This question is incomplete, the complete question is;
The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of curvature R of the hill is 15.8 m.
(a) What is the force (N) that the track must exert on the car? (positive is up)
(b) What must be the force (N) that the car exerts on a 61 kg passenger?
Answer:
a) the force (N) that the track must exert on the car is -6139.14 N
b) the force (N) that the car exerts on a 61 kg passenger is -1170.27 N
Explanation:
Given the data in the question;
Let N represent the force that the track must exerted on the car
Net force on the car Fnet = Mg + N
so
M × a = Mg + N
N = Ma - Mg
N = Ma - M(v²/R)
we substitute
N = (320kg × 9.8m/s²) - ( 320 × ((21.4m/s)² / 15.8 m) )
N = 3136 - ( 320 × 28.9848 )
N = 3136 - 9275.136
N = -6139.14 N
Therefore, the force (N) that the track must exert on the car is -6139.14 N
b) What must be the force (N) that the car exerts on a 61 kg passenger?
Let N represent the force that the car exerts on 61kg passengers
so
Net force of passengers Fnet = mg + N
Ma = Mg + N
N = Ma - Mg
N = Ma - M(v²/R)
N = (61kg × 9.8m/s²) - ( 61 × ((21.4m/s)² / 15.8 m) )
N = 597.8 - ( 61 × 28.9848)
N = 597.8 - 1768.0728
N = -1170.27 N
Therefore, the force (N) that the car exerts on a 61 kg passenger is -1170.27 N
The centripetal force of the track on the car moving in the circular path is [tex]1.465 \times 10^6 \ N[/tex].
The force (N) that the car exerts on a 61 kg passenger is 597.8 N.
Centripetal force of the track
The centripetal force of the track on the car moving in the circular path is calculated as follows;
[tex]F_c = \frac{mv^2}{r}\\\\ F_c = \frac{320 \times 21.4^2}{0.1} \\\\F_c = 1.465 \times 10^6 \ N[/tex]
Normal force of the passengerThe force (N) that the car exerts on a 61 kg passenger is equal to the force the passenger exerts on the car based on Newton's third law of motion.
F = mg
F = 61 x 9.8
F = 597.8 N
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Flying insects such as bees may accumulate a small positive electric charge as they fly. In one experiment, the mean electric charge of 50 bees was measured to be +(30±5)pC+(30±5)pC per bee. Researchers also observed the electrical properties of a plant consisting of a flower atop a long stem. The charge on the stem was measured as a positively charged bee approached, landed, and flew away. Plants are normally electrically neutral, so the measured net electric charge on the stem was zero when the bee was very far away. As the bee approached the flower, a small net positive charge was detected in the stem, even before the bee landed. Once the bee landed, the whole plant became positively charged, and this positive charge remained on the plant after the bee flew away. By creating artificial flowers with various charge values, experimenters found that bees can distinguish between charged and uncharged flowers and may use the positive electric charge left by a previous bee as a cue indicating whether a plant has already been visited (in which case, little pollen may remain). What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)?
(a) Because air is a good conductor, the positive charge on the bee’s surface flowed through the air from bee to plant.
(b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee.
(c) The plant became electrically polarized as the charged bee approached.
(d) Bees that had visited the plant earlier deposited a positive charge on the stem.
Answer:
a) True
Explanation:
There are several possible explanations for this positive charge
* The explanation of the small positive charge in the plant when the bee approaches is like a defense system of the plants,
to prevent the bees from taking the pollen, but the flowers need the bees to transport the pollen for fertilization, so this possibility is not correct
* The air is conductive so the bee indexes a charge in the nearby air, this charge must be negative and this charge induced in the air induces a charge on the flower that must be positive.
When reviewing the different statements we have
a) True, it agrees with the second explanation of the phenomenon
b) False. The earth is a deposit of negative charge
c) false. If this is the case the charge should be negative
d) False. This residual charge from the other bees is quickly neutralized by the charge from the Earth.
Answer:
Explanation:
.
what is transactive memory
Here is a copy+pasted definition...
Transactive memory is a psychological hypothesis first proposed by Daniel Wegner in 1985 as a response to earlier theories of "group mind" such as groupthink.[1] A transactive memory system is a mechanism through which groups collectively encode, store, and retrieve knowledge. Transactive memory was initially studied in couples and families where individuals had close relationships but was later extended to teams, larger groups, and organizations to explain how they develop a "group mind",[1] a memory system that is more complex and potentially more effective than that of any of its individual constituents. A transactive memory system includes memory stored in each individual, the interactions between memory within the individuals, as well as the processes that update this memory. Transactive memory, then, is the shared store of knowledge.
In simple terms...
A transactive memory is a psychological hypothesis. It is a system in the brain which allows human's to encode, store, or get information.
For example...
let's say I learn 5 colors when I'm younger. My "transactive memory" will store it and will allow my brain to "retrieve" the information to allow me to remember colors I previously "stored".
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Which describes the reaction of a substance to form one or more new substances?
a physical change
a chemical change
a change of state
How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?
This question is incomplete, the complete question is;
A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?
Answer:
the required heat energy is 16 J
Explanation:
Given the data in the question;
Lets consider the ideal gas equation;
PV = nRT
from the image, we calculate initial pressure;
Pi = ( 2000N/M × 0.06m) / 0.0008 m²
Pi = 15 × 10⁴ Pa
next we find Initial velocity
Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²
now we find the number of moles
n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K
N = 6.6 × 10⁻³ mol
next we calculate the final temperature;
Pf = ( 2000N/m × 0.08) / 0.0008 m²
Pf = 2 × 10⁵ Pa
Calculate the final Volume
Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³
we also determine the final temperature
[tex]T_{f}[/tex] = (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK
[tex]T_{f}[/tex] = 466.8 K
so change in temperature ΔT
ΔT = 466.8 K - 300K = 166.8 K
we then calculate the change in thermal energy
ΔU = nCΔT
ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K
ΔU = 13.761 J
C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic
now we calculate the work done;
W = 1/2 × K( x[tex]_{i\\}[/tex]² - x[tex]_{f\\}[/tex]² )
W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )
= - 2.8 J
and we then calculate the heat energy using the following expression;
Q = ΔU - W
we substitute
Q = 13.761 - (- 2.8 J)
Q = 13.761 + 2.8 J)
Q = 16 J
Therefore, the required heat energy is 16 J
A 680 nm laser illuminates a double slit apparatus with a slit separation distance of 7.83 μm. On the viewing screen, you measure the distance from the central bright fringe to the 2nd bright fringe to be 88.2 cm. How far away (in meters) is the viewing screen from the double slits?
Answer:
L = 5.08 10⁻¹ m = 50.8 cm
Explanation:
In a double slit experiment for constructive interference is given by
d sin θ = m λ
let's use trigonometry to find a relationship with the distance
tan θ = y / L
these experiments are very small angles
tan θ = [tex]\frac{sin \ \theta}{cos \ \theta}[/tex] = sin θ
when substituting
sin θ = y / L
substituting in the first equation
d y / L = m λ
L = [tex]\frac{d \ y}{m \ \lambda}[/tex]
let's calculate
L = [tex]\frac{ 0.783 \ 10^{-6} \ 88.2 \ 10^{-2} }{2 \ 680 \ 10^{-9}}[/tex]
L = 5.08 10⁻¹ m
The distance of viewing screen from the double slits will be L = 5.08 *10⁻¹ m = 50.8 cm
What is constructive interference?
This position, where the resulting wave is larger than either of the two original, is called constructive interference.
In a double slit experiment for constructive interference is given by
d sin θ = m λ
let's use trigonometry to find a relationship with the distance
[tex]tan\theta=\dfrac {y}{L}[/tex]
these experiments are very small angles
tan θ = = sin θ
when substituting
[tex]sin\theta = \dfrac{y}{l}[/tex]
substituting in the first equation
[tex]\dfrac{dy}{L}=m\lambda[/tex]
[tex]L=\dfrac{dy}{L\lambda}[/tex]
let's calculate
[tex]L=\frac{0.783\times 10^{-6}\times 88.2\times 10^{-2}}{2.680\times 10^{-9}}[/tex]
L = 5.08 10⁻¹ m
Hence the distance of viewing screen from the double slits will be L = 5.08 10⁻¹ m = 50.8 cm
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2. Using a giant screw, a crew does 650 J of work to drill a hole into a rock.
The screw does 65 J of work. What is the efficiency of the screw? Show your
work. Hellpppp
Answer:
42,250
Explanation:
It goes inside=
Displacemt
It does work=
Work done
To find efficiency of jule we do=
Dicplacement × Work done
650 × 65
42,250
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A rocket, with a mass of 5100 kg, has an engine that provides a net upward force of 8.0 x 10^5 N. It starts from rest and reaches a maximum speed of 900 m/s. How long does it take to reach that maximum velocity?
Answer:
5.7375 seconds
Explanation:
The computation of the time required to reach that maximum velocity is shown below:
Given that
Mass = m = 5100 kg
Net upward force F = 8 × times 10^5 N
Initial speed = V_i = 0
Maximum speed = V = 900 m.s
Based on the above information
Impluse J = m(V - V_i)
= 5100 (900 - 0)
= 459 × 10^4 kg m.s
As we know that
J = FT
So
T = J ÷ F
= (459 × 10^4) ÷ (8 × 10^5)
= 5.7375 seconds
Name the nutrients required for the body
Answer:
Explanation:
Water.
Carbohydrates.
Protein-Amino acids.
Fat.
Vitamins.
Minerals.
Omega-3 fatty acids.
Kim was adopted as a baby and raised by loving parents in an enriched environment. Studies have shown that ________
Answer:
Kim was adopted as a baby and raised by loving parents in an enriched environment. Studies have shown that the correlation between IQ
( intelligence quotient) scores of adopted children and those of their biologically unrelated family members reduce to zero as they grow into adulthood.
Explanation:
Kim was adopted as a baby and raised by loving parents in an enriched environment. Studies have shown that the correlation between IQ
( intelligence quotient) scores of adopted children and those of their biologically unrelated family members reduce to zero as they grow into adulthood.
Also, a loving environment is equally important for a child's upbringing.
A cylindrical metal rod has a resistance . If both its length and its diameter are quadrupled, its new resistance will be:________.
A. 16R
B. R/4
C. R
D. 4R
Answer:
R' = R/4
Explanation:
The resistance of a metal rod is R. It is given by the relation as follows :
[tex]R=\rho\dfrac{l}{A}[/tex]
Where
l is the length and A is the area of cross-section
[tex]A=\pi r^2=\pi (\dfrac{d}{2})^2[/tex]
If both its length and its diameter are quadrupled, it means,
l' = 4l
and d'= 4d
It means,
[tex]A'=\pi (\dfrac{4d}{2})^2[/tex]
Let new resistance be R'. So,
[tex]R'=\rho\dfrac{l'}{A'}\\\\R'=\rho\dfrac{4l}{\pi (\dfrac{4d}{2})^2}\\\\=\rho \dfrac{4l}{\pi \dfrac{16d^2}{2}}\\\\=\dfrac{4}{16}\times \dfrac{\rho l}{\pi \dfrac{d^2}{2}}\\\\=\dfrac{1}{4}\times \dfrac{\rho l}{\pi \dfrac{d^2}{2}}\\\\R'=\dfrac{R}{4}[/tex]
So, the correct option is (B) "R/4".
You are walking down a straight path in a park and notice there is another person walking some distance ahead of you. The distance between the two of you remains the same, so you deduce that you are walking at the same speed of 1.17 m/s. Suddenly, you notice a wallet on the ground. You pick it up and realize it belongs to the person in front of you. To catch up, you start running at a speed of 2.75 m/s. It takes you 10.5 s to catch up and deliver the lost wallet. How far ahead of you was this person when you started running
Answer:
16.6 m
Explanation:
Let d be the distance the other person is ahead of you. Since the other person is walking at a speed, v = 1.17 m/s, after picking the wallet, the other person moves a distance , vt in time, t = 10.5 s, the total distance covered by you till catch up is D = d + vt.
Also, you moves with a speed of v' = 2.75 m/s in time t = 10.5 s as you pick up the wallet, you covers a distance d' = v't at catch up.
At catch up, D = d'
d + vt = v't
d = v't - vt
d = (v' - v)t
Substituting the values of the variables into d, we have
d = (2.75 m/s - 1.17 m/s)10.5 s
d = (1.58 m/s)10.5 s
d = 16.59 m
d ≅ 16.6 m
So, the other person was 16.6 m ahead of you when you started running.
A simple pendulum is used to measure gravity using the following theoretical equation,TT=2ππ�LL/gg ,where L is the length of the pendulum, g is gravity, andT is the period of pendulum.Twenty measurements of T give a mean of 1.823 seconds and a standard deviation of 0.0671 s. The device used to measure time has a resolution of 0.02 s. The pendulum length is measured once to be 0.823 m (with a scale having a resolution of 0.001 m). Determine the value of g and its uncertainty (assume 90% confidence where necessary). You may use any method of uncertainty propagation that we covered in class.
Answer:
g ±Δg = (9.8 ± 0.2) m / s²
Explanation:
For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use
T = [tex]2\pi \sqrt{ \frac{L}{g} }[/tex]
T² = [tex]4\pi ^2 \frac{L}{g}[/tex]4pi2 L / g
g = [tex]4\pi ^2 \frac{L}{T^2}[/tex]
They indicate the average time of 20 measurements 1,823 s, each with an oscillation
let's calculate the magnitude
g = [tex]4\pi ^2 \frac{0.823}{1.823^2}[/tex]4 pi2 0.823 / 1.823 2
g = 9.7766 m / s²
now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation
for the period
T = t / n
ΔT = [tex]\frac{dT}{dt}[/tex] Δt + [tex]\frac{dT}{dn}[/tex] ΔDn
In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently
ΔT = Δt / n
ΔT = Δt
now let's look for the uncertainty of g
Δg = [tex]\frac{dg}{dL}[/tex] ΔL + [tex]\frac{dg}{dT}[/tex] ΔT
Δg = [tex]4\pi ^2 \frac{1}{T2}[/tex] ΔL + 4π²L (-2 T⁻³) ΔT
a more manageable way is with the relative error
[tex]\frac{\Delta g}{g} = \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}[/tex]
we substitute
Δg = g ( \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}DL / L + ½ Dt / T)
the error in time give us the stanndard deviation
let's calculate
Δg = 9.7766 ([tex]\frac{0.001}{0.823} + \frac{1}{2} \ \frac{0.671}{1.823}[/tex])
Δg = 9.7766 (0.001215 + 0.0184)
Δg = 0.19 m / s²
the absolute uncertainty must be true to a significant figure
Δg = 0.2 m / s2
therefore the correct result is
g ±Δg = (9.8 ± 0.2) m / s²
What is the cost, in dollars, of heating the hot tub, assuming 75.0% efficiency to account for heat transfer to the surroundings?
Answer:
$ 5.93
Explanation:
If we are to put into consideration the following factors even though it was not all stated in the question.
What is the cost of heating a hot tub containing 1500 kg of water from 10.0°C to 40.0°C, assuming 75.0% efficiency to take heat loss to surroundings into account? The cost of electricity is ¢8.50/kW·h.
Q = mc∆T .........1
Where m is the mass of water = 1500kg
C is the specific heat capacity of water = 4.184
∆T = temperature change
= 40° - 10°
= 30°
Now, substitute the values into equation 1
= 1500×4.184×30
= 188280
Q in kw/s
= 188280/3600
= 52.3kw/hr
Efficiency = 75%
= 75/100
= 0.75
Hence we have
52.3/0.75
= 69.73kw/hr
C = Q × cost per kw/hr
C = 69.73× $8.5/100
C = 5.92705
C =$ 5.93
Hence, the cost of heating the hot tub is $ 5.93
(It made me choose physics as the category, but the class is Astronomy)
The Earth's Escape velocity, the velocity needed to escape the Earth's
gravitational field, is
If the magnitude of the cross product of two vectors is one-half the dot product of the same vectors, what is the angle between the two vectors?
Answer:
26.57 degrees
Explanation:
Let the two vectors be a and b
The dot product of two vectors is expressed as;
a.b = |a||b|cosθ
The cross product is expressed as;
a×b = |a||b|sinθ
If the magnitude of the cross product of two vectors is one-half the dot product of the same vectors, then;
|a×b| = 1/2|a.b|
|a||b|sinθ = 1/2|a||b|cosθ
sinθ = 1/2cosθ
2sinθ = cosθ
sinθ/cosθ = 1/2
tanθ = 1/2
θ = arctan(1/2)
θ = 26.57 degrees
Hence the angle between the two vectors is 26.57 degrees
Which of the following would MOST likely slow Earth's tectonic activity?
O A. Earth's crust becomes cooler.
O B. Earth's mantle becomes warmer.
O C. Earth's mantle becomes cooler.
O D. Earth's outer core becomes warmer.
The Answer to your question is:
A.
The Earth's tectonic activity occurs by the movement of the fourteen main tectonic plates of the planet, which move over the mantle continuously, and result in the formation of mountains, earthquakes, tsunamis, volcanic activities, etc.
The decrease in the Earth's tectonic activity would therefore occur if the Earth's mantle became cooler, as the tectonic plates move over the magma, which is a paste formed by silicate of iron and magnesium and whose temperature reaches 600° and 1,200° Celsius , driven by forces from inside the planet.
Therefore the letter C is correct, as with the Earth's mantle cooler, the magma would become more solid which would decrease the Earth's tectonic activity.
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Two forces are exerted on an object in the vertical direction: a 20 N force downward and a 10 N force upward. The mass of the object is 25 kg. (1) What are some possibilities about the motion of this object? (2) Represent the motion of the object with a force diagram and a motion diagram.
Answer:
They are equal.
Explanation:
Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?
Answer:
Hello your question is poorly written below is the complete question
Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?
answer :
a) 231.48 days
b) n = 3.125 * 10^15
Explanation:
Battery moved 10,000 coulombs
current rate = 0.5 mA
A) Determine how long the clock run on the battery. use the relation below
q = i * t ----- ( 1 )
q = charge , i = current , t = time
10000 = 0.5 * 10^-3 * t
hence t = 2 * 10^7 secs
hence the time = 231.48 days
B) Determine how many electrons per second flowed
q = n*e ------ ( 2 )
n = number of electrons
e = 1.6 * 10^-19
q = 0.5 * 10^-3 coulomb ( charge flowing per electron )
back to equation 2
n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )
hence : n = 3.125 * 10^15
Can a particle move in a direction of increasing electric potential, yet have its electric potential energy decrease? Explain
Answer:
Explanation:
Yes , it is possible .
When a negative charge moves towards a positive charge , it is moving in the direction of increasing electrical potential . In the whole process , its electrical potential energy decreases and its kinetic energy increases .
Actually the potential energy of a negative charge near a positive charge is negative and it is inversely proportional to distance .
V = - Qq / R , When R decreases , the negative value of potential increases . That means potential energy decreases .
HELP PLEASE DUE IN 3 MINUTES!!!! 60 POINTS
Wind wears down rocks by blowing _________ them.
ice on
sand against
gently on
rocks toward
Answer:
i think rocks towards is correct answer
Answer:
It is gently on
Explanation:
I'm am not a 100 percent sure but try this o took that before
The resistivity of blood is related to its hematocrit, the volume fraction of red blood cells in the blood. A commonly used equation relating the hematocrit h to the blood resistivity rho (in Ω⋅m) is rho=1.32/(1−h)−0.79. In one experiment, blood filled a graduated cylinder with an inner diameter of 0.90 cm. The resistance of the blood between the 1.0 cm and 2.0 cm marks of the cylinder was measured to be 198 Ω.
Required:
What was the hematocrit for this blood?
Answer:
[tex]0.35598[/tex]
Explanation:
r = Radius = [tex]\dfrac{0.9}{2}=0.45\ \text{cm}[/tex]
R = Resistance = [tex]198\ \Omega[/tex]
A = Area = [tex]\pi r^2[/tex]
l = Length of blood in cylinder = 1 cm
h = Hematocrit of the blood
Resistivity is given by
[tex]\rho=\dfrac{1.32}{1-h}-0.79[/tex]
Resistance is given by
[tex]R=(\dfrac{1.32}{1-h}-0.79)\dfrac{l}{\pi r^2}\\\Rightarrow h=1-\dfrac{1.32}{\dfrac{R\pi r^2}{l}+0.79}\\\Rightarrow h=1-\dfrac{1.32}{\dfrac{198\times \pi\times (0.45\times 10^{-2})^2}{0.01}+0.79}\\\Rightarrow h=0.35598[/tex]
The hematocrit of the blood is [tex]0.35598[/tex].
A-10A twin-jet close-support airplane is approximately rectangular with a wingspan (the length perpendicular to the flow direction) of 17.5 m and a chord (the length parallel to the flow direction) of 3 m. The airplane is flying at standard sea level with a velocity of 200 m/s. If the flow is considered to be completely laminar, calculate the boundary layer thickness at the trailing edge and the total skin friction drag. Assume that the wing is approximated by a flat plate. Assume incompressible flow.
Solution :
Given :
Rectangular wingspan
Length,L = 17.5 m
Chord, c = 3 m
Free stream velocity of flow, [tex]$V_{\infty}$[/tex] = 200 m/s
Given that the flow is laminar.
[tex]$Re_L=\frac{\rho V L}{\mu _{\infty}}$[/tex]
[tex]$=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$[/tex]
[tex]$= 4.10 \times 10^7$[/tex]
So boundary layer thickness,
[tex]$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$[/tex]
[tex]$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$[/tex]
= 0.0024 m
The dynamic pressure, [tex]$q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$[/tex]
[tex]$ =\frac{1}{2} \times 1.225 \times 200^2$[/tex]
[tex]$=2.45 \times 10^4 \ N/m^2$[/tex]
The skin friction drag co-efficient is given by
[tex]$C_f = \frac{1.328}{\sqrt{Re_L}}$[/tex]
[tex]$=\frac{1.328}{\sqrt{4.1 \times 10^7}}$[/tex]
= 0.00021
[tex]$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$[/tex]
[tex]$=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$[/tex]
= 270 N
Therefore the net drag = 270 x 2
= 540 N
Imagin you have mixed together some sand and salt Based on the venn diagram this mixture would be placed where
Answer:
a
Explanation:
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In anticipation of a long 10o upgrade, a bus driver accelerates at a constant rate of 5 ft/s^2 while still on a level section of a highway. Knowing that the speed of the bus was 80 mph as it begins to go up the hill and that the driver does not change the setting on his throttle or shift gears, determine the distance traveled (in miles) by the bus up the hill when its speed decreased to 50 mph.
Answer:
The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles
Explanation:
The parameters of the motion of the driver are;
The upgrade of the road, θ = 10°
The rate of constant acceleration of the bus driver = 5 ft./s²
The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s
The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s
The acceleration due to gravity, g ≈ 32.1740 ft./s²
Therefore, we have;
The acceleration due to gravity down the incline plane, gₓ = g·sinθ
∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²
The net acceleration of the bus, on the incline plane, [tex]a_{Net}[/tex] = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²
The vertical component of the velocity, [tex]v_y[/tex] = v × sin(θ)
∴ [tex]v_y[/tex] = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s
vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s
The velocity of the car, v₂, on the inclined plane is given as follows;
v₂ = v₁ - [tex]a_{Net}[/tex] × t
∴ t = (v₁ - v₂)/[tex]a_{Net}[/tex] = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s
The distance covered, 's', is given as follows;
s = v₁·t - 1/2·[tex]a_{Net}[/tex]·t²
∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.
The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles
Consider the balloon and air inside the flask to be a closed system. Use the First Law of Thermodynamics to explain what happened to the balloon as heat was added by the environment.
Consider the balloon and air inside the flask to be in a closed system. Using the First Law of Thermodynamics to explain what happened to the balloon as heat was added by the environment.
Balloon will burst, due to more heat and in another flask air gets heated.
What is the first law of thermodynamics ?"The first law of thermodynamics states that energy can neither be created nor destroyed, only altered in form. For any system, energy transfer is associated with mass crossing the control boundary, external work, or heat transfer across the boundary. These produce a change of stored energy within the control volume."
What is heat ?"Heat is the energy that is transferred from one body to another as the result of a difference in temperature. If two bodies at different temperatures are brought together, energy is transferred i.e, heat flows—from the hotter body to the colder."
Know more about heat here
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What is the density of a 36 g object with a volume of 15 cm3? (Density: D = )
0.42 g/cm3
0.54 g/cm3
2.4 g/cm3
5.4 g/cm3
Answer:
density = mass/volume
so . . .
density = (36 g)/(15 cm³) = 2.4 g/cm³
Explanation:
on both sides.
F
10cm
2cm
(2 marks)
(a)
a
State the type of the lens in the box and explain your answer.
Answer:
please put pic of the questions
A baseball strikes the catcher's glove with a horizontal velocity of 40 m/s. The mass of the baseball is 0.15 kg. The displacement of the baseball due to the deformation of the catcher's glove and the movement of the catcher's hand is 8 cm from the instant it first makes contact with the glove until it stops. a. How much kinetic energy does the baseball possess just before it strikes the glove
Answer:
the kinetic energy is 120 J
Explanation:
a. The computation of the kinetic energy before it strikes the glove is as follows:
As we know that
The Kinetic energy is
= 1 ÷ 2 × m × v^2
= 1 ÷ 2 × (0.15)(40^2)
= 120 J
Hence, the kinetic energy is 120 J
Three collisions are elastic and three are inelastic. Determine which collision type took place for each collision. Support your conclusion with the data and observations from the lab.
Using your data, how are you able to determine if conservation of momentum occurs in each collision?
Write a conclusion for this lab. Also, make sure to discuss the conservation of momentum and how it applies to collisions.
Answer:
Explanation:
If objects stick together, then a collision is inelastic. When objects don't stick together, If the kinetic energy is the same, then the collision is elastic.
An object has a kinetic energy of 14 J and a mass of 17 kg , how fast is the object
moving?