Answer:
a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.
b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2
Explanation:
The expression for the angle of refraction is
n₁ sin θ₁ = n₂ sin θ₂
refractive index n₁ is for incident light and n₂ is for transmitted light.
We have two cases
a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.
b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2
Cho điện cực dưới điện cực trên . Hàm thế biến thiên theo qui luật:
Xác định sự phân bố điện tích khối và .
Answer:
Quantum theory gives the concept of
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]
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 is released from the top of the incline and slides down the incline. The maximum velocity (taken the instant before the mass reaches the bottom of the incline) is 1.06 m/s. What is the kinetic energy at that time
Answer:
0.28 J
Explanation:
Let the mass of the object is 0.5 kg
The maximum velocity of the object is 1.06 m/s.
We need to find the kinetic energy at that time. It is given by :
[tex]K=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times 0.5\times (1.06)^2\\\\K=0.28\ J[/tex]
So, the required kinetic energy is equal to 0.28 J.
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.
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.
abrief history of hand writing
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]
An empty parallel plate capacitor is connected between the terminals of a 18.8-V battery and charges up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Answer:
[tex]p.d' = 37.6 V[/tex]
Explanation:
From the question we are told that:
Potential difference [tex]p.d=18.8V[/tex]
New Capacitor [tex]C_1=C_2/2[/tex]
Generally the equation for Capacitor capacitance is mathematically given by
[tex]C=\frac{eA}{d}[/tex]
Generally the equation for New p.d' is mathematically given by
[tex]C_2V=C_1*p.d'[/tex]
[tex]p.d' = 2V[/tex]
[tex]p.d'= 2 * 18.8[/tex]
[tex]p.d' = 37.6 V[/tex]
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
3. If you change the resistance of the resistor:
a. How does the current through the circuit change? (answer, explain, evidence)
b. How does the voltage of the battery change? (answer, explain, evidence)
Answer:
Explanation:
Changing the resistance of a resistor means the resistance is either increased or decreased.
a. When the resistance of the resistor is increased, the value of current flowing through the circuit decreases.
Example: given voltage of 6V, and a resistance of 30 Ohm's. The value of current flowing in the circuit is;
V = IR
6 = I x 30
I = 0.2 A
If the resistance is changed to 50 Ohm's, then:
I = 0.12 A
(ii) When the resistance of the resistor is decreased, the value of the current flowing through the circuit increases.
In the previous example, if the resistance is changed to 5 Ohm's, then:
V = IR
6 = I x 5
I = 1.2 A
(b) The voltage of the battery does not change since it is directly proportional to the current flowing through the circuit. Consider the examples stated above.
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
b. The stream of water flowing through a hole at depth h = 10 cm in a tank holding water to height H = 40 cm. . 3 At what distance x does the stream strike the floor?
Answer:
34.64 cm
Explanation:
Given that:
The depth of the hole h = 10 cm
height of the water holding in the tank H = 40 cm
For a stream of flowing water, the distance (x) at which the stream strikes the floor can be computed by using the formula;
[tex]x = 2 \sqrt{h(H-h)}[/tex]
[tex]x = 2 \sqrt{10(40-10)}[/tex]
[tex]x = 2 \sqrt{10(30)}[/tex]
[tex]x = 2 \sqrt{300}[/tex]
[tex]x = 2 \times 17.32[/tex]
x = 34.64 cm
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
Place each description under the correct theory
Gravity is an attractive force.
Universal Law of Gravitation
General Theory of Relativity
Mass and distance affect force.
Time and space are absolute,
Time and space are relative.
Gravity is due to space-time curving.
Mass affects space-time curving.
Answer:
1) Law of Universal Gravitation Gravity is an attractive force
5) General relativity Gravity is due to the curvature of spacetime
Explanation:
In this exercise you are asked to relate the correct theory and its explanation
Theory Explanation
1) Law of Universal Gravitation Gravity is an attractive force
2) Law of universal gravitation Mass and distance affect force
3) Classical mechanics time and space are absolute
4) Special relativity Time and space are relative
5) General relativity Gravity is due to the curvature of
spacetime
6) General relativity Mass affects the curvature of space - time
Answer:
Explanation:
edge2022
If the average time it takes for the cart from point 1 to point 2 is 0.2 s, calculate the angle θ from the horizontal of the track. Assume the track is frictionless. Hint: use the definitions of acceleration and Newton’s second law.
Answer:
hehe
Explanation:
I dont know because I am a noob ant study
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]
5. For the speaker in this circuit, the voltage across it is always proportional to the current through it. Find the maximum amount of power that the circuit can deliver to the speaker.
Answer:
speaker64
--------
34x
Explanation:
64-34
x
speaker
4
2
4
788
- circuit
voltage
100000
x.34
Sorry but you have no picture shown
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.
A 2.0-kg block sliding on a rough horizontal surface is attached to one end of a horizontal spring (k = 250 N/m) which has its other end fixed. The block passes through the equilibrium position with a speed of 2.6 m/s and first comes to rest at a displacement of 0.20 m from equilibrium. What is the coefficient of kinetic friction between the block and the horizontal surface?
Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.
There are two forces performing work on the block: the restoring force of the spring and kinetic friction.
By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:
[tex]W_{\rm total}=\Delta K[/tex]
or
[tex]W_{\rm friction}+W_{\rm spring}=0-K=-K[/tex]
where K is the block's kinetic energy at the equilibrium point,
[tex]K=\dfrac12\left(2.0\,\mathrm{kg}\right)\left(2.6\dfrac{\rm m}{\rm s}\right)^2=6.76\,\mathrm J[/tex]
Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is
[tex]W_{\rm spring}=-\dfrac12\left(250\dfrac{\rm N}{\rm m}\right)(0.20\,\mathrm m)^2=-5.00\,\mathrm J[/tex]
Compute the work performed by friction:
[tex]W_{\rm friction}-5.00\,\mathrm J=-6.76\,\mathrm J \implies W_{\rm friction}=-1.76\,\mathrm J[/tex]
By Newton's second law, the net vertical force on the block is
∑ F = n - mg = 0 ==> n = mg
where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.
So we have
[tex]W_{\rm friction}=-f(0.20\,\mathrm m)[/tex]
[tex]\implies -1.76\,\mathrm J=-\mu\left(2.0\,\mathrm{kg}\right)\left(9.8\dfrac{\rm m}{\mathrm s^2}\right)(0.20\,\mathrm m)[/tex]
[tex]\implies \boxed{\mu\approx0.45}[/tex]
The coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]
What is coefficient of friction?Coefficient of friction, ratio of the frictional force resisting the motion of two surfaces in contact to the normal force pressing the two surfaces together. It is usually symbolized by the Greek letter mu (μ). Mathematically, μ = F/N, where F is the frictional force and N is the normal force.
Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.
There are two forces performing work on the block: the restoring force of the spring and kinetic friction.
By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:
[tex]W_{total}=\Delta K[/tex]
or
[tex]W_{friction}+W_{spring}=0-K=-K[/tex]
where K is the block's kinetic energy at the equilibrium point,
[tex]K=\dfrac{1}{2}(2)(2.6)^2=6.76 \ J[/tex]
Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is
[tex]W_{spring}=-\dfrac{1}{2}(250)(0.20)=-5\ J[/tex]
Compute the work performed by friction:
[tex]W_{friction}-5 =-6.76\ J=-1.76\ J[/tex]
By Newton's second law, the net vertical force on the block is
∑ F = n - mg = 0 ==> n = mg
where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.
So we have
[tex]W_{friction}=-f(0.20)[/tex]
[tex]-1.76=\mu (2)(9.8)(0.2)[/tex]
[tex]\mu =0.45[/tex]
Thus the coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]
To know more about Coefficient of friction follow
https://brainly.com/question/136431
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!
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]
5. A big wheel has a diameter of 5 m and a mass of 1500 kg when fully laden with people. a) Work out the moment of inertia of the big wheel. (Hint: which shape from the ones given on p114 would be most suitable? b) When the wheel is rotating at full speed, a person has a linear velocity of 3 m/s. What is the angular velocity of this person? c) What is the rotational kinetic energy at this speed? d) A motor takes 10 seconds to accelerate the wheel from rest to a linear velocity on the circumference of 3 m/s. What is the power of the motor?
Answer:
a) I = 3.75 10⁴ kg m², b) w = 0.6 rad / s, c) K = 6.75 10³ J, d) P = 6.75 10² W
Explanation:
This is a rotations exercise
a) the proper shape for a wheel is that of a rim where most of the weight is in the circumference plus the point weights of the people sitting on its periphery.
We are going to approximate the reda with a thin ring
I = M r²
I = 1500 5²
I = 3.75 10⁴ kg m²
b) angular and linear velocity are related
v = w r
w = v / r
w = 3/5
w = 0.6 rad / s
c) the expression for kinetic energy is
K = ½ I w²
K = ½ 3.75 10⁴ 0.6²
K = 6.75 10³ J
d) the power is
P = W / t
to find the work we use the relationship between work and the variation of kinetic energy
W = ΔK = K_f - K₀
the system part of rest wo = 0
W = K_f
W = 6.75 10³ J
we calculate
P = 6.75 10³/10
P = 6.75 10² W
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
The thermal efficiency (in %) of a system that undergoes a power cycle while receiving 1000 kJ of energy by heat transfer from a hot reservoir at 1000 K and discharging 500 kJ of energy by heat transfer to a cold reservoir at 400 K is:
Answer:
η = 0.5 = 50%
Explanation:
The efficiency of the power cycle is given by the following formula:
[tex]\eta = \frac{W}{Q_1}\\\\\eta = \frac{Q_1-Q_2}{Q_1}[/tex]
where,
where,
η = efficiency = ?
Q₁ = heat received from hot reservoir = 1000 KJ
Q₂ = heat discharged to cold reservoir = 500 KJ
Therefore,
[tex]\eta = \frac{1000\ KJ-500\ KJ}{1000\ KJ}[/tex]
η = 0.5 = 50%
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.
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]
Explain how the gravitational force between the earth and the sun changes as the earth moves from position A to B as shown in the figure. Sun Earth at position B Earth at position A
Answer:
The distance between sun & Earth at position A is less than the earth at position B. The gravitational force of two bodies is inversely proportional to the square of the distance. So At position A gravitational force is more & it decreases as it rotate towards position B.
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:
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4
Type the correct answer in the box. Use numerals instead of words.
Anne has a sample of a substance. Its volume is 20 cm and its mass is 100 grams. What is the sample's density?
The sample's density is
g/cm?
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Answer:
5g/cm
Explanation:
denisty=mass/volume
100/20
5g/cm
