Answer:
A wrecking ball has the capability to destroy a building due to the greater mass and kinetic energy than the yo-yo.
Potential energy is energy due to motion.
True or False?
Answer:
true
Explanation:
Answer:
true
Explanation:
please give brainlest need 1
Concept Simulation 4.1 reviews the central idea in this problem. A boat has a mass of 4490 kg. Its engines generate a drive force of 4520 N due west, while the wind exerts a force of 890 N due east and the water exerts a resistive force of 1210 N due east. Take west to be the positive direction. What is the boat's acceleration, with correct sign
Answer:
-0.54m/s²
Explanation:
According to Newton's second law of motion
F = ma
Force = mass * acceleration
Given
Mass m = 4490kg
Take the sum of forces
Sum of force along the east = 890+1210 = 2100N
Sum of forces along the west = -4520N
Net force = -4520+2100
Net force = -2420N
Acceleration = Net force/Mass
Acceleration = -2420/4490
Acceleration = -0.54m/s²
Hence the boat acceleration is -0.54m/s²
What is the average speed of an Olympic sprinter that runs 100 m in 9.88 s?
Answer:
speed = 10.1215 m/s
Explanation:
speed = distance / time
speed = 100 / 9.88 = 10.1215 m/s
What was the average speed in km/h of a car that travels 788 km in 7.1 h?
Answer:
Well the answer is 111 (rounded) km/h
Explanation:
788/7.1 is 110.9... so if you round it it would equal 111.
Tell types of mirros and
each
one
Answer: We can identify the different types of mirrors without touching them by looking at the image it produces. Look into each mirror, the nature of the image produced will tell you the type of mirror it is.
- A plane mirror will produce an image of the same size as your face.
- A concave mirror will produce a magnified image of your face.
- A convex mirror will produce a diminished image of your face.
MARK ME BRAINLIST
1. A wheel with spokes of length r has four masses
attached at various points where the spokes intersect the
circumference of the wheel. The wheel and spokes are
massless and spokes are all 30° apart. Determine the net
torque on the wheel.
(A) Zero (B) 22 (C) 13/2
(D) 1
A proud new Jaguar owner drives her car at a speed of 35 m/s into a corner. The coefficients of friction between the road and the tires are 0.70 (static) and 0.40 (kinetic). What is the minimum radius of curvature for the corner in order for the car not to skid
Answer:
178.6 m
Explanation:
Since the car moves in a circular path, it experiences a centripetal force, F = mv²/r where m = mass of car, v = speed of car = 35 m/s and r = radius of curvature of path.
Now, for the car not to skid, this centripetal force must be equal to the frictional force, F' acting in the opposite direction.
So, F' = μN where μ = coefficient of static friction(since the car does not move in this direction) and N = normal force = mg where m = mass of car and g = acceleration due to gravity = 9.8m/s²
F' = μmg
Since F = F'
mv²/r = μmg
dividing both sides by m, we have
v²/r = μg
multiplying both sides by r, we have
v² = μgr
dividing both sides by μg, we have
r = v²/μg
Here we use μ = coefficient of static friction(since the car does not move in this direction) = 0.70. Substituting the other variables into the equation, we have
r = v²/μg
r = (35 m/s)²/(0.70 × 9.8m/s²)
r = 1225 m²/s²/6.86m/s²)
r = 178.6 m
So, the minimum radius of curvature of the corner is 178.6 m
For the questions below, include units if applicable. If necessary, use a separate sheet of paper for 1, 6c and 7c. Tire pressure is in part a function of the temperature of the tire.
1. Based on everyday experience, state (in words) the relationship between tire pressure and temperature. Look at the data below and see if the numbers support your statement.
2. Prepare a hand-drawn plot of the two variables on the reverse side of this worksheet. Include a title, axis labels (with units), and a trendline. Estimate the tire pressure when the temperature is 18.6°C: Estimate the temperature of the air in the tire when the pressure is 37.0 psi: 3.
a. Prepare a plot using graphing software. Include a title, axis labels (with units), the equation of the best-fit Line and the R? value on the graph.
b. Re-write the equation of the best-fit line substituting "Temperature" for x and "Pressure" for y directly on the graph.
c. Attach the fully labeled graph to this worksheet.
4. What is the value of the slope for the relationship between temperature and pressure?
5. Determine the percent error using the definition of percent error: Use 0.145 psi/" for the "Actual" value of the slope. %error = Actual-Experimental % Error Actual
6. Based on your computer-generated graph,
a. visually estimate the tire pressure when the temperature is 18.6°C:
b. calculate the tire pressure at this temperature using the equation of the best fit line: the graph to ensure that this value is reasonable.
c. compare the calculated pressure to the two visually interpolated values (Steps 2 and 6a). Comment on any discrepancies.
7. Based on your computer-generated graph,
a. visually estimate the temperature of the air in the tire when the pressure is 37.0 psi:
b. calculate the temperature of the air in the tire at this pressure: Use the graph to ensure that this value is reasonable.
c. compare the calculated temperature to the two visually interpolated values (Steps 2 and 7a). Comment on any discrepancies.
Data:
Temperature (x) Tire Pressure, psi (y)
12.9°C 3.39 x 10
15.4C 34.25
-2.10 F 2.68 x 10
19.5 °C 3.50 x 10
29.6 'F 36.53
Answer:
All answer are explained below in the explanation section.
Explanation:
1. The pressure varies proportionally with the change in temperature. It can also be observed in our daily lives.
As for example, a pressure cooker uses the same principal to cook food faster. With the increasing temperature, the pressure inside the cooker increases.
Thus after a while, the excess pressure inside is released through the top nozzle. The data shown below supports that pressure and temperature varies linearly.
2. Hand drawn plot is attached in the attachment please refer to the attachment for the hand drawn plot.
Tire pressure at temperature 18.6 degree C is ~ 35 psi.
Temperature at air pressure of 37 psi is ~26.1 degree C
3. a.) Necessary values are included in the stat box. It is attached in the attachment please refer to the attachment.
3. b) The equation becomes: Pressure = 0.176 x temperature + 32.32
3. c) It is already done in part a of this question.
4. The value of the slope estimated from the linear fit is 0.176 +/- 0.094.
5. % Error = [tex]\frac{Actual - Experiment}{Actual} x 100[/tex]
Plugging in the values, we get:
Actual = 0.145, Experimental = 0.176. Thus, percentage error is given by:
% Error = 21.33%
6. a.) Visual estimation of tire pressure at t = 18.6 degree C is ~ 35 psi
6. b.) Estimation of pressure from the best fit line is given by 35.6 psi, which is consistent with the eye estimation value.
6. c.) The eye estimation and the estimation from the line fit are quite comparable. The discrepancy of +/-0.5 psi is within the percentage error calculated in 5.
7. a.) Visual estimation of temperature of the air for a tire-pressure of 37 psi is ~ 26 degree C.
7. b.) Estimation of temperature from best fit value of line is = 26.64 degree C
7 c) The values from eye estimation and evaluated from the fit are quite consistent within a random fluctuation of +/- 0.64 degree C.
Fill in the question
The name of the SI unit for magnetic field strength, such as that created around a current-carrying wire, is the
.
The name of the SI unit for magnetic field strength is Tesla. Magnetic fields are formed by moving electric charges.
What is Magnetic field strength?Magnetic field strength alludes to an actual amount that is utilized as one of the essential proportions of the power of the attractive field.
The SI unit of attractive field is tesla (T). 1 Tesla is defined as the magnetic field that carries 1C charge at the speed of 1m/s which is perpendicular to the force of 1 N.
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point.
4.
i. Explain why a particle moving with a constant speed along a circular
path has a radial acceleration.
ii. Show that the acceleration of a body moving in a circular path of
radius r with uniform speed v is → and draw a diagram to show the
direction of the acceleration.
2
iii. Show that the expression à = † is dimensionally correct.
Explanation:
Explanation: When a particle moves along a straight path, then the radius of curvature is infinitely great. This means that v2/r is zero. Explanation: When a particle moves with a uniform velocity, then dv/dt will be zero.
What current is needed in the solenoid's wires?
A researcher would like to perform an experiment in a zero magnetic field, which means that the field of the earth must be canceled. Suppose the experiment is done inside a solenoid of diameter 1.0 m, length 3.8 m , with a total of 5000 turns of wire. The solenoid is oriented to produce a field that opposes and exactly cancels the 52 μT local value of the earth's field.
What current is needed in the solenoid's wires? Express your answer with the appropriate units.
What kind of reasoning is most often used to form hypotheses?
inductive
deductive
detective
invective
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its surface. Find the electric field for the following points: (a) for all points outside the spherical shell E = keq2/r2 E = q/4πr2 none of these E = keq/r2 E = 0 (b) for a point inside the shell a distance r from the center E = keq2/r2 E = keq/r2 E = 0 E = q/4πr2 none of these
Answer:
a) E = 0
b) [tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]
Explanation:
The electric field for all points outside the spherical shell is given as follows;
a) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]
From which we have;
[tex]E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0[/tex]
E = 0/A = 0
E = 0
b) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]
[tex]E \cdot A = \dfrac{+q }{\varepsilon _{0}}[/tex]
[tex]E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}[/tex]
By Gauss theorem, we have;
[tex]E\oint dS = \dfrac{q}{\varepsilon _{0}}[/tex]
Therefore, we get;
[tex]E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}[/tex]
The electrical field outside the spherical shell
[tex]E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }[/tex]
[tex]k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }[/tex]
Therefore, we have;
[tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]
help me pls it’s a usa test prep pretty easy
Answer:
Im 99.99999% sure its c
Explanation:
i cant see the pictures too well
Assuming 84.0% efficiency for the conversion of electrical power by the motor, what current must the 13.0-V batteries of a 716 kg electric car be able to supply to climb a 3.00 x 102 m high hill in 2.00 min at a constant 22.0 m/s speed while exerting 7.00 x 102 N of force to overcome air resistance and friction
Answer:
[tex]\mathbf{ current(I) =1766.67 \ A}[/tex]
Explanation:
Given that:
The air resistance and friction = 700 N
The gravity caused force = 716 × 9.8 = 7016.8
Total force = (7016.8 + 700) N
Total force = 7716.8 N
∴
[tex]13 \times current(I) \times 0.84 = \dfrac{7716.8 \times 300}{2 \times 60}[/tex]
[tex]current(I) \times 10.92= 19292[/tex]
[tex]current(I) = \dfrac{19292}{10.92}[/tex]
[tex]\mathbf{ current(I) =1766.67 \ A}[/tex]
Even though Alice visits the wishing well frequently and always tosses in a coin for good luck, none of her wishes have come true. As a result, she decides to change her strategy and make a more emphatic statement by throwing the coin downward into the well. If the water is 5.43 m below the point of release and she hears the splash 0.85 seconds later, determine the initial speed at which she threw the coin. (Take the speed of sound to be 343 m/s.)
Answer:
Explanation:
Total time taken = 0.85 s .
Time taken by sound to travel 5.43 m + time taken by coin to fall by 5.43 m = .85
5.43 / 343 + time taken by coin to fall by 5.43 m = .85
time taken by coin to fall by 5.43 m = .85 - 5.43/343 = .834 s
Let the initial velocity of throw of coin = u
displacement of coin s = 5.43 m
time take to fall t = .834 s
s = ut + 1/2 gt²
5.43 = u x .834 + .5 x 9.8 x .834²
5.43 = u x .834 + 3.41
u x .834 = 2.02
u = 2.42 m /s .
Acceleration is the rate ot change of the velacity a -dejdt so it is the slope of the Velocity vs. Time graph Because it is dficult to drag the person in a consistent and reproducible way use the Expression Evakaator under the Special Features menu for this question lick Reset A and type in the hr on z t * t * t " t in the Expression Evaluator Click the Play button and let the simulation run roughly 5 sin ulation seconds before ressing the Pause but use the zoom buttons to a 쪄 the p s they the screen You should see 8 p at s ar l what you got in the previous question, but much smoother Look at the Postion vs Time. Velocity vs Time and Acceleration vs. Time piets h
a) the velocity is zero but the acceleration is negative
When the person is 8 to to the tight of the origin
b) the velocity is zero but the acceleration is positive
c) both the velocity and the acceleration are zero
d) both the velocity and the acceleraton are nonzero
Answer:
a) the body is changing direction,
b)the body must go to the left and the acceleration to the right
c) the movement has not started.
d) all points of the motion
Explanation:
In this exercise you are asked to find in which position you have the following characteristics of the movement
a) The velocity is zero and the acceleration is negative
This is when the body reaches the end of the travel and turns around, in this case the speed is zero and the acceleration has the opposite direction to the movement.
In this case the body moves to the right and the acceleration is to the left, therefore the speed decreases
b) The velocity is zero, but the acceleration is positive
This occurs at the points where the speed is changing direction, specifically for this case the body must go to the left and the acceleration to the right
c) Both are zero
This only occurs where the body is stopped and the movement has not started.
d) both the velocity and the relation are nonzero.
This is at all points of the motion since the velocity is constantly changing as long as there is an acceleration
How much force is needed to accelerate a 9,760 kg airplane at a rate of 3.6 m/s2?
OA. 2,711 N
OB. 35,136 N
OC. 126,490 N
OD. 9,760 N
Answer:
the answer is B: 35,136
Explanation:
force = mass × acceleration
force = 9760 × 3.6
35,136 = 9760 × 3.6
3. Two cyclists that are 500 m apart start biking toward each other. They bike at speeds of 6 and
4 m/s.
How long does it take for them to reach each other?
a.
b. How far does the slower biker travel?
Answer:
A) 50 seconds
B) 200 m
Explanation:
They are 500 metres apart.
And one of the bike loves at 6 m/s while the other loves at 4 m/s.
A) Let distance of the 6 m/s bike before they meet be x.
Thus, time = x/6
Since time = distance/speed
For the second bike at 4 m/s, his distance covered before they meet will be 500 - x
Thus, time = (500 - x)/4
Now they will meet each other at the same time. Thus;
x/6 = (500 - x)/4
Cross multiply to get;
4x = 3000 - 6x
6x + 4x = 3000
10x = 3000
x = 3000/10
x = 300 m
Thus, time will be;
t = 300/6
t = 50 seconds
B) Distance covered by the slower bike is (500 - x)
Since from a above, x = 300
Thus; distance = 500 - 300 = 200 m
A car hits a tree with a force of 45 N, the mass of the tree is 65g. What is the resulting acceleration?
a. 0.69 m/s2
b. 692 m/s2
c. 2,925 m/s2
d. 2.93 m/s2
Answer:
i think 692m/s2 is the correct answer
What is the mass of 9.11 moles of
ozone, 03?
The molecular mass of [tex]$O_{3}[/tex] is 0.43728kg
What is molecular mass?Molecular mass exists as a number equivalent to the totality of the atomic masses of the atoms in a molecule. The totality of the atomic masses of all atoms in a molecule is established on a scale in which the atomic masses of hydrogen, carbon, nitrogen, and oxygen exist 1, 12, 14, and 16, respectively.
To compute the Molecular Mass of [tex]$O_{3}[/tex]
Atomic mass of oxygen(O) = 16
As [tex]$O_{3}[/tex] contains 3 atoms,
The molecular mass of [tex]$O_{3}[/tex]
= (16 x 3) = 48g/mol
Hence the mass of 9.11 moles O3
= 9.11 mol x 48g/mol
= 437.28g
= 0.43728kg
Therefore, the molecular mass of [tex]$O_{3}[/tex] is 0.43728kg
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Atoms of which pair of elements will form covalent bonds in a compound? ASAP PLZ
A. Li and Al
B. C and O
C. Co and Fe
D. Na and F
Answer:B
Explanation:
C and O
Atoms of C and O pair of elements will form covalent bonds in a compound.Therefore the correct option is B.
What is a Chemical compound?The chemical compound is a combination of two or more either similar or dissimilar chemical elements
for example, H₂O is a chemical compound made up of two oxygen atoms and a single hydrogen atom.
These chemical compounds are formed because of different types of bonds between the constituents elements ,the chemical bonds are mainly ionic bonds, covalent bonds,s, and hydrogen bonds.
Ionic bonds are formed due to the transfer of electrons between two bond forming pairs differentiated by their electronegativity.
Covalent bonds are formed by the sharing of electrons.Generally organic compound are formed as the reason of covalent bonds.
The carbon and oxygen atoms share their valence electrons to form a covalent bond, therefore the correct option is B.
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Need help on another homework question
Rick places the blue lens of one pair of 3D glasses over the red lens of another pair. He then looks through both lenses at the same time. What color will he see?
A. blue
B. black
C. red
D. white
Answer:
a
Explanation:
what is a vector quantity?
Answer:
A quantity that has magnitude and direction. It's usually represented by an arrow whose direction is the same direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude
If you stand on a trampoline, it depresses under your weight. When you stand on a hard stone floor, __________. If you stand on a trampoline, it depresses under your weight. When you stand on a hard stone floor, __________. the floor deforms very slightly under your weight only if you are heavy enough does the floor deform at all under your weight the floor does not deform at all under your weight
Answer:
the floor deforms very slightly under your weight
Explanation:
A trampoline is made up of a large piece of strong cloth held by springs on which you jump up and down as a sport. So, If you stand on a trampoline, it depresses under your weight. However, the floor does not deform under your weight as it is too stiff.
Therefore,
when you stand on a hard stone floor, the floor deforms very slightly under your weight.
A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclined plane which makes an angle of θ degrees with the horizontal. Assume the spring has been compressed a distance d from its neutral position.
Required:
a. Set your coordinates to have the x-axis along the surface of the plane, with up the plane as positive, and the y-axis normal to the plane, with out of the plane as positive.
b. Denoting the coefficient of static friction by μs, write an expression for the sum of the forces in the x-direction just before the block begins to slide up the inclined plane. Use defined quantities and g in your expression ΣFx = 25%
c. Assuming the plane is frictionless, what will the angle of the plane be, in degrees, if the spring is compressed by gravity a distance 0.1 m?
d. Assuming θ = 45 degrees and the surface is frictionless, how far will the spring be compressed, d in meters?
Answer:
b) k Δx - W cos θ - μ mg cos θ = m a , c) θ = 86.6º, d) Δx = 1.18 m
Explanation:
a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.
F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring
b) Let's apply Newton's second law for when the spring is compressed
let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Y axis
N - W_y = 0
N = W_y
N = W cos θ
X axis
F -Wₓ -fr = ma
the force applied by the spring is given by hooke's law
F = k Δx
friction force has the expression
fr = μ N
fr = μ W cos θ
we substitute
k Δx - W cos θ - μ mg cos θ = m a ( 1)
c) If the plane has no friction, what is the angle so that Δx = 0.1m
We write the equation 1, with fr = 0 and since the system is still a = 0
k Δx - W cos θ -0 = 0
cos θ = [tex]\frac{k \Delta x}{ m g}[/tex]
cos θ = [tex]\frac{880 \ 0.1}{ 150 \ 9.8}[/tex]
cos θ = 0.0598
θ = cos⁻¹ 0.0598
θ = 86.6º
d) In this part they give the angle θ = 45º and there is no friction, they ask the compression
the acceleration is zero, we substitute in 1
k Δx - W cos θ - 0 = 0
Δx = [tex]\frac{mg \ cos \ \theta}{k}[/tex]
Δx = [tex]\frac{ 150 \ 9.8 \ cos45}{880}[/tex]
Δx = 1.18 m
m
A 3.0 kg model train going right at 2.8 bumps into another 2.0 kg model train car moving in the same
S
m
direction at 1.6 . The heavier train car has a final speed of 2.2 to the right.
S
S
What is the final speed of the lighter 2.0 kg train car?
Answer:
it’s 2.5 m/s
Explanation:
i’m too lazy but trust
This question can be solved by using the law of conservation of momentum.
The final speed of the lighter 2 kg train is " 2.5 ".
When two moving objects collide with each other, the law of conservation of momentum can be applied to them as follows:
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
where,
m₁ = mass of heavier train = 3 kg
m₂ = mass of lighter train = 2 kg
u₁ = initial speed of heavier train = 2.8
u₂ = initial speed of lighter train = 1.6
v₁ = final speed of heavier train = 2.2
v₂ = final speed of lighter train = ?
Therefore,
(3 kg)(2.8) + (2 kg)(1.6) = (3 kg)(2.2) + (2 kg)(v₂)
[tex]v_2 = \frac{5 kg}{2 kg}[/tex]
v₂ = 2.5
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The attached picture illustrates the law of conservation of momentum.
An astronaut named Sandra Bullock has drifted too far away from her spaceshuttle while attempting to repair the Hubble Space telescope. She realizes that theshuttle is moving away from her at 3 m/s. On her back is a 10 kg jetpack which consistsof an 8 kg holding tank filled with 2 kg of pressurized gas. Without the jetpack, sheand her space suit have a mass of 80 kg.
Required:
a. She is able to use the gas to propel herself in the same direction as the shuttle. The gas exits the tank at a uniform rate with a constant velocity of 100 m/s, relative to the tank (and her). After the gas in the tank has been released, what is her velocity?
b. After this, she throws her empty tank into space and relies on the conservation of momentum to increase her speed to match that of the shuttle. With what velocity (in her frame of reference!) will she have to throw the tank?
Answer:
a) v_f = 0.898 m / s, b) v₂ = -6.286 m / s
Explanation:
a) For this exercise we use the conservation of momentum, we define a system formed by the astronaut, her equipment and the expelled gases. We must also define a stationary frame of reference, let's place the system on the platform, so the speed of the subject is v = -3 m / s
Initial instant. Before you start to pass gas
p₀ = (M + Δm) v
M is the mass of the astronaut M = 80Kg and Δm the masses of the gases
Final moment. When you expel the gases
p_f = M (v + Δv) + Δm (v-v_e)
where v_e is the gas velocity v_e = 100 m / s
momentum is conserved
p₀ = p_f
M v + Δm v = Mv + M Δv + Δm v -Δm ve
0 = M Δv - Δm v_e
if we make the very small quantities Δv → dv and Δm → dm, furthermore the quantity of output gas is equal to the decrease in the total mass dm = -dM
M dv = -v_e dM
∫ dv = - v_e ∫ dM / M
We solve, between the lower limits v₀ = v with M = M₀ and the upper limit v = v_f for M = M_f
v_f - v₀ = - v_e (ln M_f - Ln M₀)
v_f - v₀ = v_e ln ([tex]\frac{M_o}{M_f}[/tex])
v_f = v₀ + v_e ln (\frac{M_o}{M_f})
let's calculate
v_f = -1.3 + 100 ln (80 + 10 + 2/80 + 10)
v_f = -1.3 +2.20
v_f = 0.898 m / s
b) launch the jetpack to increase its speed up to the speed of the platform
initial instant. Before launching the tanks
p₀ = (M + m') v_f
final instnte. After launching the tanks
p_f = M v₁ + m' v₂
indicate that the final velocity of the astronaut is the platform velocity v₁=0 m / s, since the reference system is fixed on it
p₀ = p_f
(M+ m) v_f = M v₁ + m v₂2
v₂ = [tex]\frac{ M ( v_f - v_o) + m' v_f}{m'}[/tex]
v₂ = [tex]\frac{M}{m}[/tex] (v_f -v₁) + v_f
let's calculate
v₂ = 80/10 (0.898 - 0) + 0.898
v₂ = -7.1874 + 0.898
v₂ = -6.286 m / s
As every amusement park fan knows, a Ferris wheel is a ride consisting of seats mounted on a tall ring that rotates around a horizontal axis. When you ride in a Ferris wheel at constant speed, what are the directions of your acceleration and the normal force on you (from the always upright seat) as you pass through (a) the highest point and (b) the lowest point of the ride
Answer:
Answer is explained in the explanation section below.
Explanation:
In this question, we are asked to find out the direction of acceleration and direction of the normal force acting upon us from the always upright seat.
a) You pass through the highest point:
When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the highest point, then the direction of of our acceleration will be towards the center or it will be towards downward direction.
And at the highest point on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards.
b) You pass through the lowest point of the ride:
When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the lowest point, then the direction of of our acceleration will be towards the center or it will be towards upward direction.
And at the lowest on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards again.