Answer:
Common Examples of Imagery
Taste: The familiar tang of his grandmother's cranberry sauce reminded him of his youth. Sound: The concert was so loud that her ears rang for days afterward. Sight: The sunset was the most gorgeous they'd ever seen; the clouds were edged with pink and gold.
I hope it's helpful!
Which phrase describes velocity?
u
A. A quantity with direction only
B. A quantity with magnitude only
C. A quantity with no units
D. A quantity with magnitude and direction
SUBMI
A 50kg mass is sitting on a frictionless surface. An unknown constant force pushes the mass for 2 seconds until the mass reaches a velocity of 3m/s. What was the force acting on the mass?
Answer:
75N
Explanation:
a = v/t = 3/2
F = ma = 50(3/2) = 75
Question 3 (5 points)
Yissel was going to be late to Mr. Scharff's science class. Just as the bell was about to ring. Vissel ran the last little bit of the hallway at 2.47
meters/second for 8 seconds to beat the bell. How far away was Yissel from Mr. Scharff's classroom when she started to run?
Albert Bandura emphasized the idea of __________, which is the belief one has in one’s own ability to succeed. A. operant conditioning B. determinism C. self-efficacy D. self-worth
Answer:
C
Explanation:
Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.
What is Self efficacy?
A person's self-efficacy relates to their confidence in their ability to carry out the behaviors required to achieve particular performance goals (Bandura, 1977, 1986, 1997).
The belief in one's capacity to exercise control over one's own motivation, behavior, and social environment is known as self-efficacy. The goals for which people strive, the amount of effort put out to obtain goals, and the possibility of achieving particular levels of behavioral performance are all influenced by these cognitive self-evaluations.
Self-efficacy beliefs, unlike conventional psychological notions, are anticipated to change according to the operating domain and the environment in which an action occurs.
Therefore, Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.
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Imagin you have mixed together some sand and salt Based on the venn diagram this mixture would be placed where
Answer:
a
Explanation:
ajjkiikkkkkhglutfgkitfgghiiij
If the diameter of a moose eye is 40 mm, what is the total refractive power of the anterior portion of the eye?
Answer:
-the ratio of the speed of light
in air to the speed of light in the substance.
-speed of light in air 300,000 km/sec, which decreases when it passes through a transparent substance.
-e.g.. speed of light in substance = 200,000 km/sec, R.I. = 300,000/200,000 = 1.5
Explanation:
The sides of a right triangle that has any given vector for the hypotenuse are called _____
A. Scalar
B. Component
C. Resultant
D. Vector
Answer:
They are the resultant vector.
A particle move in the xy plane so that its position vector r=bcosQi +bsinQj+ ctk, where b, Q and c are constants. show that the partial move with constant speed.
Answer:
The speed of this particle is constantly [tex]c[/tex].
Explanation:
Position vector of this particle at time [tex]t[/tex]:
[tex]\displaystyle \mathbf{r}(t) = b\, \cos(Q)\, \mathbf{i} + b\, \sin(Q) \, \mathbf{j} + c\, t\, \mathbf{k}[/tex].
Write [tex]\mathbf{r}(t)[/tex] as a column vector to distinguish between the components:
[tex]\mathbf{r}(t) = \begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}[/tex].
Both [tex]b[/tex] and [tex]Q[/tex] are constants. Therefore, [tex]b\, \cos(Q)[/tex] and [tex]b \sin (Q)[/tex] would also be constants with respect to [tex]t[/tex]. Hence, [tex]\displaystyle \frac{d}{dt}[b\, \cos(Q)] = 0[/tex] and [tex]\displaystyle \frac{d}{dt}[b\, \sin(Q)] = 0[/tex].
Differentiate [tex]\mathbf{r}(t)[/tex] (component-wise) with respect to time [tex]t[/tex] to find the velocity vector of this particle at time [tex]t\![/tex]:
[tex]\begin{aligned}\mathbf{v}(t) &= \frac{\rm d}{{\rm d} t} [\mathbf{r}(t)] \\ &=\frac{\rm d}{{\rm d} t} \left(\begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}\right) \\ &= \begin{bmatrix}\displaystyle \frac{d}{dt}[b\, \cos(Q)] \\[0.5em] \displaystyle \frac{d}{dt}[b\, \sin(Q)]\\[0.5em]\displaystyle \frac{d}{dt}[c \cdot t]\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\end{aligned}[/tex].
The speed [tex]v[/tex] (a scalar) of a particle is the magnitude of its velocity :
[tex]\begin{aligned}v(t) &= \| \mathbf{v}(t) \| \\ &= \left\|\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\right\| \\ &= \sqrt{0^2 + 0^2 + c^2} = c\end{aligned}[/tex].
Therefore, the speed of this particle is constantly [tex]c[/tex] (a constant.)
The tray dispenser in your cafeteria has broken and is not repairable. The custodian knows that you are good at design-ing things and asks you to help him build a new dispenser out of spare parts he has on his workbench. The tray dispenser supports a stack of trays on a shelf that is supported by four springs, one at each corner of the shelf. Each tray is rectangu-lar, with dimensions 45.3 cm by 35.6 cm. Each tray is 0.450 cm thick and has a mass of 580 g. The custodian asks you to design a new four-spring dispenser such that when a tray is removed, the dispenser pushes up the remaining stack so that the top tray is at the same position as the just-removed tray was. He has a wide variety of springs that he can use to build the dispenser. Which springs should he use
Answer:
you have to find 4 spring with this elastic constant k = 316 N / m
Explanation:
In this case for the design of the dispenser the four springs are placed in the four corner at the bottom, therefore we can use the translational equilibrium relationship
4 F_e -W = 0
where the elastic force is
F_e = k x
we substitute
4 kx = mg
k = [tex]\frac{mg}{4x}[/tex]
Each tray has a thickness of x = 0.450 cm = 0.450 10⁻² m, this should be the elongation of the spring so that when the tray is in position it will remain fixed.
let's calculate
k = [tex]\frac{0.580 \ 9.8}{4 \ 0.450 \ 10^{-2} }[/tex]
k = 3.1578 10² N / m
k = 316 N / m
therefore you have to find 4 spring with this elastic constant
List down the types of centripetal force?
Answer: Just a few examples are the tension in the rope on a tether ball, the force of Earth's gravity on the Moon, friction between roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge. Any net force causing uniform circular motion is called a centripetal force.
Answer:
roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge
Explanation:
A spring with a constant of 76 N/m is extended by 0.9 m. How much energy is stored in the extended spring?
Answer:
[tex]E=30.78\ J[/tex]
Explanation:
The force constant of the spring, k = 76 N/m
The extension in the spring, x = 0.9 m
We need to find the energy is stored in the extended spring. The energy stored in the spring is given by :
[tex]E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 76\times (0.9)^2\\\\E=30.78\ J[/tex]
So, 30.78 J of energy is stored in the spring.
what is the effect of divorce on females?
Answer:
Numerous studies have shown that the economic costs of divorce fall more heavily on women. After separation, women experience a sharper decline in household income and a greater poverty risk (Smock 1994; Smock and Manning
Answer:
sadness and stress...................
A diesel engine lifts the 225 kg hammer of a pile driver 20 m in 5 seconds. How much work is done on
the hammer? What is the power?
Answer:
a. Workdone = 44100 Joules
b. Power = 8820 Watts.
Explanation:
Given the following data:
Mass = 225kg
Distance = 20m
Time = 5 seconds
To find the workdone;
Workdone = force * distance
But force = mg
We know that acceleration due to gravity is equal to 9.8m/s²
Force = 225*9.8 = 2205N
Substituting the values into the equation, we have;
Workdone = 2205 * 20
Workdone = 44100 Joules
b. To find the power;
Power = workdone/time
Power = 44100/5
Power = 8820 Watts.
A liquid fueled rocket is red on a test stand. The rocket nozzle has an exit diameter of 30 cm and the combustion gases leave the nozzle at a velocity of 3800 m/s and a pressure of 100 kPa, which is the same as the ambient pressure. The temperature of the gases in the combustion area is 2400 C. Find (a) the temperature of the gases at the nozzle exit plane, (b) the pressure in the combustion area, and (c) the thrust developed. Assume that the gases have a speci c heat ratio of 1.3, and a molar mass of 9. Assume that the ow in the nozzle is isentropic.
Answer:
1. Temperature= 869.35 K
2. Pressure of combustion = 12994.043 kpa
3. Thrust = 127x10⁶N
Explanation:
this problem has been fully explained in the attachment. please use it to get a clearer explanation of the answer.
1.
The temperature = (273+2400k) - (3800)²/2(4003)
= 2673 - 14440000/8006
= 2673 - 1803.65
= 869.35 K
Approximately 869.4K
2. We first get mach number
= 3800/√1.3(923.8)(869.35)
= 3800/1021.78
= 3.719
Pressure = 100kpa[1+2.07464415]^1.3/0.3
= 12995.043kpa
C. Thrust
Pi/4(3800)²(0.3)²(100x10³)/(923.8)(869.4)
= 12678.621
= 126.781 kN
Thrust is approximately 127kN = 127x10⁶N
39. What is the change in momentum for a 5,000 kg ship in
outer space that experiences no net force over a 1 hr
period?
Answer:
Change in momentum is zero.
Explanation:
The following data were obtained from the question:
Mass (m) = 5000 kg
Time (t) = 1 h
Net force (F) = 0
Change in momentum =?
Force = Rate of change of momentum
0 = change in momentum
Change in momentum = 0
We can see from the above illustration that the net force is zero. Thus, the change in momentum is also zero.
Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1, the ball bounces off a cement floor, and in Case 2, the ball bounces off a piece of stretchy rubber. Two cases of a ball dropping and bouncing off of a surface. In case one, the ball approaches a cement floor, comes in contact, momentarily comes to rest, and then bounces off with a velocity smaller than the approach velocity. In case 2, the ball approaches a piece of stretchy rubber. When the ball comes in contact with the rubber, it deforms the rubber and comes momentarily to rest. The ball bounces off again with a velocity smaller than the approach velocity, and the rubber regains its original shape. In both cases, the balls are dropped from the same initial height and reach the same final height. In which case is the magnitude of the ball's change in momentum the greatest
Answer:
the impulse must be the same in these two cases F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
Explanation:
For this exercise we use the relationship between momentum and momentum
I = Δp
F t = m v_f - m v₀
To know the speed we use the conservation of energy
starting point. Highest point
Em₀ = U = m g h
fincla point. Just before the crash
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v = [tex]\sqrt{2gh}[/tex]
we substitute in the impulse relation
F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
therefore we can see that as in case the initial and final heights are equal, the impulse must be the same in these two cases
Two objects are electrically charged. The net charge on one object is doubled.
Therefore, the electric force _____.
reverses
doubles
quadruples
divides
4. Draw conclusions: What is the minimum energy required to break the egg?
.
Answer:
0.25 J
Explanation:
The strength of the egg shell, the size of the egg, and the force used to break it are just some of the variables that affect how much energy is needed to crack an egg. When an object hits the egg with an impact energy of 12–26 mJ, cracks occur.
What is energy?The capacity of a system or object to do work is called energy, which is a fundamental term in physics. Kinetic energy, potential energy, heat energy, electromagnetic energy, and nuclear energy are just a few of the different forms of energy.
While potential energy is the energy possessed by an object as a result of its position or position, kinetic energy is the energy of motion. While electromagnetic energy is energy carried by electromagnetic waves like light, thermal energy is energy related to the temperature of a substance. The energy stored in the nucleus of an atom is called nuclear energy.
Therefore, when an object hits the egg with an impact energy of 12–26 mJ, cracks occur.
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Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25 cm and the point-like object with charge q2 = −2.14 µC is located at x2 = −1.80 cm.
A) Determine the total electric potential (in V) at the origin.
B) Determine the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
Answer:
a) the total electric potential is 2282000 V
b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
Explanation:
Given the data in the question and as illustrated in the image below;
a) Determine the total electric potential (in V) at the origin.
We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges
so
Electric potential at p in the diagram 1 below is;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we know that; Coulomb constant, k = 9 × 10⁹ C
q1 = 4.60 uC = 4.60 × 10⁻⁶ C
r1 = 1.25 cm = 0.0125 m
q2 = -2.06 uC = -2.06 × 10⁻⁶ C
location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m
so we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )
Vp = (3312000) + ( -1030000 )
Vp = 3312000 -1030000
Vp = 2282000 V
Therefore, the total electric potential is 2282000 V
b)
the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
As illustrated in the second image;
r1² = 0.015² + 0.0125²
r1 = √[ 0.015² + 0.0125² ]
r1 = √0.00038125
r1 = 0.0195
Also
r2² = 0.015² + 0.018²
r2 = √[ 0.015² + 0.018² ]
r2 = √0.000549
r2 = 0.0234
Now, Electric Potential at P in the second image below will be;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )
Vp = 2123076.923 + ( -762962.962 )
Vp = 2123076.923 -792307.692
Vp = 1330769.23 V
Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
a) The total electric potential is 2282000 V
b) The total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
What is electric potential?The electric potential is defined as the amount of work energy needed to move a unit of electric charge from a reference point to a specific point in an electric field.
Given the data in the question and as illustrated in the image below;
a) Determine the total electric potential (in V) at the origin.
We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges
Electric potential at p in diagram 1 below is;
[tex]V_P=V_1+V_2[/tex]
[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]
we know that; the Coulomb constant, k = 9 × 10⁹ C
q1 = 4.60 uC = 4.60 × 10⁻⁶ C
r1 = 1.25 cm = 0.0125 m
q2 = -2.06 uC = -2.06 × 10⁻⁶ C
location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m
so we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )
Vp = (3312000) + ( -1030000 )
Vp = 3312000 -1030000
Vp = 2282000 V
Therefore, the total electric potential is 2282000 V
b)The total electric potential (in V) at the point with coordinates (0, 1.50 cm).
As illustrated in the second image;
[tex]r_1^2=0.015^2+0.0125^2[/tex]
[tex]r_1 = \sqrt{[ 0.015^2 + 0.0125^2 ][/tex]
[tex]r_1 = \sqrt{0.00038125}[/tex]
[tex]r_1 = 0.0195[/tex]
Also
[tex]r_2^2 = 0.015^2 + 0.018^2[/tex]
[tex]r_2 = \sqrt{0.015^2 + 0.018^2}[/tex]
[tex]r_2 = \sqrt{0.000549[/tex]
[tex]r_2 = 0.0234[/tex]
Now, Electric Potential at P in the second image below will be;
Vp = V1 + V2
[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]
we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )
Vp = 2123076.923 + ( -762962.962 )
Vp = 2123076.923 -792307.692
Vp = 1330769.23 V
Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
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A freight train has a mass of [02] kg. The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 km/h?
Answer:
t = 300.3 seconds
Explanation:
Given that,
The mass of a freight train, [tex]m=1.01\times 10^7\ kg[/tex]
Force applied on the tracks, [tex]F=7.5\times 10^5\ N[/tex]
Initial speed, u = 0
Final speed, v = 80 km/h = 22.3 m/s
We need to find the time taken by it to increase the speed of the train from rest.
The force acting on it is given by :
F = ma
or
[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.01\times 10^7\times (22.3-0)}{7.5\times 10^5}\\\\t=300.3\ s[/tex]
So, the required time is 300.3 seconds.
What is the Radiation left over from the big bang called?
Answer:
The Big Bang theory predicts that the early universe was a very hot place and that as it expands, the gas within it cools. Thus the universe should be filled with radiation that is literally the remnant heat left over from the Big Bang, called the “cosmic microwave background", or CMB.
Explanation:
The Big Bang theory suggest that the universe in early stage was at very hot place and which can be expanded, the gas within it cools. It is in an infinite universe and it has no edge.
What is big bang theory ?The Big Bang theory is a cosmological model which explain the existence of the observable universe from the earliest periods to the large-scale evolution.
The model describes the mechanism behind the universe expansion from an initial state of high density and temperature, it is very important concept as a lot of research is going on in this field to find out exactly how the universe began billions of years ago.
The universe began to cool down in order to allow the formation of particles become atoms after its initial phase of expansion, Primordial elements such as Hydrogen, Helium, and Lithium are condensed through gravity are formed early stars and galaxies.
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In the Styrofoam ball investigation, it is likely that the charges on the ball and rod are
?
A. the same
B. opposite
C. constantly changing.
Answer:
answer = opposite
they are not changing being in a place as being they-self
the same would not be true because its different then the Styrofoam and it could only be different and its make the most sense aswell.
Both ball and rod has opposite charges on their bodies because of presence of different charges on it.
What are the charges on ball and rod?In the Styrofoam ball investigation, it is likely that the charges on the ball and rod are opposite because Styrofoam ball is negatively charged due to the presence of electrons while on the other hand, the rod is positively charged because of lining of positive charges on the rod.
So we can conclude that both ball and rod has opposite charges on their bodies because of presence of different charges on it.
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1. If a wave has a wavelength of 5.5m and a frequency of 45hz, what is its speed?
Answer:
By using the most simple velocity equation, velocity = distance / time, meaning the speed would be 247.5 meters per second.
Which statement best compares potential and kinetic energy?
O Objects always have more potentiał energy than kinetic energy.
O Kinetic energy increases and potential energy decreases when the velocity of an object increases
O Only potential energy decreases when an object's height increases.
O Objects always have more kinetic energy than potential energy.
Answer:
Kinetic energy increases and potential energy decrease when velocity of an object increase.
A painter sits on a scaffold that is connected to a rope passing over a pulley. The other end of the rope rests in the hands of the painter who wants to lift the scaffold. She plans to pull downward on the loose end of the rope, thinking that the scaffold will then rise vertically with her along for the ride. The scaffold has a mass of 52 kg, and her mass is 63 kg. The painter pulls downward on the rope with a force of 600.0 N, while she and the scaffold are hanging from the other end above the ground.
Required:
a. What is the net acceleration on the system consisting of the painter and the scaffold?
b. What is the magnitude of the normal force exerted on the painter by the scaffold?
Solution :
a). From Newtons second law,
F = ma
The total tension force is 2T.
∴ 2T - (m + M)g = (m+ M)a
Then
[tex]$a=\frac{2T-(m+M)g}{m+M}$[/tex]
[tex]$a=\frac{2\times 600-(52+63)9.8}{52+63}$[/tex]
[tex]$=0.63 \ m/s^2$[/tex]
b). From the person,
F = ma
T - Mg + N = Ma
or N = Ma + Mg - T
= (63 x 9.8) + (52 x 9.8) - 600
= 617.4 + 509.6 - 600
= 527 N
QUCIK!! SOMEONE PLEASE HELP! I’LL MARK BRAINLIEST!!
Answer:
A. v = √2gh
B. No! The final velocity does not depend on the mass of the car.
C. Yes! the final velocity depends on the steepness of the hill
D. 3.28 m/s
Explanation:
A. Determination of the final velocity.
½mv² = mgh
Cancel out m
½v² = gh
Cross multiply
v² = 2gh
Take the square root of both side
v = √2gh
B. Considering the formula obtained for the final velocity i.e
v = √2gh
We can see that there is no mass (m) in the formula.
Thus, the final velocity does not depend on the mass of the car.
C. Considering the formula obtained for the final velocity i.e
v = √2gh
We can see that there is height (h) in the formula.
Thus, the final velocity depends on the steepness of the hill
D. Determination of the final velocity.
Height (h) = 0.55 m
Acceleration due to gravity (g) = 9.8 m/s²
Velocity (v) =?
v = √2gh
v = √(2 × 9.8 × 0.55)
v = √10.78
v = 3.28 m/s
What formula could be used to find distance if you know the speed an the time
Answer: d = st
Explanation:
We know that the distance is equal to the rate (speed) times the time
d = st
1. A block with mass 20 kg is
sliding up a plane (Ukinetic=0.3,
inclined at 10°) at a speed of
2 m/s to the right (positive
X-direction). How far does it
go up along the plane before
it comes to rest momentarily?
Answer: 0.435 m
Explanation:
Given
mass m=20 kg
initial speed u=2 m/s
coefficient of kinetic friction [tex]\mu_k=0.3[/tex]
deceleration which opposes the motion is given by
[tex]\Rightarrow a=g\sin \theta+\mu_kg\cos \theta\\\Rightarrow a=g(\sin \theta +\mu_k\cos \theta)[/tex]
[tex]\Rightarrow a=9.8(\sin 10^{\circ}+0.3\times \cos 10^{\circ})\\\Rightarrow a=4.59\ m/s^2[/tex]
using [tex]v^2-u^2=2as[/tex]
[tex]\Rightarrow s=\dfrac{2^2}{2\times 4.59}=0.435\ m[/tex]
Consider a uniformly charged sphere of total charge Q and radius R centered at the origin. We want to find the electric field inside the sphere (r
Answer:
Hello your question is incomplete attached below is the complete question
answer :
Total charge enclosed within the sphere : [tex]\frac{q_{r1} }{4\pi e_{0}R^3 } . r[/tex]
Total charge enclosed outside the sphere : [tex]\frac{q}{4\pi e_{0}r^2 } .r[/tex]
Explanation:
Given data:
Total charge of a uniformly charged sphere = Q
radius = R
first step : find the electric field inside and outside the uniformly charged sphere
2nd step : determine the total charge enclosed within and outside the sphere
make a sketch of the uniformly charged sphere
Attached below is a detailed solution
You are designing a new roller-coaster. The main feature of this particular design is to be a vertical circular loop-the-loop where riders will feel like they are being squished into their seats even when they are in fact upside-down (at the top of the loop). The coaster starts at rest a height of 80m above the ground, speeds up as it descends to ground level, and then enters the loop which has a radius of 20m. Suppose a rider is sitting on a bathroom scale that initially reads W (when the coaster is horizontal and at rest). What will the scale read when the coaster is moving past the top of the loop
Answer:
The reading on the scale is N = 9W
Explanation:
Since the roller coaster drops from a height, h of 80 m, the potential energy lost equals the kinetic energy gained as it enters the loop.
So, mgh = 1/2mv² where m = mass of rider, g = acceleration due to gravity = 9.8 m/s², h = initial height of roller coaster above ground level = 80 m and v = speed of roller coaster as it enters the loop.
So, mgh = 1/2mv²
v² = 2gh
v = √(2gh)
v = √(2 × 9.8 m/s² × 80 m)
v = √(1568 m²/s²)
v = 39.6 m/s
Now, as the roller coaster gets to the top of the vertical loop, the centripetal force, F and the weight W acts downwards. For the passenger not to fall off, this must equal the normal force, N
So, F = mv²/r where v = speed of roller coaster = 39.6 m/s and r = radius of vertical loop = 20 m and m = mass of rider = W/g
F = Wv²/gr
F = W(39.6 m/s)²/(9.8 m/s² × 20 m)
F = (1568 m²/s²)W/196 m²/s²
F = 8W
Since F + W = N
8W + W = N
9W = N
So, N = 9W
So, the reading on the scale is N = 9W