers, inventors, etc.
Force is directly proportional to Shift in momentum ÷ Time taken.Therefore : F is directly proportional to m×a. This relation F is directly proportional to m×a can be turned into an equation by placing in a constant k.
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What minimum volume must the slab have for a 75.0 kgkg woman to be able to stand on it without getting her feet wet
V = 0.9375 m3 is minimum volume .
What is density ?We use the word "density" to indicate how much space (or "volume") an object or substance occupies in relation to the amount of matter contained therein (its mass). Density can also be defined as the quantity of mass per unit of volume. A dense object is one that is both hefty and small.Given,
1000 kg/m3 = density of water
920 kg/m3 = density of ice
BF = Woman Weight + Slab Weight
Pvg = 75 + Pvg
V = 75 kg / (1000 - 920)
V = 75 / 80
V = 0.9375 m3
Therefore, V = 0.9375 m3 is minimum volume .
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The photographer realizes that with the lens she is currently using, she can't fit the entire landscape she is trying to photograph into her picture. Should she switch to a lens with a longer or shorter focal length
She should use shorter focal length to fit the entire landscape which she is trying to photograph into her picture.
What is focal length?
The focal length is a measure of how strongly the system converges or diverges light.
A positive focal length indicates that a system converges light, while a negative focal length indicates that the system diverges light.
For a standard rectilinear lens,
FOV = 2 arctan (x/2f)
FOV ∝ 1 / f
where x is the diagonal of the film.
Focal length (f) and field of view (FOV) of a lens are inversely proportional.
From the equation we can say that,
A shorter focal length gives you a wide angle of view which allows more view to fit in the frame.
Hence,
She should use shorter focal length to fit the entire landscape which she is trying to photograph into her picture.
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Chicago, il is currently in the e climate zone.
a) true
b) false
Answer:
false
Explanation:
what was kobes passion for sports
The density of mobile electrons in copper metal is 8.4 1028 m-3. Suppose that i = 9.0 1018 electrons/s are drifting through a copper wire. (This is a typical value for a simple circuit.) The diameter of the wire is 3.6 mm. In this case, about how many minutes would it take for a single electron in the electron sea to drift from one end to the other end of a wire 35 cm long? minutes
It would take 531 minutes for a single electron in the electron sea to drift from one end to the other end of a wire 35 cm long.
What is electric current?
A rate of electric charge flowing past a place or region is called an electric current. A net passage of electric charge through a region creates an electric current.
Here,
ρ is the density of electrons = 8.4×10²⁸ kg/m³
i is the electric current =9.08×10¹⁸ electrons/sec
d is the diameter of the wire =3.6 mm = 0.0036 m
r is radius, r = d/2, r = 1.8 mm = 0.0018 m
l is the length of wire = 34 cm = 0.34 m
v is The volume of the wire
We know that Volume:
V = π r^2* l
V = (3.14* 0.0018^2) 0.34
V = 3.45* 10^-6 m³
Now,
no of electrons= ρ × V
n = 8.4×10²⁸ * 3.45* 10^-6
n = 2.898 * 10^ 23
The time required for the drifting electron is equal to the ratio of the number of electrons to the electric current flows.
T = n/i
T = 2.898 * 10^ 23 / 9.08×10¹⁸
T = 0.3191 * 10^5 s
T = 531 minutes
Hence,
It would take 531 minutes for a single electron in the electron sea to drift from one end to the other end of a wire 35 cm long.
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One of the primary advantages of CCDs over photographic plates is the former's high efficiency in detecting light.
True
What are the advantages of CCDs over photographic plates?In place of photographic plates or films, charge coupled devices, or CCDs, are sensitive photon detectors that can be employed in telescopes to create images. CCDs were created in the late 1960s and are now found in various devices, including digital cameras and photocopiers.
CCDs are more light-efficient, catch a greater proportion of incoming photons, can store and transfer digital data, and don't require development like film does. They respond linearly and capture up to 80% of the light that strikes them.
I understand the question you are looking for is this:
One of the primary advantages of CCDs over photographic plates is the former's high efficiency in detecting light. True or false.
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15. When you cannot stop safely at a yellow traffic light before entering an intersection, ______________. A. stop in the intersection as soon as the light turns red B. accelerate so you'd cross the intersection before the light turns red C. enter the intersection carefully and continue across
Answer: When you cannot stop safely at a yellow traffic light before entering an intersection, enter the intersection carefully and continue across.
Explanation: To find the correct answer, we need to know more about the traffic signal rules.
What is the traffic signal rules?Red light- Indicator for the motorists to stop.Green-Signal for safety and word GO.Yellow- This signal let you know that the red signal is about to be displayed.when it's turned on, you can start slowing down to come to a stop in anticipation of red light.when we cannot stop safely at a yellow traffic light before entering an intersection, enter the intersection carefully and continue across.Thus, we can conclude that, the option C is correct.
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If the strength of a magnetic field at B is three units the strength of the magnetic field at A is ____?
If the strength of a magnetic field at B is three units the strength of the magnetic field at A, means that point B has three times more turns as compared to point A.
How strength of the magnetic field should be increased?We can increase the strength of the magnetic field by increasing the number of turns which increases the strength of the current flow.
So we can conclude that point B has three times more turns as compared to point A.
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A hydraulic system contains 3 pistons. The input piston has an area of 0.625 in2, one of the output pistons ( piston A ) has an unknown area and the other piston ( Output Piston ) has a diameter of 3.75 inches. If system pressure is 1,440 psi, and Piston A is supporting 20,000 lb, what is the area of Piston A
The area of Piston A is 13.88 in^2
What is a hydraulic system?
The hydraulic system functions by using pressurized fluid. The power of liquid fuel in hydraulics is significant and as a result, they are commonly used in heavy equipment.
Hydraulic system simply means a mechanical function that operates through the force of liquid pressure.
Here
A hydraulic system contains 3 pistons.
Pressure, P = 1440 psi
for piston A:
Area, A= 0.625 in^2
Force, F = 20,000 lb
We know that,
Force = Pressure x Area
F = P x A
Area ( piston A) = F / P
A = 20,000 / 1440
A = 13.88 in^2
Hence,
The area of Piston A is 13.88 in^2
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Which type of motorcycle features rear-positioned footrests, high power-to-weight ratio, and a forward leaning ride
Sport-type motorcycles typically feature a higher than average power-to-weight ratio.
Sport-style motorcycles often have a power-to-weight ratio that is above average. Fuel refers to opening the fuel supply valve, if available. Set the choke and squeeze the clutch lever together is referred known as "choke and clutch." Even though some motorcycles lack a choke, the choke should be ON for a cold engine.What is scrambler motorcycle?
Scrambler motorbikes, also referred to as dirt bikes, have special features designed for off-road activities and sports. Off-road, to be clear, is any route that isn't paved or that follows a set pattern. Scrambler bicycles feature relatively straightforward designs, long suspensions, and high ground clearance.Learn more about Scrambler motorbikes
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Tuning forks with a frequency of 500 to 1000 hz are most commonly used to measure:________
Tuning forks with a frequency of 500 to 1000 hz are most commonly used to measure: hearing loss
A person is said to have hearing loss if they are not able to hear as well as someone with normal hearing, meaning hearing thresholds of 20 dB or better in both ears. It can be mild, moderate, moderately severe, severe or profound, and can affect one or both ears.
High-pitched tuning fork with frequency of 500 to 1000 hertz (Hz) should be used to estimate hearing loss in range of normal speech (300 to 3000 Hz).
The 512-Hz Rinne tuning fork test could be very effective at detecting conductive hearing losses when performed by an experienced tester and when masking was used . Tuning forks are used to carry Rinne and Weber tests. These tests provide a quick easy method of determining whether a hearing loss is situated in the middle ear or inner ear
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A satellite with a mass of 110 kg and a kinetic energy of 3.0 ´ 109 J must be moving at a speed of
Answer: A satellite with a mass of 110 kg and a kinetic energy of 3.08×10^9 J must be moving at a speed of 7483 m/s.
Explanation: To find the answer we need to know about the kinetic energy of a body.
How to solve the problem the equation of kinetic energy?We have the expression for kinetic energy of a body as,[tex]KE=\frac{1}{2}mv^2[/tex]
Given that,[tex]m=110kg\\KE=3.08*10^9J\\[/tex]
We have to find the speed of the satellite,[tex]v=\sqrt{\frac{2KE}{m} } =\sqrt{\frac{2*3.08*10^9}{110} } =7.483*10^3 m/s[/tex]
Thus, we can conclude that, the velocity of the satellite will be 7438m/s.
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In order to move at a speed of 7483 m/s, a satellite with a mass of 110 kg and a kinetic energy of 3.08× 10^-9 J must be travelling.
Understanding a body's kinetic energy is necessary in order to determine the solution.
How can the kinetic energy equation be solved?The phrase "kinetic energy of a body" might be used as,[tex]KE=\frac{1}{2}mV^2[/tex]
Given that,[tex]m=110kg\\KE=3.08*10^9J[/tex]
We must determine the satellite's speed.[tex]V=\sqrt{\frac{2KE}{m} }=7.48km/s[/tex]
Thus, we can infer that the satellite will move at a speed of 7438 m/s.
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How many outer-orbital electrons are found in an atom of:a) Na?b) P?c) Br?d) I?e) Te?f) Sr?
Answer:
Na=1P=5Br=7I=7Te=6Sr=2The requried number of outer-orbital electrons in an atom of Na is 1, P is 5, Br is 7, I is 7, Te is 6, and Sr is 2.
What is orbital?An orbital is a region of space around the nucleus of an atom where there is a high probability of finding an electron. The electrons in an atom are arranged in specific energy levels or shells, and each shell contains one or more subshells, which in turn contain one or more orbitals.
Here,
To determine the number of outer-orbital electrons in an atom, we need to look at the position of the element in the periodic table. The number of outer-orbital electrons is equal to the group number of the element for groups 1, 2, and 13 to 18. For groups 3 to 12, we need to subtract the group number from 10.
a) Na is in group 1, so it has 1 outer-orbital electron.
b) P is in group 15, so it has 5 outer-orbital electrons.
c) Br is in group 17, so it has 7 outer-orbital electrons.
d) I is in group 17, so it has 7 outer-orbital electrons.
e) Te is in group 16, so it has 6 outer-orbital electrons.
f) Sr is in group 2, so it has 2 outer-orbital electrons.
Therefore, the number of outer-orbital electrons in an atom of Na is 1, P is 5, Br is 7, I is 7, Te is 6, and Sr is 2.
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The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s. Find m.
Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s. Then the mass m is 0.625kg.
Explanation: To find the answer, we need to know more about the simple harmonic motion.
What is simple harmonic motion?A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.We have the equation of time period of a SHM as,[tex]T=2\pi \sqrt{\frac{m}{k} }[/tex]
Where, m is the mass of the body and k is the spring constant.How to solve the problem?Given that,[tex]T_1=2s\\m_1=m\\m_2=m+2kg\\T_2=3s[/tex]
We have to find the value of m,[tex]T_1=2\pi \sqrt{\frac{m}{k} } \\T_2=2\pi \sqrt{\frac{m+2}{k} } \\\frac{T_1}{T_2} =\sqrt{\frac{m}{m+2} }\\\frac{2}{3} =\sqrt{\frac{m}{m+2} }\\\\[/tex]
[tex]m=\frac{5}{8} =0.625kg[/tex]
Thus, we can conclude that, the mass m will be 0.625kg.
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A small mailbag is released from a helicopter that is descending steadily at 1.53 m/s. (a) After 4.00 s, what is the speed of the mailbag? v = m/s (b) How far is it below the helicopter? d = m (c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.53 m/s?
Answer: A small mailbag is released from a helicopter that is descending steadily at 1.53 m/s. Then,
a) after 4s, the speed of the mailbox will be, 40.73m/s downwards.
b)78.4m downwards.
c) v=37.67m/s downward. and d=78.4m downward.
Explanation: To find the answer, we need to know more about the equations of uniformly accelerated motion.
What are the equations of uniformly accelerated motion?The equations of uniformly accelerated motion under gravity are,[tex]v=u+at\\S=ut+\frac{1}{2}at^2\\ v^2-u^2=2aS\\S=\frac{v+u}{2}t[/tex]
How to solve the problem?a) speed of the mailbag after 4s,[tex]v=u+gt=-1.53-(9.8*4)=-40.73 m/s^2[/tex]
b) How far it is below the helicopter,[tex]d_1=ut+\frac{1}{2}gt^2=(-1.53*4)+\frac{-9.8*16}{2} =-84.52 m\\d_2=vt=-40.73*4=-162.92 m\\d=-162,92+84.52=-78.4m[/tex]
c) If the helicopter is rising steadily, then v and d will be,[tex]v=u+gt=1.53-(9.8*4)=-37.67 m/s[/tex]
[tex]d_1=ut+\frac{1}{2}gt^2=(1.53*4)+\frac{-9.8*16}{2} =-72.28 m\\ d_2=vt=-37.67*4=-150.68 m\\d=-150.68+72.28=-78.4 m[/tex]
Thus, we can conclude that, the answers to the question are, a) after 4s, the speed of the mailbox will be, 40.73m/s downwards.
b)78.4m downwards.
c) v=37.67m/s downward. and d=78.4m downward.
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There are many muscles in our body. our body uses muscles to move parts of our body. a part that needs to move a lot will have a higher number of muscles. which of the following parts is most likely to have the most muscles? a. ear lobe b. stomach c. teeth d. brain
Option b) Stomach is most likely to have the most muscles.
There are five main muscles in stomach pyramidalis, rectus abdominus, external obliques, internal obliques, and transversus abdominis.
Muscles circulate frame parts by contracting after which relaxing. muscular tissues can pull bones, however, they cannot push them to return to their unique role. so they paintings in pairs of flexors and extensors. The flexor contracts to bend a limb at a joint.
The muscle groups that circulate our frame elements are called skeletal muscles, and they're a form of striated muscle. We will actively manipulate those with our brains. another sort of striated muscle is those that maintain our hearts pumping, which we are not able to actively control.
Muscles pull at the joints, allowing us to move. a couple of muscular tissues are used for any sort of motion of a bone. Although we are sitting flawlessly nevertheless, muscle tissues at some point in the frame are constantly shifting.
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we observe stars orbiting at distances of 25000 light years from the centers of galaxy x and galaxy y for this orbital distance stars in galaxy x have higher orbital speeds tan stars in galaxy y what can we conclude
Galaxy X contains more total mass within 25,000 light-years of its center than Galaxy Y.
Light-yearsA light-year, a significant unit of length used to express astronomical distances, is roughly 9.46 trillion kilometers, or 5.88 trillion miles. A light-year is the distance that light travels in a vacuum in one Julian year, according to the International Astronomical Union. The name "light-year" is occasionally misunderstood as a unit of time since it contains the time measurement word "year". When expressing distances to stars and other distances on a galactic scale, the light-year is most frequently employed, especially in non-specialized situations and popular science publications. The parsec, which comes from astrometry and is the distance at which one astronomical unit subtends an angle of one second of arc, is the unit that is most frequently used in professional astronomy.
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When starting a foot race, a 64 kilogram sprinter exerts an average force of 693 newtons backward on the ground for 0.59 seconds. how far does he travel in meters during this time?
The distance traveled by the sprinter in meters is determined as 1.88 m.
Acceleration of the sprinterThe acceleration of the sprinter is the rate of change of velocity of the sprinter with time.
The acceleration of the sprinter is calculated as follows;
Apply Newton's second law of motion as follows;
F = ma
a = F/m
where;
F is the applied force by the sprinterm is mass of the sprintera is acceleration of the sprintera = 693 N / 64 kg
a = 10.83 m/s²
Distance traveled by the sprinterThe distance traveled by the sprinter is calculated as follows;
s = ut + ¹/₂at²
where;
u is initial velocity = 0s = ¹/₂at²
where;
t is time of motiona is accelerations = (0.5)(10.83)(0.59²)
s = 1.88 m
Thus, the distance traveled by the sprinter in meters is determined as 1.88 m.
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A gymnast initially has a moment of inertia of 6 kgm^2 and rotation rate of 12rad/s. His rotation rate slows to 8rad/s because he extends his arms and legs. What is his new moment of inertia
The new moment of inertia of the gymnast is 9kgm^2.
To find the answer, we need to know about the conservation of angular momentum.
What's conservation of angular momentum?It says the angular momentum of a rotating body remains conserved until an external torque is applied on it. Angular momentum= moment of inertia × angular velocity = I×wWhat's the final moment of inertia of the gymnast, if his initial moment of inertia, initial angular velocity and final angular velocity are 6 kgm^2, 12rad/s and 8rad/s respectively?Let I₁= 6 kgm^2, w₁= 12rad/s, w₂= 8rad/s and I₂= final angular velocityFrom conservation of angular momentum,I₁×w₁= I₂×w₂
I₂= I₁×w₁/w₂
= (6×12)/8
= 9kgm^2
Thus, we can conclude that the new moment of inertia of the gymnast is 9kgm^2.
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A firm current ratio is 1. 0 and its quick ratio is 1. 0. If current liabilities are 12300, what are its inventories?
A firm current ratio is 1. 0 and its quick ratio is 1. 0. If current liabilities are 12300 then its inventories will be 12300
Inventory is the accounting of items, component parts and raw materials that a company either uses in production or sells
The quick and current ratios are liquidity ratios that help investors and analysts gauge a company's ability to meet its short-term obligations. The current ratio divides current assets by current liabilities. The quick ratio only considers highly-liquid assets or cash equivalents as part of current assets.
current ratio = current assets / current liabilities
current assets = current ratio * current liabilities
= 1 * 12300 = 12300
since , inventory is a current asset for accounting purpose , hence inventories will be 12300
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In addition to ozone, what four other greenhouse gases or groups of greenhouse gases are included in nearly all the climate models?
In addition to ozone, the four other greenhouse gases or groups of greenhouse gases are included in nearly all the climate models are Carbon dioxide, water vapor, and methane, nitrous oxide.
Carbon dioxide, absorbs energy at a variety of wavelengths, a range that overlaps with that of infrared energy. As CO2 soaks up this infrared energy, it vibrates and re-emits the infrared energy back in all directions. About half of that energy goes out into space, and about half of it returns to Earth as heat, contributing to the ‘greenhouse effect.’
As methane is emitted into the air, it reacts in several ways. It reacts and after oxidization methane forms water vapor and carbon dioxide. So, not only does methane contribute to global warming directly but also, indirectly through the release of carbon dioxide.
Nitrous oxide enhances the greenhouse effect by capturing reradiated infrared radiation from the Earth’s surface and subsequently warming the troposphere . It is chemically inert in the troposphere and stays in the troposphere for about 120 years before moving into the stratosphere where it ultimately leads to destruction of stratospheric ozone.
Water vapor does absorb longwave radiation and radiates it back to the surface, thus contributing to warming.
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Two boys with masses of 40 kg and 60 kg stand on a horizontal frictionless surface holding the ends of a light 10-m long rod. The boys pull themselves together along the rod. When they meet the 40-kg boy will have moved what distance?.
when they meet the 40-kg boy will have moved a distance of 6 m.
What is friction force:Friction is the force that resists motion when the surface of one object comes in contact with the surface of another.
The principle of center mass.It is the average position of all the parts of the system, weighted according to their masses.
here,
masses of the boys
M=40 kg,
m=60 kg
x1 = 10 m
The displacement of the 40 kg boy is calculated from the principle of center mass.
XM = ( mx1 + Mx2) / ( m + M )
X(40) = (60 x 10 + 40 x 0) / (40 kg + 60 kg)
X(40) = (600)/(100)
X(40) = 6 m
X(60) = (60 x 0 + 40 x 10 m) / (40 kg + 60 kg)
X(60) = (400)/(100)
X(60) = 4 m
Thus,
when they meet the 40-kg boy will have moved a distance of 6 m.
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A bicycle with 20 -inch diameter wheels is traveling at 16 mi/h. Find the angular speed of the wheels in rad/min.
The angular speeds of the wheels is 1,647.28 rad/min
What is Angular Velocity?angular velocity is the speed at which the angle between two bodies changes as an item rotates or revolves around an axis.
The diameter of a bicycle, d = 20inches
Radius, r = 10 inches
The bicycle is travelling at 16 mi/h.
We need to find the angular speed of the wheels in rad/min.
We know that,
1 mi/h = 0.0166667 mi/min
It means,
16 mi/h = 0.26 mi/min
Also,
10 inch = 0.0001578356 miles
The angular speeds of the wheels is given by :
v = rω
ω = v/r
ω = 0.26/0.0001578356
ω = 1,647.28 rad/min
So, the angular speeds of the wheels is 1,647.28 rad/min
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What is constant in the horizontal direction for a projectile?
Answer:
initial velocity is the constant in horizontal projectile
Water falls without splashing at a rate of 0.400 L/s from a height of 2.20 m into a 0.720-kg bucket on a scale. If the bucket is originally empty, what does the scale read 3.60 s after water starts to accumulate in it
The reading of the scale after 3.6 seconds is 23.79 N.
Velocity of the water
v = √2gh
v = √(2 x 9.8 x 2.2)
v = 6.57 m/s
Rate of change of momentum of the water after impactmv/t = F = 0.4 x 6.57 = 2.628 N
Weight of the bucketWb = 0.72 x 9.8
Wb = 7.056 N
Weight of the accumulated waterWa = (RT)g
Wa = (0.4 x 3.6) x 9.8
Wa = 14.11 N
Reading of the scaleR = Wa + Wb + F
R = 14.11 N + 7.056 N + 2.628 N
R = 23.79 N
Thus, the reading of the scale after 3.6 seconds is 23.79 N.
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Three charges, qA is +6.0 μC, qB is –5.0 μC, and qc is +6.0 μC, are located at the corners of a square with each side length at 5.0 cm as shown in the diagram. Calculate the electric field at point D.
The net electric field at point D is determined as 3.95 x 10⁷ N/C.
Electric field at D due to charge A
E = kq/r²
where;
r is the distance between A and Dq is charge AE(AD) = (9 x 10⁹ x 6 x 10⁻⁶)/(0.05²)
E(AD) = 2.16 x 10⁷ i N/C
Electric field at D due to charge B
E = kq/r²
where;
r is the distance between A and Bq is charge Br² = 5² + 5²
r² = 50
r = √50
r = 7.07 cm
E(BD) = (9 x 10⁹ x 5 x 10⁻⁶)/(0.0707²)
E(BD) = 9 x 10⁶ N/C
in x - direction = 9 x 10⁶ N/C x cos(45) = 6.36 x 10⁶ i N/C
in y - direction = 9 x 10⁶ N/C x sin(45) = 6.36 x 10⁶j N/C
Electric field at D due to charge CE = kq/r²
where;
r is the distance between C and Dq is charge CE(CD) = (9 x 10⁹ x 6 x 10⁻⁶)/(0.05²)
E(CD) = 2.16 x 10⁷ j N/C
Net electric field in x directionEi = 2.16 x 10⁷ i N/C + 6.36 x 10⁶ i N/C
Ei = 2.796 x 10⁷ i N/C
Net electric field in y directionEj = 2.16 x 10⁷ j N/C + 6.36 x 10⁶j N/C
Ej = 2.796 x 10⁷ j N/C
Resultant electric field at DE = √Ei² + Ej²
E = √[(2.796 x 10⁷)² + (2.796 x 10⁷)²]
E = 3.95 x 10⁷ N/C
Thus, the net electric field at point D is determined as 3.95 x 10⁷ N/C.
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