Answer:
a) T = 0.68 s, b) f = 1.47 Hz, c) λ = 54 m
Explanation:
Simple harmonic motion is described by the expression
x = A cos (wt + Ф)
a) angular velocity and period are related
w = 2π / T
The period is the time it takes for the movement to repeat itself, in this case they indicate the time to reach two consecutive zeros, this corresponds to half a period
T / 2 = 0.34 s
T = 2 0.34
T = 0.68 s
b) period and frequency are inverse
f = 1 / T
f = 1 / 0.68
f = 1.47 Hz
c) the amplitude of the movement is proportional to the energy carried by the wave, which cannot be calculated with the data provided.
With the length data we can find the wavelength
λ / 2 = 27 m
λ = 54 m
Can an electron be diffracted? Can it exhibit interference?
Answer:
Yeah, it can be diffracted. Though it depends on a diffracting medium.
It must have some magnetic fields .
Forexample; X-ray diffraction where electrons are diffracted to the target filament.
A monochromatic light falls on two narrow slits that are 4.50 um apart. The third destructive fringes which are 35° apart were formed at a screen 2m from the light source. The light source is 0.30 m from the slits. () Calculate Ym. (4 marks) Compute the wavelength of the light. (4 marks)
Answer:
y = 1.19 m and λ = 8.6036 10⁻⁷ m
Explanation:
This is a slit interference problem, the expression for destructive interference is
d sin θ = m λ
indicate that for the angle of θ = 35º it is in the third order m = 3 and the separation of the slits is d = 4.50 10⁻⁶ m
λ = d sin θ / m
let's calculate
λ = 4.50 10⁻⁶ sin 35 /3
λ = 8.6036 10⁻⁷ m
for the separation distance from the central stripe, we use trigonometry
tan θ= y / L
y = L tan θ
the distance L is measured from the slits, it indicates that the light source is at x = 0.30 m from the slits
L = 2 -0.30
L = 1.70 m
let's calculate
y = 1.70 tan 35
y = 1.19 m
Suppose a 42-turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.125 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.40 T
Answer:
EMF = 73.5 volts
Explanation:
Given that,
The number of a coil, N = 42
The area of the coil, A = 0.125 m²
It is stretched to have no area in 0.100 s
The magnetic field strength is 1.4 T.
We need to find the average induced emf in the coil. We know that,
[tex]\epsilon=N\dfrac{d\phi}{dt}[/tex]
[tex]\epsilon=N\dfrac{BA}{dt}\\\\\epsilon=42\times \dfrac{1.4\times 0.125}{0.1}\\\\=73.5\ V[/tex]
So, the induced emf in the coil is 73.5 volts.
The USDA recommends that women consume about 2000 Calories per day and men consume about 2500 Calories per day. How much average power use (in Watts) does this imply for a human body
Answer:
1500 W to 2200 W
Explanation:
Every person does work in his daily day to day life. A person needs energy in order to perform work. And the energy consumed by an individual while performing a daily work is directly responsible to the mount of oxygen consumed by the person.
The USDA is the federal agency which looks after the food and agriculture matters of the US government. It deals with and formulates different policies and laws for the country and it s people. It recommends about 2000 calories per day for women and for men it recommends about 2500 calorie per days of food intake.
Accordingly, the average power required by a human body for doing regular work is in the range of 1500 W to 2200 W.
1. A positive electric charge in a region of changing electric potential will: A. move in the direction of decreasing potential B. move in the direction of increasing potential C. move in an undetermined way; we need more information D. remain at rest
Answer:
The correct option is (B).
Explanation:
The electric potential is the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field. The electric potential due to a point charge is given by :
[tex]V=\dfrac{kQ}{r}[/tex]
Where
k is the electrostatic constant
Q is the electric charge
r is the distance from the charge
So, a positive electric charge in a region of changing electric potential will move in the direction of increasing potential.
In a calorimetry experiment, three samples A, B, and C with TA> TB> Tc are placed in thermal contact. When the samples have reached thermal equilibrium at a common temperature T, which one of the following must be true?
a. QA > QB >QC
b. QA< 0, QB <0, and Qc > 0
c. T> TB
d. T
e. TA > T> Tc
Answer:
e. TA>T>Tc
Explanation:
a) In this case, we cannot say for sure QA>QB>QC. This is because the magnitude of the heat flow will depend on the specific heat and the mass of each sample. Due to the equation:
[tex]Q=mC_{p}(T_{f}-T_{0})[/tex]
if we did an energy balance of the system, we would get that>
QA+QB+QC=0
For this equation to be true, at least one of the heats must be negative. And one of the heats must be positive.
We don't know either of them, so we cannot determine if this statement is true.
b) We can say for sure that QA<0, because when the two samples get to equilibrum, the temperatrue of A must be smaller than its original temperature. Therefore, it must have lost heat. But we cannot say for sure if QB<0 because sample B could have gained or lost heat during the process, this will depend on the equilibrium temperature, which we don't know. So we cannot say for sure this option is correct.
c) In this case we don't know for sure if the equilibrium temperature will be greater or smaller than TB. This will depend on the mass and specific heat of the samples, just line in part a.
d) is not complete
e) We know for sure that A must have lost heat, so its equilibrium temperature must be smaller than it's original temperature. We know that C must have gained heat, therefore it's equilibrium temperature must be greater than it's original temperature, so TA>T>Tc must be true.
The distance between the two object is fixed at 5.0 m. The uncertainty distance measurement is? The percentage error in the distance is?
Two distinct systems have the same amount of stored internal energy. 500 J are added by heat to the first system and 300 J are added by heat to the second system. What will be the change in internal energy of the first system if it does 200 J of work? How much work will the second system have to do in order to have the same internal energy?
Answer:
The change in the internal energy of the first system is 300 J
The second system will do zero work in order to have the same internal energy.
Explanation:
Given;
heat added to the first system, Q₁ = 500 J
heat added to the second system, Q₂ = 300 J
work done by the first system, W₁ = 200 J
The change in the internal energy of the system is given by the first law of thermodynamics;
ΔU = Q - W
where;
ΔU is the change in internal energy of the system
The change in the internal energy of the first system is calculated as;
ΔU₁ = Q₁ - W₁
ΔU₁ = 500 J - 200 J
ΔU₁ = = 300 J
The work done by the second system to have the same internal energy with the first.
ΔU₁ = Q₂ - W₂
W₂ = Q₂ - ΔU₁
W₂ = 300 J - 300 J
W₂ = 0
The second system will do zero work in order to have the same internal energy.
• Explain how sound travels
Sound is a type of energy made by vibrations. These vibrations create sound waves which move through mediums such as air, water and wood. When an object vibrates, it causes movement in the particles of the medium. This movement is called sound waves, and it keeps going until the particles run out of energy.
Sound is a type of energy made by vibrations. These vibrations create sound waves which move through mediums such as air, water and wood. When an object vibrates, it causes movement in the particles of the medium. This movement is called sound waves, and it keeps going until the particles run out of energy.
A kugel fountain, also known as a floating sphere fountain, is a sculpture that suspends a large, typically stone, sphere on a thin plane of water. One of three in Columbus, is on campus in the Chadwick Arboretum on the corner of Lane Ave. and Olentangy. Kugel fountains have very little friction between the stone sphere and the hollow in which they sit, so it takes very little torque to start one moving from rest.
Required:
a. The kugel fountain in the Richard and Annette Bloch Cancer Survivor's Plaza in the Lane Ave Gardens has a granite sphere with a mass of approximately 1,800 kg, and a radius of about half a meter. You use your hand to apply a tangential force on the sphere to start it rotating from rest, and it takes 12 seconds to complete a quarter revolution. What is the average torque you applied to the sphere?
b. What is the average magnitude of the tangential force you applied?
c. How does this compare to the weight of a cellphone?
Answer:
Explanation:
Moment of inertia of sphere I = 2 /5 x m R²
m is mass and R is radius of the sphere .
I = 2/5 x 1800 x 0.5²
= 180 kg m².
For rotational motion starting from rest
θ = 1/2 α t²
Here θ = π /2 = 1.57 radians
t = 12 s
1.57 radians = 1/2 x α x 12²
Angular acceleration α = .022 rad /s²
Torque applied = I α
= 180 kg m² x .022 rad /s²
= 3.96 N.m
b ) Torque = F x r
F = Torque / r
= 3.96 / .5 = 7.92 N .
c )
Average weight of a cellophone is 1.3 N .
So the force applied is a little more than weight of a cellophone.
Consider a uniform electric field of 50 N/C directed toward the east. If the voltage measured relative to ground at a given point in the field is 80 V, what is the voltage at a point 1.0 m directly east of the point
Answer:
30 V
Explanation:
Given that:
The uniform electric field = 50 N/C
Voltage = 80 V
distance = 1.0 m
The potential difference of the electric field = Δ V
E_d = V₁ - V₂
50 × 1 = 80V - V₂
50 - 80 V = - V₂
-30 V = - V₂
V₂ = 30 V
A crate with a mass of 161.5 kg is suspended from the end of a uniform boom with a mass of 72.3 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall. Calculate the tension in the cable.
Answer:
The correct answer is "2205.72 N".
Explanation:
Given:
m₁ = 161.5 kg
m₂ = 72.3 kg
By taking moment about point A, we get
⇒ [tex]m_1 g Cos \theta.\frac{l}{2 Cos \theta} + m g Cos \theta.\frac{l}{ Cos \theta} -T Cos(90-2 \theta).\frac{l}{2 Cos \theta} = 0[/tex]
By substituting the values, we get
⇒ [tex]\frac{161.5\times 9.8\times 13}{2}+72.3\times 9.8\times 13-T.2 Sin \theta.l =0[/tex]
⇒ [tex]10287.55+9211.02-T 2\times 0.34\times 13=0[/tex]
⇒ [tex]T\times 8.84=19498.57[/tex]
⇒ [tex]T= 2205.72 N[/tex]
A long, current-carrying solenoid with an air core has 1550 turns per meter of length and a radius of 0.0240 m. A coil of 200 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system
Answer:
[tex]M=7.05*10^{-4}[/tex]
Explanation:
From the question we are told that:
Coil one turns N_1=1550 Turns/m
Radius [tex]r=0.0240m[/tex]
Turns 2 [tex]N_2=200N[/tex]
Generally the equation for area is mathematically given by
[tex]A=\pi*r^2[/tex]
[tex]A=\pi*0.024^2[/tex]
[tex]A=\1.81*10^{-3} m^2[/tex]
Therefore
The mutual inductance of this system is
[tex]M=\mu*N_1*N_2*A[/tex]
[tex]M=(4 \pi*10^{-7})*1550*200*1.81*10^{-3}[/tex]
[tex]M=7.05*10^{-4}[/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
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]
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
Welcome to this IE. You may navigate to any page you've seen already using the IE Outline tab on the right. A particle beam is made up of many protons each with a kinetic energy of 3.25 x 10-15J. A proton has a mass of 1.673x10-27kg and a charge of 1.602x10-19C. What is the magnitude of a uniform electric field that will stop these protons in a distance of 2 m
Answer:
the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Explanation:
Given the data in the question;
Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J
Mass of proton = 1.673 × 10⁻²⁷ kg
Charge of proton = 1.602 × 10⁻¹⁹ C
distance d = 2 m
we know that
Kinetic Energy = Charge of proton × Potential difference ΔV
so
Potential difference ΔV = Kinetic Energy / Charge of proton
we substitute
Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )
Potential difference ΔV = 20287.14 V
Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;
E = Potential difference ΔV / distance d
we substitute
E = 20287.14 V / 2 m
E = 10143.57 V/m or 1.01 × 10⁴ V/m
Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
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
What fraction of the total energy of a SHO is kinetic when the displacement is one third the amplitude
Answer:
The fraction of kinetic energy to the total energy is [tex]\frac{K}{T}=\frac{8}{9}[/tex].
Explanation:
displacement is one third of the amplitude.
Let the amplitude is A.
x= A/3
The kinetic energy of the body executing SHM is
[tex]K = 0.5 mw^2(A^2 - x^2)\\\\K = 0.5 m w^2 \left ( A^2 -\frac{A^2}{9} \right )\\\\K = 0.5 mw^2\times \frac{8A^2}{9}......(1)[/tex]
The total energy is
[tex]T =0.5 mw^2A^2 ..... (2)[/tex]
Divide (1) by (2)
[tex]\frac{K}{T}=\frac{8}{9}[/tex]
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 cylinder within a piston expands from a volume of 1.00 L to a volume of 2.00 L against an external pressure of 1.00 atm. How much work (in Joule) was done by the expansion?
Answer:
1.671L
Explanation:
In the given case the pressure is constant therefore it is an isobaric process, the process in which the pressure remains constant is called as isobaric process.
Work done= external pressure× change in volume.
288= 2×( final volume-intial volume)×101.32
144÷101.32=(finalvolume-0.250)
1.421=final volume-0.250
final volume=1.671L
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]
A stone is dropped from a cliff 64 feet above the ground. Answer the following, using -32 ft/sec2 as the acceleration due to gravity. Show all work and submit to D2L.
a. Find functions that represent the acceleration, velocity, and position of the stone above the ground at time t.
b. How long does it take the stone to reach the ground?
c. With what velocity does the stone hit the ground?
Answer:
(a) v = 32 t
h = 16 t^2
g = 32 ft/s^2
(b) 64 ft/s
Explanation:
height, h = 64 feet
g = - 32 ft/s^2
(a) Let the time is t .
Let the velocity after time t is v.
Use first equation of motion
v = u + at
- v = 0 - 32 t
v = 32 t
Let the distance is h from the top.
Use second equation of motion
[tex]h = u t + 0.5 at^2 \\\\- h = 0 - 0.5\times 32 \times t^2\\\\h = 16 t^2[/tex]
The acceleration is constant for entire motion.
(b) Let the velocity is v as it hits the ground. Use third equation of motion
[tex]v^2 = u^2 + 2 a s \\\\v^2 = 0 + 2 \times 32\times 64\\\\v = 64 feet/s[/tex]
A 2600 kg truck travelling at 72 km/h slams on the brakes and skids to a stop. The frictional force from the road is 8200 N. Use the relationship between kinetic energy and mechanical work to determine the distance it takes for the truck to stop.
Answer:
Approximately [tex]63\; \rm m[/tex].
Explanation:
Convert the initial speed of this truck to standard units:
[tex]\begin{aligned} v &= 72\; \rm km \cdot h^{-1} \times \frac{1\; \rm h}{3600\; \rm s} \times \frac{1000\; \rm m}{1\; \rm km} \\ &= 20\; \rm m \cdot s^{-1}\end{aligned}[/tex].
Calculate the initial kinetic energy of this truck:
[tex]\begin{aligned}\text{KE} &= \frac{1}{2} \, m \cdot v^{2} \\ &= \frac{1}{2} \times 2600\; \rm kg \times (20\; \rm m \cdot s^{-1})^{2} \\ &= 5.2 \times 10^{5} \; \rm J\end{aligned}[/tex].
The kinetic energy of the truck would be [tex]0[/tex] when the truck is not moving. Hence, the friction on the truck would need to do [tex](-5.2 \times 10^{5}\; \rm J)[/tex] of work on the truck to bring the truck to a stop.
Calculate the displacement required to achieve [tex](-5.2 \times 10^{5}\; \rm J)[/tex] of work at [tex]8200\; \rm N[/tex]:
[tex]\begin{aligned}x &= \frac{W}{F} \\ &= \frac{-5.2 \times 10^{5}\; \rm J}{8200\; \rm N} \approx -63\; \rm m\end{aligned}[/tex].
This result is negative because the direction of friction is opposite to the direction of the motion of the truck.
Hence, the truck would come to a stop after skidding for approximately [tex]63\; \rm m[/tex].
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
How far from a 10 µC point charge will the potential be 1000 V?
a.
10 m
b.
90 m
c.
80 m
d.
60 m
• Point charge (Q) = 10 μC = 10 × 10⁻⁶ C
• Potential (V) = 1000 V
• Distance (r) = ?
[tex]\implies V = \dfrac{KQ}{r} \\ [/tex]
[tex]\implies r= \dfrac{KQ}{V}[/tex]
[tex]\implies r= \dfrac{9 \times {10}^{9} \times 10 \times {10}^{ - 6} }{1000} \\ [/tex]
[tex]\implies r= \dfrac{90 \times {10}^{3} }{1000} \\ [/tex]
[tex]\implies r= \dfrac{90 \times {10}^{3} }{ {10}^{3} } \\ [/tex]
[tex]\implies r= 90 \times {10}^{3} \times {10}^{ - 3} \\ [/tex]
[tex]\implies\bf r= 90\:m \\ [/tex]
Hence,the option B) 90 m is the correct answer.
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
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
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
