The percentage by which the water density increased is 4.1[tex]\mathbf{\overline 6}[/tex] %
The known values are;
The increase in pressure per 10 meter increase in depth = 1.0 atm
The depth of the deepest ocean = 12 km = 12,000 m
The compressibility of the ocean = 5.0 × 10⁻⁵ 1/atm
The unknown
The percentage the density of water increased in the deepest ocean
Strategy;
Find the pressure at the deepest point of the deepest ocean and apply the compressibility
We have;
[tex]\mathbf{Compressibility = \dfrac{1}{V} \times \dfrac{\partial V}{\partial p}}[/tex]
The change in pressure, [tex]\partial p[/tex] = (12,000 m/(10 m)) × 1.0 atm = 1,200 atm
Therefore, we have for one cubic meter of water
[tex]\mathbf{5.0 \times 10^{-5} \ atm^{-1} = \dfrac{1}{1 \, m^3} \times \dfrac{\partial V}{1,200 \, atm}}[/tex]
Therefore;
[tex]\mathbf{\partial}[/tex]V = 5.0 × 10⁻⁵ atm⁻¹ × 1 m³ × 1,200 atm = 0.06 m³
The new volume = V - [tex]\mathbf{\partial}[/tex]V
∴ The new volume = 1 m³ - 0.06 m³ = 0.94 m³
The initial density = mass/(1 m³)
The new density = mass/(0.96 m³)
The percentage increase in density, [tex]\partial[/tex]ρ%, is given as follows;
[tex]\mathbf{\partial p \% = \dfrac{ \dfrac{Mass}{0.96 \ m^3} - \dfrac{Mass}{1 \ m^3} }{ \dfrac{Mass}{1 \ m^3}} \times 100 = \dfrac{25}{6} \% = 4.1 \overline 6 \%}[/tex]
∴ [tex]\mathbf{\partial}[/tex]ρ% = 4.1[tex]\mathbf {\overline 6}[/tex] %
The percentage by which the water density increased, [tex]\partial[/tex]ρ% = 4.1[tex]\mathbf{\overline 6}[/tex] %
Learn more about compressibility here;
https://brainly.com/question/18746977
A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is n = 1.50, how thick would you make the coating?
Answer:
[tex]t=0.50cm[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lamda=3c[/tex]m
Refraction Index [tex]n=1.50[/tex]
Generally the equation for Destructive interference for Normal incidence is mathematically given by
[tex]2nt=m(\frac{1}{2})\lambda[/tex]
Since Minimum Thickness occurs at
At [tex]m=0[/tex]
Therefore
[tex]t=\frac{\lambda}{2}[/tex]
[tex]t=\frac{3}{4(1.50)}[/tex]
[tex]t=0.50cm[/tex]
An eagle flying at 35 m/s emits a cry whose frequency is 440 Hz. A blackbird is moving in the same direction as the eagle at 10 m/s. (Assume the speed of sound is 343 m/s.)
(a) What frequency does the blackbird hear (in Hz) as the eagle approaches the blackbird?
Hz
(b) What frequency does the blackbird hear (in Hz) after the eagle passes the blackbird?
Hz
Answer:
a) [tex]F=475.7Hz[/tex]
b) [tex]F'=410.899Hz[/tex]
Explanation:
From the question we are told that:
Velocity of eagle [tex]V_1=35m/s[/tex]
Frequency of eagle [tex]F_1=440Hz[/tex]
Velocity of Black bird [tex]V_2=10m/s[/tex]
Speed of sound [tex]s=343m/s[/tex]
a)
Generally the equation for Frequency is mathematically given by
[tex]F=f_0(\frac{v-v_2}{v-v_1})[/tex]
[tex]F=440(\frac{343-10}{343-35})[/tex]
[tex]F=475.7Hz[/tex]
b)
Generally the equation for Frequency is mathematically given by
[tex]F'=f_0(\frac{v+v_2}{v+v_1})[/tex]
[tex]F'=440(\frac{343+10}{343+35})[/tex]
[tex]F'=410.899Hz[/tex]
Question 4 of 5
How can the Fitness Logs help you in this class?
O A. They can't; the Fitness Logs are only useful to your teacher.
B. They show your parents how much you're learning.
C. They let you keep track of your thoughts, feelings, and progress.
D. They help you evaluate yourself for your final grade.
SUBMIT
Answer:
C is the right answer
Explanation:
fitness logs is a great way to track your progress. You can easily look back and see how you have progressed over time. In addition, it can help you plan and prepare for future workouts, as well as identify patterns of what seems to work well for you and when you have the most success
hope it was useful for you
A 0.20 kg mass on a horizontal spring is pulled back a certain distance and released. The maximum speed of the mass is measured to be 0.30 m/s. If, instead, a 0.40 kg mass were used in this same experiment, choose the correct value for the maximum speed.
a. 0.40 m/s.
b. 0.20 m/s.
c. 0.28 m/s.
d. 0.14 m/s.
e. 0.10 m/s.
Answer:
b. 0.20 m/s.
Explanation:
Given;
initial mass, m = 0.2 kg
maximum speed, v = 0.3 m/s
The total energy of the spring at the given maximum speed is calculated as;
K.E = ¹/₂mv²
K.E = 0.5 x 0.2 x 0.3²
K.E = 0.009 J
If the mass is changed to 0.4 kg
¹/₂mv² = K.E
mv² = 2K.E
[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 0.009}{0.4} } \\\\v = 0.21 \ m/s\\\\v \approx 0.20 \ m/s[/tex]
Therefore, the maximum speed is 0.20 m/s
In a robotics circuit, a voltage source of 75V is supplying a current, I to a series circuit of 5
resistances. Resistance, R1 = 5 KΩ and R2 = 10 KΩ. The voltage drops across 3 black boxes of
resistances R3 , R4 and R5 are 15V, 20V and 25V respectively. The current through the black
box of resistance, R5 is measured as 1mA. Calculate the voltage V1 and V2 across the
resistance R1 and R2 using the Voltage Divider Rule.
Answer:
In the given circuit, R
2
,R
6
and R
4
are in series. So,
R
1
′
=7+5+12=24Ω
Now R
1
′
and R
5
′
are in parallel. So,
R
2
′
1
=
8
1
+
24
1
=
24
3+1
=
24
4
=
6
1
R
2
′
=6ohm.
Now R
2
′
,R
1
and R
3
are in series. So,
R=R
2
′
+R
1
+R
3
=6+3+2=11ohm.
We know i=
R+r
E
=
11+1
6
=
12
6
=
2
1
i=0.5amp.
Two positive charges ( 8.0 mC and 2.0 mC) are separated by 300 m. A third charge is placed at distance r from the 8.0 mC charge in such a way that the resultant electric force on the third charge due to the other two charges is zero. The distance r is
Answer:
[tex]r=200m[/tex]
Explanation:
From the question we are told that:
Charges:
[tex]Q_1=8.0mC[/tex]
[tex]Q_2=2.0mC[/tex]
[tex]Q_3=8.mC[/tex]
Distance [tex]d=300m[/tex]
Generally the equation for Force is mathematically given by
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Therefore
[tex]F_{32}=F_{31}[/tex]
[tex]\frac{q_2}{(300-r)^2}=\frac{q_1}{r^2}[/tex]
[tex]\frac{2*10^{-3}}{(300-r)^2}=\frac{8*10^{-3}}{r^2}[/tex]
[tex]r=2(300-r)[/tex]
[tex]r=200m[/tex]
A 50.0 kg person is walking horizontally with constant acceleration of 0.25 m/s² inside an elevator. The elevator is also accelerating downward at a rate of 1.0 m/s². Sketch the path of the man as it is observed from someone on the ground. Explain your choice.
Answer:
The acceleration is in 2 D as in between east and south.
Explanation:
mass, m = 50 kg
acceleration, a = 0.25 m/s^2 horizontal
acceleration of elevator, a' = 1 m/s^2 downwards
When a person on the ground the resultant acceleration of the person with respect to the ground is between east and south direction so the path os parabolic in nature. It graph is shown below:
The driver provides a constant force on the engine through the foot pedal. Eventually the van stops accelerating and reaches a constant speed.
c Explain why the van reaches a constant speed if the driver provides a constant driving force to the van.
It follows from Newton's second law that there is some counteractive force that cancels out the force exerted by the engine - it's most likely drag due to air resistance in combination with static friction between the tires and the road. The car is moving at constant speed past a certain point, so the net force on the car is
∑ F = (force from engine) - (resistive forces) = 0
Consider an airplane with a total wing surface of 50 m^2. At a certain speed the difference in air pressure below and above the wings is 4.0 % of atmospheric pressure.
Required:
Find the lift on the airplane.
Answer:
[tex]F=202650N[/tex]
Explanation:
From the question we are told that:
Area [tex]a=50m^2[/tex]
Difference in air Pressure [tex]dP=4.0\% atm=>0.04*101325=>4035Pa[/tex]
Generally the equation for Force is mathematically given by
[tex]F=dP*A[/tex]
[tex]F=4053*50[/tex]
[tex]F=202650N[/tex]
A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 15-m-long cable. You are (unwisely!) standing directly beneath the point from which the wrecking ball is hung when you notice that the ball has just been released and is swinging directly toward you. How much time do you have to move out of the way? answer in seconds.
Answer:
Time to move out of the way = 1.74 s
Explanation:
Time to move out of the way is one fourth of period = 6.95/4 = 1.74 seconds.
Time to move out of the way = 1.74 s
A wheel rotates at an angular velocity of 30rad/s. If an acceleration of 26rad/s2 is applied to it, what will its angular velocity be after 5.0s
True or false: Increasing the Young’s modulus of a beam in bending will cause it to deflect less.
Answer:
false?
Explanation:
The higher the modulus, the more stress is needed to create the same amount of strain; an idealized rigid body would have an infinite Young's modulus.
Answer:
I think the answer is False.
Put the balloon near (BUT NOT TOUCHING) the wall. Leave about as much space as the width of your pinky finger between the balloon and wall. Does the balloon move, if so which way
Answer:
Move towards the wall.
Explanation:
When the balloon is kept near to the wall not touching the wall, there is a force of electrostatic attraction so that the balloon moves towards the wall and stick to it.
As there is some charge on the balloon and the wall is uncharged so the force is there due to which the balloon moves towards the wall.
In which type of mixture do the physically distinct component parts each have distinct properties?
Answer:
In heterogeneous mixture do the physically distinct component parts each have distinct properties.
A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm. Fwithout belt
This question is incomplete, the complete question is;
Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.
1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?
Fwith belt =
2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.
Fwithout belt =
Answer:
1) The Net force on the driver with seat belt is 10.3 KN
2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN
Explanation:
Given the data in the question;
from the equation of motion, v² = u² + 2as
we solve for a
a = (v² - u²)/2s ----- let this be equation 1
we know that, F = ma ------- let this be equation 2
so from equation 1 and 2
F = m( (v² - u²)/2s )
where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.
1)
Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.
i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m
so we substitute the given values into the equation;
F = 70( ((0)² - (18)²) / 2 × 1.1 )
F = 70 × ( -324 / 2.4 )
F = 70 × -147.2727
F = -10309.09 N
F = -10.3 KN
The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.
Fwith belt = 10.3 KN
Therefore, Net force of the driver is 10.3 KN
2)
No sit belt,
m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m
we substitute
F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )
F = 70 × ( -324 / 0.022 )
F = 70 × -14727.2727
F = -1030909.08 N
F = -1030.9 KN
The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.
Fwithout belt = 1030.9 KN
Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN
Two objects moving with a speed v travel in opposite directions in a straight line. The objects stick together when they collide, and move with a speed of v/2 after the collision.
Required:
a. What is the ratio of the final kinetic energy of the system to the initial kinetic energy?
b. What is the ratio of the mass of the more massive object to the mass of the less massive object?
Answer:
Explanation:
Let the mass of objects be m₁ and m₂ .
Total kinetic energy = 1/2 m₁ v² + 1/2 m₂ v²= 1/2 ( m₁ + m₂ ) v²
Total kinetic energy after collision= 1/2 ( m₁ + m₂ ) v² / 4 = 1/2 ( m₁ + m₂ ) v² x .25
final KE / initial KE = 1/2 ( m₁ + m₂ ) v² x .25 / 1/2 ( m₁ + m₂ ) v²
= 0.25
b )
Applying law of conservation of momentum to the system . Let m₁ > m₂
m₁ v - m₂ v = ( m₁ + m₂ ) v / 2
m₁ v - m₂ v = ( m₁ + m₂ ) v / 2
m₁ - m₂ = ( m₁ + m₂ ) / 2
2m₁ - 2 m₂ = m₁ + m₂
m₁ = 3m₂
m₁ / m₂ = 3 / 1
Answer:
(a) The ratio is 1 : 4.
(b) The ratio is 1 : 3.
Explanation:
Let the mass of each object is m and m'.
They initially move with velocity v opposite to each other.
Use conservation of momentum
m v - m' v = (m + m') v/2
2 (m - m') = (m + m')
2 m - 2 m' = m + m'
m = 3 m' .... (1)
(a) Let the initial kinetic energy is K and the final kinetic energy is K'.
[tex]K = 0.5 mv^2 + 0.5 m' v^2 \\\\K = 0.5 (m + m') v^2..... (1)[/tex]
[tex]K' = 0.5 (m + m') \frac{v^2}{4}.... (2)[/tex]
The ratio is
K' : K = 1 : 4
(b) m = 3 m'
So, m : m' = 3 : 1
Galaxies that are 400 million light years away have a red shift of 0.03 approximately. A radio wave coming from one of these galaxies has an observed wavelength of 125 meters. What is the emitted wavelength in meters and the observed frequency in Megahertz
Answer:
The correct answer is "121.36 meters and 2.40 MHz".
Explanation:
Given:
Red shift,
[tex]\frac{v}{c}=\frac{\lambda - \lambda_0}{\lambda_0} = 0.03[/tex]
Wavelength,
[tex]\lambda = 125 \ meters[/tex]
The observed frequency will be:
⇒ [tex]f = \frac{c}{\lambda}[/tex]
[tex]=\frac{3\times 10^8}{125}[/tex]
[tex]=24\times 10^5[/tex]
[tex]= 2.40\times 10^6[/tex]
[tex]=2.40 \ MHz[/tex]
hence,
The Emitted wavelength will be:
⇒ [tex]\frac{125-\lambda_0}{\lambda_0} =0.03[/tex]
[tex]\frac{125}{\lambda_0}-1 =0.03[/tex]
[tex]125=1.03 \ \lambda_0[/tex]
[tex]\lambda_0=\frac{125}{1.03}[/tex]
[tex]=121.36 \ m[/tex]
Hi,A body changes its velocity from 60 km/hr to 72 km/hr in 2 sec.Find the acceleration and distance travelled.
Answer:
Initial velocity, u = 60 km/h = 16.7 m/s
Final velocity, v = 72 km/h = 20 m/s
time, t = 2 sec
From first equation of motion:
[tex]{ \bf{v = u + at}}[/tex]
Substitute the variables:
[tex]{ \tt{20 = 16.7 + (a \times 2)}} \\ { \tt{2a = 3.3}} \\ { \tt{acceleration = 1.65 \: {ms}^{ - 2} }}[/tex]
A hockey ball is flicked of the ground with initial velocity of 2.0m/s upwards and 10m/s horizontally. Calculate the distance travelled from the point where the ball is flicked and to the point where the ball hits the ground.
Answer:
imma try and fail again and again
A pulse traveled the length of a stretched spring the pulse transferred...A)energy only B)mass only C)both energy and mass D) neither energy nor mass
Answer:
A
Explanation:
So a pulse is a part of a mechanical wave, and mechanical waves are energy transfer trough some medium, in this case a stretched spring. So the correct answer is (A) energy only. The pulse cant be transferred into mass.
g A CD is spinning on a CD player. You open the CD player to change out the disk and notice that the CD comes to rest after 15 revolutions with a constant deceleration of 120 r a d s 2 . What was the initial angular speed of the CD
Answer:
[tex]\omega_1=150rads/sec[/tex]
Explanation:
From the question we are told that:
Number of Revolution [tex]N=15=30\pi[/tex]
Deceleration [tex]d= -120 rads/2[/tex]
Generally the equation for initial angular speed [tex]\omega_1[/tex] is mathematically given by
[tex]\omega_2^2=\omega_1^2 +2(d)(N)[/tex]
[tex]0=\omega_1^2 +2(-120)(20 \pi)[/tex]
[tex]\omega_1^2=7200 \pi[/tex]
[tex]\omega_1=150rads/sec[/tex]
An infinite plane lies in the yz-plane and it has a uniform surface charge density.
The electric field at a distance x from the plane
a.) decreases as 1/x^2
b.) increases linearly with x
c.) is undertermined
d.) decreases linearly with x
e.) is constant and does not depend on x
Answer:
So the correct answer is letter e)
Explanation:
The electric field of an infinite yz-plane with a uniform surface charge density (σ) is given by:
[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]
Where ε₀ is the electric permitivity.
As we see, this electric field does not depend on distance, so the correct answer is letter e)
I hope it helps you!
If you tethered a space station to the earth by a long cable, you could get to space in an elevator that rides up the cable much simpler and cheaper than riding to space on a rocket. There's one big problem, however: There is no way to create a cable that is long enough. The cable would need to reach 36,000 km upward, to the height where a satellite orbits at the same speed as the earth rotates; a cable this long made of ordinary materials couldn't even support its own weight. Consider a steel cable suspended from a point high above the earth. The stress in the cable is highest at the top; it must support the weight of cable below it.
What is the greatest length the cable could have without failing?
Answer:
[tex]l=12916.5m[/tex]
Explanation:
Distance [tex]d=3600km[/tex]
Since
Density of steel [tex]\rho=7900kg/m^3[/tex]
Stress of steel [tex]\mu= 1*10^9[/tex]
Generally the equation for Stress on Cable is mathematically given by
[tex]S=\frac{F}{A}[/tex]
[tex]S=\frac{\rho Alg}{A}[/tex]
Therefore
[tex]l=\frac{s}{\rhog}[/tex]
[tex]l=\frac{ 1*10^9}{7900kg/m^3*9.8}[/tex]
[tex]l=12916.5m[/tex]
An artificial satellite circling the Earth completes each orbit in 125 minutes. (a) Find the altitude of the satellite.(b) What is the value of g at the location of this satellite?
Answer:
(a) Altitude = 1.95 x 10⁶ m = 1950 km
(b) g = 5.9 m/s²
Explanation:
(a)
The time period of the satellite is given by the following formula:
[tex]T^2 = \frac{4\pi^2r^3}{GM_E}[/tex]
where,
T = Time period = (125 min)([tex]\frac{60\ s}{1\ min}[/tex]) = 7500 s
r = distance of satellite from the center of earth = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
[tex]M_E[/tex] = Mass of Earth = 6 x 10²⁴ kg
Therefore,
[tex](7500\ s)^2 = \frac{4\pi^2r^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}\\\\r^3 = \frac{(7500\ s)^2(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}{4\pi^2}\\\\r = \sqrt[3]{5.7\ x\ 10^{20}\ m^3} \\[/tex]
r = 8.29 x 10⁶ m
Hence, the altitude of the satellite will be:
[tex]Altitude = r - radius\ of\ Earth\\Altitude = 8.29\ x\ 10^6\ m - 6.34\ x\ 10^6\ m[/tex]
Altitude = 1.95 x 10⁶ m = 1950 km
(b)
The weight of the satellite will be equal to the gravitational force between satellite and Earth:
[tex]Weight = Gravitational\ Force\\\\M_sg = \frac{GM_EM_s}{r^2}\\\\g = \frac{GM_E}{r^2}\\\\g = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}{(8.23\ x\ 10^6\ m)^2}[/tex]
g = 5.9 m/s²
A soap bubble, when illuminated at normal incidence with light of 463 nm, appears to be especially reflective. If the index of refraction of the film is 1.35, what is the minimum thickness the soap film can be if it is surrounded by air
Answer:
the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm
Explanation:
Given the data in the question;
wavelength of light; λ = 463 nm = 463 × 10⁻⁹ m
Index of refraction; n = 1.35
Now, the thinnest thickness of the soap film can be determined from the following expression;
[tex]t_{min[/tex] = ( λ / 4n )
so we simply substitute in our given values;
[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 4(1.35)
[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 5.4
[tex]t_{min[/tex] = ( 463 × 10⁻⁹ m ) / 4(1.35)
[tex]t_{min[/tex] = 8.574 × 10⁻⁸ m
[tex]t_{min[/tex] = 85.74 × 10⁻⁹ m
[tex]t_{min[/tex] = 85.74 nm
Therefore, the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm
When an external magnetic flux through a conducting loop decreases in magnitude, a current is induced in the loop that creates its own magnetic flux through the loop. How does that induced magnetic flux affect the total magnetic flux through the loop
Answer:
Len's law
Explanation:
We can explain this exercise using Len's law
when the magnetic flux decreases, a matic flux appears that opposes the decrease, thus maintaining the value of the initial luxury.
The mass of a hot-air balloon and its occupants is 381 kg (excluding the hot air inside the balloon). The air outside the balloon has a pressure of 1.01 x 105 Pa and a density of 1.29 kg/m3. To lift off, the air inside the balloon is heated. The volume of the heated balloon is 480 m3. The pressure of the heated air remains the same as that of the outside air. To what temperature in kelvins must the air be heated so that the balloon just lifts off
Answer:
In order to lift off the ground, the air in the balloon must be heated to 710.26 K
Explanation:
Given the data in the question;
P = 1.01 × 10⁵ Pa
V = 480 m³
ρ = 1.29 kg/m³
M = 381 kg
we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol
let F represent the force acting upward.
Now in a condition where the hot air balloon is just about to take off;
F - Mg - m[tex]_g[/tex]g = 0
where M is the mass of the balloon and its occupants, m[tex]_g[/tex] is the mass of the hot gas inside the balloon.
the force acting upward F = Vρg
so
Vρg - Mg - m[tex]_g[/tex]g = 0
solve for m[tex]_g[/tex]
m[tex]_g[/tex] = ( Vρg - Mg ) / g
m[tex]_g[/tex] = Vρg/g - Mg/g
m[tex]_g[/tex] = ρV - M ------- let this be equation 1
Now, from the ideal gas law, PV = nRT
we know that number of moles n = m[tex]_g[/tex] / μ
where μ is the molecular mass of air
so
PV = (m[tex]_g[/tex]/μ)RT
solve for T
μPV = m[tex]_g[/tex]RT
T = μPV / m[tex]_g[/tex]R -------- let this be equation 2
from equation 1 and 2
T = μPV / (ρV - M)R
so we substitute in our values;
P = 1.01 × 10⁵ Pa
V = 480 m³
ρ = 1.29 kg/m³
M = 381 kg
we know that; R = 8.31 J/mol.K and the molecular mass of air μ = 29 × 10⁻³ kg/mol
T = [ (29 × 10⁻³) × (1.01 × 10⁵) × 480 ] / [ (( 1.29 × 480 ) - 381)8.31 ]
T = 1405920 / 1979.442
T = 710.26 K
Therefore, In order to lift off the ground, the air in the balloon must be heated to 710.26 K
The temperature required for the air to be heated is 710.26 K.
Given data:
The mass of a hot air-balloon is, m = 381 kg.
The pressure of air outside the balloon is, [tex]P = 1.01 \times 10^{5} \;\rm Pa[/tex].
The density of air is, [tex]\rho = 1.29 \;\rm kg/m^{3}[/tex].
The volume of heated balloon is, [tex]V = 480 \;\rm m^{3}[/tex].
The condition where the hot air balloon is just about to take off is as follows:
[tex]F-mg - m'g =0[/tex]
Here,
m' is the mass of hot gas inside the balloon and g is the gravitational acceleration and F is the force acting on the balloon in upward direction. And its value is,
[tex]F = V \times \rho \times g[/tex]
Solving as,
[tex](V \times \rho \times g)-mg - m'g =0\\\\ m'=(V \rho )-m[/tex]
Now, apply the ideal gas law as,
PV = nRT
here, R is the universal gas constant and n is the number of moles and its value is,
[tex]n=\dfrac{m'}{M}[/tex]
M is the molecular mass of gas. Solving as,
[tex]PV = \dfrac{m'}{M} \times R \times T\\\\\\T=\dfrac{P \times V\times M}{m'R}\\\\\\T=\dfrac{P \times V\times M}{(V \rho - m)R}[/tex]
Since, the standard value for the molecular mass of air is, [tex]M = 29 \times 10^{-3} \;\rm kg/mol[/tex]. Then solve for the temperature as,
[tex]T=\dfrac{(1.01 \times 10^{5}) \times 480\times 381}{(480 \times (1.29) - 381)8.31}\\\\\\T = 710.26 \;\rm K[/tex]
Thus, we can conclude that the temperature required for the air to be heated is 710.26 K.
Learn more about the ideal gas equation here:
https://brainly.com/question/18518493
A child weighing 200 N is being held back in a swing by a horizontal force of 125 N, as shown in the image. What is the tension T in the rope that supports the swing in units of Newtons? Note: Please enter only the numerical answer. If you include any units in your answer, your answer will be counted as incorrect. T F= 125 N Weight = 200 N
Answer:
75
Explanation:
i am not sure but if 200N boy is being held back then the force that's holding him back must be equal to or greater than his weight. if 125N is already exerted then the tension will be:
T=200-125
= 75
When a car's starter is in use, it draws a large current. The car's lights draw much less current. As a certain car is starting, the current through the battery is 54.0 A and the potential difference across the battery terminals is 9.18 V. When only the car's lights are used, the current through the battery is 2.10 A and the terminal potential difference is 12.6 V. Find the battery's emf.
Answer:
12.74 V
Explanation:
We are given that
Current, I1=54 A
Potential difference, V1=9.18V
I2=2.10 A
V2=12.6 V
We have to find the battery's emf.
[tex]E=V+Ir[/tex]
Using the formula
[tex]E=9.18+54r[/tex] ....(1)
[tex]E=12.6+2.10r[/tex] .....(2)
Subtract equation (1) from (2)
[tex]0=3.42-51.9r[/tex]
[tex]3.42=51.9r[/tex]
[tex]r=\frac{3.42}{51.9}=0.0659ohm[/tex]
Using the value of r in equation (1)
[tex]E=9.18+54(0.0659)[/tex]
[tex]E=12.74 V[/tex]
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