Answer:
Explanation:
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 bus is moving at a speed of 20 m/s, when it begins to slow at a constant rate of 5 m/s2 in order to stop at a bus stop. If it comes to rest at the bus stop, how far away was the bus from the stop?
Answer:
40 m
Explanation:
Applying,
v² = u²+2as............... Equation 1
Where v = final velocity, u = initial velocity, a = deceleration, s = distance
make s the subject of the equation
s = (v²-u²)/2a............ Equation 2
From the question,
Given: v = 0 m/s (comes to rest), u = 20 m/s, a = -5 m/s² (begins to slow)
Substitute these values into equation 2
s = (0²-20²)/[2(-5)]
s = -400/-10
s = 40 m
Hence the bus was 40 m from the stop
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]
An electron starts from rest and falls through a potential rise of 100V. What is its final
speed?
Answer:
The final speed of electron=[tex]5.93\times 10^6m/s[/tex]
Explanation:
We are given that
Initial velocity, u=0
Potential, V=100 V
We have to find the final speed.
Mass of electron, [tex]m=9.1\times 10^{-31} kg[/tex]
Charge on electron, q=[tex]1.6\times 10^{-19}C[/tex]
We know that
[tex]qV=\frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]
Using the formula
[tex]1.6\times 10^{-19}\times 100=\frac{1}{2}\times 9.1\times 10^{-31} v^2-0[/tex]
[tex]v^2=\frac{2\times 1.6\times 10^{-19}\times 100}{9.1\times 10^{-31}}[/tex]
[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 100}{9.1\times 10^{-31}}}[/tex]
[tex]v=5.93\times 10^6m/s[/tex]
Hence, the final speed of electron=[tex]5.93\times 10^6m/s[/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
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
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
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]
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 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
1) Define Mechanical Advantage?
2) What factor affect the mechanical advantage of a machine?
3) Define ideal machine?
4) What are output work and input work?
5) What is moment?
Answer:
1) ans: The ratio of load to effort in a simple machine is called Mechanical Advantage.
2) ans: Frictuon produced in Simple machine affect the mechanical advantage of a machine.
3) ans: The machine whose efficiency is 100% is called ideal machine.
4) ans: The work done by the machine is called output work.
ans: The work done in the machine is called input work.
5) ans: The turning effect of force is called moment.
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 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.
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 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].
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
abrief history of hand writing
A spaceship travels at a constant speed from earth to a planet orbiting another star. When the spacecraft arrives, 13 years have elapsed on earth, and 7.9 years have elapsed on board the ship. How far away (in meters) is the planet, according to observers on earth
Answer:
[tex]L=9.76*10^{16}m[/tex]
Explanation:
From the question we are told that:
Time on earth [tex]T_e= 13yrs[/tex]
Time on ship [tex]T_s= 7.9yrs[/tex]
Therefore
[tex]r=\frac{t_e}{t_s}[/tex]
[tex]r=\frac{13}{7.9}[/tex]
[tex]r=1.65[/tex]
Generally the equation for Constant Velocity is mathematically given by
[tex]V=C\sqrt{1-\frac{1}{r^2}}[/tex]
[tex]V=3*10^8\sqrt{1-\frac{1}{1.64^2}}[/tex]
[tex]V=2.38*10^8m/s[/tex]
Therefore
[tex]L=V*t[/tex]
Where
[tex]t=(13*365.25*24*3600)s[/tex]
[tex]t=4.1*10^8[/tex]
[tex]L=2.38*10^8m/s*4.1*10^8[/tex]
[tex]L=9.76*10^{16}m[/tex]
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.
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 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]
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 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
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.
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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]
During the 5 minute period described in question 4, the water in the insulated vessel undergoes a temperature increase of 20 C. Assuming all of the electrical energy dissipated by the resistor was transferred to the water as heat, what is the mass of the water
Answer:
Please find the complete question in the attached file.
Explanation:
[tex]V=12\ V\\\\I=1.2\ A\\\\T=5\times 60=300\ second\\\\[/tex]
Calculating the electrical energy dissipated:
[tex]w=p\cdot t=V\cdot I \cdot t\\\\[/tex]
[tex]=12\times 1.2 \times 300 \ J\\\\=4320\ J[/tex]
[tex]\Delta T=20^{\circ}\ C\\\\W=m\cdot c\cdot \Delta T\\\\4320=m(4186 \times 20)\\\\m=\frac{4320}{4186 \times 20}=51.6 \ grams=0.516 \ kg\\\\[/tex]
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]
Ann and Bob are carrying a 18.5 kg table that is 2.25 m long. A 8.33 kg box sits on the table 0.750 m from Ann. How much lift force does Ann exert? Use 9.80 for gravity and answer in Newtons
Answer:
F = 118 N
Explanation:
Assume Ann and Bob lift at their respective ends of the table
Sum moments about Bob's position to zero.
Let F be Ann's upward force
F[2.25] - 18.5(9.80)[2.25 / 2] - 8.33(9.80)[0.750] = 0
F = 117.86133333...
Given:
Mass of table, [tex]m_t = 18.5 \ kg[/tex]Mass of box, [tex]m_b = 8.33 \ kg[/tex]Length of table, [tex]2.25 \ m[/tex]Length of box, [tex]0.75 \ m[/tex]The weight of table will be:
→ [tex]W_t = m_t g[/tex]
[tex]= 18.5\times 9.8[/tex]
[tex]= 181.3 \ N[/tex]
Now,
→ [tex]\sum M_A_{nn} = -81.634\times 0.750-181.3\times 1.125+R_{bob}\times 1.125[/tex]
or,
→ [tex]R_{bob} = \frac{61.2255+203.9625}{2.25}[/tex]
[tex]= \frac{265.188}{2.25}[/tex]
[tex]= 117.9 \ N[/tex]
Thus the above answer is right.
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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]
