The angular speed of the merry-go-round reduced more as the sandbag is
placed further from the axis than increasing the mass of the sandbag.
The rank from largest to smallest angular speed is presented as follows;
[m = 10 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 20 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 0.5·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 1.0·R]
Reasons:
The given combination in the question as obtained from a similar question online are;
1: m = 20 kg, r = 0.25·R
2: m = 10 kg, r = 1.0·R
3: m = 10 kg, r = 0.25·R
4: m = 15 kg, r = 0.75·R
5: m = 10 kg, r = 0.5·R
6: m = 40 kg, r = 0.25·R
According to the principle of conservation of angular momentum, we have;
[tex]I_i \cdot \omega _i = I_f \cdot \omega _f[/tex]
The moment of inertia of the merry-go-round, [tex]I_m[/tex] = 0.5·M·R²
Moment of inertia of the sandbag = m·r²
Therefore;
0.5·M·R²·[tex]\omega _i[/tex] = (0.5·M·R² + m·r²)·[tex]\omega _f[/tex]
Given that 0.5·M·R²·[tex]\omega _i[/tex] is constant, as the value of m·r² increases, the value of [tex]\omega _f[/tex] decreases.
The values of m·r² for each combination are;
Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²
Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²
Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²
Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²
Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²
Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²
Therefore, the rank from largest to smallest angular speed is as follows;
Combination 3 > Combination 1 > Combination 5 = Combination 6 >
Combination 2
Which gives;
[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =
10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].
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How is heat transferred by convection?
Answer:
Convective heat transfer is the transfer of heat between two bodies by currents of moving gas or fluid. In free convection, air or water moves away from the heated body as the warm air or water rises and is replaced by a cooler parcel of air or water.
Explanation:
Hope this helped and have a great day
Heat can be transferred by convection through liquids. Heat travels where warmer liquid rises and cooler liquid sinks. An example of this can be heating up a pot of water. Hope this helps!
HELP PLS!!
A 3 kg mass is raised a distance of 14 m above the earth by a vertical force of 93 N.
a
The gain in potential energy of the mass, to 3 significant figures, is:
Hi there!
We know that:
U (Gravitational Potential Energy) = mgh
Where:
g = acceleration due to gravity (m/s²)
m = mass (kg)
h = height/displacement (m)
Plug in the values:
U = 3 × 9.8 × 14 = 412 J
Elevations on the tongue are called
sulci
taste buds
papillae
gyri
Answer:
Papillae is correct
Explanation:
hope it helps you
Answer:
Papillae is the correct answer of this question
A farm tractor starting from the rest attains a speed of 36000 m/s after covering a distance of 2000 m. Work out the magnitude of the net force the tractor weighs 5000 kg.
Answer:
the answer is 3,888.7
Explanation:
Hope this answer helped!:)
a load of 600 N is lifted using a first class lever applying an effort of 350 N. If the distance between the fulcrum and the effort is 60 cm and the distance between the load and the fulcrum is 30cm, calculate it's efficiency
Explanation:
(a) A machine is a device by which we can either overcome a large resistive force at some point by applying a small force at a convenient point and in a desired direction or by which we can obtain a gain in the speed.
(b) An ideal machine is a machine whose parts are weightless and frictionless so that which there is no dissipation of energy in any manner. Its efficiency is 100%, i.e. the work output is equal to work input.
1. How far does a train travel in the fourth second if it starts from rest with a uniform
acceleration of 2.0 m/s2?
Answer:
Explanation:
d = ½(2.0)4² - ½(2.0)3² = 7 m
Nikolai is using a hand-operated grain mill to grind wheatberries into flour. The mill is operated by spinning a fly-wheel with radiusR= 23 cm, which has a handle attachedto the outer edge. After grinding for a few minutes at a con-stant angular speedωi, he lets go of the handle and allows themechanism to come to rest as it undergoes constant angularacceleration. This happens over the course oft= 0.50 s, andthe flywheel undergoes a quarter of a rotation during this time.What is the linear tangential accelerationaof the handle as itcomes to rest? For the limits check, investigate what happenstoaas the time required to stop the flywheel becomes small(t→0).
Answer:
Explanation:
α = Δω/t = (0 - ωi)/0.50 = -2ωi rad/s²
ωf² = ωi² + 2αθ
θ = (ωf² - ωi²) / 2α
2π/4 = (0² - ωi²) / (2(-2ωi))
2π/4 = ωi / 4
ωi = 2π rad/s
α = -2(2π) = -4π rad/s²
a = rα = 0.23(-4π) = 0.92π m/s² ≈ -2.89 m/s²
as the time to stop approaches zero, acceleration goes toward infinity.
How do I resolve moments about the point P?
Answer:
By applying the definition of torques ( [tex]\vec \tau = \vec r \times \vec F[/tex] ) and them remembering a few tricks.
Namely: if you wrap your RIGHT hand fingers around something and stick your thumb out, the direction your finger wraps gives you the verse of rotation and the thumb the orientation of the torque. Bottom force (4N) will give a counterclockwise rotation, torque is pointing up; top force (3N) will give a clockwise rotation and its torque its pointing down (read up and down as if the sheet the image is printed on is on your table).
In terms of magnitude the trick is easy: You want to multiply the intensity of the force (3N and 4N) by the distance between the point and the line the force it is applied to (that is, you don't care about the length of r itself, but the distance at a right angle, which is 0.9 and 0.8m respectively.
At this point, assuming "upwards" (relative to the plane of the sheet that is) torques positive, the 3N force gives you a torque of [tex]- 3N \times 0.9m = - 2.7N\cdot m[/tex] and the 4N force provides [tex]+4N\times 0.8 m = +3.2 N\cdot m[/tex]
Use a trigonometric equation to
determine the leg of this triangle.
10 m
[?]
m
30°
Answer:
5m
Explanation:
x=?
[tex]sin30^{0} =\frac{x}{10}[/tex]
[tex]x=10sin30^{0} =10(0.5)=5[/tex]
Hope this helps
help me for a physics project please
Mister Brainly Please Help Me
Write 10 Information's About Sound
-cant travel through space since there's no molecules to travel through
-sound travels 4.3 times faster in water than air
-sounds are waves that pass through our ears via vibrations and travel by vibrations of molecules
-different types of sound like audible, inaudible, infrasonic, ultrasonic,
-sounds waves are either longitudinal, mechanical and pressure waves
-sound travels at 767 miles per hour
What is the voltage if the current is 4 A and the resistance is 10 Ω?
Answer:
40 volts
Explanation:
Use the equation [tex]V=IR[/tex]
[tex]V=(4)(10)[/tex]
[tex]V=40[/tex]
A car is driving 12m/sec, has to stop suddenly because a pedestrian dashes out in front of the car. If the coefficient of kinetic friction between the tires and parking lot is ∪=60
what is the time, after the breaks are applied, before the car comes to a stop? Sketch the velocity time graph for the car's motion from the instant the breaks are applied until the car comes to a stop.
Answer:
Approximately [tex]2\; \rm s[/tex], assuming that the floor of this parking lot is level, [tex]\mu_{\rm k} = 0.60[/tex], and [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].
Explanation:
Let [tex]m[/tex] denote the mass of this vehicle. Weight of this vehicle: [tex]m\, g[/tex].
If the floor of this parking lot is level, the normal force on this vehicle would be equal to its weight: [tex]N = m \, g[/tex].
Given that [tex]\mu_{\rm k}[/tex], the kinetic friction between this vehicle and the ground would be consistently [tex]\mu_{\rm k} \, N = \mu_{\rm k} \, m \, g[/tex] until the vehicle comes to a stop.
Assuming that all forces on this vehicle other than friction are balanced. The net force of this vehicle during braking would be [tex](-\mu_{\rm k} \, m \, g)[/tex] (negative because this force is opposite to the direction of the motion.)
By Newton's second law of motion, the acceleration of this vehicle would be:
[tex]\begin{aligned}a &= \frac{F_\text{net}}{m} \\ &= \frac{-\mu_{\rm k} \, m \, g}{m} \\ &= -\mu_{\rm k}\, g \\ &= -0.60 \times 9.81\; \rm m\cdot s^{-2} \\ &= -5.886\; \rm m\cdot s^{-2}\end{aligned}[/tex].
In other words, braking would reduce the velocity of this vehicle by a constant [tex]5.886\; \rm m\cdot s^{-1}[/tex] every second until the vehicle comes to a stop. Calculate the time it would take to reduce the velocity of this vehicle from [tex]v_{0} = 12\; \rm m\cdot s^{-1}[/tex] to [tex]v_{1} = 0\; \rm m\cdot s^{-1}[/tex]:
[tex]\begin{aligned}t &= \frac{v_{1} - v_{0}}{a} \\ &= \frac{0\; \rm m\cdot s^{-1} - 12\; \rm m\cdot s^{-1}}{-5.886\; \rm m \cdot s^{-2}} \\ &\approx 2.0\; \rm s \end{aligned}[/tex].
Acceleration is the slope of the velocity-time graph. Since the acceleration here is constant, the velocity-time graph of this vehicle would be a line with a negative slope.
A racing car on the straight accelerates from 100 km/h to 316 km/h in three seconds.
What is its acceleration?
40m/s2
30m/s2
20m/s2
72m/s2
Answer:
[tex]20m/s^2[/tex]
Explanation:
Solution is attached. I apologize if it is a little messy.
Please help me with this problem
Answer:
Summertime
Explanation:
the sun never sets south of the Antarctic circle in the summertime.
It velocity of light scalar or vector equality
Answer:
velocity is a vector quantity
Explanation:
velocity is a vector quantity because it has a mass and a direction
Which of the following can cause an object to accelerate?
Select one:
a. Force
b. Inertia
c. Mass
d. Kinetic Energy
I think force can accelerate
Explanation:
i think force of an object
F (N)
4
* 0
3
A
2
FIGURE 2
t(s)
5
0
1
2
3
4
3) A force of magnitude Fx acting in the x-direction on a 2.00 kg particle varies in time as shown
in FIGURE 2. Find
a) The impulse of the force
b) The final velocity of the particle if it is initially at rest
c) The final velocity of the particle if it is initially moving along the x-axis with velocity
of -2.00 ms -1
Answer:
Mark me as brainlist please.
Students were asked to create roller coasters for marbles. The only requirement is that the roller coaster include at least one hill that the marble must roll over in order for the roller coaster to be considered a success. Students are building their designs with marbles and foam tubing so there will be some air resistance and friction. What should the students keep in mind if they want to create a successful roller coaster?
Question 23 options:
1)
Without an extra push at the bottom of the first hill, there is no way the car will make it back up a hill of any height because the marble doesn't have enough mass or velocity
2)
The hill should be at least a little lower than the starting height because some of the kinetic energy at the bottom of the first hill will be converted to other types of energy due to friction so it will not have as much potential energy at the top of the next hill
3)
The hill should be taller than the starting height because the marble will pick up speed on the downward hill and the increased velocity will allow it to travel higher on the next hill
4)
The hill should be the exact same height as the starting height because 100% of the kinetic energy at the bottom of the first hill will be converted back to potential energy at the top of the next hill
Since energy is lost in the roller coaster due to friction, the hill should little lower than the starting height since some of the kinetic energy at the bottom of the first hill is lost due to friction so it will not have as much potential energy at the top of the next hill.
A roller coaster is a good way to demonstrate the principle of conservation of energy. Recall that energy is neither created nor destroyed but can be converted from one form to another.
In a roller coaster, all the heights are not the same because energy is lost along the line. Therefore, the students must bear in mind that the hill should be at least a little lower than the starting height because some of the kinetic energy at the bottom of the first hill will be converted to other types of energy due to friction so it will not have as much potential energy at the top of the next hill.
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What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1000 m/sm/s
Explanation:
Since the neutron is only moving at 1000 m/s, we are going to ignore the relativistic effects on its mass and energy. The mass of a neutron in [tex]m_n = 1.67×10^{-27}\:\text{kg}[/tex] so its kinetic energy KE is
[tex]KE = \frac{1}{2}m_nv^2[/tex]
[tex]\:\:\:\:\:\:\:\:= \frac{1}{2}(1.67×10^{-27}\:\text{kg})(10^3\:\text{m/s})^2[/tex]
[tex]\:\:\:\:\:\:\:\:= 8.35×10^{-22}\:\text{J}[/tex]
A photon's energy E is defined as
[tex]E = h\nu[/tex]
where [tex]\nu[/tex] is the photon's frequency and h is the Planck's constant. Solving for the frequency, we get
[tex]\nu = \dfrac{E}{h} = \dfrac{8.35×10^{-22}\:\text{J}}{6.63×10^{-34}\:\text{J-s}}[/tex]
[tex]\:\:\:\:\: = 1.26×10^{12}\:\text{Hz}[/tex]
which is right around the infrared radiation range.
The angle of incidence (5 points)
must equal the angle of reflection
is always less than the angle of reflection
is always greater than the angle of reflection
may be greater than, less than, or equal to the angle of reflection
Answer:
D.) the same.
They are traveling in the same vacuum so there is no resistance and no outside influences. They will travel at the same speed as each other will little to no variance in their speed.
Explanation:
hope this helps. . . <3
good luck! uωu
A 1.5 kg cart is attached to a spring with spring constant of 5 N/m. The cart & spring is pulled to stretch the spring by 3 meters.
What is the SPE?
22.5 J
Explanation:
Given:
x = 3 m
[tex]k = 5\:\text{N/m}[/tex]
The spring potential energy [tex]PE_s[/tex] is
[tex]PE_s = \frac{1}{2}kx^2 = \frac{1}{2}(5\:\text{N/m})(3\:\text{m})^2[/tex]
[tex]\:\:\:\:\:\:\:=22.5\:\text{J}[/tex]
Five ramps lead from the ground to the second floor of a workshop, as sketched below. All five ramps have the same height; ramps B, C, D and E have the same length; ramp A is longer than the other four. You need to push a heavy cart up to the second floor and you may choose any one of the five ramps.Assuming no frictional forces on the cart, which ramp would require you to do the least work?
The mechanical advantage of ramp A is greater than others and it will require the least force to move the load to greater distance.
Let the height of the ramp = hLet the length of ramp B, C, D and E = LLet the length of the ramp A = 2LThe mechanical advantage of the ramp is calculated as follows;
[tex]M.A = \frac{L}{h}[/tex]
The mechanical advantage of the ramp B, C, D and E is calculated as;
[tex]M.A = \frac{L}{h} \\\\[/tex]
The mechanical advantage of the ramp A is calculated as follows;
[tex]M.A = \frac{2L}{h} \\\\M.A = 2(\frac{L}{h} )[/tex]
Since the length of the ramp A is greater than other ramps, the mechanical advantage will be greater and it will require the least force to move the load to greater distance.
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Explain why a tiny 1.5 V cell can operate a calculator for a year, while a much larger 1.5 V cell burns out in a few hours in a tiny robot. I WILL CHOOSE BRAINIEST!! PLEASE HELP
The reason why a tiny 1.5 V cell can operate a calculator for a year, but would burn out in a few hours in a tiny robot is because the power demands of the calculator are way less than that of the tiny robot.
What controls how long a cell lasts ?Cell capacity is measured by the amount of Amps and Amp Hour (A.H.) capacity for how long a battery can endure.
On the label or in the user's manual, batteries mention their reserve capability, which specifies the approximate amount of time they can operate between charges. You might notice a shorter or longer battery life if your circuit uses more or less power than this hypothetical circuit. Calculate the battery's entire capacity and divide it by the power of your circuit to see how long it can survive.
Load current, which is influenced by the power of the linked item, influences how quickly the battery's electrical capacity will be used up.
The reason the 1.5 V cell would operate the calculator for a year and yet die in a few hours with the robot is that the robot has a much higher load current than the calculator.
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What are the 7 different states of matter in Chemistry?How many states of matter are there?
Answer:
The 7 states of matter are solid, loquid, gas, fermionoc condensate, quark gluton plasm, bose einetein condensate amd ionised plasm but its usually only 3 they teach you
Answer:
7
Explanation:
solid, liquid,gas,fermionoc condensate,quark glutton plasm,bose einetein condensate amd ionised plasm.
A cubical box with sides of length 0.368 m contains 1.980 moles of neon gas at a temperature of 298 K. What is the average rate (in atoms/s) at which neon atoms collide with one side of the container? The mass of a single neon atom is 3.35x10-26 kg.
The average rate at which the neon atoms collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].
The given parameters;
length of the cubical box, L = 0.368number of moles of the gas, n = 1.98temperature, T = 298 Kmass of the gas, m = 3.35 x 10⁻²⁶ kgThe average kinetic energy of the gas molecules is calculated as follows;
[tex]K = \frac{3}{2} \frac{R}{N_a} T\\\\K = \frac{3 \times 8.314\times 298}{2\times 6.022 \times 10^{23}} \\\\K = 6.17\times 10^{-21} \ J/atoms[/tex]
The average speed of the gas molecules is calculated as follows;
[tex]K = \frac{1}{2}mv_{rms}^2\\\\v_{rms} = \sqrt{\frac{2K}{m} } \\\\v_{rms} = \sqrt{\frac{2\times 6.17 \times 10^{-21}}{3.35\times 10^{-26}} } \\\\v_{rms} = 607 \ m/s[/tex]
The time of collision of the gas molecules with the walls of the container is calculated as follows;
[tex]t = \frac{2d}{v} \\\\t = \frac{2\times 0.368}{607} \\\\t = 0.0012 \ s[/tex]
The average rate at which the gases collide with a single wall out of the 3 identical walls is calculated as follows;
[tex]rate =\frac{1}{3} \times \frac{n \times N_a}{t} \\\\rate = \frac{1.98 \times 6.02 \times 10^{23} \ atoms}{3 \times 0.0012 \ s} \\\\rate = 3.31 \times 10^{26} \ atoms/s[/tex]
Thus, the average rate at which the gases collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].
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Rachel drops a ball from a hot–air balloon while her friend Lisa is watching her from the ground. Which statement about the ball's motion is true from Lisa's point of view?
Assume that there is no air resistance and the hot–air balloon is moving horizontally.
A. The ball drops to the ground along a straight–line path.
B.When the ball lands, the hot–air balloon will be ahead of it.
C. When the ball lands, the hot–air balloon will be behind it.
D. When the ball lands, the hot–air balloon will be directly above it.
Answer:
According Lisa, both the ball and the balloon have the same forward velocity of Vx.
(D) is correct
Please Help
A projectile fired over level ground has an initial total velocity of 41.3 m/s. It is in the air for 5.1 s. What is the x-component of the projectile's initial velocity?
Answer:
Explanation:
In the vertical analysis assuming launch from ground level.
0 = 0 + (41.3sinθ)(5.1) + ½(-9.8)5.1²
(41.3sinθ)(5.1) = ½(9.8)5.1²
(41.3sinθ) = ½(9.8)5.1
sinθ = ½(9.8)5.1/41.3
sinθ = 0.60508...
θ = 37.235°
vx = 41.3cos37.235
vx = 32.881452...
vx = 32.9 m/s
how does the structure of compounds determines the properties of the compounds?
Answer:
The chemical structure of the molecule is responsible for each of these characteristics. The chemical structure is comprised of the bonding angle, the kind of bonds, the size of the molecule, and the interactions that occur among the molecules. Even little changes in the chemical structure of a molecule may have a significant impact on the characteristics of the substance.
Explanation:
Hope it helps:)
What does Newton's third law describe?
A) the tendency of stationary objects to remain at rest.
B) How the force applied to an object is related to a change in its motion.
C) The four fundamental forces of nature.
D) The forces between two objects
Thanks!!
Answer:
What does Newton's third law describe?
D) The forces between two objectsA ball is dropped from an 80.0 m building. What is the ball's velocity after 3.00 s? Use an order-of-magnitude estimation to identify the correct choice.
A. -2.9 m/s
B. -29.4 m/s
C -8.8 m/s
D. -88.3 m/s
Answer:b
Explanation:
-29.4 m/s
The velocity of the ball dropped from 80 m if it reaches the ground within 3 seconds is 26.6 m/s. If it is in midway within this time, then the velocity will be 29.4 m/s.
What is velocity ?Velocity of a moving object is the measure of its distance travelled per unit time. Velocity is a vector quantity having both magnitude and direction. Acceleration is the rate of change in velocity.
Given that, height of the building = 80 m
the ball is moving downwards by acceleration due to gravity g = 9.8 m/s².
Then after 3 seconds, the velocity of the ball is calculated as follows:
velocity = acceleration × time
v = 9.8 m/s² × 3 s = -29.4 m/s
If the ball reaches the ground within the time of 3 s, then, the velocity is:
v = 80 m/3s = 26.6 m/s.
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