We know that
• The mass is m = 30.0 kg.
,• The vertical height is h = 10.0 m.
(a)We have to use the conservation of energy theorem, which states that mechanical energy is constant all the time. Also, halfway down means a height of 5.0 m. It's important to know that at the top the total energy is potential, while halfway is distributed as kinetic and potential, the expression below shows this
[tex]E_{p1}=E_{k1}+E_{p2}[/tex]Then, using the definition of each energy, we have
[tex]mgh_1=\frac{1}{2}mv^2+mgh_2[/tex]Now, we use the given values to find the speed.
[tex]\begin{gathered} \text{mgh}_1=m(\frac{1}{2}v^2+gh_2) \\ gh_1=\frac{1}{2}v^2+gh_2 \\ 9.81m/s^2\cdot10m=\frac{1}{2}v^2+9.81m/s^2\cdot5m \\ 98.1m^2/s^2=\frac{1}{2}v^2+49.05m^2/s^2 \\ 98.1m^2/s^2-49.05m^2/s^2=\frac{1}{2}v^2 \\ 2\cdot49.05m^2/s^2=v^2 \\ v=\sqrt[]{98.1m^2/s^2} \\ v\approx9.9m/s \end{gathered}[/tex]Therefore, the speed of the child halfway down is 9.9 meters per second.(b)In this case, we just have to use as the second height of the equation the magnitude 2.5 meters because that's 3/4 of the way down. So, let's use the same process and expression
[tex]\begin{gathered} gh_1=\frac{1}{2}v^2+gh_2 \\ 9.81m/s^2\cdot10m=\frac{1}{2}v^2+9.81m/s^2\cdot2.5m \\ v=\sqrt[]{2(98.1m^2/s^2-24.53m^2/s^2)} \\ v\approx12.1m/s \end{gathered}[/tex]Therefore, the speed of the child 3/4 of the way down is 12.1 meters per second.A planet has mass M and radius 2R.
a) Derive an expression for the escape speed from the planet.
b) A projectile of mass m is shot directly away from the surface of the planet at ⅓ of the escape
speed from the planet. Derive an expression for the maximum distance from the center of the
planet the projectile reaches, in terms of R. Simplify as far as possible. (Ignore the existence of
all other celestial objects.)
The escape velocity of the planet is [tex]\sqrt{gR\\}\\[/tex] when mass is M and radius is 2R.
In astronomy and space research, escape velocity refers to the speed at which a body can leave a gravitational field without being further accelerated. It is given be the expression ,[tex]v_{e} = \sqrt{2gR}[/tex]
Where, Ve is the escape velocity, g is gravity , and R is the radius of the earth.
If a planet has mass M and radius 2R then an expression for the escape velocity from the planet can be calculated as,
[tex]v_{e} = \sqrt{(2GM)/R[/tex]
Here, G is the gravitational force of the planet therefore Newton's law of gravitational force can be used as,
g = GM/R²
gR²= GM
Thus, with the help of above equations the escape velocity of the planet can be calculated when mass is M and radius is 2R i.e.
[tex]v_{e} = \sqrt{(2GM)/2R[/tex]
[tex]v_{e} = \sqrt{(2GR)/2R[/tex]
[tex]v_{e} = \sqrt{gR[/tex]
Therefore , the escape velocity of planet is equal to the root of gravity g and radius R .
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What is thermodynamics Movement of heatMeasure of heatIt’s magic None of them
To find
What is thermodynamics
Explanation
Thermodynamic study about the transfer of energy to and fro from a system.
Thus thermodynamics is the movement of heat
Conclusion
The correct option is movement of heat
A 7 lb bowling ball accelerates at a rate of 7 m/s2.
According to the given statement The new acceleration is 21 m/s².
The correct option is D.
What is an acceleration defined as?Any process where the velocity varies is referred to as acceleration. There are only two ways to speed up: changing your speed or changing your direction, or changing both. This happens because velocity entails both a speed and a direction.
Briefing:The initial force is F = mass x acceleration
F = 3.17515 kg x 7 m/s²
F = 22.23 N
When the net force is tripled, the new acceleration will be
3 x F =m x a'
3 x 22.23 = 3.17515 x a'
a' = 21.003 m/s²
The new acceleration is 21 m/s². The correct option is d.
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The complete question is -
A 7 lb bowling ball accelerates at a rate of 7 m/s². If the net force on the bowling ball is tripled, what will be its new acceleration?
a. 2.3 m/s²
b. 7 m/s²
c. 14 m/s²
d. 21 m/s²
Calculate the force between charges 4 x 10^-8 C and 1.8 x 10^-6 C if they are 3.5 m apart.
Explanation
Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.
[tex]F=k\frac{q_1\cdot q_2}{r^2}[/tex]where k is the coulomb constant
[tex]k=8.9875\cdot10^9\text{ N }\frac{m^2}{C^2}[/tex]Step 1
let
[tex]\begin{gathered} q_1=4\cdot10^{-8}C \\ q_2=1.8\cdot10^{-6} \\ r=3.5\text{ m} \\ k=8.9875\cdot10^9\text{ N }\frac{m^2}{C^2} \end{gathered}[/tex]now, replace.
[tex]\begin{gathered} F=k\frac{q_1\cdot q_2}{r^2} \\ F=8.9875\cdot10^9\text{ N }\frac{m^2}{C^2}(\frac{4\cdot10^{-8}C\cdot1.8\cdot10^{-6}}{(3.5m)^2}) \\ F=5.2824\cdot10^{-5}\text{ N} \end{gathered}[/tex]I hope this helps you
10. A punter kicks the ball high enough that it spends 5.2 seconds in the air. How fast was it
going when it left his foot?
With the use of first equation of motion, the ball will be going with a velocity of - 51 m/s
What is Velocity ?Velocity is the distance travel in a specific direction per time taken. It is a vector quantity and it is measured in m/s
Given that a punter kicks the ball high enough that it spends 5.2 seconds in the air. To know how fast was it, let us assumed that the initial velocity u of the ball is zero. The ball will be moving under the influence of gravity.
To find the velocity, let us use first equation of motion. That is,
v = u - gt
where
Final velocity v = ?Initial velocity u = 0Acceleration due to gravity g = 9.8 m/s²Time t = 5.2 sSubstitute all the parameters into the equation
v = - 9.8 × 5.2
v = - 50.96 m/s
Therefore, the ball will be going as fast as -51 m/s when it left his foot.
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29. [-/10 Points]DETAILSSERCPWA11 4.WA.010.An object of mass 0.77 kg is initially at rest. When a force acts on It for 2.9 ms It acquires a speed of 15.5 m/s. Find the magnitude (In N) of the average force acting on the object during the2.9 ms time interval.
We have:
m = mass = 0.77 kg
t = time = 2.9 ms = 0.0029 s
vf = final speed = 15.5 m/s
vi= initial speed = 0 m/s
Apply:
• F = m (vf-vi) / t
Replacing:
F = 0.77 (15.5 - 0 )/0.0029
F= 4,155.5 N
why do surface waves compared to body waves have a high amplitude?
You pull a box 23 m horizontally, If the rope tension is 120 N , and if the rope does 2500 J of work on the box, what angle θ does the rope make with the horizontal?
The angle θ made by the rope with the box is equal to 25.17 degrees.
If the rope is pulled at an angle A and the tension in the rope was T, then the rope will have two components of forces,
One Vertical which is TsinA and one along horizontal which is TcosA,
And we know,
W = F.s.cosα
Where,
F is the force applied,
s is the Displacement along force,
α is the angle between them,
We know, the box is moving forward, so the angle between the vertical component of tension and displacement is 90 degrees, so the vertical component of tension will not do any work, and the angle between displacement and the horizontal component is 0 degrees,
So, the work done by the horizontal component is 0 degrees,
So, we cam write,
W = TcosA.D
where D is the displacement of the box, which is given to be 23m,
T is the tension in the string which is 120N,
W is the work done by the rope on the box which is 25000J,
so, putting all the values,
2500 = 120.cosA.23
cosA = 2500/120x23
cos A = 0.9
So, A is 25.17 degrees.
The angle made by the rope is 25.17 degrees,
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A hollow, spherical shell with mass 2.00 kg rolls without slipping down a 33.0 ∘ slope.
Find the acceleration.
Find the friction force.
The acceleration of the spherical shell is determined as 6.37 m/s².
The friction force is 12.74 N.
What is the coefficient of friction of the surface?
The coefficient of friction of the surface is calculated as follows;
Fsinθ = μFncosθ
where;
F is the applied force on the spherical shellFn is the weight of the spherical shellmg(sin θ) = (μmg) cosθ
(sin θ) = μ cosθ
μ = sinθ/cos θ
where;
g is acceleration due to gravityθ is angle of inclination of the slopeμ = tan θ
μ = tan(33)
μ = 0.65
The acceleration of the spherical shell is calculated as;
a = μg
a = 0.65 x 9.8 m/s²
a = 6.37 m/s²
The frictional force is calculated as follows;
Ff = μmg
Ff = 0.65 x 2 x 9.8
Ff = 12.74 N
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A 58 V AC voltage source is connected across a 3.7 H inductor. What is the current through the inductor if the frequency is 63 Hz?
We will have the following:
First:
[tex]x_l=2\pi(63Hz)(3.7H)\Rightarrow x_l=466.2\pi\Omega[/tex]Then:
[tex]\begin{gathered} I=\frac{58V}{466.2\pi\Omega}\Rightarrow I=0.03960097254...A \\ \\ \Rightarrow I\approx0.04A \end{gathered}[/tex]So, the current is approximately 0.04 A.
The process of charging a capacitor consists of transferring charge from the plate at lower ____ to the plate at higher one
The process of charging a capacitor consists of transferring charge from the plate at lower potential to the plate at higher one (higher potential).
What steps are involved in charging a capacitor?When a battery is connected in series with a resistor and capacitor, a large initial current flows when the battery moves charge from one capacitor plate to the other. As the capacitor charges to the battery voltage, the charging current asymptotically decreases until it is zero.
Therefore, a unidirectional flow of electric charge is known as direct current. When the voltage from the power source matches the voltage at the capacitor terminals, the capacitor is fully charged. When the electrical circuit's current stops flowing, the charging phase of the capacitor is complete.
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How do magnification and resolution compare between electron and light microscopes?
magnification and resolution compare between electron and light microscopes by they produce an photograph of a specimen with the aid of using the usage of a beam of electrons instead of a beam of mild.
Electron microscopes range from mild microscopes in that they produce an photograph of a specimen with the aid of using the usage of a beam of electrons instead of a beam of mild. Electrons have a whole lot a shorter wavelength than seen mild, and this lets in electron microscopes to supply higher-decision photos than popular mild microscopes. Electron microscopes may be used to observe now no longer simply complete cells, however additionally the subcellular systems and cubicles inside them.
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Explain how a $10.00 compact fluorescent light bulb (15 W) can be an overall money saver compared to an incandescent light bulb (60 W) that costs $0.50.
Answer:
Explanation:
We would consider the kilowatt hour for both bulbs. It is the energy consumed by an appliance per hour.
Recall, 1 kw = 1000w
15w = 15/1000 = 0.015 kw
60w = 60/1000 = 0.06kw
In 1 hour,
1)
The 15 W bulb would consume 0.015 kw.
For 10 hours per day, the consumption is 0.015 x 10 = 0.15kw
For a month of 30 days, the consumption is 0.15 x 30 = 4.5
Let's assume that the charge per KW is $0.2
Consumption cost per month = 0.2 x 4.5 = $0.9 per month
2)
The 60 W bulb would consume 0.06 kw.
For 10 hours per day, the consumption is 0.06 x 10 = 0.6kw
For a month of 30 days, the consumption is 0.6 x 30 = 18
Let's assume that the charge per KW is $0.2
Consumption cost per month = 0.2 x 18 = $3.6 per month
Thus, the 15W bulb would be an overall money saver as time goes on.
2x6573858493.8404? please
Answer:
13,147,716,987.681
if you multiply 2 by 6573858493.8404, you get 13,147,716,987.681
Help I am so confused on what to do with theses steps to get the answer.
Step 1 convert all the dimensions of the box from cm to m
Step 2 calculate the maximum buoyant force if the box was completely submerged. Fb = pfluid V g
Step 3 calculate the Fg of the box Fg = mg
Step 4 Subtract the Fg from the Fb to get the Force of gravity from extra mass
Step 5. Use the Fg from extra mass =mg equation to solve for mass. Using
The buoyant force if the box was completely submerged is 93.84 N.
The length of the box is 40.8 cm.
The height of the box is 10.4 cm and the depth of the box is 19 cm.
Now we know,
1 m = 100 cm
So,
1 cm = 0.01 m
Therefore,
Length, L = 40.8 cm = 40.8 × 0.01 = 0.408 m
Height, H = 10.4 cm = 10.4 × 0.01 = 0.104 m
Depth, D = 19 cm = 19 × 0.01 = 0.19 m
The density of the fluid is 1187 kg/m³.
ρ = 1187 kg/m³ = 1187 × 10³ g/m³
The mass of the box, m = 2325.4 g
The maximum buoyant force will be:
F = ρgV
Here g = 1 g cc
Therefore,
F = 1187 × 1 × ( 0.408 × 0.104 × 0.19 )
F = 93.846590 N
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A bicycle and rider going 15 m/s approach a hill. Their total mass is 66 kg.
(a) What is their kinetic energy?
7425 J
(b) If the rider coasts up the hill without pedaling, how high above its starting level will the bicycle be when it finally rolls to a stop?
Need help with B Please help!!
a)The kinetic energy is 7425 J.
b) Theight at which the bicycle starts is 1147 m.
For the given problem, we are dealing with kinetic and potential energy, where Kinetic energy is the energy an object has since its movement. On the off chance that we need to accelerate an object, at that point we must apply a force. Applying a force requires us to do work. After work has been done, energy has been transferred to the object, and the object will be moving with a new constant speed,However when its position is modified from its regular equilibrium position, the bow is able to store energy by virtue of its position. This put-away energy of position is alluded to as potential energy. Potential energy is the stored energy of position had by an object.
Since we were given a mass which is 66 kg and a velocity of 15 m/s
Since the formula of kinetic energy is:
K.E= (1/2)mv², where m is the mass of the object and v is the velocity of the object moving .
= 1/2 (66)(15 m/s)²
=7425 J
For the second part the kinetic energy is converted to the potential energy, so the formula for the potential energy is
P.E = mgh, where m is the mass , g is the acceleration due to gravity and h is the height at which the object.
=7425 J =mgh
=>h= 7425/(mg)
=>h =7425/66*15
=>h =11.47 m
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A football punter accelerates a football from rest to a speed of 13 m/s during the time in which his toe is in contact with the ball (about 0.24 s). If the football has a mass of 0.49 kg, what average force does the punter exert on the ball?_____ N
Given:
The initial speed of the football is: vi = 0 m/s (as initially, the football is at rest)
The final speed of the football is: vf = 13 m/s
The time for which the football accelerates is: t = 0.24 s
The mass of the football is: m = 0.49 kg
To find:
The force exerted on the ball
Explanation:
The acceleration of the ball can be determined by using the following equation.
[tex]a=\frac{v_f-v_i}{t}[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} a=\frac{13\text{ m/s}-0\text{ m/s}}{0.24\text{ s}} \\ \\ a=\frac{13\text{ m/s}}{0.24\text{ s}} \\ \\ a=54.17\text{ m/s}^2 \end{gathered}[/tex]The force exerted by the punter on the ball can be determined by using Newton's second law of motion.
According to Newton's second law of motion,
[tex]F=ma[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} F=0.49\text{ kg}\times54.17\text{ m/s}^2 \\ \\ F=26.54\text{ N} \end{gathered}[/tex]Final answer:
The force exerted by the punter on the ball is 26.54 Newtons (N).
A 1800 kg object moving to the east at 13 m/s, collided with a 200 kg object that was moving to the north at 32 m/s. What was the angle of motion
Explain the flow of energy needed for the deer to eat.
Energy flow in an needed for the deer to eat starts at the sun which produces all the energy needed for life. Then comes the producers, the autotrophs(plants). These organisms harness the energy of the sun and convert it into sugars. Then deer which eat the autotrophs get the next most energy.
What is energy flow?One of the key components that enables the survival of such a large number of organisms is the energy flow within the ecosystem. Solar energy is the main source of energy for almost all living things on Earth.
It is interesting to learn that less than 50% of the sun's actual radiation reaches Earth. The radiation that plants can use to perform photosynthesis is the radiation that is considered to be effective.
Through the food chain and food web, energy is transferred. Being the energy producers, plants use their chloroplasts to absorb sunlight, part of which is converted into chemical energy during the process of photosynthesis.
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. A circular loop of wire of area 10 cm2 carries a current of 25 A. At a particular instant, the loop lies in the xy-plane and is subjected to a magnetic field B=(2.0iˆ+6.0jˆ+8.0kˆ)×10−3T. As viewed from above the xy-plane, the current is circulating clockwise. (a) What is the magnetic dipole moment of the current loop? (b) At this instant, what is the magnetic torque on the loop?
The magnetic dipole moment of the current loop is 0.025 Am².
The magnetic torque on the loop is 2.5 x 10⁻⁴ Nm.
What is magnetic dipole moment?The magnetic dipole moment of an object, is the measure of the object's tendency to align with a magnetic field.
Mathematically, magnetic dipole moment is given as;
μ = NIA
where;
N is number of turns of the loopA is the area of the loopI is the current flowing in the loopμ = (1) x (25 A) x (0.001 m²)
μ = 0.025 Am²
The magnetic torque on the loop is calculated as follows;
τ = μB
where;
B is magnetic field strengthB = √(0.002² + 0.006² + 0.008²)
B = 0.01 T
τ = μB
τ = 0.025 Am² x 0.01 T
τ = 2.5 x 10⁻⁴ Nm
Thus, the magnetic dipole moment of the current loop is determined from the current and area of the loop while the magnetic torque on the loop is determined from the magnetic dipole moment.
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What is the magnification of a curved mirror if a 10.0 cm tall object is placed 8.00 cmfrom the mirror and produces an image 12.00 cm in front of the mirror?-0.6671.50-1.500.667
We will have that the magnification will be given by:
[tex]m=-\frac{v}{u}[/tex]That is:
[tex]\begin{gathered} m=-\frac{8cm}{12cm}\Rightarrow m=-\frac{2}{3} \\ \\ \Rightarrow m\approx-0.667 \end{gathered}[/tex]So, the magnification is approximately -0.667.
If the actual release height is 2 h , calculate the normal force exerted by the track at the bottom of the loop.
The normal force exerted by the track at the bottom of the loop is F = m(v²/h + g).
What is the force exerted at the bottom of the loop?
The normal force at the bottom of a loop, is determined from two opposite forces.
One is reaction to weight and other is from reaction of centripetal force which is due to the curvature of the track.
The normal force exerted at the bottom of a circular loop is calculated as follows;
F = ma + mg
where;
a is the centripetal acceleration of the objectm is mass of the objectg is acceleration due to gravityF = m(v²/r) + mg
where;
v is the velocity of the objectr is the radius of the loopif the release height of the object = 2h
then, the radius of the track = 2h/2 = h
F = m(v²/h) + mg
F = m(v²/h + g)
Thus, the normal force exerted by the track at the bottom of the loop depends on the weight of the object and centripetal force.
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What is the average velocity of a car driving down the highway if it’s displacement is 123m west during a time period of 14.0s
The average velocity of a car driving down the highway if it’s displacement is 123 m west during a time period of 14.0 s is 8.79 m/sec.
What is Velocity?The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity. Unit of velocity is m/sec.
Given in the question a car driving down the highway it’s displacement is 123 m west during a time period of 14.0 s then the average velocity is given as,
Average velocity = displacement / time
= 123/14
= 8.79 m/sec
The average velocity of a car driving down the highway if it’s displacement is 123 m west during a time period of 14.0 s is 8.79 m/sec.
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Satellite orbits at a distance from earths center of about 6.6 earth radii and takes 24h to go around once. What distance (in meters) does the satellite travel in one day? What is the orbital velocity (in m/s)?
We know that the radii of the orbit is 6.6 the earth one. To get the distance travel in one day we calculate the circunference of that orbit:
[tex]\begin{gathered} C=(6.6)(6371)(2\pi) \\ C=264,199.14 \end{gathered}[/tex]Therefore the distance travel by the satellite is 264199.14 km per day.
Now, to get the orbital velocity we need to use the equation:
[tex]v=\sqrt[]{\frac{MG}{r}}[/tex]where M is the mass of the object at the center (in this case the earth), G is the gravitational constant and r is the radius of the orbit, then we have:
[tex]\begin{gathered} v=\sqrt[]{\frac{(5.9722\times10^{24})(6.67\times10^{-11})}{(6.6)(6.371\times10^6)}} \\ v=3077.89 \end{gathered}[/tex]Therefore the orbital velocity is 3077.83 m/s.
The diagram shows what happens when a polythene rod is rubbed with a dry cloth. Select the statements that are true
When a polythene rod is rubbed with a dry cloth, the dry cloth loses electrons, the polythene rod gains electrons and the dry cloth is left with positive charge.
The electrons on the dry cloth's surface move to the polyethene rod when it is brushed against a polythene rod. It is considered to be negatively charged because more electrons have accumulated within the polythene rod as a result of the rod being electron rich and the electrons moving there.
The amount of electrons inside the cloth has dropped, making the cloth electron-deficient. It is referred to as positively charged because of this.
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Each set of protons and electrons represents a different atom. Place the atoms in order of their overall charge. Order them from most positive to most negative.swap_vert22 protons, 18 electronsswap_vert12 protons, 10 electronsswap_vert17 protons, 10 electrons
Protons have positive charge
Electrons have negative charge
• 22 protons (+22) + 18 electrons (-18 )
22 - 18 = 4 (positive)
• 12 -10 = 2
• 17 - 10 = 7
• 2 - 1 = 1
From most positive to most negative overall charges.
7 - 4 - 2 - 1
A car of mass m = 1090 kg is traveling down a θ = 11 degree incline. When the car's speed is v0 = 16 m/s, a mechanical failure causes all four of its brakes to lock. The coefficient of kinetic friction between the tires and road is μk = 0.45.
Calculate the distance the car travels down the hill L in meters until it comes to a stop at the end.
The distance travelled by the car along the plane is 29.56 m.
What is the acceleration of the car down the incline?The acceleration of the car down the incline is calculated by applying Newton's second law of motion as shown below;
F - Ff = ma
where;
F is the applied force on the car = 0Ff is the frictional force acting on the car, trying to stop it m is the mass of the cara is the acceleration of the car down the inclinefrictional force acting on the car, Ff = μmg cosθ
0 - μmg cosθ = ma
-μmg cosθ = ma
-μg cosθ = a
where;
μ is coefficient of kinetic frictionθ is the angle of inclination of the planea = -(0.45 x 9.8 x cos11)
a = -4.33 m/s²
The distance travelled by the car along the plane is calculated as follows;
v² = u² + 2aL
where;
L is the length of the incline = distance travelled by the carv is the final velocity of the car when it stops = 0u is the initial velocity of the car = 16 m/sa is the acceleration of the car0 = (16²) + 2(-4.33)L
0 = 256 - 8.66L
8.66L = 256
L = 256/8.66
L = 29.56 m
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Please list two materials that are conductors and two materials that are insulators.
Two materials that are conductors - silver, copper
Two materials that are insulators - rubber, dry wood
A ball is orbits in a circle of radius 10m with a speed 50 m/s. What is its angular velocity?
The angular velocity of the ball orbiting in circle will be 5 rad/s.
What is angular velocity?Angular velocity describes the rate at which an object rotates.
Given is a ball that orbits in a circle of radius 10m with a speed of 50 m/s.
The relation between the linear and angular velocity of a body is given by-
v = rω
where -
v is the linear velocity
r is the radius
ω is the angular velocity
On substituting the values, we get -
50 = 10 x ω
ω = 50/10
ω = 5 rad/s
Therefore, the angular velocity of the ball orbiting in circle will be
5 rad/s.
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. What are the velocities of sound waves in air, helium, and carbon dioxide? Show your data. Which was the fastest? Why do sound waves travel at different speeds in each gas? Explain. Note that the values for helium will be incomplete in these frequency ranges.
Answer:
Taking into account the table we get that, for n = 2, the velocities of sound waves are
Velocity on air: 267.3 m/s
Velocity on helium: 260 m/s
Velocity of carbon dioxide: 829.3 m/s
The velocity of the sound travels at different speeds in each gas because the velocity depends on the properties of the medium like the density. Since the air, helium, and carbon dioxide have different densities, the sound speed will be different.