In order to solve this equation, we will need to use the formula
[tex]f=\frac{c}{\lambda}[/tex]where f = frequency, c is the speed of light and lamda is wavelength
c = 3x10^8 m/s
lamda = 2.55x10^-3 m
f = (3x10^8)/(2.55x10^-3) = 1.18x10^11 1/s
The average radius of Earth is 6,371 km. If the average thickness of oceanic crust is 7.5 km and the average thickness of continental crust is 35 km, what fraction of Earth's radius is each type of crust? Please show how to solve .
If the average thickness of the oceanic crust is 7.5 km and the average thickness of the continental crust is 35 km, then the fraction of the Earth's radius is each type of crust would be 0.00117720922 and 0.005493 respectively.
What is the percentage of a number?It is the relative value that represents the hundredth part of any number for example 2% of any number represents, 2 multiplied by the 1/100th of that number.
As given in the problem average thickness of the oceanic crust is 7.5 kilometers and the average thickness of the continental crust is 35 km,
The fraction of the earth's radius as oceanic crust = 7.5 / 6371
= 0.00117720922
The fraction of the earth's radius as continental crust = 35 / 6371
= 0.005493
Thus, the fraction of the Earth's radius is each type of crust would be 0.00117720922 and 0.005493 respectively.
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Example of a balanced force
what is the mass on grams of 0.56 moles of NaCl
Answer:
1 mole of Na = 23 g
1 mole of Cl = 35 g
1 mole of NaCl = 58 g
.56 * 58 g = 32.5 g
A basketball player jumps for a rebound and reaches a maximum height of 1.5 m. with what speed did he jump off the floor? How long was he in the air?
The final speed of the player can be given as,
[tex]v^2=u^2-2gh[/tex]At the maximum height, the final speed of player is zero.
Plug in the known values,
[tex]\begin{gathered} (0m/s)^2=u^2-2(9.8m/s^2)(1.5\text{ m)} \\ u^2=29.4m^2s^{-2} \\ u=5.42\text{ m/s} \end{gathered}[/tex]Thus, the speed with which it jump off the floor is 5.42 m/s.
The time for which the player was in player is,
[tex]t=\sqrt[]{\frac{2h}{g}}[/tex]Plug in the known values,
[tex]undefined[/tex]A 228-turn, 24.506-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 27 degrees away from vertical increases from 0.807 T to 4.68 T in 13.843 s. Determine the emf induced in the coil.
Given:
• Number of turns, N = 228
,• Diameter, d = 24.506 cm
,• θ = 27 degrees
,• Initial Magnetic field, B1 = 0.807 T
,• Final, B2 = 4.68 T
,• Time , t = 13.843 s
Let's find the induced emf in the coil.
To find the induced EMF, apply Faraday's law:
[tex]\begin{gathered} E=N\frac{d}{dt}(B*A) \\ \\ E=N*Acos\theta\frac{d}{dt}(B) \\ \\ E=N*(\pi r^2)cos\theta(\frac{B_2-B_1}{t}) \end{gathered}[/tex]Where:
A is the area in meters.
Rewrite the diameter from cm to meters.
Where:
100 cm = 1 meters
24.056 cm = 0.24506 m
Now, the radius will be:
radius = diameter/2 = 0.24506/2 = 0.12253 m
Now, plug in the values and solve for E:
[tex]\begin{gathered} E=228*(\pi *(0.12253)^2)cos(27)*(\frac{4.68-0.807}{13.843}) \\ \\ E=228*0.0471666*cos(27)*0.27978 \\ \\ E=2.6\text{ volts} \end{gathered}[/tex]Therefore, the EMF induced in the coil is 2.6 volts.
ANSWER:
2.6 v
if you had only one telescope and wanted to take both visible-light and ultraviolet pictures of stars, where should you locate your telescope?
If we just had one telescope and wanted to photograph stars in both visible and ultraviolet light, we should put it in space.
While visible light is observable from Earth, ultraviolet light can only be seen from space. Indeed, Hubble's ability to observe ultraviolet light gives it a major advantage over larger ground-based observatories.
Rank the visible light hues from left to right according to the altitude in the atmosphere where they are totally absorbed. All visible light wavelengths reach the Earth's surface, which is why we can see all colors and why visible-light telescopes perform well on the ground.
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5. Draw a transverse wave with two wavelengths and label amplitude, crest, trough, and
equilibrium position.
A wave is considered to be transverse if its oscillations run counterclockwise to the wave's direction of advance. A longitudinal wave, on the other hand, moves in the direction of its oscillations. Transverse waves include water waves.
A waveform signal's wavelength, which is the distance between two identical locations (adjacent crests) in the succeeding cycles, determines whether it is sent through space or via a wire. This length is typically defined in wireless systems in metres (m), centimetres (cm), or millimetres (mm).
The largest displacement or distance made by a point on a wave or vibrating body relative to its equilibrium position is its amplitude. It is equal to one-half of the vibration path.
The crest and trough of a wave, respectively, are its highest and lowest surface portions. The wave height is the vertical distance between the peak and trough. The wavelength is the horizontal separation between two consecutive crests or troughs.
The horizontal line at the wave's centre stands in for balance. A period is the length of time it takes to complete a cycle, which includes travelling from one peak to another, from one trough to another, or from one equilibrium point to another (both equilibrium points same direction).
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Sodium has a density of 1.95 g/cm3. What is the volume of 56.2 g of sodium?
Answer: couldn't type in that little 3 haha
Explanation:
A card is drawn at random from a standard deck. Determine whether the events are mutually exclusive or not mutually exclusive. Then find each probability (2 or black card).
Consider that the event are no mutually exclusive because you can obtain a black card and 2 in one card.
In a standard deck you have 52 cards. There are 26 black cards and 4 cards with number 2.
Consider that in the 26 black cards there are two black cards with number 2.
Then, to get the probability consider that the required result can be obtained for 26 + 2 = 28 cards (26 black cards and two cards with number 2).
The probability is the quotient between the number of options over the number of cards:
[tex]p=\frac{28}{52}=0.54[/tex]Hence, the probability is 0.54
Austin does his Power lifting every morning to stay in shape. He lifts a 90 kg barbell, 2.3 m above the ground.a) How much energy does it have when it was on the ground? Jb)How much energy does it have after being lifted 2.3 m? Jc) What kind of energy does it have after being lifted? d) How much work did Austin do to lift the barbell? Je) If he lifted it in 1.9s, what was his power? W
ANSWER
[tex]\begin{gathered} (a)0J \\ (b)2030J \\ (c)\text{ Potential energy} \\ (d)2030J \\ (e)1070W \end{gathered}[/tex]EXPLANATION
Parameters given:
Mass of barbell, m = 90 kg
Height above ground, h = 2.3 m
(a) We want to find the energy the barbell has on the ground. #
When it is on the ground, the barbell is stationary, which means its velocity is 0 m/s, hence, its kinetic energy is also 0 J, since kinetic energy is given as:
[tex]\begin{gathered} KE=\frac{1}{2}mv^2 \\ KE=\frac{1}{2}\cdot m\cdot0=0J \end{gathered}[/tex]Also, on the ground, it is at a height of 0 m, hence, its potential energy is 0 J:
[tex]\begin{gathered} PE=mgh \\ PE=m\cdot g\cdot0=0J \end{gathered}[/tex]where g = acceleration due to gravity
Therefore, on the ground, the energy the barbell had was 0 J.
(b) After it had been lifted 2.3 m, its height above the ground became 2.3 m.
Now, we can find the potential energy possessed by the barbell:
[tex]\begin{gathered} PE=90\cdot9.8\cdot2.3 \\ PE=2028.6J\approx2030J \end{gathered}[/tex]After it is lifted, it is once again stationary, hence, it has no kinetic energy.
Therefore, the energy the barbell has after it has been lifted 2.3 m is 2070J.
(c) As stated in (b) above, after being lifted, the barbell only possesses potential energy since it is at a height above the ground and it is not moving.
(d) The work done in lifting the barbell is equal to the force applied multiplied by the height moved by the barbell.
That is:
[tex]W=F\cdot d[/tex]The force applied is equal to the weight of the barbell:
[tex]\begin{gathered} F=W=mg \\ F=90\cdot9.8 \\ F=882N \end{gathered}[/tex]Therefore, the work done is:
[tex]\begin{gathered} W=882\cdot2.3 \\ W=2028.6J\approx2030J \end{gathered}[/tex](e) He lifted the barbell in 1.9 seconds. To find his power, we have to divide the work done by the time taken to do the work.
That is:
[tex]\begin{gathered} P=\frac{W}{t} \\ P=\frac{2030}{1.9} \\ P=1068.4W\approx1070W \end{gathered}[/tex]That was his power.
What is troubling about the circumstance when a coil is stationary and a magnet moves as compres to when a magnet is stationary and the coil moves?
When the magnet is moved, the galvanometer needle will deflect. It shows that the current is flowing in the coil. When the magnet moves into the coil, the needle deflects into one way, and if the magnet moves out of the coil, the needle deflects into the other way.
When the magnet is held stationary near, or even inside, the coil, no current will flow through the coil.
Determine the speed of the Earth in its motion around the Sun using Newton's Law of Universal Gravitation and centripetal force. Look up the values of the Earth's mass, the Sun's mass, and the average distance of Earth from the Sun; other than G, nothing else is needed
In order to determine the speed of the Earth, proceed as follow:
Consider that the centripetal force must be equal to the gravitational force between the Earth and the Sun (because guarantees the stability of the system):
[tex]F_g=F_c[/tex]Fg is the gravitational force and Fc the centripetal force. The expressions for each of these forces are:
[tex]\begin{gathered} F_g=\text{G}\frac{\text{mM}}{r^2} \\ F_c=ma_c=m\frac{v^2}{r} \end{gathered}[/tex]where,
G: Cavendish's constant = 6.67*10^-11 Nm^2/kg^2
m: Earth's mass = 5.97*10^24 kg
M: Sun's mass = 1.99*10^30kg
v: speed of Earth around the Sun = ?
r: distance between the center of mass of Earth and Sun = 1.49*10^8km = 1.49*10^11 m
Equal the expressions for Fg and Fc, solve for v, replace the previous values of the parameters and simplify:
[tex]\begin{gathered} \text{G}\frac{\text{mM}}{r^2}=m\frac{v^2}{r} \\ v^{}=\sqrt[]{\frac{GM}{r}} \\ v=\sqrt[]{\frac{(6.67\cdot10^{-11}N\frac{m^2}{\operatorname{kg}^2})(1.99\cdot10^{30}kg)}{1.49\cdot10^{11}m}} \\ v\approx29846.7\frac{m}{s} \end{gathered}[/tex]Hence, the speed of the Earth around the Sun is approximately 29846.7m/s
If a copper wire with a 0.300 mm diameter is to have a resistance of 0.500 Ω at 20.0ºC, how long should it be? The resistivity of copper is 1.68 * 10^-8 Ω*m
Given:
• Diameter = 0.300 mm
,• Resistance = 0.500 Ω
,• Temperature = 20.0ºC
,• Resistivity = 1.68 * 10⁻⁸ Ω*m
Let's find the length of the copper wire.
To find the length of the wire, apply the formula:
[tex]\begin{gathered} R=\frac{\rho L}{A} \\ \\ R=\frac{\rho L}{\pi r^2} \end{gathered}[/tex]Where:
R is the resistance = 0.500 Ω
p is the resistivity = 1.68 * 10⁻⁸ Ω*m
L is the length
r is the radius = diameter/2 = 0.300mm/2 = 0.15 mm
Convert the radius to meters, where:
1 m = 1000 mm
0.15mm = 0.15 x 10⁻³ m
Now, input values into the formula and solve fo L:
[tex]0.500=\frac{1.68\times10^{-8}\times L}{\pi\times(0.15\times10^{-3})^2}[/tex]Solving further, rewrite the equation for the length L:
[tex]\begin{gathered} L=\frac{0.500\pi\times(0.15\times10^{-3})^2}{1.68\times10^{-8}} \\ \\ L=\frac{3.534\times10^{-8}}{1.68\times10^{-8}} \\ \\ L=2.1\text{ m} \end{gathered}[/tex]Therefore, the length of the copper wire is 2.1 meters.
ANSWER:
2.1 m
What physical property is described in the following statement?: The asphaltis smooth.O A. TextureO B. ShapeO C. HardnessO D. SizeSUBMIT
A. Texture
Texture is the feel or appearence of a surface.
Modern roller coasters have vertical loops like the one shown in the figure. The radius of curvature is smaller at the top than on the sides so the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats.
1. What is the speed of the roller coaster, in meters per second, at the top of the loop if the radius of curvature there is 14 m and the downward acceleration of the car is 1.1g? Note that g here is the acceleration due to gravity.
2. The beginning of this roller coaster is at the top of a high hill. If it started from rest at the top of this hill, how high, in meters, above the top of the loop is this initial starting point? You may assume there is no friction anywhere on the track.
3. If it actually starts 7.5 m higher than your answer to the previous part (yet still reaches the top of the loop with the same velocity), how much energy, in joules, did it lose to friction? Its mass is 1800 kg.
1 ) The speed of the roller coaster = 12.28 m / s
2 ) Height of the hill above the top of the loop = 7 m
3 ) Energy lost due to friction = 132 KJ
1 ) The speed of the roller coaster,
[tex]a_{c}[/tex] = v² / r
[tex]a_{c}[/tex] = Centripetal acceleration
v = Linear velocity
r = Radius
r = 14 m
[tex]a_{c}[/tex] = 1.1 g = 1.1 * 9.8
[tex]a_{c}[/tex] = 10.78 m / s²
v² = [tex]a_{c}[/tex] * r
v² = 10.78 * 14
v² = 150.9
v = 12.28 m / s
2 ) Initial starting point,
Considering hill as 1 and the loop as 2,
v1 = 0
h2 = 2 r = 2 * 14
h2 = 28 m
∑ [tex]F_{y}[/tex] = m [tex]a_{c}[/tex]
[tex]F_{N}[/tex] + [tex]F_{g}[/tex] = m [tex]a_{c}[/tex]
0 + m g = m v2² / r
v2² = g r
According to law of conservation of energy,
E1 = E2
m g h1 + 1 / 2 m v1² = m g h2 + 1 / 2 m v2²
m g h1 + 0 = 28 m g + 1 / 2 m g r
h1 = 28 + 1 / 2 ( 14 )
h1 = 35 m
Height of the hill above the top of the loop = h1 - h2
Height of the hill above the top of the loop = 35 - 28
Height of the hill above the top of the loop = 7 m
3 ) Energy lost due to friction,
h1 = 35 + 7.5
h1 = 42.5 m
m = 1800 kg
v2² = g r
v2² = 9.8 * 14
v2² = 137.2 m / s
Since energy is lost due to friction,
KE1 + U1 = KE2 + U2 + W
0 + m g h1 = 1 / 2 m v2² + m g h2 + W
( 1800 * 9.8 * 42.5 ) = ( 0.5 * 1800 * 137.5 ) + ( 1800 * 9.8 * 28 ) + W
749700 = 123750 + 493920 + W
W = 749700 - 617670
W = 132030 J
W = 132 KJ
Therefore,
1 ) The speed of the roller coaster = 12.28 m / s
2 ) Height of the hill above the top of the loop = 7 m
3 ) Energy lost due to friction = 132 KJ
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Your engineering team has created a 46.9 kg spider robot that moves along a strand of web for
Halloween. The spider begins at rest and moves straight down the strand increasing its speed
at a constant rate. It covers 3.04 m in a time of 6.08 s. What is the Tension, in Newtons, in
the strand of web?
The fabric of the spiderwebs is so thin that it adheres easily to the hook side of a Velcro strip. Push Velcro onto the object you want to attach your webs to, then remove the adhesive backing. Grab a few of the strands and put them onto the Velcro to keep the webs in place.
Explain about the web for Halloween?A black spider spread between two slices of buttered bread is said to offer a witch great power. Spiders were rumoured to help witches cast charms. Old customs claim that if you see a spider on Halloween, the ghost of a departed loved one is purportedly watching over you.
The spiderwebs stick in the hook side of a Velcro strip with ease because they are constructed of a very thin fabric. When you wish to attach your webs to something, push Velcro onto the surface and peel off the sticky backing. To ensure that the webs stay in place, grab a handful of the strands and push them onto the Velcro.
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What is the displacement of the particle in the time interval 7 seconds to 8 seconds?OA. O metersОВ.1.5 metersOC. 3 metersOD. 7 meters
Given,
A velocity-time graph.
The area under the curve of a velocity-time graph of an object gives us the displacement of the object.
From the graph, we can see that the area under the curve from 7 seconds to 8 seconds is a triangle.
The height of the triangle is h=6 m/s.
And the base of the triangle is b=1 s.
The area of a triangle is given by,
[tex]A=\frac{1}{2}bh[/tex]On substituting the known values,
[tex]\begin{gathered} A=0.5\times6\times1 \\ =3\text{ m} \end{gathered}[/tex]Therefore the displacement of the particle in the time interval 7 s to 8 s is 3 meters.
Thus, the correct answer is option C.
You exert a force of 5.3 N on a book to slide it across a table. If you do 2.5 J of work in the process, how far did the book move?
We will have the following:
[tex]2.5J=5.3N\cdot x\Rightarrow x=\frac{2.5J}{5.3N}[/tex][tex]\Rightarrow x=\frac{25}{53}m\Rightarrow x\approx0.47m[/tex]So, the book moved 25/53 meters, that is approximately 0.47 meters.
A car travels 400 km in the first 4.5 hours of a trip. It stops for an hour and then travels final 300 km in 2.5 hours. Find the average speed of the car.
Given data:
Distance traveled by car in t_1=4.5 hr is s_1=400 km.
Distance traveled by car in t_2=1 hr is s_2=0 km (as the car was stopped).
Distance traveled by car in t_3=2.5 hr is s_3=300 km.
The average speed is given as,
[tex]\begin{gathered} v_{avg}=\frac{\text{ total distance traveled}}{\text{total time taken}} \\ =\frac{s_1+s_2+s_3}{t_1+t_2+t_3} \end{gathered}[/tex]Substitute all known values,
[tex]\begin{gathered} v_{avg}=\frac{(400\text{ km})+(0\text{ km})+(300\text{ km})}{(4.5\text{ hr})+(1\text{ hr})+(2.5\text{ hr})} \\ =87.5\text{ km/h} \end{gathered}[/tex]Therefore, the average speed of the car is 87.5 km/h.
As the speed of an object falling toward Earth increases, the gravitational potential energy of the object with respect to EarthA. IncreasesB. DecreasesC. Remains the same
Answer:
B. Decreases
Explanation:
When an object is falling toward Earth, the height of the object decreases, and the speed increases. Then, the gravitational potential energy decreases, and the kinetic energy increase because the first one depends on the height and the second one depends on the speed. Therefore, the answer is:
B. Decreases
g) 0.35 oz to mgh) 75 mL to gali) 54 mi to kmj) 1789 ft to km
In order to covnert the given quantities, use the correct covnersion factor, as follow:
g) 0.35 oz to mg
[tex]0.35oz\cdot\frac{28.3495g}{1oz}\cdot\frac{1000mg}{1g}=9922.325mg[/tex]h) 75 mL to gal
[tex]75mL\cdot\frac{1L}{1000mL}\cdot\frac{0.2641722gal}{1L}\approx0.02gal[/tex]i) 54 mi to km
[tex]54mi\cdot\frac{1.61\operatorname{km}}{1mi}=86.94\operatorname{km}[/tex]j) 1789 ft to km
[tex]1789ft\cdot\frac{0.3048m}{1ft}\cdot\frac{1\operatorname{km}}{1000m}\approx0.54\operatorname{km}[/tex]A fireman standing on a 14 m high ladderoperates a water hose with a round nozzle ofdiameter 2.65 inch. The lower end of the hose(14 m below the nozzle) is connected to thepump outlet of diameter 3.49 inch. The gaugepressure of the water at the pump isCalculate the speed of the water jet emerg-ing from the nozzle. Assume that water is anincompressible liquid of density 1000 kg/m3and negligible viscosity. The acceleration ofgravity is 9.8 m/s?Answer in units of m/s.
Given data,
The height, H = 14 m
The diameter, D = 2.65 inch
The gauge pressure, P = 317.84 kPa
We need to calculate the speed of the water jet emerging from the nozzle.
Using Bernoulli's equation,
[tex]\begin{gathered} \frac{1}{2}\rho(v^2_n-v^2_p)=P_{gauge\text{ }}-\rho gh \\ (v^2_n-v^2_p)=(\frac{2}{\rho})P_{gauge}-2gh \\ v^2_n-(\frac{A_n}{A_p})^2v^2_n=(\frac{2}{\rho})P_{gauge}-2gh \\ v^2_n-(\frac{r_n}{r_p_{}})^4v^2_n=(\frac{2}{\rho})P_{gauge}-2gh \end{gathered}[/tex]Further solved as,
[tex]\begin{gathered} v_n=\sqrt[]{\frac{(\frac{2}{\rho})P_{gauge}-2gh}{1-(\frac{r_n}{r_p})^4}} \\ v_n=\sqrt[]{\frac{(\frac{2}{1000})\times317.84\times10^3-2\times9.8\times14}{1-(\frac{1.325_{}}{1.745_{}})^4}} \\ v_n=\sqrt[]{\frac{635-274.4}{0.667}} \\ v_n=\sqrt[]{540.62} \end{gathered}[/tex]Thus, the speed of the water jet is
[tex]v=23.25\text{ m/s}[/tex]The velocity-time table represents the motion of a rightward-
moving motorcycle.
Magnitude =
Time (s)
0.0
0.5
1.0
1.5
2.0
What is the magnitude (i.e., value) and direction of the
acceleration?
Direction =
Velocity (m/s)
24.0, right
22.0, right
20.0, right
18.0, right
16.0, right
m/s/s
(No -
sign.)
(Tap field to change.)
NE
The direction of motion of a body or object depends on its velocity. Speed can be thought of as a scalar quantity in its simplest form. In essence, velocity is a vector quantity. It is the speed at which distance is changing.
What is velocity?Galileo Galilei, an Italian physicist, is credited with being the first to calculate speed by dividing it by the required amount of time and the distance traveled.
Speed, according to Galileo, is the distance traveled in a set amount of time.
The speed at which an object is moving is referred to as its velocity.
Examples of fast motion include an automobile driving north on a highway or a rocket blasting off.
The equation v = u + at, where v signifies the ultimate speed, can be used to express an object's final velocity, which is equal to its starting velocity plus acceleration times the distance traveled.
Therefore, the direction of motion of a body or object depends on its velocity. Speed can be thought of as a scalar quantity in its simplest form. In essence, velocity is a vector quantity. It is the speed at which distance is changing.
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ine? Bir10. Two people are pulling on opposite ends of a rope so that ithas a tension of 150 newtons. If the rope is not moving, withwhat pulling force is each of the two people pulling?
ANSWER
150 N
EXPLANATION
The rope is not moving, so the net force on it is 0. The force that each person exerts on the rope is equal and opposite to the tension on the rope, so the sum of the forces acting on it is zero.
Hence, the force that each of the two people is exerting on the rope while pulling is 150 Newtons.
Identify as many different ways as you can for giving energy to a basketball? (Select all that apply)
To answer this question we need to remember each kind of energy:
• Potential energy is the energy held by an object because of its position relative to other objects.
,• Kinetic energy is the energy held by an object due to is motion.
,• Internal energy is the energy due to the movement of the molecules of the object.
With this in mind we conclude that the following are ways of giving energy to a basketball:
• You can give a basketball kinetic energy by pushing on it with your hand, as in throwing or dribbling.
• You can give a basketball kinetic energy by spinning it on your finger.
• You can give a basketball potential energy by lifting it upward with your hand, as when shooting a free throw.
,• You can give a basketball internal energy by heating it.
which of the following are independent of the mass of an object falling freely near earth's surface: (may have more than 1 answer) 1) acceleration of the object 2) gravitational force acting on the object 3) gravitational force acting on the object 4) magnitude of the gravitational field
As the object is falling freely, the acceleration of the object will be equal to the acceleration due to gravity.
It is given as,
[tex]g=\frac{GM}{R^2}[/tex]Here, G is the univarshal gravitational constant and M is the mass of the Earth.
means acceleration of the object is constant and independent of the mass of the object.
so option 1 is correct.
now the gravitational force on that object is,
[tex]F=\frac{GMm}{R^2}[/tex]here this is dependent on the mass of the object(m).
NOw the gravitational field means the force per unit mass and is given by,
[tex]E=\frac{GM}{R^2}[/tex]Here we can se that this gravitational field is also independent of the mass of the object.
So, option 1 and 4 are correct.
Energy transformations always produce a wasteful amount of energy called ?
Energy transformation always produce a wasteful amount of energy called heat energy.
Hence, the answer is heat energy.
A dentist causes the bit of a high speed drill to accelerate from an angular speed of 1.76 x 10^4 rads to an angular speed of 4.61 x 10^4 rat. In the process, the bit turns through 1.97 x 10 ^4 rad. Assuming a constant angular acceleration, how long would it take the reach its maximum speed of 7.99 x 10^4 rads starting from rest?
The time taken for the bit to reach the maximum speed is 1.35 seconds.
What is the angular acceleration of the bit?The angular acceleration of the bit is determined by applying the following kinematic equation as shown below.
ωf² = ωi² + 2αθ
where;
ωf is the final angular speedωi is the initial angular speedθ is the angular displacementα is the angular accelerationα = (ωf² - ωi²)/2θ
α = (46,100² - 17,600²) / (2 x 19,700)
α = 46,077.4 rad/s²
The time taken for the bit to reach the maximum speed is calculated as follows;
ωf = ωi + αt
t = (ωf - ωi) / α
t = (79,900 - 17,600) / (46,077.4)
t = 1.35 seconds
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If Hubble's constant had a value of 95 km/s/Mpc, what would be the age of the Universe?
The age of universe will be 0.0103 million years.
Hubble's law states that the rate at which any galaxy is receding from another galaxy is proportional to its distance from the galaxy. In simple form,
v = H₀ × d
where v is the velocity of the galaxy, d is its distance, and H₀ is the Hubble's constant.
Given: Hubble's constant, H₀ = 95 km/s/Mpc.
The term 1 / H₀ is called the Hubble's time and gives the age of the universe. That is, if 't' is the age of the universe, then
t = 1 / H₀.
Substitute the given value.
⇒ t = 1 / (95 km/s/Mpc)
(1 pc = 3.086 ×[tex]10^{13}[/tex] km)
⇒ t = 1 / (95/3.086 ×[tex]10^{13}[/tex] ) s
⇒ t = (1 / 30.784) ×[tex]10^{13}[/tex] s
⇒ t = 0.0325 ×[tex]10^{13}[/tex] s
(1 year = 3.154 ×[tex]10^{7}[/tex] s)
⇒ t = 10305.68 years
⇒ t = 0.0103 million years
Therefore, the age of universe will be 0.0103 million years.
Learn more about Hubble's law here, https://brainly.com/question/13705068.
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A person is at the top of a slide with 200 J of potential energy. Explain what happens to the energy as they slide down.
Answer:
the person changes from potential energy , to kinetic energy back to potential