Using the principle of buoyancy and the ideal gas law equation, we can calculate the temperature needed inside a hot air balloon to provide the required lift. The maximum altitude of the balloon is limited by air temperature, pressure, and other environmental factors.
Explanation:The principle of buoyancy used by hot air balloons to create lift is based on Archimedes' Principle, which states that the buoyant or lifting force exerted on a body immersed in a fluid equals the weight of the fluid the body displaces. In order for a hot air balloon to gain lift, the air inside the balloon must be heated to a temperature that makes it less dense than the surrounding cooler air. This difference in air density is what actually provides the uplift.
To calculate the required temperature inside the balloon, we can use the ideal gas law equation P₁V₁ / P₂V₂. Given the values in the question and the atmospheric pressure, we can derive the temperature inside the balloon required to provide the needed lift.
The maximum altitude attainable by a hot air balloon is limited by the ambient air temperature and pressure, which decrease as altitude increases. If the balloon's lift decreases (because it can no longer heat the air inside it sufficiently to provide lift), the balloon will stop ascending. Wind currents and weather conditions can also affect the maximum altitude.
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When a honeybee flies through the air, it develops a charge of +18pC. How many electrons did it lose in the process of acquiring this charge?
The honeybee lost about 112 million electrons to acquire a charge of +18pC. This calculation involves converting the charge from picocoulombs to coulombs and then dividing by the charge of a single electron.
Explanation:To calculate the number of electrons a honeybee lost to acquire a charge of +18pC, we first need to understand that the fundamental unit of charge, often represented as e, is +1.602 x 10-19 C for a proton and -1.602 x 10-19 C for an electron.
The number of electrons lost (n_e) is the total charge divided by the charge per electron. Therefore, we convert the charge of the honeybee from picocoulombs (pC) to coulombs (C) by multiplying by 10-12, because 1pC = 10-12C. The +18pC charge is thus equivalent to 18 x 10-12 C.
In relation to the charge of an electron, the honeybee's charge is -18 x 10-12 C / -1.602 x 10-19 C/e-, which gives approximately 1.12 x 108 or 112,000,000 electrons.
So, the honeybee lost about 112 million electrons to get a positive charge of +18pC.
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A 6.1-m-long solid circular metal rod (E = 150 GPa) must not stretch more than 10 mm when a load of 109 kN is applied to it. Determine the smallest diameter rod that should be used. State your answer in mm.
To solve this problem we will apply the concepts related to the deformation of a body and the normal effort, from which we will obtain the area. From this area applied to the geometric concept of a circular bar we will find the diameter.
The deformation equation in a rod tells us that
[tex]\delta = \frac{PL}{AE}[/tex]
Here,
P = Load
L = Length
A = Cross-sectional area
E = Elastic Modulus
Rearranging the Area,
[tex]A = \frac{PL}{\delta E}[/tex]
Replacing we have that the area is,
[tex]A = \frac{(109*10^{3})(6.1)}{(10*10^{-3})(150*10^9)}[/tex]
[tex]A = 0.000443266m^2[/tex]
[tex]A = 44.32*10^{-6}m^2[/tex]
Using the geometric expression for the Area we have,
[tex]A = \frac{\pi}{4} d^2[/tex]
[tex]d = \sqrt{\frac{4A}{\pi}}[/tex]
[tex]d = \sqrt{\frac{4(44.32)}{\pi}}[/tex]
[tex]d = 7.51mm[/tex]
Therefore the smalles diameter rod is 7.51mm
The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, increasing its kinetic energy by 4.80×10−19 JJ . In what direction and through what potential difference Vb−VaVb−Va does the particle move?
1) Potential difference: 1 V
2) [tex]V_b-V_a = -1 V[/tex]
Explanation:
1)
When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by
[tex]\Delta U=q\Delta V[/tex]
where
q is the charge's magnitude
[tex]\Delta V[/tex] is the potential difference between the initial and final position
In this problem, we have:
[tex]q=4.80\cdot 10^{-19}C[/tex]is the magnitude of the charge
[tex]\Delta U = 4.80\cdot 10^{-19}J[/tex] is the change in kinetic energy of the particle
Therefore, the potential difference (in magnitude) is
[tex]\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V[/tex]
2)
Here we have to evaluate the direction of motion of the particle.
We have the following informations:
- The electric potential increases in the +x direction
- The particle is positively charged and moves from point a to b
Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)
This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:
[tex]V_b-V_a = - 1V[/tex]
Final answer:
The particle moves in the direction of decreasing potential and through a potential difference of 3.0 V.
Explanation:
The potential difference through which the particle moves can be calculated using the formula for work done by the electric force: W = q(Vb - Va), where W is the change in kinetic energy, q is the charge of the particle, and (Vb - Va) is the potential difference. Rearranging the formula, we have Vb - Va = W/q. Substituting the given values, Vb - Va = (4.80×10-19 J) / (-1.60×10-19 C) = -3.0 V.
This implies that the particle moves in the direction of decreasing potential. Therefore, the particle moves from point a to point b through a potential difference of 3.0 V in the direction of decreasing potential.
In a double-slit interference experiment, interference fringes are observed on a distant screen. The width of both slits is then doubled without changing the distance between their centers.
a. What happens to the spacing of the fringes?
b. What happens to the intensity of the bright fringes?
Final answer:
If the width of both slits in a double-slit interference experiment is doubled without changing the distance between their centers, the spacing of the fringes decreases and the intensity of the bright fringes decreases.
Explanation:
In a double-slit interference experiment, if the width of both slits is doubled without changing the distance between their centers:
a. The spacing of the fringes will decrease.
b. The intensity of the bright fringes will decrease.
When the width of the slits is increased, the interference fringes become wider and less compact. This means that the spacing between the fringes becomes smaller. Additionally, the intensity of the bright fringes decreases because spreading out the light over a wider area results in less light reaching a specific point on the screen.
A jet fighter pilot wishes to accelerate from rest at a constant acceleration of 5 g to reach Mach 3 (three times the speed of sound) as quickly as possible. Experimental tests reveal that he will black out if this acceleration lasts for more than 5.0 s. Use 331 m/s for the speed of sound. (a) Will the period of acceleration last long enough to cause him to black out? (b) What is the greatest speed he can reach with an acceleration of 5g before he blacks out?
Final answer:
The jet fighter pilot will black out during the acceleration period as it lasts longer than 5.0 s. The greatest speed the pilot can reach with an acceleration of 5g before blacking out is 245 m/s.
Explanation:
To determine whether the jet fighter pilot will black out, we need to calculate the time of acceleration. The speed of sound is v = 331 m/s. The speed the pilot wants to reach is 3 times the speed of sound, so the desired speed is 3 * 331 m/s = 993 m/s. We can use the equation:
v = u + at
Where v is the final velocity, u is the initial velocity (0 m/s), a is the acceleration (5 g = 5 * 9.8 m/s^2 = 49 m/s^2), and t is the time.
Substituting the values, we get:
993 m/s = 0 m/s + 49 m/s² x t
Rearranging this equation to solve for t, we get:
t = 993 m/s / 49 m/s^2 = 20.265 s
Since the acceleration lasts for longer than 5.0 seconds, the pilot will black out during the acceleration period.
To calculate the greatest speed the pilot can reach before blacking out, we can use the same equation, but rearrange it to solve for v:
v = u + at
Substituting the values, we get:
v = 0 m/s + 49 m/s^2 x 5.0 s = 245 m/s
Therefore, the greatest speed the pilot can reach with an acceleration of 5 g before blacking out is 245 m/s.
A muon is produced by a collision between a cosmic ray and an oxygen nucleus in the upper atmosphere at an altitude of 50 . It travels vertically downward to the surface of the Earth where it arrives with a total energy of 178 . The rest energy of a muon is 105.7 . What is the kinetic energy of a muon at the surface
Answer:
72.3 MeV
Explanation:
E = Total energy of muon = 178 MeV
[tex]E_0[/tex] = Rest energy of a muon = 105.7 MeV
Kinetic energy of the muon at the surface of the Earth is given by the difference of the total energy and the rest energy of the muon
[tex]K=E-E_0\\\Rightarrow K=178-105.7\\\Rightarrow K=72.3\ MeV[/tex]
The kinetic energy of a muon at the surface is 72.3 MeV
If a 430 mL ordinary glass beaker is filled to the brim with ethyl alcohol at a temperature of 6.00°C, how much (in mL) will overflow when their temperature reaches 22.0°C?
Answer : The volume of ethyl alcohol overflow will be, 7.49 mL
Explanation :
To calculate the volume of ethyl alcohol overflow we are using formula:
[tex]\Delta V=V_o(\alpha \Delta T)\\\\\Delta V=V_o\times \alpha \times (T_2-T_1)[/tex]
where,
[tex]\Delta V[/tex] = volume expand = ?
[tex]\alpha[/tex] = volumetric expansion coefficient = [tex]0.00109/^oC[/tex]
[tex]V_o[/tex] = initial volume = 430 mL
[tex]T_2[/tex] = final temperature = [tex]22.0^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]6.00^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta V=(430mL)\times (0.00109/^oC)\times (22.0-6.00)^oC[/tex]
[tex]\Delta V=7.49mL[/tex]
Thus, the volume of ethyl alcohol overflow will be, 7.49 mL
Answer:
7.3094 ml
Explanation:
Initial volume of the glass, Vo = 430 ml
Initial temperature, T1 = 6°C
final temperature, T2 = 22°C
Temperature coefficient of glass, γg = 27.6 x 10^-6 /°C
Temperature ethyl alcohol, γa = 0.00109 /°C
Use the formula of expansion of substances
Expansion in volume of glass
ΔVg = Vo x γg x ΔT
ΔVg = 430 x 27.6 x 10^-6 x 16 = 0.1898 ml
Expansion in volume of ethyl alcohol
ΔVa = Vo x γa x ΔT
ΔVa = 430 x 0.00109 x 16 = 7.4992 ml
The amount of volume over flow is
ΔV = ΔVa - ΔVg
ΔV = 7.4992 - 0.1898
ΔV = 7.3094 ml
Thus, the amount of ethyl alcohol over flow is 7.3094 ml.
A steel piano wire is 0.7 m long and has a mass of 5 g. It is stretched with a tension of 500 N. What is the speed of transverse waves on the wire? To reduce the wave speed by a factor of 2 without changing the tension, what mass of copper wire would have to be wrapped around the wire?
The speed of transverse waves on the steel piano wire can be calculated using the formula √(Tension / (Mass per unit length)). To reduce the wave speed by a factor of 2 without changing the tension, we can solve for the new wave speed, and then calculate the difference in mass per unit length with the copper wire.
Explanation:The speed of transverse waves on a steel piano wire can be calculated using the formula:
Speed = √(Tension / (Mass per unit length))
Where the tension in the wire is 500 N and the mass per unit length is calculated by dividing the mass of the wire by its length. Therefore, the mass per unit length is 5 g / 0.7 m = 7.14 g/m.
To reduce the wave speed by a factor of 2 without changing the tension, we can use the equation:
New wave speed = √(New tension / (Mass per unit length))
We can solve this equation for the new tension by rearranging it as:
New tension = (New wave speed)^2 * (Mass per unit length)
Since we want to reduce the wave speed by a factor of 2, the new wave speed is half the original speed. Substituting these values into the equation, we have:
(0.5 * Old wave speed)^2 * (Mass per unit length) = 500 N
Solving for the new mass per unit length gives:
New mass per unit length = 500 N / (0.5 * Old wave speed)^2
The difference between the new mass per unit length and the mass per unit length of copper wire is the mass of copper wire that needs to be wrapped around the steel wire. However, we would need additional information to calculate this difference.
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Assuming the same initial conditions as described in FNT 2.2.1-1, use the energy-interaction model in two different ways (parts (a) and (b) below) to determine the speed of the ball when it is 4 meters above the floor headed down:
a) Construct a particular model of the entire physical process, with the initial time when the ball leaves Christine’s hand, and the final time when the ball is 4 meters above the floor headed down.
b) Divide the overall process into two physical processes by constructing two energy-system diagrams and applying energy conservation for each, one diagram for the interval corresponding to the ball traveling from Christine’s hand to the maximum height, and then one diagram corresponding to the interval for the ball traveling from the maximum height to 4 meters above the floor headed down.
c) Did you get different answers (in parts (a) and (b)) for the speed of the ball when it is 4 meters above the floor headed down?
Answer:
(a). Vf = 7.14 m/s
(b). Vf = 7.14 m/s
(c). same answer
Explanation:
for question (a), we would be applying conservation of energy principle.
but the initial height is h = 1.5 m
and the initial upward velocity of the ball is Vi = 10 m/s
Therefore
(a). using conservation law
Ef = Ei
where Ef = 1/2mVf² + mghf ........................(1)
also Ei = 1/2mVi² + mghi ........................(2)
equating both we have
1/2mVf² + mghf = 1/2mVi² + mghi
eliminating same terms gives,
Vf = √(Vi² + 2g (hi -hf))
Vf = √(10² + -2*9.8*2.5) = 7.14 m/s
Vf = 7.14 m/s
(b). Same process as done in previous;
Ef = Ei
but here the Ef = mghf ...........(3)
and Ei = 1/2mVi² + mghi ...........(4)
solving for the final height (hf) we relate both equation 3 and 4 to give
mghf = 1/2mVi² + mghi ..............(5)
canceling out same terms
hf = hi + Vi²/2g
hf = 1.5 + 10²/2*9.8 = 6.60204m ............(6)
recalling conservation energy,
Ef = Ei
1/2mVf² + mghf = mghi
inputting values of hf and hi we have
Vf = √(2g(hi -hf)) = 7.14 m/s
Vf = 7.14 m/s
(c). From answer in option a and c, we can see there were no changes in the answers.
What is the current in amperes if 1400 Na+ ions flow across a cell membrane in 3.3 μs ? The charge on the sodium is the same as on an electron, but positive.
Answer:
Current, [tex]I=6.78\times 10^{-11}\ A[/tex]
Explanation:
In this case, we need to find the current in amperes if 1400 Na+ ions flow across a cell membrane in 3.3 μs.
Charge, [tex]q=1400\times 1.6\times 10^{-19}=2.24\times 10^{-16}\ C[/tex]
Time taken, [tex]t=3.3\ \mu s=3.3\times 10^{-6}\ s[/tex]
Let I is the current. It is given by total charge per unit time. It is given by :
[tex]I=\dfrac{q}{t}[/tex]
[tex]I=\dfrac{2.24\times 10^{-16}}{3.3\times 10^{-6}}[/tex]
[tex]I=6.78\times 10^{-11}\ A[/tex]
So, the current of [tex]6.78\times 10^{-11}\ A[/tex] is flowing across a cell membrane. Hence, this is the required solution.
The position x, in meters, of an object is given by the equation:
x = A + Bt + Ct^2,
where t represents time in seconds.
1. What are the SI units of A, B, and C?
A. m, s, s
B. m, m/s, m/s^2
C. m, m, m
D. m/s, m/s^2, m/s^3
E. m, s, s^2
Answer:
The SI units of A, B and C are :
[tex]m,\ m/s\ and\ m/s^2[/tex]
Explanation:
The position x, in meters, of an object is given by the equation:
[tex]x=A+Bt+Ct^2[/tex]
Where
t is time in seconds
We know that the unit of x is meters, such that the units of A, Bt and [tex]Ct^2[/tex] must be meters. So,
[tex]A=m[/tex][tex]bt=m[/tex][tex]b=\dfrac{m}{s}=m/s[/tex]
[tex]Ct^2=m[/tex][tex]C=m/s^2[/tex]
So, the SI units of A, B and C are :
[tex]m,\ m/s\ and\ m/s^2[/tex]
So, the correct option is (B).
The SI units of A, B, and C is option B [tex]m, m/s, m/s^2[/tex]
The calculation is as follows;The position x, in meters, of an object is provided by the equation:
[tex]x = A + Bt + Ct^2[/tex]
Here
t should be t in seconds
As We know that the unit of x should be in meters, in such a way that the units of A, Bt and must be meters.
So,
A = m
bt = m
[tex]b = m\div s = m/s\\\\ Ct^2 = m\\\\ C = m/s^2[/tex]
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Can you find a vector quantity that has a magnitude of zero but components that are not zero? Explain. Can the magnitude of a vector be less than the magnitude of any of its components? Explain.
It is not possible to find a vector quantity of magnitude zero but components different from zero
The magnitude can never be less than the magnitude of any of its components
A capacitor stores an energy of 8 Joules when there is a voltage of 51 volts across its terminals. A second identical capacitor of the same value is stores an energy of 2 Joules. What is the voltage across the terminals of the second capacitor?
Answer:
6.38 V
Explanation:
Note: Since the capacitors are identical, The same Charge flows through them.
For the first capacitor,
The Energy stored in a capacitor is given as
E = 1/2qv ............... Equation 1
Where E = Energy stored by the capacitor, q = charge in the capacitor v = Voltage.
make q the subject of the equation
q = 2E/v ................ Equation 2
Given: E = 8 J, v = 51 v.
Substitute into equation 2
q = 8/51
q = 0.3137 C.
For the second capacitor,
v = 2E/q ................... Equation 3
Given: q = 0.3137 C, E = 2 J.
Substitute into equation 3
v = 2/0.3137
v = 6.38 V
Hence the voltage across the second capacitor = 6.38 V
Why are jovian planets so much larger than terrestrial planets?
Explanation:
The temperature in the inner solar system was too high for light gases to condense, while in the outer solar system, the temperature was much lower, which allowed the Jovian planets to form, which grew enough to accumulate and retain the hydrogen gas that remained in the solar nebula, which led to its high levels of hydrogen and large size.
Two identical stones, A and B, are thrown from a cliff from the same height and with the same initial speed. Stone A is thrown vertically upward, and stone B is thrown vertically downward. Which of the following statements best explains which stone has a larger speed just before it hits the ground, assuming no effects of air friction?
a. Both stones have the same speed; they have the same change in Ugand the same Ki
b. A, because it travels a longer path.
c. A, because it takes a longer time interval.
d. A, because it travels a longer path and takes a longer time interval.
e. B, because no work is done against gravity.
Answer:
Option A
Explanation:
This can be explained based on the conservation of energy.
The total mechanical energy of the system remain constant in the absence of any external force. Also, the total mechanical energy of the system is the sum of the potential energy and the kinetic energy associated with the system.
In case of two stones thrown from a cliff one vertically downwards the other vertically upwards, the overall gravitational potential energy remain same for the two stones as the displacement of the stones is same.
Therefore the kinetic energy and hence the speed of the two stones should also be same in order for the mechanical energy to remain conserved.
Answer:
b. A, because it travels a longer path.
Explanation:
If the stone A is thrown is thrown vertically upwards and another stone is dropped down directly from the same height above the ground then the stone A will hit the ground with a higher speed because it falls down from a greater height above the earth surface.This can be justified by the equation of motion given below:
[tex]v^2=u^2+2\times a\times s[/tex]
where:
[tex]v=[/tex] final velocity
[tex]u=[/tex] initial velocity
[tex]a=[/tex] acceleration = g (here)
[tex]s=[/tex] displacement of the body
Now we know that at the maximum height the speed of the object will be zero for a moment. So for both the stones A and B the initial velocity is zero, stone B is also dropped from a height with initial velocity zero.Acceleration due to gravity is same for the stones so the only deciding factor that remains is s, displacement of the stones. Since stone A is thrown upwards it will attain a greater height before falling down.In what ways do observed extrasolar planetary systems differ from our own solar system?
Answer:
Extrasolar solar system differ from our solar system in many ways such as of mass, size and shape of the planet, as well as temperature or amount of heat received in each planet.
Explanation:
An extrasolar planet is a planet outside the Solar System, while the Solar System orbit around the sun as a result of the gravitational pull of the sun.
Thus, we can say that the major difference between extrasolar planetary systems and solar system is that in solar system, planets orbit around the Sun, while in extrasolar planetary systems, planets orbit around other stars.
All of the planets in our solar system orbit around the Sun. Planets that orbit around other stars are called exoplanets or extrasolar.
Extrasolar solar system differ from our solar system in many ways such as of mass, size and shape of the planet. They also differ in terms of temperature, because the temperature in each planet in solar system depends on its distance from the sun while that of the extrasolar depends on the activities of the star.
The volume of a fluid in a tank is 0.25 m3 . of the specific gravity of the fluid is 2.0. Determine the mass of the fluid. Given the density of water is 1000 kg/m3. Express the answer in Kg.
Answer:
Explanation:
Given
volume of Tank [tex]V=0.25\ m^3[/tex]
Specific gravity [tex]=2[/tex]
specific gravity is the defined as the ratio of density of fluid to the density of water
Density of water [tex]\rho _w=1000\ kg/m^3[/tex]
Density of Fluid [tex]\rho =2\times 1000=2000\ kg/m^3[/tex]
We know mass of a fluid is given by the product of density and volume
[tex]m=\rho \times V[/tex]
[tex]m=2000\times 0.25[/tex]
[tex]m=500\ kg[/tex]
Calculate the initial (from rest) acceleration of a proton in a 5.00 x 10^6 N/C electric field (such as created by a research Van de Graaff). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.
Answer:
Acceleration of the proton will be equal to [tex]4.79\times 10^{14}m/sec^2[/tex]
Explanation:
We have given electric field [tex]E=5\times 10^6N/C[/tex]
Mass of proton is equal to [tex]m=1.67\times 10^{-27}kg[/tex]
And charge on proton is equal to [tex]e=1.6\times 10^{-19}C[/tex]
Electrostatic force will be responsible for the motion of proton
Electrostatic force will be equal to [tex]F=qE=1.6\times 10^{-19}\times 5\times 10^6=8\times 10^{-13}N[/tex]
According to newton law force on the proton will be equal to F = ma, here m is mass of proton and a is acceleration
This newton force will be equal to electrostatic force
So [tex]1.67\times 10^{-27}\times a=8\times 10^{-13}[/tex]
[tex]a=4.79\times 10^{14}m/sec^2[/tex]
So acceleration of the proton will be equal to [tex]4.79\times 10^{14}m/sec^2[/tex]
Final answer:
The initial acceleration of a proton in a [tex]5.00 * 10^6 N/C[/tex] electric field is approximately [tex]4.79 * 10^1^5 m/s^2[/tex], calculated by applying the electric force formula and Newton's second law.
Explanation:
Calculating Proton Acceleration in an Electric Field
To calculate the initial acceleration of a proton in a [tex]5.00 * 10^6 N/C[/tex] electric field, we follow the steps in the Problem-Solving Strategy for electrostatics:
Identify the known quantities. The electric field (E) is [tex]5.00 * 10^6 N/C[/tex], and the charge of a proton (q) is approximately [tex]1.60 * 10^-^1^9[/tex] C.
Write down the formula for electric force (F = qE).
Calculate the force on the proton: F = [tex](1.60 * 10^-^1^9 C)(5.00 * 10^6 N/C) = 8.00 * 10^-^3 N.[/tex]
Use Newton's second law (F = ma) to find the acceleration (a), knowing the mass of a proton (m) is approximately [tex]1.67 * 10^-^2^7 kg[/tex].
Solve for acceleration: a = F/m =[tex](8.00 * 10^-^3 N) / (1.67 * 10^-^2^7 kg) = 4.79 * 10^1^5 m/s^2[/tex].
Thus, the initial acceleration of the proton is approximately[tex]4.79 * 1015 m/s^2.[/tex]
Starting with the definition 1 in. = 2.54 cm, find the number of (a) kilometers in 1.00 mile and (b) feet in 1.00 km.
Answer :
(a) [tex]1.00\text{ mile}=1.61km[/tex]
(b) [tex]1.00km=3.28\times 10^3ft[/tex]
Explanation :
As are given:
1 inch = 2.54 cm
(a) Now we have to convert the number of kilometers in 1.00 mile.
Conversions used:
1 mile = 5280 ft
1 ft = 12 inch
1 inch = 2.54 cm
1 cm = 10⁻⁵ km
Thus,
[tex]1.00\text{ mile}=5280ft\times \frac{12in}{1ft}\times \frac{2.54cm}{1in}\times \frac{10^{-5}km}{1cm}[/tex]
[tex]1.00\text{ mile}=1.61km[/tex]
(b) Now we have to convert the number of feet in 1.00 km.
[tex]1.00km=10^5cm\times \frac{1in}{2.54cm}\times \frac{1ft}{12inch}[/tex]
[tex]1.00km=3.28\times 10^3ft[/tex]
The unit 1 mile is equivalent to 1.61 km.
Part BThe unit 1 km is equivalent to 3280 ft.
How do you convert the miles into km and km into feet?Part A
The unit 1 miles will convert into km. We know that 1 mile = 5280 ft and 1 ft = 12 inch.
Given that 1 inch = 2.54 cm. Hence
1 mile = 5280 ft.
1 mile = [tex]5280 \times 12[/tex] inch
1 mile = [tex]63360 \times 2.54[/tex] cm
1 mile = 160934.4 cm
Now, 1 cm = 10^-5 km. Then
1 mile = [tex]160934.4 \times 10^{-5}[/tex] km
1 mile = 1.61 km
The unit 1 mile is equivalent to 1.61 km.
Part B
The unit 1 km will convert into feet. We know that 1 km = 10^5 cm
Given that 1 inch = 2.54 cm
1 km = [tex]10^5[/tex] cm
1 km = [tex]10^5 \times \dfrac {1}{2.54}[/tex] inch
1 km = [tex]10^5 \times \dfrac {1}{2.54}\times \dfrac{1}{12}[/tex] ft.
1 km = 3280 ft.
The unit 1 km is equivalent to 3280 ft.
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Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a 750-ft ramp at a high speed v0 and travels 540 ft in 6 s at constant deceleration before its speed is reduced to v0/2. Assuming the same constant deceleration, determine (a) the additional time required for the truck to stop, (b) the additional distance traveled by the truck.
Final answer:
By using the equations of motion under constant acceleration, we can calculate the additional time and distance required for a truck, decelerating on a ramp, to come to a complete stop after having its speed reduced to half of its initial value.
Explanation:
A truck enters a 750-ft ramp at high speed v0 and travels 540 ft in 6 s at constant deceleration before its speed is reduced to v0/2. To solve for both the additional time required for the truck to stop and the additional distance traveled, we use the equations of motion under constant acceleration.
Given:
Initial distance traveled: 540 ft
Time taken: 6 seconds
Initial speed: v0
Final speed at this stage: v0/2
Solution:
Calculate the constant deceleration using the formula: v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Using the determined deceleration, calculate the additional time required for the truck to stop using the formula: t = (v - u) / a.
To find the additional distance traveled, we use the formula: s = ut + 0.5at2.
Through these calculations, we can determine the additional time required for the truck to stop and the additional distance it will travel.
A ball is released from rest at the top of an incline. It is measured to have an acceleration of 2.2. Assume g=9.81 m/s2. What is the angle of the incline in degrees?
Answer:
θ=12.7°
Explanation:
Lets take the mass of the ball = m
The acceleration due to gravity = g = 9.81 m/s²
The acceleration of the block = a
a= 2.2 m/s²
Lets take angle of incline surface = θ
When block slide down :
The gravitational force on the block = m g sinθ
By using Newton's second law
F= m a
F=Net force ,a acceleration ,m=mass
m g sinθ = ma
a= g sinθ
Now by putting the values in the above equation
2.2 = 9.81 sinθ
[tex]sin\theta =\dfrac{2.2}{9.81}\\sin\theta=0.22\\\theta = 12.7\ degrees[/tex]
θ=12.7°
What are (a) the lowest frequency, (b) the second lowest frequency, (c) the third lowest frequency of transverse vibrations on a wire that is 10.0m long, has a mass of 100g, and is stretched under tension of 250 N?
Answer:f1 = 7.90Hz, f2= 15.811Hz, f3 = 23.71Hz.
Explanation: length of string = 10m, mass of string = 100g = 0.1kg, T= 250N
We need the velocity of sound wave in a string when plucked with a tension T, this is given below as
v = √T/u
Where u = mass /length = 0.1/ 10 = 0.01kg/m
Hence v = √250/0.01, v = √25,000 = 158.11
a) at the lowest frequency.
At the lowest frequency, the length of string is related to the wavelength with the formulae below
L = λ/2, λ= 2L.
λ = 2 * 10
λ = 20m.
But v = fλ where v = 158.11m/s and λ= 20m
f = v/ λ
f = 158.11/ 20
f = 7.90Hz.
b) at the first frequency.
The length of string and wavelength for this case is
L = λ.
Hence λ = 10m
v = 158.11m/s
v= fλ
f = v/λ
f = 158.11/10
f = 15.811Hz
c) at third frequency
The length of string is related to the wavelength of sound with the formulae below
L =3λ/2, hence λ = 2L /3
λ = 2 * 10 / 3
λ = 20/3
λ= 6.67m
v = fλ where v = 158.11m/s, λ= 6.67m
f = v/λ
f = 158.11/6.67
f = 23.71Hz.
You spot a plane that is 1.37 km north, 2.71 km east, and at an altitude 4.65 km above your position. (a) How far from you is the plane? (b) At what angle from due north (in the horizontal plane) are you looking? °E of N (c) Determine the plane's position vector (from your location) in terms of the unit vectors, letting î be toward the east direction, ĵ be toward the north direction, and k be in vertically upward. ( km)î + ( km)ĵ + ( km) k (d) At what elevation angle (above the horizontal plane of Earth) is the airplane?
Answer:
Explanation:
a ) Position of the plane with respect to observer ( origin ) is
R = 2.71 i + 1.37 j + 4.65 k
magnitude of R = √ (2.71² + 1.37² + 4.65²)
√(7.344 + 1.8769 + 21.6225)
=√30.8434
= 5.55 km
b ) angle with north
cos Ф = 1.37 / 5.55
= .2468
Ф = 75°
c )
R = 2.71 i + 1.37 j + 4.65 k
=
Aristarchus measured the angle between the Sun and the Moon when exactly half of the Moon was illuminated. He found this angle to be A greater than 90 degrees. B exactly 90 degrees. C less than 90 degrees by an amount too small for him to measure. D less than 90 degrees by an amount that was easy for him to measure.
Answer:
when the Sun illuminates half of the Moon it must be at 90°
Explanation:
The Moon has a circular motion around the Earth and the relative position of the sun, the earth and the moon create the lunar phases.
For this case when the Sun illuminates half of the Moon it must be at 90°, this angle changes with the movement of the moon, it is zero degree for the new moon and 180° for the full moon
the boiling point of sulfur is 444.6 celsius .sulfur melting point is 586.1 fahrenheit lower than its boiling?
point.
a. Determine the melting point of sulfur is degrees celsius.
b. FInd the melting and boiling Points in degrees fahrenheit.
c. FInd the melting and boiling points in kelvins
Answer:
(a) Melting point is 136.8°C
(b) Melting point is 278.24°F
Boiling point is 832.28°F
(c) Melting point is 409.8K
Boiling point is 717.6K
Explanation:
(a) 586.1°F = 5/9(586.1 - 32)°C = 307.8°C
Melting point = 444.6°C - 307.8°C = 136.8°C
(b) Melting point = 136.8°C = (9/5×136.8) + 32 = 278.24°F
Boiling point = 444.6°C = (9/5×444.6) + 32 = 832.28°F
(c) Melting point = 136.8°C = 136.8 + 273 = 409.8K
Boiling point = 444.6°C = 444.6 + 273 = 717.6K
To determine the melting point of sulfur in degrees Celsius and Fahrenheit, and in Kelvin.
Explanation:a. To determine the melting point of sulfur in degrees Celsius, we can subtract the Fahrenheit value from the boiling point of sulfur. Given that the melting point of sulfur is 586.1 Fahrenheit lower than its boiling point, we can calculate the melting point as follows:
Melting point in degrees Celsius = Boiling point in degrees Celsius - 586.1
b. To convert the melting and boiling points from degrees Celsius to Fahrenheit, we can use the formula:
Degrees Fahrenheit = (Degrees Celsius × 9/5) + 32
c. To convert the melting and boiling points from degrees Celsius to Kelvin, we can use the formula:
Kelvin = Degrees Celsius + 273.15
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A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/m
Part A What work is done by the electric force when the charge moves a distance of 0.480 m to the right?
Part B What work is done by the electric force when the charge moves a distance of 0.660 m upward?
Part C What work is done by the electric force when the charge moves a distance of 2.50 m at an angle of 45.0° downward from the horizontal?
A) The work done by the electric field is zero
B) The work done by the electric field is [tex]9.1\cdot 10^{-4} J[/tex]
C) The work done by the electric field is [tex]-2.4\cdot 10^{-3} J[/tex]
Explanation:
A)
The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).
The work done by a force is given by the equation
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force
d is the displacement of the particle
[tex]\theta[/tex] is the angle between the direction of the force and the direction of the displacement
In this problem, we have:
The force is directed vertically upward (because the field is directed vertically upward)The charge moves to the right, so its displacement is to the rightThis means that force and displacement are perpendicular to each other, so
[tex]\theta=90^{\circ}[/tex]
and [tex]cos 90^{\circ}=0[/tex]: therefore, the work done on the charge by the electric field is zero.
B)
In this case, the charge move upward (same direction as the electric field), so
[tex]\theta=0^{\circ}[/tex]
and
[tex]cos 0^{\circ}=1[/tex]
Therefore, the work done by the electric force is
[tex]W=Fd[/tex]
and we have:
[tex]F=qE[/tex] is the magnitude of the electric force. Since
[tex]E=4.30\cdot 10^4 V/m[/tex] is the magnitude of the electric field
[tex]q=32.0 nC = 32.0\cdot 10^{-9}C[/tex] is the charge
The electric force is
[tex]F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N[/tex]
The displacement of the particle is
d = 0.660 m
Therefore, the work done is
[tex]W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J[/tex]
C)
In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is
[tex]\theta=90^{\circ}+45^{\circ}=135^{\circ}[/tex]
Moreover, we have:
[tex]F=1.38\cdot 10^{-3} N[/tex] (electric force calculated in part b)
While the displacement of the charge is
d = 2.50 m
Therefore, we can now calculate the work done by the electric force:
[tex]W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J[/tex]
And the work is negative because the electric force is opposite direction to the displacement of the charge.
Learn more about work and electric force:
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In uniform circular motion, the acceleration is perpendicular to the velocity at every instant. Is this true when the motion is not uniform—that is, when the speed is not constant?
Answer:
No.
Explanation:
Through uniform We can presume you mean constant in both that is in magnitude as well as in direction ⠀
When you apply a steady, non-zero acceleration to a resting body, the velocity can be parallel to the acceleration and not always perpendicular at any time later.
We can do the same for an object with an initial velocity in a direction which is different than that of acceleration, and the direction of its velocity will asymptotically approach to that of acceleration over time.
In both cases the motion of a object undergoing a non-zero persistent acceleration can not be uniform That is the concept of acceleration: the velocity change over time.
A ball player catches a ball 3.0 sec after throwing it vertically upward. With what speed did he throw it, and what height did it reach?
Answer:
14.715 m/s
11.03625 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = a
Time taken to go up will be [tex]\dfrac{3}{2}=1.5\ s[/tex]. This is also equal to the time taken to go down.
[tex]v=u+at\\\Rightarrow v=0+9.81\times 1.5\\\Rightarrow v=14.715\ m/s[/tex]
The speed of the ball when it reaches the player is 14.715 m/s
This is equal to the speed at which the player threw the ball
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{14.715^2-0^2}{2\times 9.81}\\\Rightarrow s=11.03625\ m[/tex]
The ball reached a height of 11.03625 m
An object with a mass of 49.9 pounds is moving with a uniform velocity of 54.4 miles per hour. Calculate the kinetic energy of this object in joules.
Answer:
6698.03 J
Explanation:
Kinetic Energy: This is a form of mechanical energy that is due to a body in motion. The S.I unit of Kinetic Energy is Joules (J).
The formula for kinetic energy is given as
Ek = 1/2mv².......................... Equation 1
Where Ek = kinetic Energy, m = mass of the object, v = velocity of the object.
Given: m = 49.9 pounds, v = 54.4 miles per hours.
Firstly, we convert pounds to kilogram.
If 1 pounds = 0.454 kg,
Then, 49.9 pounds = (0.454×49.9) kg = 22.655 kg.
Secondly, we convert miles per hours to meters per seconds.
If 1 miles per hours = 0.447 meter per seconds,
Then, 54.4 miles per hours = (0.447×54.4) = 24.3168 meters per seconds.
Substitute the value of m and v into equation 1
Ek = 1/2(22.655)(24.3168)²
Ek = 6698.03 J.
Thus the Kinetic energy of the object = 6698.03 J
Given two vectors A⃗ = 4.20 i^+ 7.00 j^ and B⃗ = 5.70 i^− 2.60 j^ , find the scalar product of the two vectors A⃗ and B⃗ .
Applying the concept of scalar product. We know that vectors must be multiplied in their respective corresponding component and then add the magnitude of said multiplications. That is, those corresponding to the [tex]\hat {i}[/tex] component are multiplied with each other, then those corresponding to the [tex]\hat {j}[/tex] component and so on. Finally said product is added.
The scalar product between the two vectors would be:
[tex]\vec{A} \cdot \vec{B} = (4.2\hat{i}+7\hat{j})\cdot (5.7\hat{i}-2.6\hat{j})[/tex]
[tex]\vec{A} \cdot \vec{B} = (4.2*5.7) +(7*(-2.6))[/tex]
[tex]\vec{A} \cdot \vec{B} = 5.74[/tex]
Therefore the scalar product between this two vectors is 5.74
Final answer:
The scalar product of vectors A = 4.20 i + 7.00 j and B = 5.70 i - 2.60 j is calculated by multiplying corresponding components and adding them up, resulting in a scalar product of 5.74.
Explanation:
To find the scalar product (also known as the dot product) of two vectors, you multiply the corresponding components of the vectors and then add these products together. Given two vectors A = 4.20 i + 7.00 j and B = 5.70 i - 2.60 j, the scalar product A cdot B is calculated as follows:
Multiply the x-components together: (4.20)(5.70)
Multiply the y-components together: (7.00)(-2.60)
Add these two products together to get the scalar product.
Now let's do the calculations:
(4.20)(5.70) = 23.94
(7.00)(-2.60) = -18.20
23.94 + (-18.20) = 5.74
Therefore, the scalar product of vectors A and B is 5.74.