Answer:
The radius of the bubble when it reaches the surface at 30 ºC is 1.015 centimeters.
Explanation:
Let suppose that air bubble behaves as ideal gas, whose equation of state is:
[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex] (Eq. 1)
Where:
[tex]P[/tex] - Pressure of the bubble, measured in kilopascals.
[tex]V[/tex] - Volume of the bubble, measured in cubic meters.
[tex]n[/tex] - Molar amount of the bubble, measured in kilomoles.
[tex]T[/tex] - Temperature, measured in Kelvin.
[tex]R_{u}[/tex] - Ideal gas constant, measured in kilopascal-cubic meter per kilomole-Kelvin.
Then, we eliminate the molar amount and the ideal gas constant by constructing the following relationship:
[tex]\frac{P_{A}\cdot V_{A}}{T_{A}} = \frac{P_{B}\cdot V_{B}}{T_{B}}[/tex] (Eq. 2)
Where:
[tex]P_{A}[/tex], [tex]P_{B}[/tex] - Pressure of the bubble at bottom and surface, measured in kilopascals.
[tex]V_{A}[/tex], [tex]V_{B}[/tex] - Volume of the bubble at bottom and surface, measured in cubic meters.
[tex]T_{A}[/tex], [tex]T_{B}[/tex] - Temperature of the bubble at bottom and surface, measured in Kelvin.
The pressure experimented by the bubble at bottom and surface are, respectively:
[tex]P_{A} = 101.325\,kPa+\left(1027\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{kg}{m^{3}} \right)\cdot (25\,m)\cdot \left(\frac{1}{1000}\,\frac{kPa}{Pa} \right)[/tex]
[tex]P_{A} = 353.120\,kPa[/tex]
[tex]P_{B} = 101.325\,kPa[/tex]
If we know that [tex]P_{A} = 353.120\,kPa[/tex], [tex]P_{B} = 101.325\,kPa[/tex], [tex]V_{A} = 1.20\times 10^{-6}\,m^{3}[/tex], [tex]T_{A} = 290.15\,K[/tex] and [tex]T_{B} = 303.15\,K[/tex], then the volume of the bubble at surface is:
[tex]\frac{(353.120\,kPa)\cdot (1.20\times 10^{-6}\,m^{3})}{290.15\,K} = \frac{(101.325\,kPa)\cdot V_{B}}{303.15\,K}[/tex]
[tex]1.460\times 10^{-6} = 0.334\cdot V_{B}[/tex]
[tex]V_{B} = 4.372\times 10^{-6}\,m^{3}[/tex]
[tex]V_{B} = 4.372\,cm^{3}[/tex]
And the volume of the air bubble is determined by this formula:
[tex]V_{B} = \frac{4\pi\cdot R^{3}}{3}[/tex] (Eq. 3)
Where [tex]R[/tex] is the radius of the air bubble, measured in centimeters.
If we know that [tex]V_{B} = 4.372\,cm^{3}[/tex], then the radius of the air bubble is:
[tex]4.372 = \frac{4\pi\cdot R^{3}}{3}[/tex]
[tex]R^{3} = 1.044[/tex]
[tex]R \approx 1.015\,cm[/tex]
The radius of the bubble when it reaches the surface at 30 ºC is 1.015 centimeters.
What would happen if the intake stroke never worked?
Answer:
you will die by it
Explanation:
Answer:
If the intake stroke never occurred, then the whole four stoke enegine process would not work
Explanation:
What does the acronym BMI stand for?
O A Body Median isotopes
O B. Beginning Mean Index
O . Body Mass Index
O D. Big Mass Inside
Answer:
body mass index
Explanation:
Answer:
[tex] \boxed{ \sf{ \: Option \: C \: is \: correct.}}[/tex]
Explanation:
BMI stands for Body Mass Index .
Hope I helped!
Best regards! :D
~TheAnimeGirl
2. Heather and Matthew walk with an average velocity of +0.87 m/s eastward.
If it takes them 27 minutes to walk to the store, what is their
displacement? (include direction)
(5 points)
A fish is hung by two spring scales as shown in the diagram. Assume the spring scales are massless. if the bottom scale reads 5N, what will the top scale read?
Question 3 options:
10 N
15 N
5 N
0 N
Answer:
I'm going to guess 15 N
Explanation:
Answer my last post with all needed work for 75 POINTSSSS vist my profile to see it its on projectile motion !!!!!!!!!!
Answer:
ok
Explanation:
ok ok I will thanks for free
Review Question 1
A group of students collected the data shown below while attempting to
measure the coefficient of static friction (of course, it looks like this group
varied the amount of mass sitting on the block with each trial - this is not
recommended). Nonetheless, what is their average coefficient of static friction?
Mass of block (g)
Hanging mass (kg)
Trial
1
105
0.053
2
165
0.081
3
220
0.118
4
280
149
5
315
0.180
6
385
0.198
Answer:
The average coefficient of static friction = 0.130
Explanation:
The average coefficient of static friction = [tex]\frac{sum of coefficient of static friction}{number of trials}[/tex]
From the given question, the coefficient of static friction given are:
0.053, 0.081, 0.118, 0.149, 0.180, 0.198
Thus,
the sum of the given coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198
= 0.779
The average coefficient of static friction = [tex]\frac{0.779}{6}[/tex]
= 0.129833
= 0.130
Therefore, the average coefficient of static friction is 0.130.
Jeffrey is using a pulley to get water from a well. He loops a rope over the top of a pulley wheel, then ties the rope to a bucket. He lowers the bucket into the water. To bring the bucket and water up from the well, Jeffrey pulls down on his end of the rope. Why does Jeffrey apply a pulling force to the rope? A. because adding his force to the force generated by the pulley will get the job done faster B. to test the tension in the rope before he uses it to climb down into the well C. to signal to the pulley to start drawing water from the well D. because work must be done on the pulley before it can do work on the bucket
Answer:
D
Explanation:
Bob roller skates with a constant speed of 6 mi/hr. How long will it take him to travel 6 mi.? t=d/s
Answer:
1 hour
Explanation:
t = 6/6
6/6 = 1
You just said that his speed enables him to cover 6 miles in each hour. From this information, I calculated that it will take him one hour to travel 6 miles.
you a viewing a read out on an oscilloscope that shows you when different voltages in a circuit maximize. If the period of the circuit is 0.04 seconds, how much time is there between when the capacitor voltage maximizes and the inductor voltage maximizes ?
Answer:
0.01 s
Explanation:
When the capacitor voltage is maximum, the inductor voltage is at a minimum, and when the capacitor voltage is minimum, the inductor voltage is at a maximum.
The time difference between minimum voltage and maximum voltage for the capacitor or inductor is the same as that for maximum voltage of capacitor to maximum voltage of inductor and this is t = T/4 where T = period.
Since T = 0.04 s, t = T/4
= 0.04 s/4
= 0.01 s
So, the amount of time is there between when the capacitor voltage maximizes and the inductor voltage maximizes is 0.01 s
Taranga lifts 400N load using first class lever. if Taranga applies effort of 100N at 30 cm from the fulcrum what should be the load distance to balance it?
Answer:
d = 7.5 [cm]
Explanation:
To be able to solve this problem we must understand well the concept of a simple lever of First Class, which consists of a support point (fulcrum) and a beam where a force is applied at one end, while at the other end the force exerted is multiplied.
In the attached image we can see an image that will help us better understand the concept of this first-class lever.
The moment created by the effort and the distance is equal to:
M = 100*30 = 3000 [N*cm]
Now the balance is created because we have the same moment exerted by the effort and the one created by the load.
3000 = 400*d
d = 7.5 [cm]
you know,dinner is taking,like,forever to cook,and i'm sterving
Answer:
OK?
Explanation:
Answer:
It happens and they might not even cook what you feel like eating.
A wire with a linear mass density of 1.17 g/cm moves at a constant speed on a horizontal surface and the coefficient of kinetic friction between the wire and the surface is 0.250. If the wire carries a current of 1.24 A westward and moves horizontally to the south, determine the magnitude and direction of the smallest magnetic field that can accomplish this.
Answer:
The value is [tex]B = 0.2312 \ T[/tex]
The direction is into the surface
Explanation:
From the question we are told that
The mass density is [tex]\mu =\frac{m}{L} = 1.17 \ g/cm =0.117 kg/m[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.250[/tex]
The current the wire carries is [tex]I = 1.24 \ A[/tex]
Generally the magnetic force acting on the wire is mathematically represented as
[tex]F_F = F_B[/tex]
Here [tex]F_F[/tex] is the frictional force which is mathematically represented as
[tex]F_F = \mu_k * m * g[/tex]
While [tex]F_B[/tex] is the magnetic force which is mathematically represented as
[tex]F_B = BILsin(\theta )[/tex]
Here [tex]\theta =90^o[/tex] is the angle between the direction of the force and that of the current
So
[tex]F_B = BIL[/tex]
So
[tex]BIL = \mu_k * m * g[/tex]
=> [tex]B = \mu_k * \frac{m}{L} * [\frac{g}{I} ][/tex]
=> [tex]B = 0.25 * 0.117 * [\frac{9.8}{1.24} ][/tex]
=> [tex]B = 0.2312 \ T[/tex]
Apply the right hand curling rule , the thumb pointing towards that direction of the current we see that the direction of the magnetic field is into the surface as shown on the first uploaded image
A 200 – g object is tied to the end of a cord and it is turning in horizontal circle of radius 1.20 Cm at the constant 3 rev/sec. Calculate the centripetal acceleration of the object.
At angular speed of 3 rev/s, the object moves a distance equal to 3 times the circumference of the circle each second, or a distance of 3 • 2π (1.20 cm) ≈ 22.6 cm.
So, with a linear speed of 22.6 cm/s = 0.226 m/s, the object has a centripetal acceleration a of
a = (0.226 m/s)² / (0.012 m) ≈ 18.8 m/s²
directed toward the center of the circle.
The temperature of a 2.0-kg block increases by 5°C when 2,000 J of thermal energy are added to the block. What is the specific heat of the block?
2. How much time will it take to move 120 m if an object
moves 300 m/s?
Answer:
2.5 seconds
Explanation:
Time is equal to distance over speed
300 divide by 120 u get 2.5 seconds
How many neutrons does protium have
Answer:
I believe none
Explanation:
HELP ASAP !!!!!!!!!!!!!!
Answer:
big bang theory
Explanation:
Edwin Hubble is credited for the initial development of the Big Bang theory, an idea which helps to explain the formation of the universe over 15 billion years ago.
NEED HELP ASP
WILL GIVE POINTS OR WHATEVER
Answer:
you need to multiply the momentum and the mass
Find the magnitude of the gravitational force (in N) between a planet with mass 6.00 X 10^24 kg and its moon, with mass 2.50 X 10^22 kg, if the average distance between their centers is 2.70 X 10^8 m. What is the moon's acceleration (in m/s2) toward the planet? (Enter the magnitude.) What is the planet's acceleration (in m/s2) toward the moon? (Enter the magnitude.)
Answer:
Magnitude of gravitational force between this planet and its moon: approximately [tex]1.37 \times 10^{20}\; \rm N[/tex].
Acceleration of this moon towards the planet: approximately [tex]5.49 \times 10^{-3}\; \rm m \cdot s^{-2}[/tex].
Acceleration of this planet towards its moon: approximately [tex]2.29 \times 10^{-5}\; \rm m \cdot s^{-2}[/tex].
Explanation:
Look up the gravitational constant, [tex]G[/tex]:
[tex]G \approx 6.67 \times 10^{-11}\; \rm m^3\cdot kg^{-1} \cdot s^{-2}[/tex].
Assume that both this planet and its moon are spheres of uniform density. When studying the gravitational interaction between this planet and its moon, this assumption allows them to be considered as two point masses.
The formula for the size of gravitational force between two point masses [tex]m_1[/tex] and [tex]m_2[/tex] with a distance of [tex]r[/tex] in between is:
[tex]\displaystyle F = \frac{G \cdot m_1 \cdot m_2}{r^2}[/tex],
where [tex]G[/tex] is the gravitational constant.
Let [tex]m_1[/tex] and [tex]m_2[/tex] denote the mass of this planet and its moon, respectively.
Calculate the size of gravitational force between this planet and its moon:
[tex]\begin{aligned} F &= \frac{G \cdot m_1 \cdot m_2}{r^2} \\ &\approx \frac{6.67 \times 10^{-11}\; \rm m^3 \cdot kg^{-1} \cdot s^{-2} \times 6.00 \times 10^{24}\; \rm kg \times 2.50 \times 10^{22}\; \rm kg}{{\left(2.70 \times 10^{8}\; \rm m\right)}^2} \\ &\approx 1.37 \times 10^{20}\; \rm N\end{aligned}[/tex].
Assume that other than the gravitational force between this planet and its moon, all other forces (e.g., gravitational force between this planet and the star) are negligible. The magnitude of the net force on the planet and on the moon should both be approximately [tex]1.37 \times 10^{20}\; \rm N[/tex].
Apply Newton's Second Law of motion to find the acceleration of this planet and its moon:
[tex]\displaystyle \text{acceleration} = \frac{\text{net force}}{\text{mass}}[/tex].
For this moon:
[tex]\displaystyle \frac{1.37 \times 10^{20}\; \rm N}{2.50 \times 10^{22}\; \rm kg} \approx 5.49\times 10^{-3}\; \rm m \cdot s^{-2}[/tex].
For this planet:
[tex]\displaystyle \frac{1.37 \times 10^{20}\; \rm N}{6.00 \times 10^{24}\; \rm kg} \approx 2.29 \times 10^{-5}\; \rm m \cdot s^{-2}[/tex].
A 30 kg box is being pulled with a force of 125 N. The coefficient of static friction between the box and the floor is 0.35. What is the minimum downward force on the box that will keep it from slipping?
Answer:
The minimum downward force on the box that will keep it from slipping is 63.14 N
Explanation:
Given;
mass of the object, m = 30 kg
applied force, f = 125 N
coefficient of static friction, μ = 0.35
Normal reaction (R) is acting upwards, weight of the box (mg) is acting downwards and the minimum downward force (F) on the box that will keep it from slipping is also acting downwards.
The net vertical forces on the box is given by;
R - mg - F = 0
F = R - mg
Now, determine normal reaction, R
f = μR
R = f / μ
R = 125 / 0.35
R = 357.14 N
Finally, determine the minimum downward force on the box that will keep it from slipping;
F = R - mg
F = 357.14 - (30 x 9.8)
F = 357.14 - 294
F = 63.14 N
Therefore, the minimum downward force on the box that will keep it from slipping is 63.14 N
A mother and daughter press their hands together and then push apart while ice skating. Immediately after they push away from each other, how does the motion of the mother and daughter change?
Answer:
J
Explanation:
The daughter moves with greater acceleration backwards because of her weight.
Both mother and daughter move backward, but the daughter moves with greater acceleration. Therefore, option (J) is correct.
What is Newton's second law of motion?This law states that the acceleration of a body depends on two factors. The first one is the mass of the body and the net force acting on the body. The acceleration is inversely proportional to the mass and is directly proportional to the net force acting on the body and
Newton’s second law can be written in an expression as follows:
F = ma
or, a = F/m
According to Newton's third law of motion, there is an equal and opposite reaction for every action. So when the two skaters push each other. They both will move in a backward direction.
But the mass of the daughter is 50 Kg while the mass of the mother is 100 Kg. As the mass of the daughter is less. Therefore daughter moves in a backward direction with greater acceleration.
Therefore, both move backward but the daughter moves with greater acceleration in comparison to the mother.
Learn more about Newton's second law of motion, here:
brainly.com/question/13447525
#SPJ2
Catching a wave, a 77 kg surfer starts with a speed of 1.3 m/s, drops through a height of 1.65 m, and ends with a speed of 8.2 m/s. How much non-conservative work was done on the surfer?
Answer:
Explanation:
The total work done by the wave is expressed as;
Workdone = Potential energy + Kinetic energy
Workdone = mgh + 1/2mv²
m is the mass = 77kg
g is the acceleration due to gravity = 9.8m/s²
v is the velocity = 8.2m/s
h is the height = 1.65m
Substitute into the formula;
Workdone = 77(9.8)(1.65) + 1/2(77)8.2²
Workdone = 1245.09 + 2588.74
Workdone = 3833.83Joules
Hence the amount of non conservative work done on the sofa is 3833.83Joules
Given:
Velocity, v = 8.2 m/sHeight, h = 1.65 mMass, m = 77 kgWe know,
→ [tex]Work \ done = Potential \ energy +Kinetic \ energy[/tex]
or,
[tex]= mgh +\frac{1}{2} mv^2[/tex]
By putting the values,
[tex]= 77\times 9.8\times 1.65+\frac{1}{2}\times 77\times (8.2)^2[/tex]
[tex]= 1245.09+2588.74[/tex]
[tex]= 3833.83 \ Joules[/tex]
Thus the above approach is right.
Learn more about work done here:
https://brainly.com/question/24230840
Give real-world examples of evidence that supports the evolution of Earth in each category:
Deposition -
Chemical weathering -
Volcanic eruption -
Answer:
real world examples
Explanation:
concave wave rock- The wave rock is formed by weathering of the surrounding area, which helps in proving the deposition part, as the wave rock was below the ground, occurred due to deposition of rock years over years. It is made from very tough material from its surrounding. The weathering reduced the surrounding terrain, while the bedrock remained to witness for the history. --Also, volcanic eruptions have changed earth greatly. EX- Ash and sulfur went into Earth's atmosphere because of the Tambora eruption which dimmed incoming sunlight. It lowered global temperatures by about 3°F. The Mount Pinatubo of 1991 that erupted in the Philippines cooled the planet by about 1°F.
A tuning fork vibrates at 15,660 oscillations every minute. What is the period (in seconds) of one back and forth vibration of the tuning fork?
We are given:
The tuning fork vibrates at 15660 oscillations per minute
Period of one back-and forth movement:
the given data can be rewritten as:
1 minute / 15660 oscillations
60 seconds / 15660 oscillations (1 minute = 60 seconds)
dividing the values
0.0038 seconds / Oscillation
Therefore, one back and forth vibration takes 0.0038 seconds
A soccer ball is kicked with an initial velocity of 35 m/s at an angle of 60 degrees from the horizontal. How far down the field will the ball hit the ground
Answer:
The ball will hit the ground 108.25 m down the field
Explanation:
To determine how far down the field the ball will hit the ground, that is the Range of the ball. From formula to calculate Range in projectile motion,
[tex]R = \frac{u^{2} sin2\theta }{g}[/tex]
Where R is the Range
u is the initial velocity
θ is the angle of projection
and g is acceleration due to gravity (Take g = 9.8 m/s²)
From the question,
Initial velocity, u = 35 m/s
Angle, θ = 60°
Putting these values into the equation, we get
[tex]R = \frac{35^{2} ( sin2(60)) }{9.8}[/tex]
[tex]R = \frac{1225 \times sin(120)}{9.8}[/tex]
[tex]R = \frac{1225 \times 0.8660}{9.8}[/tex]
[tex]R = \frac{1060.85}{9.8}[/tex]
[tex]R = 108.25 m[/tex]
Hence, the ball will hit the ground 108.25 m down the field.
what evidence do you that suggest water waves are transverse wave
Answer:
If you throw a pebble into a pond, ripples
spread out from where it went in. These
ripples are waves travelling through the
water. The waves move with a transverse
motion.
Explanation:
Just before it landed on the moon, the Apollo 12 lunar lander had a mass of 1.4×104 kg. What rocket thrust was necessary to have the lander touch down with zero acceleration?
Answer:
[tex]\mathbf{F_{thrust} \simeq 2.3 \times 10^4 \ N}[/tex]
Explanation:
For a lunar lander acceleration in an upward direction, the net force that acts on it can be expressed as:
[tex]F_{net} = F_{thrust} - F_{g(moon)}[/tex]
The equation for the net force that acts on the lunar lander is:
[tex]F_{net} = m_La[/tex]
For the lander touch down to be zero acceleration, it implies that the acceleration of the moving rocket is zero(a free body fall)
i.e. a = 0
We can now regard the Apollo 12 lunar as a freely falling body
However; the force of gravity as a result of the moon acting on the lunar rocket is:
[tex]F_{g(moon)} = m_Lg_m[/tex]
Then; the equation for the thrust force of the lunar rocket is:
[tex]F_{net} = F_{thrust} - F_{g(moon)}[/tex]
[tex]m_La = F_{thrust}-m_Lg_m[/tex]
[tex]m_L(0)= F_{thrust}-m_Lg_m[/tex]
[tex]0= F_{thrust}-m_Lg_m[/tex]
[tex]-F_{thrust}= -m_Lg_m[/tex]
[tex]F_{thrust}= m_Lg_m[/tex]
Finally; the thrust force of the lunar rocket can be determined as:
[tex]F_{thrust}= m_Lg_m[/tex]
Acceleration due to gravity ot the surface of the moon = 1.625 m/s²
[tex]F_{thrust}=(1.4 \times 10^{4} \ kg) (1.625 \ m/s^2)[/tex]
[tex]F_{thrust}=2.275 \times 10^4 \ N[/tex]
[tex]\mathbf{F_{thrust} \simeq 2.3 \times 10^4 \ N}[/tex]
A thrust force of 22680N will be necessary to to have the lander touch down with zero acceleration.
Given the data in the question;
Mass of Apollo 12 lunar lander; [tex]m_L = 1.4 * 10^4 kg[/tex]
Using the expression for the net-force that acts on the lander when it is accelerated in the upward direction:
[tex]F_{net} = F_{thrust} - F_{g(moon)}[/tex] ------- Let this be equation 1
Also, the net-force that acts on the lunar lander can be expressed as:
[tex]F_{net} = m_L*a[/tex] ------- Let this be equation 2
Where [tex]m_L[/tex] is the mass of the lunar lander and a is the acceleration ( 0 ),
( the acceleration of the moving rocket is zero and its freely falling )
Also, Force of gravity due to moon that acts on the lunar lander is expressed as:
[tex]F_{g(moon)} = m_L * g_m[/tex] ------- Let this be equation 3
Where [tex]m_L[/tex] is the mass of the lunar lander and [tex]g_m[/tex] is the moon's gravity ( [tex]1.62m/s^2[/tex] )
Lets substitute equation 2 and 3 into 1
[tex]m_L * a = F_{thrust} - (m_L * g_m)\\\\m_L * 0= F_{thrust} - (m_L * g_m)\\\\0=F_{thrust} - (m_L * g_m)\\\\ F_{thrust} = m_L * g_m[/tex]
We substitute our values into the equation
[tex]F_{thrust} = (1.4*10^4kg) * 1.62m/s^2\\\\F_{thrust} = 22680 kg.m/s^2\\\\F_{thrust} = 22680N[/tex]
Therefore, a thrust force of 22680N will be necessary to to have the lander touch down with zero acceleration.
Learn more: https://brainly.com/question/7602665
Two balls collide – one heavy and one light. Why will the light one roll farther back than the heavy one?
Answer:
During a collision, some of the ball's energy is converted into heat. As no energy is added to the ball, the ball bounces back with less kinetic energy and cannot reach quite the same height. The heavy ball, on the other hand, is left behind with little energy and does not move much.
Your friend clams that objects do not have to be thouching for a magnetic force to cause motion. How would you support your friends claim? Due today
Help me guys please with this question
Answer:
[tex]\mid \vec C\mid=31.9[/tex]
Explanation:
Consider the vectors:
[tex]\vec A=9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}}[/tex]
[tex]\vec B=-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}}[/tex]
Calculate the magnitude of
[tex]\vec C=-2\vec B-\vec A[/tex]
Substitute the values of the vectors:
[tex]\vec C=-2(-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}})-(9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}})[/tex]
Operate and remove parentheses:
[tex]\vec C=19\mathbf{\hat{i}}+26.8\mathbf{\hat{j}}-9.4\mathbf{\hat{i}}+3.6\mathbf {\hat{j}}[/tex]
Operating both components separately:
[tex]\vec C=9.6\mathbf{\hat{i}}+30.4\mathbf{\hat{j}}[/tex]
Now find the magnitude of C:
[tex]\mid \vec C\mid=\sqrt{9.6^2+30.4^2}[/tex]
[tex]\mid \vec C\mid=\sqrt{1016.32}[/tex]
[tex]\mathbf{\mid \vec C\mid=31.9}[/tex]
