Answer:
m ω^2 R = centripetal force
G M m / R^2 = gravitational force
ω^2 = G M / R^3 for the forces to be equal
d (density) = M / V = M / (4/3 R^3) = 3/4 M / R^3
or M / R^3 = 4 d / 3
Then
ω^2 = G 4 d / 3
d = 3.7 g/cm^2 = 3700 kg/m*3 converting
ω^2 = 6.67E-11 * 3700 * 4 / 3 = 32.9E-8
ω = 5.74E-4
P = 1 / f = 2 pi / ω = 6.28 / 5.74E-4 = 1.09E4 sec
10900 / 3600 = 3.04 hrs for the minimum rotation period
1. Wha' is the relationship between potential and kinetic energy?
As potential energy increases, kinetic energy increases.
b. As potential energy increases, kinetic energy decreases.
C. As potential energy decreases, kinetic energy decreases.
d. Potential and kinetic energy are two separate things and have no
relationship.
Answer:
B
Explanation:
The kinetic energy in an object is converted into potential energy. This makes the kinetic decrease, while the potential increases.
K
Mission CG9: Weightlessness
Consider the several locations along a roller coaster
track. In which location(s) would the riders feel less
than their normal weight? Select all that apply.
Location A
Location B
Location C
a
=-10 m/s/s, dn
--2 m/s/s, up
a--6 m/s/s, dn
Location D
Location E
x=-12 m/s/s, dn
---6 m/s/s, up
The locations where the riders feel less than their normal weight are Location A, Location C and Location D.
The given parameters;
Location A, a = 10 m/s² downLocation B, a = 2 m/s² upLocation C, a = 6 m/s² downLocation D, a = 12 m/s² downLocation E, a = 6 m/s² upThe normal weight of the riders is calculated by applying Newton's second law of motion as follows;
W = mg
W = 9.8m
The apparent weight of the riders for the upward acceleration is calculated as follows;
[tex]R = m(g + a)[/tex]
The apparent weight of the riders for the downward acceleration is calculated as follows;
[tex]R = m(g - a)[/tex]
The apparent weight of the riders at location A is calculated as follows;
[tex]R_ A = m(9.8 - 10)\\\\R_ A = -0.2 m[/tex]
The apparent weight of the riders at location B is calculated as follows;
[tex]R_B = m(9.8 + 2)\\\\R_B = 11.8 m[/tex]
The apparent weight of the riders at location C is calculated as follows;
[tex]R_C = m(9.8 - 6)\\\\R_C = 3.8 m[/tex]
The apparent weight of the riders at location D is calculated as follows;
[tex]R_D = m(9.8 - 12)\\\\R_D = -2.2 m[/tex]
The apparent weight of the riders at location E is calculated as follows;
[tex]R_E = m(9.8 + 6)\\\\R_E = 15.8 m[/tex]
Thus, the locations where the riders feel less than their normal weight are;
Location ALocation CLocation D.Learn more about Newton's second law and reading of a scale here: https://brainly.com/question/11603452
A pendulum with a length of 2 m has a period of 2.8 s. What is the period of a pendulum with a length of 8 m
Answer:
P = 2 pi (L / g)^1/2
P2 / P1 = (8 / 2)^1/2 = 2
The period would be twice as long or 5.6 sec.
True or False: When looking at kinetic vs. thermodynamic products the kinetic product predominates at low temperature.
Answer:
true answer this question
A car move at an initial velocity of 240m and reach at the final velocity of 540m in 8hours. calculate its acceleration.
Answer:
a = 0.01m/s²
Explanation:
V_f = V_0+a*t
V_f = Velocity final
V_0 = Velocity initial
a = acceleration
t = time
a = (V_f-V_0)/t
a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))
a = 0.01m/s²
How much power does it take to lift 30.0 N 10.0 m high in 10.00 s?
Answer:
60w
Explanation:
The power required is 30 Watt.
Let us recall that power is defined as the rate of doing work. Hence, we can write as follows;
Power = Work done/ time taken
Now;
work done = Force × distance
Force = 30.0 N
Distance = 10.0 m
work done = 30.0 N × 10.0 m = 300 J
The power expended = 300 J/10.00 s = 30 Watt
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When shopping for margarine what information should you look for on the label
Answer:
When buying a margarine, people should look for trans fat and saturated fat content. If it is over 2%, then it is not healthy. If it is lower that means it is healthy and fresh!
Explanation:
Why does the earth stay in orbit around the sun instead of drifting away from it into space?
A Electric force between the Sun and Earth
B Magnetic force between the Sun and Earth
Gravitational force between the Sun and Earth
Answer:
b it is b part answer i think so
If the velocity and frequency of a wave are both doubled, how does the wavelength change?
The wavelength will remain unchanged.
Explanation:
The velocity [tex]v[/tex] of a wave in terms of its wavelength [tex]\lambda[/tex] and frequency [tex]\nu[/tex] is
[tex]v = \lambda\nu[/tex] (1)
so if we double both the velocity and the frequency, the equation above becomes
[tex]2v = \lambda(2\nu)[/tex] (2)
Solving for the wavelength from Eqn(2), we get
[tex]\lambda = \dfrac{2v}{2\nu} = \dfrac{v}{\nu}[/tex]
We would have gotten the same result had we used Eqn(1) instead.
Answer:
the wavelength increases
Explanation:
What is work? A. What happens when energy is destroyed B. The number of molecules in a substance C. The total size of an object divided by its mass D. What happens when a force causes an object to move
Answer:
Which extended definition would be most helpful to add to this body paragraph?
Laughter Yoga, a new form of yoga, is becoming increasingly popular.
Laughter Yoga, which is becoming more and more popular, was created in the 1990s.
Laughter Yoga is modeled after traditional yoga but includes laughter exercises.
Laughter Yoga is a form of yoga that was created by Dr. Madan Kataria in the 1990s.
Explanation:
c
The security alarm on a parked car goes off and produces a frequency of 769 Hz. The speed of sound is 343 m/s. As you drive toward this parked car, pass it, and drive away, you observe the frequency to change by 69.5 Hz. At what speed are you driving
Answer:
Explanation:
ASSUMING your speed is constant
f₀ = f(v + vo)/(v + vs)
Δf = f approach - f depart
69.5 = (769(343 + vo)/(343 + 0)) - (769(343 - vo)/(343 + 0))
69.5 = 769(2vo/343)
vo = 15.5 m/s
The speed of car driving is 15.5 m/s as the car is parked and drive away.
What is speed?Speed is defined as a measurement of the length of time it takes for an object to travel a certain distance. You can determine an object's speed if you know how far it moves in a given amount of time. Time does not move, hence there is no concept of a speed of time. Time refers to how we move through the temporal realm. Speed is a unit of measurement for how quickly something is moving. A change in velocity results in a change in speed.
To calculate the speed we use the formula
f₀ = f (v + vo) / (v + vs)
Δf = f approach - f depart
69.5 = (769(343 + vo) / (343 + 0)) - (769(343 - vo)/(343 + 0))
69.5 = 769(2vo/343)
vo = 15.5 m/s
Thus, the speed of car driving is 15.5 m/s as the car is parked and drive away.
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The more matter in an object, the _____.(1 point)
lower its weight and mass
lower its weight and mass
more likely it is to have gravity
more likely it is to have gravity
greater its gravitational attraction to Earth
greater its gravitational attraction to Earth
less likely it is to have weight on the moon
Answer:
more likely it is to have gravity
A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:
The wavelengths of the constituent travelling waves CANNOT be 400 cm.
The given parameters:
Length of the string, L = 100 cmThe wavelengths of the constituent travelling waves is calculated as follows;
[tex]L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}[/tex]
for first mode: n = 1
[tex]\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm[/tex]
for second mode: n = 2
[tex]\lambda = \frac{2L}{2} = L = 100 \ cm[/tex]
For the third mode: n = 3
[tex]\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm[/tex]
For fourth mode: n = 4
[tex]\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50 \ cm[/tex]
Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.
The complete question is below:
A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:
A. 400 cm
B. 200 cm
C. 100 cm
D. 67 cm
E. 50 cm
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22 Newton force is working on a 1,901 gram object. What is the acceleration in
meter/s^2 unit
Answer:
11.573
Explanation:
f = m*a
where f is the force in Newtons, m is the mass of the object (in kg) and a is the acceleration
so, we solve for a
a = f/m
a = 22/1.901
a = 11.573
the turns ratio for a transformer with 225 turns of wire in its primary winding and 675 turns in the secondary is: n
The ratio of the primary turns to the secondary turns is 1/3
The correct answer to the question is Option A. 1/3
From the question given above, the following data were obtained:
Primary turn (Nₚ) = 225 turnsSecondary turn (Nᵣ) = 675 turns Ratio of primary to secondary =?Ratio = Nₚ/Nᵣ
Nₚ/Nᵣ = 225 / 675
Nₚ/Nᵣ = 1/3
Therefore, the ratio of the primary turns to the secondary turns is 1/3
Complete question:
See attached photo
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For northern hemisphere observers, which celestial object would be above the horizon for the greatest
amount of time: one that is on the celestial equator, one that is 30° above the celestial equator, one that is
70° above the celestial equator, or one that is 40" below the celestial equator? Which one would be above
the horizon the greatest amount of time for southern hemisphere observers? Explain your answer.
Answer:
Explanation:
For a person at about 20° North latitude, an object 70° above the celestial equator would never set. It's arc path would touch the horizon be never sink below it. Observers north of 20° see it all night. Observers south of 20° an object 70° above the celestial equator would spend the greatest amount of time above the horizon.
For southern hemisphere observers, the object 40" below the celestial equator will spend the most time above the horizon. Nearly 12 hours per day. Did you mean 40°? 40 seconds is very close to the equator itself. However, the result is the same.
For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.
What is the equator?The Equator is an imaginary line passing through the middle of a globe. It is equidistant from the North Pole and the South Pole, Its is a horizontal line residing at 0 degrees latitude.
For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.
One that is 40" below the celestial equator would be above the horizon for the greatest amount of time for southern hemisphere observers.
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What activity in orienteering can improve a player's math skills?
homework
or just play some 2012 math game
Answer:
I think calculating distance (if thats an option)
Explanation:
Really hope this helped!
Scenario 3: You are driving through a rain storm talking to your family but you can only hear every other word. What 2 medias are the waves passing in this scenario?
Help me outtttt jejjejejeje
Answer:
do it got a picture
Explanation:
A cyclist rides in a circle with speed 8.1 m/s. What is his centripetal
acceleration if the circle has a radius of 27 m?
Explanation:
We know that the tangent velocity is 8.1 m/s. We also know that the tangent velocity can be written in the following way:
Vt = ωr with ω being the angular velocity.
We now calculate ω:
ω = Vt/r = 8.1 m/s / 27m = 0.3 rad/s
Now that we have ω we can calculate the centripetal aceleration:
a = ω^2 * r = ( 0.3 )^2 * 27 = 2.43 m/s^2
please help 9.2.1 project in science just ned an example
Answer:
Give me what kind of example you need please so I can help you. Put it in the comments.
Explanation:
how far from the lens is the image of the house if the house is 16 ft from the thin convex lens with a focal length of 8 ft
This question involves the concepts of the thin lens formula, focal length, and image distance.
The image of the house is "16 ft" away from the lens.
According to the thin lens formula:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
where,
f = focal length = 8 ft
p = object distance = 16 ft
q = image distance = ?
Therefore,
[tex]\frac{1}{8\ ft}=\frac{1}{16\ ft}+\frac{1}{q}\\\\\frac{1}{q}=\frac{1}{8\ ft}-\frac{1}{16\ ft}\\\\\frac{1}{q}=0.125\ ft^{-1}-0.0625\ ft^{-1}\\\\q=\frac{1}{0.0625\ ft^{-1}}\\\\[/tex]
q = 16 ft
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At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11, how much time does it take the puck to come to rest?
At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11, time it take the puck to come to rest is 4.51 sec.
What is speed?The speed of an item, which is a scalar quantity in everyday usage and kinematics, is the size of the change in that object's position over time or the size of the change in that object's position per unit of time.
Given in the question at the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11,
force = ma = μmg, putting the value,
a = - 1.175 m/sec²
now using equation of motion
v = u + at
t = 4.51 sec
So, time taken by the puck to come to rest is 4.51 sec.
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A car slams on its brakes creating an acceleration of -4.7 m/s^2. It comes to rest after traveling a distance of 235 m. What was its velocity before it began to accelerate?
Answer:
Explanation:
v² = u² + 2as
0² = u² + 2(-4.7)(235)
u² = 2209
u = 47 m/s
A ball is launched as a projectile with initial speed v at an angle θ above the horizontal. using conservation of energy, find the maximum height hmax of the ball's flight. express your answer in terms of v, g, and θθ.
Answer:
Explanation:
I will us θ as the angle above horizontal as I don't know how to make your chosen symbol for angle.
KEy = PE
½m(vsinθ)² = mgh
h = (vsinθ)²/2g
The qualitative equivalent of external validity is:
A- Credibility
B- Dependability
C- Transformability
D- Confirmability
Which statement describes the particles in a liquid?
A. They do not move.
B. They are far apart.
C. They are close together.
D. They are locked in position.
Question No. 1 Marks = = 5 +5 +2 = 12
The driver of a 2.0 × 103 kg red car traveling on the highway at 45m/s slams on his brakes to avoid striking a second yellow car in front of him, which had come to rest because of blocking ahead as shown in above Fig. After the brakes are applied, a constant friction force of 7.5 × 103 N acts on the car. Ignore air resistance.
(a) Determine the least distance should the brakes be applied to avoid a collision with the other vehicle?
(b) If the distance between the vehicles is initially only 40.0 m, at what speed would the collision occur?
(c) Write your conclusive observations on the result obtained from this numerical. i.e. the importance of Physics in daily life.
Question No. 2 Marks = 8
Consider an automobile moving at v mph that skids d feet after its brakes lock. Calculate how far it would skid if it was moving at 2v and the brakes were locked.
The kinematics and Newton's second law we can find the results for the questions about the braking movement of the car are;
Question 1.
a) The stopping distance is: x = 270 m
b) The initial velocity is: v₀ = 17.3 m / s
c) Concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.
Question 2.
The stopping distance is: x = 4d
Given parameters
Mass of the red carriage m1 = 2,0 10³ kg Red car speed vo = 45 m / s Friction force fr = 7.5 10³ N.To find
Question 1.
a) Minimum braking distance.
b) If the distance is x = 40.0 m, what speed should the vehicles have?
c) Conclusive importance of physics in daily life.
Question 2.
The distace to stop.
Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.
v² = v₀² - a2 x
Where v and v₀ are the current and initial velocity, respectively, at acceleration and x the distance traveled.
Newton's second law states that the net force is proportional to the mass and the acceleration of the body.
F = ma
Where F is force, m is mass and acceleration.
In the attachment we see a diagram of the forces in the system. Let's look for the acceleration of the body
fr = m a
a =[tex]\frac{fr}{m}[/tex]
a = [tex]\frac{7.5 \ 10^3}{2.0 \ 10^3 }[/tex]
a = 3.75 m / s²
This acceleration is in the opposite direction to the speed.
Let's find the distance needed to stop, the final speed is zero.
0 = v₀² - 2 ax
x = [tex]\frac{v_o^2 }{ 2a}[/tex]
Let's calculate.
x = [tex]\frac{45^2 }{2 3.75}[/tex]
x = 270 m
This is the minimum distance that the two vehicles must separate to avoid a collision.
b) We look for speed.
v₀ = [tex]\sqrt{2ax}[/tex]
v₀ = [tex]\sqrt{2 \ 3.75 \ 40.0}[/tex]
v₀ = 17.3 m / s
c) The concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.
2) They indicate that the initial velocity is v and the distance traveled to stop is d, let's find the acceleration.
0 = v₀² - 2ax
Let's substitute.
a = [tex]\frac{v^2}{2d}[/tex]
They ask the distance traveled if this car traveled from an initial speed 2v.
0 = v² - 2 a x
x = [tex]\frac{v^2}{2a}[/tex]
We substitute
x = [tex]\frac{(2v)^2 }{2} \ (\frac{2d}{v^2})[/tex]
x = 4 d
In conclusion, using the kinematic relations and Newton's second law we can find the results for the questions about the braking movement of the car are;
Question 1
a) The stopping distance is: x = 270 m
b) The initial velocity is: v₀ = 17.3 m / s
c) concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.
Question 2.
The stopping distance is x = 4d
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Shorter the vibrating part more will be the pitch. How?
Answer:When the length of a string is changed, it will vibrate with a different frequency.Shorter strings have higher frequency and therefore higher pitch.
I can learn new things but I cannot change how good I am at math. A strongly agree B. agree C disagree D. strongly disagree
Mathematics is a broad subject that anyone who constantly practices by learning from first principle and from example will grow in perfection of the subject as time goes by. Hence, I strongly disagree
During my schools days, I struggled learning mathematics, at the time it was difficult, I began practising and learning from text examples, with time I got a hang of it and my grade and confidence level in the subject increased.
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