Answer:
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
Explanation:
Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration ([tex]a[/tex]), measured in meters per square second, is estimated by this kinematic formula:
[tex]a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s }[/tex] (1)
Where:
[tex]\Delta s[/tex] - Travelled distance, measured in meters.
[tex]v_{o}[/tex], [tex]v[/tex] - Initial and final speeds of the spaceship, measured in meters.
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v = 10968\,\frac{m}{s}[/tex] and [tex]\Delta s = 212.8\,m[/tex], then the acceleration experimented by the spaceship is:
[tex]a = \frac{\left(10968\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (212.8\,m)}[/tex]
[tex]a = 282652.782\,\frac{m}{s^{2}}[/tex]
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
A researcher who does not want to manipulate or interfere with the behavior of research subjects would conduct a __________ study.
A.
cross-sectional
B.
naturalistic observation
C.
case
D.
longitudinal
Please select the best answer from the choices provided
A
B
C
D
Answer:
b
Explanation:
A researcher who does not want to manipulate or interfere with the behavior of research subjects would conduct a Naturalistic observation study.
What is observational research?Observational technique in the research field is usually, " watching something attentively in a scientific manner". If we are observing research technique and its phenomena naturally it's our interest to see in our own subjects. The subject will be known only when we observe the things. It's all about understanding, analyzing, predicting and manufacturing in a unique manner.
The types of observation techniques are:
Naturalistic observation.Participant observation.Structured observation.Naturalistic observation:
This is the observation which can be observed naturally and the phenomenon will be interacted only with the researchers not with the participants.
Some of the observational research methods are:
Thematic analysisGrounded theoryHence, Option B is the correct answer.
Learn more about the Naturalistic observation,
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Sam heaves a 16lb shot straight upward, giving it a constant upward acceleration from rest of 35 m/s^2 for 64.0 cm. He releases it 2.20m above the ground. You may ignore air resistance.
(a) What is the speed of the shot when Sam releases it?
(b) How high above the ground does it go?
(c ) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?
Answer:
6.69 m/s
4.483 m
1.42s
Explanation:
Given that:
Initial Velocity, u = 0
Final velocity, v =?
Acceleration, a = 35m/s²
1.) using the relation :
v² = u² + 2as
v² = 0 + 2(35) * 64*10^-2m
v² = 70 * 0.64
v = sqrt(44.8)
v = 6.693
v = 6.69 m/s
B.) height from the ground, h0 = 2.2
How high ball went , h:
Using :
v² = u² + 2as
Upward motion, g = - ve
0 = 6.69² + 2(-9.8)*(h - 2.2)
0= 6.69² - 19.6(h - 2.2)
44.7561 + 43.12 - 19.6h = 0
19.6h = 44.7561 - 43.12
h = 87.8761 / 19.6
h = 4.483 m
C.)
vt - 0.5gt² = h - h0
6.69t - 0.5(9.8)t²
6.69t - 4.9t² = 1.83 - 2.2
-4.9t² + 6.69t + 0.37 = 0
Using the quadratic equation solver :
Taking the positive root:
1.4185 = 1.42s
Suppose it takes 1.4 s for the monkey to catch the coconut and the initial upward speed of the coconut is 2.9 m/s. Assume the acceleration of gravity is 9.8 m/s 2 . Determine the y coordinate of the location where the monkey catches the coconut. Answer in units of m.
Answer:
y-coordinate where the monkey catches the coconut is 13.664 m.
Explanation:
Given;
time taken for the monkey to catch the coconut, t = 1.4 s
initial upward speed of the coconut, Uy = 2.9 m/s
acceleration due to gravity, g = 9.8 m/s²
The y-coordinate where the monkey catches the coconut is calculated as;
[tex]h_y = U_yt +\frac{1}{2} gt^2\\\\h_y = (2.9\times 1.4) +\frac{1}{2} (9.8)(1.4^2)\\\\h_y = 4.06 + 9.604\\\\h_y = 13.664 \ m[/tex]
Therefore, y-coordinate where the monkey catches the coconut is 13.664 m.
Cite an example of an issue showing the power of Media and information to affect change? sentence pls :)
Answer:
The influence of Media and Information over elections, politics and, governance.
Explanation:
Once we used to have politicians travel to interact with citizens and meet them face to face. That is slowly fading away in place of townhalls that take place online.
Video conferencing, video calls, social media apps, and technology now make it easy to connect with a lot of people at once, across various timezones, without changing location.
Once upon a time, the Television media, Radio Media, Cable Television (all categorized as traditional media) used to be the custodian of information, hence, the chief influencers of politics and governance.
Today, the internet has made it possible for individuals or a group of people without much tech-spend to set up and garner a following that has enough power to influence governance.
Online petitions and Wikileaks, the proliferation of terrorist organizations that have strong information and technology as well as social media skills have modified forever, the way public office holders and other stakeholders involved used to look at governance.
All types of mass movement are caused by the force of
Answer:
Mass and Acceleration
Explanation:
The typical Force equation is:
F = ma
where m = mass, and a=acceleration.
Answer:
Gravity is the main force responsible for mass movements. Gravity is a force that acts everywhere on the Earth's surface, pulling everything in a direction toward the center of the Earth
Explanation:
A particle with charge Q and mass M has instantaneous speed uy when it is at a position where the electric potential is V. At a later time, the particle has moved a distance R away to a position where the electric potential is V2 ) Which of the following equations can be used to find the speed uz of the particle at the new position?
a. 1/2M(μ2^2-μ1^2)=Q (v1-v2)
b. 1/2M(μ2^2-μ1^2)^2=Q(v1-v2)
c. 1/2Mμ2^2=Qv1
d. 1/2Mμ2^2=1/4πx0 (Q^2/R)
Answer:
A
Explanation:
Ke = 1/2 MV^2
If a cyclist travels 30 km in 2 h, What is her average speed?
Answer:
15km/h
Explanation:
→ Speed = Distance ÷ Time
30 ÷ 2 = 15km/h
What is the displacement of an object during a specific unit of time.
Answer:
velocity
Explanation:
the displacement of an object during a specific unit of time.
A power source is going to shake a string and create waves 5.0 cm in amplitude. The string has a linear mass density of 4.0x10-2 kg/m and is pulled under a tension of 100 N. The source can deliver a maximum power of 300 W. What is the highest frequency f at which the source can shake without exceeding the maximum power?
Answer:
55.13 Hz
Explanation:
Given that :
Amplitude, A =. 5cm = 0.05m
Density, d = 4 * 10^-2
Tension, T = 100N
Power = 300 W
Recall:
Power = 0.5dA²w²v
v = velocity = √T/d ; w = Angular velocity = 2πf
Rewriting our equation :
P = 0.5 * d * A² * (2πf)² * √T/d
300 = 0.5 *(4*10^-2)* 0.05² * 4π² * f² * sqrt(100/0.04)
300 = 0.5*(4*10^-2) * 0.05² * 4π² * f² * sqrt(2500)
300 = 0.02 * 50 * 0.05² * 39.478417 * f²
300 = 0.0986960 * 0.05² * f²
f² = 300 / 0.0986960
f² = 3039.6355
f = sqrt(189.97722)
f = 55.132889
= 55.13 Hz
In which number are the zeros not significant?
100.0
O 0.0003
O 4.00005
O 1.0004
Answer:
0.0003
Explanation:
In the rules of Sig Figs, all zeros before with decimals are not sigificant. I.E. 0.00000000000000009. Despite how many 0's there are, only the 9 is significant. Zeros before a number is not significant. In 100, only the one is signficant in 100. with a dot at the end, the one and the two zeros are significant. hope this helps.
Answers:
the second option
Explanation:
Un autobús en una autopista lleva una magnitud de la velocidad de 95 km/h, el conductor observa que debido a un derrumbe la carretera está cerrada, en ese instante acciona los frenos, deteniendo el autobús después de recorrer 60 m. a) ¿Cuál es el valor de la aceleración en el autobús?
Answer:
La aceleración del autobús es -5.80 m/s².
Explanation:
Podemos encontrar la aceleración del autobús usando la siguiente ecuación:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
Where:
[tex]v_{f}[/tex]: es la velocidad final = 0 (se detiene al final)
[tex]v_{0}[/tex]: es la velocidad inicial = 95 km/h
d: es la distancia recorrida = 60 m
Por lo tanto, la aceleración es:
[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} = \frac{0 - (95 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2*60 m} = -5.80 m/s^{2} [/tex]
El signo negativo se debe a que el autobús está desacelerando (hasta que se detiene).
Entonces, la aceleración del autobús es -5.80 m/s².
Espero que te sea de utilidad!
What is average acceleration due to gravity on Earth for a 2000 kg boulder, in proper SI units?
Answer:
9.8m/s²
Explanation:
The average acceleration due to gravity on Earth for a 2000kg boulder is 9.8m/s².
Every object on earth is accelerated towards the center by a rate of change of velocity with time value of 9.8m/s².
The acceleration due to gravity on earth is a constant value from places to places.
For other planetary bodies, the value varies and it differs.
But on earth every object is accelerated at 9.8m/s².
Which element has the same number of valence electrons as krypton (Kr)?
fluorine (F)
chlorine (Cl)
neon (Ne)
selenium (Se)
The element that has the same number of valence electrons as krypton (Kr) is known as Neon (Ne). Thus, the correct option for this question is C.
What are Valence electrons?Valence electrons may be defined as the number of electrons present in the outermost shell of any atom during the time of electronic configuration. For example, Sodium has an atomic number of 11, its electronic configuration is 2, 8, and 1. It means that the valence electron on sodium is 1.
Neon and Krypton being the members of the same family which is known as noble gases or inert gases occupy the same valence electrons of eight in their outermost shell. It is group 8A of the periodic table. Apart from Ne, and Kr, other elements like argon, Xenon, and Radon also belong to the same family and occupy the same valence electron except for helium.
Therefore, the element that has the same number of valence electrons as krypton (Kr) is known as Neon (Ne). Thus, the correct option for this question is C.
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Question 7 of 25
A sound wave with a wavelength of 26 m passes through glass, and then it
passes into air, where its wavelength changes to 1.6 m. If the sound wave has
a constant frequency of 220 Hz, what is its approximate speed as it passes
through the air? (The equation for the speed of a wave is v= f *1.)
O A. 3,500 m/s
B. 350 m/s
O C. 5,700 m/s
O D. 570 m/s
Answer: 350 ms
Explanation:
Just took the quiz:)
Answer:
B. 350
Explanation:
You are playing with a yoyo . If Potential energy of the yoyo is 18 J , and the total mechanical energy is 20 J , how much kinetic energy does the yoyo have ?
Answer:
the kinetic energy of the yoyo is 2 J.
Explanation:
Given;
potential energy of yoyo, P.E = 18 J
total mechanical energy, M.E = 20 J
The kinetic energy of yoyo is calculated as;
M.E = K.E + P.E
where;
K.E is the kinetic energy
K.E = M.E - P.E
K.E = 20 J - 18 J
K.E = 2 J
Therefore, the kinetic energy of the yoyo is 2 J.
How is energy transferred when
hitting a nail?
Answer:
kinetic energy
kinetic energy
A primary coil has 360 turns, while secondary has 120 turns in a transformer. The output voltage is 220V. What is the input voltage, and what type of transformer is it
Answer:
550V
Step - Down transformer
Explanation:
Given parameters:
Number of turns in primary coil = 300 turns
Secondary turns = 120 turns
Output voltage = 220V
Unknown:
Input voltage = ?
Type of transformer = ?
Solution:
To solve this problem, we use the expression below:
[tex]\frac{V_{out} }{V_{in} }[/tex] = [tex]\frac{Ns}{Np}[/tex]
So insert the parameters and find Vin;
[tex]\frac{220}{Vin}[/tex] = [tex]\frac{120}{300}[/tex]
120Vin = 220 x 300
Vin = [tex]\frac{220 x 300}{120}[/tex] = 550V
Since the input voltage is greater than the output voltage, this is step - down transformer.
Two children, each with a mass of 25.4 kg, are at fixed locations on a merry-go-round (a disk that spins about an axis perpendicular to the disk and through its center). One child is 0.78 m from the center of the merry-go-round, and the other is near the outer edge, 3.14 m from the center. With the merry-go-round rotating at a constant angular speed, the child near the edge is moving with translational speed of 11.5 m/s.
a. What is the angular speed of each child?
b. Through what angular distance does each child move in 5.0 s?
c. Through what distance in meters does each child move in 5.0 s?
d. What is the centripetal force experienced by each child as he or she holds on?
e. Which child has a more difficult time holding on?
Answer:
a) ω₁ = ω₂ = 3.7 rad/sec
b) Δθ₁ = Δθ₂ = 18.5 rad
c) d₁ = 14.5 m d₂ = 57.5 m
d) Fc1 = 273.9 N Fc2 = 1069.8 N
e) The boy near the outer edge.
Explanation:
a)
Since the merry-go-round is a rigid body, any point on it rotates at the same angular speed.However, linear speeds of points at different distances from the center, are different.Applying the definition of angular velocity, and the definition of angle, we can write the following relationship between the angular and linear speeds:[tex]v = \omega*r (1)[/tex]
Since we know the value of v for the child near the outer edge, and the value of r for this point, we can find the value of the angular speed, as follows:[tex]\omega = \frac{v_{out} }{r_{out} } = \frac{11.5m/s}{3.14m} = 3.7 rad/sec (2)[/tex]
As we have already said, ωout = ωin = 3.7 rad/secb)
Since the angular speed is the same for both childs, the angle rotated in the same time, will be the same for both also.Applying the definition of angular speed, as the rate of change of the angle rotated with respect to time, we can find the angle rotated (in radians) as follows:[tex]\Delta \theta = \omega * t = 3.7 rad/sec* 5.0 sec = 18.5 rad (3)[/tex]⇒ Δθ₁ = Δθ₂ = 18.5 rad.
c)
The linear distance traveled by each child, will be related with the linear speed of them.Knowing the value of the angular speed, and the distance from each boy to the center, we can apply (1) in order to get the linear speeds, as follows:[tex]v_{inn} = \omega * r_{inn} = 3.7 rad/sec * 0.78 m = 2.9 m/s (4)[/tex]
vout is a given of the problem ⇒ vout = 11. 5 m/s
Applying the definition of linear velocity, we can find the distance traveled by each child, as follows:[tex]d_{inn} = v_{inn} * t = 2.9m/s* 5.0 s = 14.5 m (5)[/tex]
[tex]d_{out} = v_{out} * t = 11.5 m/s* 5.0 s = 57.5 m (6)[/tex]
d)
The centripetal force experienced by each child is the force that keeps them on a circular movement, and can be written as follows:[tex]F_{c} = m*\frac{v^{2}}{r} (7)[/tex]
Replacing by the values of vin and rin, since m is a given, we can find Fcin (the force on the boy closer to the center) as follows:[tex]F_{cin} = m*\frac{v_{in}^{2}}{r_{in}} = 25.4 kg* \frac{(2.9m/s)^{2} }{0.78m} = 273.9 N (8)[/tex]
In the same way, we get Fcout (the force on the boy near the outer edge):[tex]F_{cout} = m*\frac{v_{out}^{2}}{r_{out}} = 25.4 kg* \frac{(11.5m/s)^{2} }{3.14m} = 1069.8 N (9)[/tex]
e)
The centripetal force that keeps the boys in a circular movement, is not a different type of force, and in this case, is given by the static friction force.The maximum friction force is given by the product of the coefficient of static friction times the normal force.Since the boys are not accelerated in the vertical direction, the normal force is equal and opposite to the force due to gravity, which is the weight.As both boys have the same mass, the normal force is also equal.This means that for both childs, the maximum possible static friction force, is the same, and given by the following expression:[tex]F_{frs} = \mu_{s} * m* g (10)[/tex]If this force is greater than the centripetal force, the boy will be able to hold on.So, as the centripetal force is greater for the boy close to the outer edge, he will have a more difficult time holding on.How does the magnetic force move throughout the magnet
Answer:
its always active
Explanation:
Your friend, a world-class long jumper, is trapped on the roof of a burning building. His only escape route is to jump to the roof of the next building. Fortunately for him, he is in telephone contact with you, a Physics 131 student, for advice on how to proceed. He has two options. He can jump to the next building by using the long-jump technique where he jumps at 45o to the horizontal. Or, he can take his chances by staying where he is in the hopes that the fire department will rescue him. You learn from the building engineers that the next building is 10 m away horizontally and the roof is 3 m below the roof of the burning building. You also know that his best long-jump distance is 7.9 m . What do you advise him to do
Answer:
y = 7.33 m, x= 3 m, t = 1.608 s
it is still higher than the second building, which indicates that if it jumps it will be saved
Explanation:
Let's use the projectile launch ratios, let's start with the range ratio
R = v₀² sin² 2θ / g
in this case the range is R = 7.9m and the angle is 45º, let's find the initial velocity
v₀² = R g / sin² 2θ
let's calculate
v₀ = [tex]\sqrt{ \frac{ 7.9 \ 9.8}{ 1} }[/tex]
v₀ = 8.80 m / s
Let's find the components of the initial velocity
v₀ₓ = v₀ cos 45 = 8.80 cos 45
[tex]v_{oy}[/tex] = v₀ sin 45 = 8.80 sin 45
v₀ₓ = 6.22 m / s
v_{oy} = 6.22 m / s
To save yourself, you have to be at the same time as the other building or higher.
x = v₀ₓ t
t = x / v₀ₓ
t = 10 / 6.22
t = 1.608 s
let's see how much it has descended in this time
y =y₀ + v_{oy} t - ½ g t²
y = 10+ 6.22 1.608 - ½ 9.8 1.608²
y = 7.33 m
therefore it is still higher than the second building, which indicates that if it jumps it will be saved
what would happen to the surface temperature on Earth if large amounts of carbon dioxide were removed from the atmosphere?
Answer:
If carbon emissions stopped, the oceans catch up with the atmosphere, the Earth's temperature would rise about another 1.1F (0.6C).
Explanation:
The small spheres that are moving through the circuit are the electric current. Current is the flow or movement of electrons. Describe how the current moves in the simulation.
Answer:
When the circuit is connected properly, the current starts flowing from one end of the battery to the other end.
Explanation:
hope this helps!
Answer:
When the circuit is connected properly, the current starts flowing from one end of the battery to the other end.
Explanation:
Edmentum answer.
An empty parallel plate capacitor is connected between the terminals of a 5.12-V battery and charges up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Answer:
V₂ = 2* V₁ = 10.24 V
Explanation:
By definition, the capacitance is given by the following relationship between the charge on one of the plates (for a parallel plate capacitor) and the voltage between them:[tex]C = \frac{Q}{V} (1)[/tex]
Applying (1) to the capacitor, once fully charged and disconnected from the battery, V = V₁ = 5.12 V.Now, we know that for a parallel plate capacitor, the capacitance is independent from the voltage applied, as follows:[tex]C = \frac{\epsilon_{0} * A}{d} (2)[/tex]
where ε₀ = 8.85*10⁻¹² F/m, A is the area of one of the plates, and d is the spacing between plates.If we double the spacing between the plates d, the capacitance will be reduced to half, due to d₂ = 2* d₁Once disconnected from the battery, due to the principle of conservation of the charge, the charge Q must remain constant, i.e.,Q₁ = Q₂ = Q
Since the only variable that can be modified is the voltage V, if the capacitance reduces to the half of the original value, the voltage must be doubled in order to keep C₂ = C₁/2, true.⇒ Q/V₂ = Q/V₁ * 1/2 ⇒ V₂ = 2* V₁ = 2* 5.12 V = 10.24 V
a plane passes over Point A with a velocity of 8,000 m/s north. Forty seconds later it passes over Point B with a velocity of 10,000 m/s north. Which is the plane's acceleration from A to B ?
Acceleration = (change in velocity) / (time for the change)
Change in velocity = (ending velocity) - (starting velocity)
Change in the plane's velocity = (10,000 m/s north) - (8,000 m/s north)
Change in the plane's velocity = 2,000 m/s north
Time for the change = 40 seconds
Acceleration = (2,000 m/s north) / (40 seconds)
Acceleration = 50 m/s² north
The law of ____ states that energy in a system can change forms but can never be created or destroyed.
Answer:
The Law of Conversation
7 x 7 x 7 x.........x 7 = n (30 times)
7 is multiplied 30 times to get a number n. Which of the following will be a factor of n?
A 30 kg male emperor penguin under a clear sky in the Antarctic winter loses very little heat to the environment by convection; its feathers provide very good insulation. It does lose some heat through its feet to the ice, and some heat due to evaporation as it breathes; the combined power is about 12 W. The outside of the penguin's body is a chilly −22∘C, but its surroundings are an even chillier −38∘C. The penguin's surface area is 0.56 m2, and its emissivity is 0.97. What is the rate of energy loss by radiation
Answer:
Rate of energy loss by radiation is 28.31 Watt
Explanation:
Given that;
m = 30 kg
power p = 12 W
emissivity e = 0.97
Surface Area A = 0.56 m²
outside of the penguin's body T = −22°C
surroundings Temperature Ts = -38°C
the rate of energy loss by radiation = ?
Now, using Stefan-Boltzmann law;
P = σeA [ T⁴ - Ts⁴ ]
Stefan's constant σ = 5.67 × 10⁻⁸
so we substitute
P = 5.67 × 10⁻⁸ × 0.97 × 0.56 [ (-22 + 273 k)⁴ - (-38 + 273 k )⁴]
= 3.079944 × 10⁻⁸ [ 919325376]
= 28.31 Watt
the rate of energy loss by radiation is 28.31 Watt
What is displacement?
a. The distance an object travels.
b. The distance between the starting point and the ending point of an object's
journey.
C. The amount of time it takes an object to travel to a destination.
d. The path in which an object travels.
Answer:
displacement is the distance between the starting point and the ending point of an object's journey
An 8.0-kg object moving with an initial velocity of 8.0 m/s on a surface comes to rest due to friction after it travels a horizontal distance of 11 m. What is the coefficient of kinetic friction between the object and the surface
Answer:
μk = 0.3
Explanation:
According the work-energy theorem, the net work done on an object, is equal to the change in the kinetic energy of the object.In this case, assuming that the object was not accelerating when moving at 8.0 m/s, the only force doing work on it, is the kinetic friction force, acting through the 11 m distance before causing the object to come to rest.The kinetic friction force, is given by the following expression:[tex]f_{frk} = \mu_{k} * F_{n} (1)[/tex]
where μk = coefficient of kinetic friction, Fn = normal force.Assuming that the surface is horizontal, since the normal force is always perpendicular to the surface, and the object is not accelerating in the vertical direction, this means that the normal force must be equal and opposite to the force that gravity exerts on the object, as follows:[tex]F_{n} = m*g (2)[/tex]
The friction force and the horizontal displacement produced by it have opposite directions, so the angle between the force and the displacement is 180º.So, we can express the work done by the kinetic friction force, as follows:[tex]W_{ffr} = -F_{fr} * d = - \mu_{k}* m*g*d (3)[/tex]
We have already said that (3) must be equal to the change in the kinetic energy, ΔK, as follows:[tex]\Delta K = K_{f} - K_{o} (4)[/tex]
Since the object comes to rest, Kf = 0.Replacing K₀ = 1/2*m*v₀² in (4), we have:[tex]\Delta K = 0 - \frac{1}{2} *m * v_{o} ^{2} = - \frac{1}{2} *m * v_{o} ^{2} (5)[/tex]
From (3) and (5), since they are equal each other, we get:[tex]W_{ffr} = \mu_{k}* g*d = \frac{1}{2} * v_{o} ^{2} (6)[/tex]
Replacing by the givens, and solving for μk, we get:[tex]\mu_{k} = \frac{1}{2} * v_{o} ^{2} *\frac{1}{g*d} =\frac{(8.0m/s)^{2}}{2*9.8m/s2*11m} = 0.3 (7)[/tex]
Earth has a mass of 5.98 1024 kg and a radius of 6.38 106 m. Assume it is a uniform solid sphere. The distance of Earth from the Sun is 1.50 1011 m. (Assume Earth completes a single rotation in 24.0 hours and orbits the Sun once every 365 Earth days.) (a) What is the rotational kinetic energy of Earth on its axis
Answer:
Explanation
Rotational kinetic energy of the earth = 1/2 Iω²
where I is moment of inertia of the earth and ω is angular velocity .
I = 2/5 m R² , m is mass and R is radius of the earth .
I = 2/5 x 5.98 x 10²⁴ x ( 6.38 x 10⁶ )²
=97.36 x 10³⁶
ω = 1 / T
T = 24 x 60 x 60 = 86400 s
ω = 1 / 86400
= 11.57 x 10⁻⁶ rad / s
Rotational Kinetic energy = 1/2 Iω²
= .5 x 97.36 x 10³⁶ x (11.57 x 10⁻⁶ )²
= 6516.54 x 10²⁴ J