Answer:
Divide then multiply or multiply then divide
Explanation:
to get the answer of a and b
A satellite of mass m orbits a moon of mass M in uniform circular motion with a constant tangential speed of v. The satellite orbits at a distance R from the center of the moon. Write down the correct expression for the time T it takes the satellite to make one complete revolution around the moon?
The gravitational force exerted by the moon on the satellite is such that
F = G M m / R ² = m a → a = G M / R ²
where a is the satellite's centripetal acceleration, given by
a = v ² / R
The satellite travels a distance of 2πR about the moon in complete revolution in time T, so that its tangential speed is such that
v = 2πR / T → a = 4π ² R / T ²
Substitute this into the first equation and solve for T :
4π ² R / T ² = G M / R ²
4π ² R ³ = G M T ²
T ² = 4π ² R ³ / (G M )
T = √(4π ² R ³ / (G M ))
T = 2πR √(R / (G M ))
The correct expression for the time T it takes the satellite to make one complete revolution around the moon is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].
We can find the period T (the time it takes the satellite to make one complete revolution around the moon) from the gravitational force:
[tex] F = \frac{GmM}{R^{2}} [/tex] (1)
Where:
G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²
R: is the distance between the satellite and the center of the moon
m: is the satellite's mass
M: is the moon's mass
The gravitational force is also equal to the centripetal force:
[tex] F = ma_{c} [/tex] (2)
The centripetal acceleration ([tex]a_{c}[/tex]) is equal to the tangential velocity (v):
[tex] a_{c} = \frac{v^{2}}{R} [/tex] (3)
And from the tangential velocity we can find the period:
[tex] v = \omega R = \frac{2\pi R}{T} [/tex] (4)
Where:
ω: is the angular speed = 2π/T
By entering equations (4) and (3) into (2), we have:
[tex] F = m\frac{v^{2}}{R} = m\frac{(\frac{2\pi R}{T})^{2}}{R} = \frac{mR(2\pi)^{2}}{T^{2}} [/tex] (5)
By equating (5) and (1), we get:
[tex] \frac{mR(2\pi)^{2}}{T^{2}} = \frac{GmM}{R^{2}} [/tex]
[tex] T^{2} = \frac{R^{3}(2\pi)^{2})}{GM} [/tex]
[tex] T = \sqrt{\frac{R^{3}(2\pi)^{2})}{GM}} [/tex]
[tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex]
Therefore, the expression for the time T is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].
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You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90. The true value is 9.81. Calculate the mean value of your results to three significant digits. ________
Answer: The mean value = 9.85m/s².
Explanation:
Mean = [tex]\dfrac{\text{Sum of n observations}}{n}[/tex]
The given measurements the acceleration of gravity (units of m/s²): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90.
Number of measurements =9
Sum of measurements = 88.69
Mean = [tex]\dfrac{88.69}{9}=9.85444444\approx9.85[/tex]
Hence, the mean value = 9.85m/s².
Agnes makes a round trip at a constant speed to a star that is 16 light-years distant from Earth, while twin brother Bert remains on Earth. When Agnes returns to Earth, she reports that she has celebrated 20 birthdays during her journey. (a) What was her speed during her journey
Answer:
Speed of Agnes during her journey was 0.848c
Explanation:
Given that;
Age of Agnes t₀ = 20 years
distance d = 2 × distance of star from Earth = 2 × 16 light-years
= 32 light-years
so get her speed speed; we use the following expression
Yvt₀ = d
( v / √(1 - ([tex]\frac{v}{c}[/tex])²) )² = ( 32 light-years / 20 yrs )²
v² / (1 - ( v²/c²)) = ( 32 × c / 20 )²
v² / (1 - ( v²/c²)) = 2.56 × c²
v² / c²-v²/c² = 2.56 × c²
v²c² / c² - v² = 2.56 × c²
v² / c² - v² = 2.56
v² = 2.56 (c² - v²)
v² = 2.56 (c² - v²)
v² = 2.56c² - 2.56v²
v² + 2.56v² = 2.56c²
3.56v² = 2.56c²
v² = (2.56/3.56)c²
v = √((2.56/3.56)c²)
so v = 0.848c
Therefore, Speed of Agnes during her journey was 0.848c
A 10 kg remote control plane is flying at a height of 111 m. How much
potential energy does it have?
Answer:
10.88kJ
Explanation:
Given data
mass= 10kg
heigth= 111m
Applying
PE= mgh
assume g= 9.81m/s^2
substitute
PE= 10*9.81*111
PE=10889.1 Joules
PE=10.881kJ
Hence the potential energy is 10.88kJ
what is the result of mixing 15 garm of water 80 degree celsius with 10 gram of ice -10 degree Celsius ? give specific heat capctiy of ice 0.5 calorie per gram Celsius and letent heat of fusion of ice 80 calorie per gram.
Answer:
50
Explanation:
Current Attempt in Progress The atomic radii of a divalent cation and a monovalent anion are 0.52 nm and 0.125 nm, respectively. (a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another). Enter your answer for part (a) in accordance to the question statement N (b) What is the force of repulsion at this same separation distance
Answer:
a) F = 1.70 10⁻⁹N, F = 1.47 10⁻⁸ N,
b) * the electronegative repulsion, from the repulsion by quantum effects
Explanation:
a) The atraicione force comes from the electric force given by Coulomb's law,
F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]
divalent atoms
In this case q = 2q₀ where qo is the charge of the electron -1,6 10⁻¹⁹ C and the separation is given
F = k q² / r²
F = [tex]2 \ 10^9 \ \frac{2 (1.6 \ 10^{-19} )^2}{ (0.52 10^{-9} )^2 }[/tex]
F = 1.70 10⁻⁹N
monovalent atoms
in this case the load is q = q₀
F = 2 \ 10^9 \ \frac{ (1.6 \ 10^{-19} )^2}{ (0.125 10^{-9} )^2 }
F = 1.47 10⁻⁸ N
b) repulsive forces come from various sources
* the electronegative repulsion of positive nuclei
* the electrostatic repulsion of the electrons when it comes to bringing the electron clouds closer together
* from the repulsion of electron clouds, by quantum effects
Hlo
what is a force........??
Explanation:
In physics, a force is any interaction that, when unopposed, will change the motion of an object. A force can cause an object with mass to change its velocity, i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity.
Formula
Newton's Second Law
F = m * a
F = force
m = mass of an object
a = acceleration
I don’t know what to do
Answer:
So increasing the voltage increases the charge in direct proportion to the voltage. If the voltage exceeds the capacitors rated voltage, the capacitor may fail due to breakdown of the dielectric between the two plates that make up the capacitor.
Explanation:
A option.
Now complete your visual overview by identifying the known variables and the variables you must find. Assume charge 1 is located at the origin of the x axis and the positive x axis points to the right. Let x1, x2, and x3 denote the positions of charge 1, charge 2, and charge 3, respectively. Determine which of the following quantities are known and which are unknown.
Answer:
Explanation:
From the given information;
If we assume that charge 1 is located at the origin;
Then, using the visual overview for identification, we will realize that the known quantities are:
[tex]\mathbf{= q_1, \ q_2 , \ q_3, \ x_1 \ and \ x_2}[/tex]
However, provided that we do not know the exact location of [tex]x_3[/tex],
Then, the unknown quantity is [tex]\mathbf{ x_3}[/tex]
3N
3 N
What is the net force of the box?
What is unusual about the material that Emily invented?
Answer:
The material that Emily invented can be easily repaired by shining ultraviolet light on it.
Explanation:
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Answer:
The material that Emily invented can be easily repaired by shining ultraviolet light on it.
Explanation:
A quarterback, Patrick, throws a football down the field in a long arching trajectory to wide receiver, Tyreek. The football and Tyreek are traveling in the same direction, started at the same spot, and the football was thrown at the same instant that Tyreek began running. Furthermore, both Tyreek and the football have horizontal components of speed of 22.6 mph. Under these circumstances, no matter what angle the football is thrown at, it will land on Tyreek (whether he catches it or not). True or false
Answer:
the statement is False [tex]\frac{x_{ball} }{ x_{rec} }[/tex] = cos θ
Explanation:
Let's analyze this problem, the ball and the receiver leave the same point and we want to know if at the same moment they reach the same point, for this we must have both the ball and the receiver travel the same distance.
Let's start by finding the time it takes for the ball to reach the ground
y = [tex]v_{oy}[/tex] t - ½ g t²
when it reaches the ground its height is y = 0
0 = vo sin θ - ½ g t²
0 = t (vo sin θ - ½ g t)
The results are
t = 0 exit point
t = 2 v₀ sin θ/g arrival point
at this point the ball traveled
[tex]x_{ball}[/tex]= v₀ₓ t
x_{ball} = v₀ cos θ 2v₀ sin θ / g
x_{ball}= 2 v₀² cos θ sin θ/ g
Now let's find that distantica traveled the receiver in time
[tex]x_{rec}[/tex] = v₀ t
x_{rec} = v₀ (2 v₀ sin θ / g)
x_{rec} = 2 v₀² sin θ / g
without dividing this into two distances
[tex]\frac{x_{ball} }{ x_{rec} }[/tex] = cos θ
therefore the distances are not equal to the ball as long as behind the receiver
Therefore the statement is False
Convert 20 C to F
-40 C to F
40C to F
Answer:
20 C to F
Ans: 68F
-40 C to F
Ans:-40F
40C to F
Ans:104F
a 70 kg student stands on top of a 5.0 m platform diving board . how much gravitational potential energy does the student have?
how much work did it take for the student to travel from the ground to the top of the platform diving board?
Answer:
a. P.E = 3430Joules.
b. Workdone = 3430Nm
Explanation:
Given the following data;
Mass = 70kg
Distance = 5m
We know that acceleration due to gravity is equal to 9.8m/s²
To find the potential energy;
Potential energy = mgh
P.E = 70*9.8*5
P.E = 3430J
b. To find the workdone;
Workdone = force * distance
But force = mass * acceleration
Force = 70*9.8
Force = 686 Newton.
Workdone = 686 * 5
Workdone = 3430Nm