Answer:
1) c. Helium
2) Iron
3) False.
Explanation:
1. Red dwarf is the smallest and the coolest star on the sequence. These are common stars in the milky way. Red dwarfs contains metals and the elements with higher atomic number. It is found that Helium is produced in red dwarf stars.
2. Iron is the highest atomic number element that is produced in cores of largest stars. The highest mass stars can make all elements up to iron, which is the heaviest element they can produce.
3. The end of stars life is dependent on the mass they are born with. It is not necessary that all red dwarf stars will become white dwarf stars faster than sun like star.
select the example that best describes a renewable resource.
A.after a shuttle launch, you can smell the jet fuel for hours.
B.solar panels generate electricity that keeps the satellites running.
C.tractor trailers are large trucks that run on diesel fuel.
D. we use our barbeque every night; it cooks with propane.
Answer:
B.solar panels generate electricity that keeps the satellites running.
Explanation:
Solar panels are a renewable resource because they take energy from the sun.
Two particles, of charges q1 and q2, are separated by a distance d on the x-axis with q1 at the origin and q2 in the positive direction. The net electric field due to the particles is zero at x = d/4. With V = 0 at infinity, locate (in terms of d) any point on the x-axis (other than infinity) at which the electric potential due to the two particles is zero.
Answer:
No point on the x-axis
Pls see attached file
A 18.0 kg electric motor is mounted on four vertical springs, each having a spring constant of 24.0 N/cm. Find the period with which the motor vibrates vertically.
Answer:
Explanation:
Total mass m = 18 kg .
Spring are parallel to each other so total spring constant
= 4 x 24 = 96 N/cm = 9600 N/m
Time period of vibration
[tex]T=2\pi\sqrt{\frac{m}{k} }[/tex]
Putting the given values
[tex]T=2\pi\sqrt{\frac{18}{9600} }[/tex]
= .27 s .
A flashlight is held at the edge of a swimming pool at a height h = 1.6 m such that its beam makes an angle of θ = 38 degrees with respect to the water's surface. The pool is d = 1.75 m deep and the index of refraction for air and water are n1 = 1 and n2 = 1.33, respectively.
Required:
What is the horizontal distance from the edge of the pool to the bottom of the pool where the light strikes? Write your answer in meters.
one of the answers that i found was 5.83 m i did some more research and it showed the same answer again. good luck with it. hope i was able to help you.
A block of mass m is suspended by a vertically oriented spring. If the mass of a block is increased to 4m, how does the frequency of oscillation change, if at all
Answer:
The frequency will be reduced by a factor of √2/2
Explanation:
Pls see attached file
The new frequency of oscillation will be half the original frequency of oscillation of spring-block system.
Let the initial mass of block be m.
And new mass is, 4m.
The frequency of oscillating motion is defined as the number of complete oscillation made during the time interval of 1 second. The mathematical expression for the frequency of oscillation of block-spring system is given as,
[tex]f = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
Here,
k is the spring constant.
If the mass of block increased to 4m, then the new frequency of oscillation of spring will be,
[tex]f' = \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{4m}}\\\\\\f' =\dfrac{1}{2} \times \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{m}}\\\\\\f' =\dfrac{1}{2} \times f[/tex]
Thus, we can conclude that the new frequency of oscillation will be half the original frequency of oscillation of spring-block system.
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A narrow copper wire of length L and radius b is attached to a wide copper wire of length L and radius 2b, forming one long wire of length 2L. This long wire is attached to a battery, and a current is flowing through it. If the electric field in the narrow wire is E, the electric field in the wide wire is
Answer:
electric field in the wide wire is
E₂ =[tex]\frac{E}{4}[/tex]
Explanation:
given
length of the copper wire = L
radius of the copper wire r₁ = b
length of the second copper wire = L
radius of the second copper wire r₂ = 2b
electric field in the narrow wire = E₁=E
recall
resistance R = ρL/A
where ρ is resistivity of the copper wire, L is the length, and A is the cross sectional area.
Resistance of narrow wire, R₁
R₁ = ρL/A
where A = πb²
R₁ = ρL/πb²---------- eqn 1
Resistance of wide wire, R₂
R₂ = ρL/A
where A = π(2b)²
R₂ = ρL/π(2b)²
R₂ = ρL/4πb²-------------- eqn 2
R₂ = ¹/₄(ρL/πb²)
comparing eqn 1 and 2
R₁ = 4R₂
calculating the current in the wire,
I = E/(R₁ + R₂)
recall
R₁ = 4R₂
∴ I = E/(4R₂ + R₂)
I = E/5R₂
calculating the potential difference across R₁ & R₂
V₁ = IR₁
I = E/5R₂
∴ V₁ = ER₁/5R₂
R₁ = 4R₂
V₁ = 4ER₂/5R₂
∴V₁ = ⁴/₅E
potential difference for R₂
V₂= IR₂
I = E/5R₂
∴ V₂ = ER₂/5R₂
V₂ = ER₂/5R₂
∴V₂ = ¹/₅E
so, electric field E = V/L
for narrow wire E₁ = V₁/L ----------- eqn 3
for wide wire, E₂ = V₂/L------------ eqn 4
compare eqn 3 and 4
E₂/E₁ = V₂/V₁( L is constant)
E₂/E₁ = ¹/₅E/⁴/₅E
E₂ = E₁/4
note E₁ = E
∴E₂ =[tex]\frac{E}{4}[/tex]
A solenoid 26.0 cm long and with a cross-sectional area of 0.580 cm^2 contains 490 turns of wire and carries a current of 90.0 A.
Calculate:
(a) the magnetic field in the solenoid;
(b) the energy density in the magnetic field if the solenoid is filled with air;
(c) the total energy contained in the coil’s magnetic field (assume the field is uniform);
(d) the inductance of the solenoid.
Answer:
A.21.3T
B.1.8x 10^6J/m^3
C.0.27x10^2J
D.6.6x10^-3H
Explanation:
Pls see attached file
Calculate the maximum kinetic energy of electrons ejected from this surface by electromagnetic radiation of wavelength 236 nm.
Answer:
Explanation:
Using E= hc/wavelength
6.63 x10^-34 x3x10^8/ 236nm
19.86*10^-26/236*10^-9
=0.08*10^-35Joules
n oscillator is driven by a sinusoidal force. The frequency of the applied force A : must be less than the natural frequency of the oscillator. B : is independent of the natural frequency of the oscillator. C : becomes the natural frequency of the oscillator. D : must be equal to the natural frequency of the oscillator. E : must be greater than the natural frequency of the oscillator
Answer:
B : is independent of the natural frequency of the oscillator
Explanation:
You can apply any force you like to a natural oscillator. It is independent of the natural frequency of the oscillator.
The result you get will depend on how the frequency of the applied force and the natural frequency relate to each other. It will also depend on the robustness of the oscillator with respect to the applied force.
Clearly, if the force is small enough, it will have no effect on the oscillator. If it is large enough, it will overpower any motion the oscillator may attempt. For forces in the intermediate range, there will be some mix of natural oscillation and forced behavior. One may modulate the other, for example.
An air conditioner connected to a 103 V rms AC line is equivalent to a 20 resistance and a 1.68 inductive reactance in series. a) What is the impedance of the air conditioner
Answer:
20.07ohms
Explanation:
Impedance is defined as the opposition to the flow of current through the elements of the circuit.
Impedance for R-L AC circuit is expressed as Z = √R²+XL²
R is the resistance
XL is the inductive reactance.
Given resistance of the air condition = 20 ohms
Inductive reactance XL = 1.68 ohms
Z = √20²+1.68²
Z = √400+2.8224
Z = √402.8224
Z = 20.07 ohms
Hence the impedance of the air conditioner is 20.07ohms
The force required to compress a spring with elastic constant 1500N / m, with a distance of 30 cm is
Explanation:
F = kx
F = (1500 N/m) (0.30 m)
F = 450 N
Gravitational Force: Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and the separation of the balls, what will now be the magnitude of the attractive force on each one
Answer:
F' = F
Hence, the magnitude of the attractive force remains same.
Explanation:
The force of attraction between two bodies is given by Newton's Gravitational Law:
F = Gm₁m₂/r² --------------- equation 1
where,
F = Force of attraction between balls
G = Universal Gravitational Constant
m₁ = mass of first ball
m₂ = mass of 2nd ball
r = distance between balls
Now, we double the masses of both balls and the separation between them. So, the force of attraction becomes:
F' = Gm₁'m₂'/r'²
here,
m₁' = 2 m₁
m₂' = 2 m₂
r' = 2 r
Therefore,
F' = G(2 m₁)(2 m₂)/(2 r)²
F' = Gm₁m₂/r²
using equation 1:
F' = F
Hence, the magnitude of the attractive force remains same.
A capacitor that is initially uncharged is connected in series with a resistor and an emf source with E=110V and negligible internal resistance. Just after the circuit is completed, the current trhough the resistor is 6.5*10^-5A. the time constant for the circuit is 4.8s.
1) What is the resistance of the resistor?
2) What is the capacitance of the capacitor?
Answer:
1
[tex]R = 1.692*10^{6} \Omega[/tex]
2
[tex]C = 2.837 *10^{-6} \ F[/tex]
Explanation:
From the question we are told that
The voltage is [tex]E = 110 \ V[/tex]
The current is [tex]I = 6.5 *10^{-5} \ A[/tex]
The time constant is [tex]\tau = 4.8 \ s[/tex]
The resistance of resistor is mathematically evaluated as
[tex]R = \frac{E}{I}[/tex]
substituting values
[tex]R = \frac{ 110 }{ 6.5*10^{-5}}[/tex]
[tex]R = 1.692*10^{6} \Omega[/tex]
The capacitance of the capacitor is mathematically represented as
[tex]C = \frac{\tau}{R}[/tex]
substituting values
[tex]C = \frac{ 4.8}{ 1.692*10^{6}}[/tex]
[tex]C = 2.837 *10^{-6} \ F[/tex]
A boat floating in fresh water displaces 16,000 N of water. How many newtons of salt water would it displace if it floats in salt water of specific gravity 1.10
Answer:
It will displace the same weight of fresh water i.e.16000N. The point is the body 'floats'- which is the underlying assumption here, and by Archimedes Principle, for this body or vessel or whatever it may be, to float it should displace an equal weight of water
Explanation:
If the car decelerates uniformly along the curved road from 27 m/s m/s at A to 13 m/s m/s at C, determine the acceleration of the car at B
Answer:
0.9m/s²
Explanation:
See attached files
Rick spends four hours researching on the internet and does 1090 J of work. In the process, his internal energy decreases by 2190 J. Determine the value of Q, including the algebraic sign.
Answer:
Q = -3280J
Explanation:
From the First Law of Thermodynamics, energy cannot be created nor destroyed but it can be converted from one form to another with the interaction of heat. Mathematically, this can be expressed as:
ΔU = Q + W ----------(i)
Where;
ΔU = total change in internal energy of a system.
Q = heat exchanged between the system and the surrounding
W = work done by or on the system.
If heat is lost into the surrounding, then Q = -ve, else Q = +ve
If work is done on the system, then W = -ve, else W = -ve
=> From the question, Rick is the system and does a work of
W = +1090J [since Rick does the work, W = +ve]
=>Also, the internal energy decreases by 2190J, therefore,
ΔU = -2190J [since there is a decrease in internal energy]
Substitute the values of W and ΔU into equation (i) as follows;
-2190 = Q + 1090
=> Q = -2190 - 1090
=> Q = -3280J
Therefore, the value of Q = -3280J
Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies. Express your answer in terms of Ki and Kf.
The net work done (W) on a particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies is equal to [tex]W = K_f - K_i[/tex]
The net work done (W) can be defined as the work done in moving an object by a net force, which is the vector sum of all the forces acting on the object.
According to Newton's Second Law of Motion, the net work done (W) on an object or physical body is equal to the change in the kinetic energy possessed by the object or physical body.
Mathematically, the net work done (W) on an object or physical body is given by the formula:
[tex]W =\Delta K_E\\\\W = K_f - K_i[/tex]
Where:
W is the net work done.[tex]K_f[/tex] is the initial kinetic energy.[tex]K_i[/tex] is the final kinetic energy.Read more: https://brainly.com/question/22599382
At t=0 a 2150kg rocketship in outer space fires the engine which exerts a force=At2, and F(1.25s)=781.25N in the x direction. Find the impulse J during the interval t=2.00s and t=3.5s
Answer:
5.81 X 10^3 Ns
Explanation:
Given that
F = At² and F at t = 1.25 s is 781.25 N ?
A = F/t² at t = 1.25 s => F = 781.25/(1.25)² = 500 N/s²
d(Impulse) = Fdt
Impulse = ∫Fdt =∫At²dt evaluated in the interval 2.00 s ≤ t ≤ 3.50 s
Impulse = At³/3 = (500/3)(t³) = 166.7t³ between t = 2.00 s and t = 3.50 s
Impulse = 166.7[3.5³ - 2³] = 166.7[42.875 - 8] = 166.7[34.875] = 5813.7 Ns
5.81 X 10^3 N.s
which example describes a nonrenewable resource?
A. everyone in our neighborhood uses solar panels to generate electricity to run their pool pumps.
B. once up and running, the power plant will convert the energy from tides and waves into electricity.
C. there is a long stretch of land in the desert with many windmills that are able to generate enough electricity to run the town.
D. there are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity.
Answer:
D. There are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity
Explanation:
Solar panels are a renewable resource because the sun will not run out. The power plant uses water, so it is also a renewable resource. Windmills use wind, and wind will not run out so it is a renewable resource. However, natural gas and oil are not renewable resources because they will run out one day.
A resistor, capacitor, and switch are all connected in series to an ideal battery of constant terminal voltage. Initially, the switch is open and the capacitor is uncharged. What is the voltage across the resistor and the capacitor at the moment the switch is closed
Answer:
The voltage across the resistor is zero, and the voltage across the capacitor is equal to the terminal voltage of the battery.
Explanation:
This is because when a capacitor is charged no current or voltage flows through it so it will have a voltage equal to the terminal voltage of the battery
Receiver maxima problem. When the receiver moves through one cycle, how many maxima of the standing wave pattern does the receiver pass through
The number of maxima of the standing wave pattern is two.
Maxima problem:At the time when the receiver moves via one cycle so here two maximas should be considered. At the time when the two waves interfere by traveling in the opposite direction through the same medium so the standing wave pattern is formed.
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A circular loop of wire of radius 10 cm carries a current of 6.0 A. What is the magnitude of the magnetic field at the center of the loop
Answer:
3.77x10^-5T
Explanation:
Magnetic field at center of the loop is given as
B=uo*I/2r =(4pi*10-7)*6/2*0.1
B=3.77*10-5Tor 37.7 uTi
A string on the violin has a length of 24.20 cm and a mass of 0.0992 g. The fundamental frequency of the string is 659.3 Hz.
Required:
a. What is the speed of the wave on the string?
b. What is the tension in the string?
Answer:
a. The speed of the wave is 319.1m
b. The tension in the string is 41.74N
Explanation:
Please see the attachments below
In the 1980s, the term picowave was used to describe food irradiation in order to overcome public resistance by playing on the well-known safety of microwave radiation. Find the energy in MeV of a photon having a wavelength of a picometer.
Answer:
E = 1.24MeV
Explanation:
The photon travels at the speed of light, 3.0 × [tex]10^{8}[/tex] m/s, and given that its frequency = 1 picometer = 1.0 × [tex]10^{-12}[/tex] m.
Its energy can be determined by;
E = hf
= (hc) ÷ λ
where E is the energy, h is the Planck's constant, 6.626 × [tex]10^{-34}[/tex] Js, c is the speed of the light and f is its frequency.
Therefore,
E = (6.626 × [tex]10^{-34}[/tex]× 3.0 × [tex]10^{8}[/tex]) ÷ 1.0 × [tex]10^{-12}[/tex]
= 1.9878 × [tex]10^{-25}[/tex] ÷ 1.0 × [tex]10^{-12}[/tex]
E = 1.9878 × [tex]10^{-13}[/tex] J
But, 1 eV = 1.6 × [tex]10^{-19}[/tex] J. So that;
E = [tex]\frac{1.9878*10^{-13} }{1.6*10^{-19} }[/tex]
= 1242375 eV
∴ E = 1.24MeV
The energy of the photon is 1.24MeV.
Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
1. Blood flow through the aorta is approximately 11.2 cm/s
2. The diameter of the aorta is approximately 3.0 cm
3. Assume the heart pumps its own volume with each beat
4. Assume a pulse rate of 67 beats per minute.
Answer:
Explanation:
radius of aorta = 1.5 cm
cross sectional area = π r²
= 3.14 x 1.5²
= 7.065 cm²
volume of blood flowing out per second out of heart
= a x v , a is cross sectional area , v is velocity of flow
= 7.065 x 11.2
= 79.128 cm³
heart beat per second = 67 / 60
= 1.116666
If V be the volume of heart
1.116666 V = 79.128
V = 70.86 cm³.
1. The frequency of a wave defines
O A. the minimum height of a wave.
O B. the maximum height of a wave.
O C. how fast the wave is moving in cycles per second.
D. the height of the wave at a given point.
Answer:
The answer is CExplanation:
Frequency, in physics, the number of waves that pass a fixed point in unit time; also, the number of cycles or vibrations undergone during one unit of time by a body in periodic motion. ... See also angular velocity; simple harmonic motion.
A string passing over a pulley has a 3.85-kg mass hanging from one end and a 2.60-kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.5 cm and mass 0.79 kg .
A. If the bearings of the pulley were frictionless, what would be the acceleration of the two masses?
B. In fact, it is found that if the heavier mass is given a downward speed of 0.20 m/s , it comes to rest in 6.4 s . What is the average frictional torque acting on the pulley?
Answer:
Explanation:
Let the acceleration be a of the system
T₁ and T₂ be the tension in the string attached with 3.85 and 2.6 kg of mass
for motion of 3.85 kg , applying newton's law
3.85g - T₁ = 3.85 a
for motion of 2.6 kg
T₂ - 2.6g = 2.6 a
T₂ - T₁ + 1.25 g = 6.45 a
T₁ - T₂ = 1.25 g - 6.45 a
for motion of pulley
(T₁ - T₂ ) x R = I x α where R is radius of pulley , I is its moment of inertia and α is angular acceleration
(T₁ - T₂ ) x R = 1 /2 m R² x a / R
(T₁ - T₂ ) = m x a / 2 = .79 x a / 2 = . 395 a
1.25 g - 6.45 a = .395 a
1.25 g = 6.845 a
a = 1.79 m /s²
B )
When heavier mass is given speed of .2 m /s , it comes to rest in 6.4 s
Average deceleration = .2 / 6.4 = .03125 m /s²
Total deceleration created by frictional torque = 1.79 + .03125
= 1.82125 m /s²
If R be the average frictional torque acting on the pulley
angular deceleration of pulley = a / R
= 1.82125 / .045
= 40.47 rad /s²
Now R = I x 40.47 , I is moment of inertia of pulley
= 1 /2 x .79 x .045² x 40.47
= .0323 N.m
Torque created = .0323 Nm
The acceleration of the two masses hanging from ends of the pulley is 31 m/s².
The average frictional torque acting on the pulley is 0.55 Nm.
The given parameters;
mass of the pulley, = M = 0.79 kgfirst mass, m₁ = 3.85 kgsecond mass, m₂ = 2.6 kgradius, R = 4.5 cm = 0.045 mThe acceleration of the two masses is determined by taking net torque acting on the pulley;
[tex]\tau _{net} = I \alpha[/tex]
[tex]T_1R - T_2R = I \alpha\\\\[/tex]
where;
T is the tension on both stings suspending the masses = mgI is the moment of inertia of the pulley [tex]= \frac{MR^2}{2}[/tex]α is the angular acceleration[tex]R(T_1 - T_2) = (\frac{MR^2}{2} )(\frac{a}{R} )\\\\T_1 - T_2 = (\frac{MR^2}{2} )(\frac{a}{R} ) \times \frac{1}{R} \\\\T_1 - T_2 = \frac{M}{2} \times a\\\\a = \frac{2}{M} (T_1 - T_2)[/tex]
Substitute the given parameters, to solve for the acceleration of the masses;
[tex]a = \frac{2}{M} (m_1g - m_2 g)\\\\a = \frac{2g}{M} (m_1 - m_2)\\\\a = \frac{2 \times 9.8}{0.79} (3.85 - 2.6)\\\\a = 31 \ m/s^2[/tex]
The average frictional torque acting on the pulley when the heavier mass speeds down by 0.2 m/s and stop by 6.4 s.
[tex]a = \frac{v}{t} = \frac{0.2}{6.4} = 0.031 \ m/s^2 \\\\ a_t = 31 m/s^2+ 0.031 m/s^2 = 31.031 m/s^2 \\\\\tau = I \alpha\\\\\tau = (\frac{MR^2}{2} )(\frac{a_t}{R} )\\\\\tau = (\frac{0.79 \times 0.045^2 }{2} ) (\frac{31.031}{0.045} )\\\\\tau = 0.55 \ Nm[/tex]
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A 900 kg roller coaster car starts from rest at point A. rolls down the track, goes
around a loop (points B and C) and then flies off the inclined part of the track (point D),
Figure 2.
The dimensions are: H =80 m.
r= 15m, h=10m and theta =9.30°
Calculate the
(a) gravitational potential energy at point A.
(b) velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J.
c) distance of the car land (in the horizontal direction) from point D if given the
velocity at point D is 37.06 m/s
I
Answer:
gravitational potential energy at point A.
A) The gravitational potential energy at point A is; 705600 J
B) The velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J is; v = 31.295 m/s
A) Formula for gravitational potential energy is;
PE = mgh
At point A;
mass; m = 900 kg
height; h = 80 m
Thus;
PE = 900 × 9.8 × 80
PE = 705600 J
B) Kinetic energy of the roller coaster at point C is given as;
KE = PE - W
We are given Workdone; W = 264870 J
Thus;
KE = 705600 - 264870
KE = 440730 J
Thus, velocity at point C is gotten from the formula of kinetic energy;
KE = ½mv²
v = √(2KE/m)
v = √(2 × 440730/900)
v = 31.295 m/s
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At 30.0 m below the surface of the sea (density = 1 025 kg/m3), where the temperature is 5.00°C, a diver exhales an air bubble having a volume of 0.95 cm3. If the surface temperature of the sea is 20.0°C, what is the volume of the bubble just before it breaks the surface?
Answer:
The volume is [tex]V_a = 1.510 *10^{-5} m^3[/tex]
Explanation:
From the question we are told that
The depth below the see is [tex]d_1 = 30.0 \ m[/tex]
The density of the sea is [tex]\rho_s = 1025 \ kg /m^3[/tex]
The temperature at this level is [tex]T_d = 5.00 ^oC = 278 \ K[/tex]
The volume of the air bubble at this depth is [tex]V_d = 0.95 \ cm^3 = 0.95 *0^{-6}\ m[/tex]
The temperature at the surface is [tex]T_a = 20^oC =293\ K[/tex]
Generally the pressure at the given depth is mathematically evaluated as
[tex]P_d = P_o + \rho_s * g * d[/tex]
Where [tex]P_o[/tex] is the atmospheric pressure with a constant value
[tex]P_o = 1.013 *10^{5} \ Pa[/tex]
substituting values
[tex]P_d = 1.013 * 10^{5} * + (1025 * 9.8 * 30 )[/tex]
[tex]P_d = 4.02650 * 10^{5} \ Pa[/tex]
According to the combined gas law
[tex]\frac{P_a * V_a }{T_a } = \frac{P_d * V_d }{T_d }[/tex]
=> [tex]V_a = \frac{4.026650 *10^{5} * 0.95 *10^{-6} * 293 }{278 * 1.013*10^{5} }[/tex]
=> [tex]V_a = 1.510 *10^{-5} m^3[/tex]
Velocity of a Hot-Air Balloon A hot-air balloon rises vertically from the ground so that its height after t sec is given by the following function.
h=1/2t2+1/2t
(a) What is the height of the balloon at the end of 40 sec?
(b) What is the average velocity of the balloon between t = 0 and t = 30?
ft/sec
(c) What is the velocity of the balloon at the end of 30 sec?
ft/sec
Answer:
Explanation:
Given the height reached by a balloon after t sec modeled by the equation
h=1/2t²+1/2t
a) To calculate the height of the balloon after 40 secs we will substitute t = 40 into the modeled equation and calculate the value of t
If h(t)=1/2t²+1/2t
h(40) = 1/2(40)²+1/2 (40)
h(40) = 1600/2 + 40/2
h(40) = 800 + 20
h(40) = 820 feet
The height of the balloon after 40 secs is 820 feet
b) Velocity is the change of displacement of a body with respect to time.
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
when v = 0sec
v(0) = 0 + 1/2
v(0) = 1/2 ft/sec
at v = 30secs
v(30) = 30 + 1/2
v(30) = 30 1/2 ft/sec
average velocity = v(30) - v(0)
average velocity = 30 1/2 - 1/2
average velocity of the balloon between t = 0 and t = 30 = 30 ft/sec
c) Velocity is the change of displacement of a body with respect to time.
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
The velocity of the balloon after 30secs will be;
v(30) = 30+1/2
v(30) = 30.5ft/sec
The velocity of the balloon after 30 secs is 30.5 feet/sec
A) The height of the balloon at the end of 40 sec is 820 feet.
B) The average velocity of the balloon is 30 ft/sec.
C) The velocity of the balloon at the end of 30 sec is
VelocityGiven :
h=1/2t²+1/2tPart A)
The height of the balloon after 40 secs is :
h(t)=1/2t²+1/2t
h(40) = 1/2(40)²+1/2 (40)
h(40) = 1600/2 + 40/2
h(40) = 800 + 20
h(40) = 820 feet
The height of the balloon after 40 secs is 820 feet
Part B)
The average velocity of the balloon is :
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
when v = 0 sec
v(0) = 0 + 1/2
v(0) = 1/2 ft/sec
When at v = 30secs
v(30) = 30 + 1/2
v(30) = 30 1/2 ft/sec
average velocity = v(30) - v(0)
average velocity = 30 1/2 - 1/2
average velocity of the balloon = 30 ft/sec
The average velocity of the balloon is 30 ft/sec.
Part C)
The velocity of the balloon at the end of 30 sec is :
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
The velocity of the balloon after 30secs will be;
v(30) = 30+1/2
v(30) = 30.5ft/sec
The velocity of the balloon after 30 secs is 30.5 feet/sec.
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