Answer:
x = 1382.9 km
Explanation:
The speed of the wave is constant, so we can use the uniform motion relationships
p wave
[tex]v_p[/tex] = x / t₁
t₁ = x /v_p
S wave
v_s = x / t₂
t₂ = x / v_s
indicate that the time difference between the two waves is
t₂ - t₁ = 3.25 min (60 s / 1 min)
t₂ -t₁ = 195 s
let's substitute
[tex]\frac{x}{v_s} - \frac{x}{v_p}[/tex] = 195
x ([tex]\frac{1}{v_s} - \frac{1}{v_p}[/tex] = 195
let's calculate
x [tex]( \frac{1}{4.2} - \frac{1}{10.3} )[/tex] = 195
x (0.1410) = 195
x = 195 /0.141
x = 1382.9 km
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
The sound from a trumpet travels at 351 m/s in the frequency of the note is 294 Hz, what is the wavelength of the sound wave?
Answer:
1.19m
Explanation:
Given parameters:
Speed of the trumpet sound = 351m/s
Frequency of the note = 294Hz
Unknown:
Wavelength of the sound = ?
Solution:
To solve this problem, we use the wave - velocity equation.
V = F x ∧
V is the velocity of the body
F is the frequency
∧ is the wavelength
So;
351 = 294 x ∧
∧ = [tex]\frac{351}{294}[/tex] = 1.19m
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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Two satellites are approaching the Earth from opposite directions. According to an observer on the Earth, satellite A is moving at a speed of 0.648c and satellite B at a speed of 0.795c. What is the speed of satellite A as observed from satellite B
Answer:
Explanation:
Satellite A and satellite B are approaching the earth from opposite directions , that means they are approaching each other . The velocity of satellite A and B are .648c and .795c respectively . Their velocities are comparable to velocity of light so they will follow relativistic laws .
Their relative velocity will be given by the following relation .
[tex]V_r=\frac{u+v}{1+\frac{uv}{c^2} } }[/tex]
where u and v are velocities of vehicles coming from opposite direction and c is velocity of light .
[tex]V_r=\frac{.795c+.648c}{1+\frac{.648c\times .795c}{c^2} } }[/tex]
[tex]V_r=\frac{1.443c}{1+.515 } }[/tex]
= .952c
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
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]
A racecar reaches 24 m/s in 6 seconds at the start of a race. What is the acceleration of the car?
Answer:
4m/s^2
Explanation:
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:
The only force acting on a 3.1 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a velocity of 4.4 m/s in the positive x direction and some time later has a velocity of 6.4 m/s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time
Answer:
33.5J
Explanation:
Given:
Mass of the canister= 3.1 kg
Initial velocity of canister= v(i)= 4.4i m/s
Final velocity of canister= v(f)= 6.4j m/s
Force magnitude( xy plane)= 5 N
The magnitude of vector V'= Vxi + Vyj + Vzk
|V|= √( Vx^2 + Vy^2 + Vz^2
From Kinectic energy and work theorem.
Net work = Kinectic energy of the canister
ΔK= W
(Kf - Ki)= W
Where Kf= final Kinectic energy
= 1/2 mv^2
If we input the given values we have,
= 1/2 × 3.1 ×√(4.4^2 + 0^2 + 0^2)^2 = 30J
Ki= initial Kinectic energy
= 1/2 mv^2
If we substitute the given values we have
=1/2 × 3.1 ×√(0^2 + 6.4^2 + 0^2)^2 = 63.5 J
Work done by canister = (final Kinectic energy - initial energy)
= 63.5- 30
=33.5J
Hence, work done on the canister 33.5J
Jogger A has a mass m and a speed v, jogger B has a mass m/2 and a speed 3v, jogger C has a mass 3m and a speed v/2, and jogger D has a mass 4m and a speed v/2. Rank the joggers in order of increasing kinetic energy. Indicate ties when appropriate. Show your work.
Answer: B>A=D>C
Explanation:
Kinetic Energy is the product of mass and square of the velocity
For Jogger A
[tex]K.E._a=\frac{1}{2}mv^2[/tex]
For Jogger B
[tex]K.E._b=\frac{1}{2}\times\frac{m}{2}\times(3v)^2=\frac{9}{4}mv^2\\K.E._b=2.25mv^2[/tex]
For Jogger C
[tex]K.E._c=\frac{1}{2}\times3m\times(\frac{v}{2})^2=\frac{3}{8}mv^2\\K.E._c=0.375mv^2[/tex]
For Jogger D
[tex]K.E._d=\frac{1}{2}\times4m\times(\frac{v}{2})^2=\frac{1}{2}mv^2[/tex]
Kinetic Energy of Joggers in increasing order
B>A=D>C
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.
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
Which characterictic of motion could change without changing the velocity of an object
Answer:
The direction could change
1.
Atennis ball is shot straight up with an initial velocity of 34 m/s. What is its velocity two seconds after launch?
Answer:
The speed (magnitude of the velocity) is 14.4 m/s
Explanation:
Vertical Launch Upwards
It occurs when an object is launched vertically up without taking into consideration any kind of friction with the air.
If vo is the initial speed and g is the acceleration of gravity, the speed vf at any time is calculated by:
[tex]v_f=v_o-g.t[/tex]
A tennis ball is launched vertically up with an initial speed of vo=34 m/s. At time t=2 s, its speed is:
[tex]v_f=34-9.8*2[/tex]
[tex]v_f=34-19.6[/tex]
[tex]v_f=14.4\ m/s[/tex]
The speed (magnitude of the velocity) is 14.4 m/s
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.
what Species can change over time to adapt to their environment
Answer:
Explanation:
Most environments have many niches. If a niche is "empty" (no organisms are occupying it), new species are likely to evolve to occupy it. This happens by the process of natural selection. By natural selection, the nature of the species gradually changes to become adapted to the niche.
Astronaut 1 has a mass of 75 kg. Astronaut 2 has a mass of 80 kg. Astro 1 and 2 want to travel to separate planets, but they want to experience the same weight (in N). Astro 1 visits a planet with gravitational acceleration 12 m/s2. What must be Astro 2 planet's ag to equal Astro 1's weight
Answer:
weight = 900 N
acceleration = 11.25 m/s²
Explanation:
given data
mass m1 = 75 kg
mass m2 = 80 kg
gravitational acceleration = 12 m/s²
solution
As we know weight of a mass that is
weight of mass = Mass × Acceleration due to gravity .................1
so Astro 1 weight is
weight = 75kg × 12 m/s²
weight = 900 N
and
so, when Astro 2 needs this much weight the planet on which he is will have the acceleration
acceleration = Weight ÷ Mass of Astro 2 .....................2
acceleration = 900 ÷ 80 m/s²
acceleration = 11.25 m/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
3N
3 N
What is the net force of the box?
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
The dotted lines and arrows represent
The audio power of the human voice is concentrated at about 300 Hz. Antennas of the appropriate size for this frequency are impracticably large, so that to send voice by radio the voice signal must be used to modulate a higher (carrier) frequency for which the natural antenna size is smaller. a. What is the length of an antenna one-half wavelength long for sending radio at 300 Hz
Answer:
the length of an antenna one-half wavelength long for sending radio at 300 Hz is 500 km
Explanation:
Given the data in the question;
we know that wave length is;
λ = c/f
c is the speed of voice ( 3 × 10⁸ m/s )
frequency f = 300 Hz
so we substitute
λ = 3 × 10⁸ / 300
λ = 1000000 m
we know that; 1 km = 1000 m
so
λ = 1000000 m / 1000
λ = 1000 km
hence, an antenna one-half wavelength will be;
λ /2
= 1000 km / 2
= 500 km
Therefore, the length of an antenna one-half wavelength long for sending radio at 300 Hz is 500 km
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:
a) Two workers are trying to move a heavy crate. One pushes onthe crate with a force A, which has amagnitude of 445 newtons and is directed due west. The other pusheswith a force B, which has a magnitude of 325newtons and is directed due north. What are the magnitude anddirection of the resultant force A + Bapplied to the crate?
b) Suppose the second worker applies a force of-B instead of B. What then arethe magnitude and direction of the resultant force A -B applied to the crate? In both cases express thedirection relative to due west.
Answer:
Divide then multiply or multiply then divide
Explanation:
to get the answer of a and b
If the grasses have 100 units of energy, how much can be passed to the grasshoppers?
After an initial race George determines that his car loses 35 percent of its acceleration due to air resistance travelling at 38 m/s on flat ground. Assuming that his car travels with a constant acceleration, calculate the maximum speed (Vm), in meters per second, his car can reach on flat ground?\
Answer:
64.2 m/s
Explanation:
We are given that
Speed ,v=38 m/s
We have to find the maximum speed when his car reach on flat ground.
Using dimensional analysis
[tex]F_{res}\propto v^2[/tex]
If 35% acceleration reduced by F(res) at 38 m/s
Then, 100% acceleration can be reduced by F(res) at v' m/s
[tex]\frac{F_1}{F_2}=\frac{v^2}{v'^2}[/tex]
[tex]v'^2=\frac{F_2}{F_1}v^2[/tex]
[tex]v'=v\sqrt{\frac{F_2}{F_1}}[/tex]
Substitute the values
[tex]v'=38\times \sqrt{\frac{100}{35}}[/tex]
[tex]v'=64.2 m/s[/tex]
Hence, the maximum speed when his car can reach on flat ground=64.2 m/s
The maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.
What is the relation between resistance and speed?
The air resistance is directly proportional to the square of the velocity of an object.
R ∝ v²
The speed of the car was reduced by 35 % at 38 m/s.
So, the speed of the car was reduced by 100% at v' m/s.
The relationship can be given by,
[tex]\dfrac {R_1}{R_2} = \dfrac {v^2}{v'^2}[/tex]
Put the values in the formula and calculate for [tex]v'[/tex],
[tex]v' = \sqrt {\dfrac {R_2 v^2}{R_1 }}[/tex]
[tex]v' = \sqrt {\dfrac { 100\times 38 ^2}{35 }}\\\\v' = 62.2 \rm \ m/s[/tex]
Therefore, the maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.
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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
The lawn outside your neighbor's house has an approximate area of 175 m2One night it snows so that the snow on the lawn has a uniform depth of 25.5 cm. What volume of snow is on the lawn, in cubic m
The volume of snow on the lawn, in cubic m, is 44.63 or 44.63 m³ if the lawn outside your neighbor's house has an approximate area of 175 m².
What is volume?It is defined as a three-dimensional space enclosed by an object or thing.
It is given that:
The lawn outside your neighbor's house has an approximate area of 175 m². One night it snows so that the snow on the lawn has a uniform depth of 25.5 cm.
As we know, unit conversion can be defined as the conversion from one quantity unit to another quantity unit followed by the process of division, and multiplication by a conversion factor.
25.5 cm = 0.255 m
Volume = area×depth
Volume = 175×0.255
Volume = 44.63 cubic meters
Thus, the volume of snow on the lawn, in cubic m is 44.63 or 44.63 m³ if the lawn outside your neighbor's house has an approximate area of 175 m².
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