Answer:
45.89m/s²
Explanation:
Given
Distance S = 305m
Time t = 3.64s
To get the acceleration during this run, we will apply the equation of motion:
S = ut+1/2at²
Substitute the given parameters into the formula and calculate the value of a
305 = 0+1/2 a(3.64)²
304 = 1/2(13.2496)a
304 = 6.6248a
a = 304/6.6248
a = 45.89m/s²
Hence the average acceleration during this run is 45.89m/s²
Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m57.0 m . If the track is completely flat and the race car is traveling at a constant 24.5 m/s24.5 m/s (about 55 mph55 mph ) around the turn, what is the race car's centripetal (radial) acceleration
Answer:
10.53m/s²
Explanation:
Centripetal acceleration is the acceleration of an object about a circle. The formula for calculating centripetal acceleration is expressed by:
[tex]a = \frac{v^2}{r}[/tex]
v is the velocity of the car = 24.5m/s
r is the radius of the track = 57.0m
Substitute the given values into the formula:
[tex]a = \frac{24.5^2}{57} \\\\a = \frac{600.25}{57}\\ \\a = 10.53m/s^{2}[/tex]
Hence the centripetal acceleration of the race car is 10.53m/s²
Please help with this science thank you
Answer:acbd
Explanation:
When adding the components of a vector, the resultant magnitude is always the sum of the two component magnitudes.
True or false?
Answer:False
Explanation:
Answer:
False
Explanation:
4. Tires on the road show how friction produces
a. lubrication
b. heat
c. gravity
d. force
help :(
I think it is D. Force
Answer:
Explanation:
Do you mean tire tracks?
If you do then the tracks show that there must have been a lot of heat involved. When the tires skid, the friction produced must be awfully hot to stop and 800 kg vehical (which is pretty small). The tire tracks that you see are bits of melted rubber taken from the tire.
Force is involved, but heat is the better answer.
gravity is necessary for friction to take place. If you were out in space, you could not get enough friction to leave tire tracks. (Force is awfully small).
Lubrication is there to give you a 4th choice. The exact opposite is what takes place. Lubrication reduces friction.
Apply the general results obtained in the full analysis of motion under the influence of a constant force in Section 2.5 to answer the following questions. You hold a small metal ball of mass a height above the floor. You let go, and the ball falls to the floor. Choose the origin of the coordinate system to be on the floor where the ball hits, with up as usual. Express all results in terms of , , and . Just after release, what are and
Answer:
y(i) = h
v(y.i) = 0
Explanation:
See attachment for elaboration
A 66-N ⋅ m torque acts on a wheel with a moment of inertia 175 kg ⋅ m2. If the wheel starts from rest, how long will it take the wheel to make one revolution?
Answer:
t = 5.77 s
Explanation:
This exercise will use Newton's second law for rotational motion
τ = I α
α = τ / I
α = 66/175
α = 0.3771 rad/s²
now we can use the rotational kinematics relations, remember that all angles must be in radians
θ = 1 rev = 2π radians
θ = w₀ t + ½ α t²
as the wheel starts from rest w₀ = 0
t = √ (2θ/α)
let's calculate
t = √ (2 2π / 0.3771)
t = 5.77 s
An airline employee tosses two suitcases in rapid succession with a horizontal velocity of 7.2 ft/s onto a 50-lb baggage carrier which is initially at rest. Problem 14.003.a Conservation of momentum: two colliding suitcases Knowing that the final velocity of the baggage carrier is 4.8 ft/s and that the first suitcase the employee tosses onto the carrier has a weight of 30 lb, determine the weight of the other suitcase. (You must provide an answer before moving on to the next part.) The weight of the other suitcase is lb.
Answer:
m₁ = 70 lb
Explanation:
Here we will use the law of conservation of momentum:
m₁u₁ + m₂u₂ + m₃u₃ = m₁v₁ + m₂v₂ + m₃v₃
where,
m₁ = mass of first suitcase = ?
m₂ = mass of second suitcase = 30 lb
m₃ = mass of baggage carrier = 50 lb
u₁ = initial speed of first suitcase = 7.2 ft/s
u₂ = initial speed of second suitcase = 7.2 ft/s
u₃ = initial speed of baggage carrier = 0 ft/s
v₁ = Final speed of first suitcase = 4.8 ft/s
v₂ = Final speed of second suitcase = 4.8 ft/s
v₃ = Final speed of baggage carrier = 4.8 ft/s
because after collision all three will have same speed
Therefore,
(m₁)(7.2 ft/s) + (30 lb)(7.2 ft/s) + (50 lb)(0 ft/s) = (m₁)(4.8 ft/s) + (30 lb)(4.8 ft/s) + (50 lb)(4.8 ft/s)
(m₁)(7.2 ft/s) + (216 lb ft/s) + (0 lb ft/s) = (m₁)(4.8 ft/s) + (144 lb ft/s) + (240 lb ft/s)
(m₁)(7.2 ft/s) - (m₁)(4.8 ft/s) = 168 lb ft/s
m₁ = (168 lb ft/s)/(2.4 ft/s)
m₁ = 70 lb
How have the owners of the game reserve invested in the local community?
Answer:They have made community members shareholders so they get a share of the profits, which they use for schools and healthcare clinics
Explanation: Edmentum
Answer:
shareholders of the community get profits and that is used for schools and healthcare clinics. cs.
Explanation:
A single living thing.
Answer:
What do you mean ma´am/sir?
Explanation:
What is the resultant velocity of a plane that is traveling at 245 m/s North and encounters a tailwind of 55 m/s North?
Answer:
b
Explanation:
Determine the accelerations that result when a 12-N net force is applied to a 3-kg object and then to a 6-kg object.
ans: 4m/s²
and 2m/s²
step:
f=m.a
in first condition
f=12N , M=3kg
so,
12=3.a
a=12/3
a=4m/s².
similarly
in second condition
f=12N ,M=6Kg
so,
12=a.6
a=12/6
a=2m/s²
QUESTION 10
An archer fires an arrow towards a tree with initial speed 65 m/s and angle 25 degrees above the horizontal. If the arrow takes 0.85
seconds to hit the tree, calculate the horizontal distance between the archer and the tree.
QUESTION 11
A monkey throws a banana from a tree into a nearby river. The banana has initial speed 7.6 m/s, is angled 40 degrees below the
horizontal, and takes 0.75 seconds to land in the river. Calculate the speed of the banana when it hits the water.
Answer:
10) The distance between the archer and the tree is 50.074 meters.
11) The speed of the banana when it hits the water is approximately 13.554 meters per second.
Explanation:
10) The arrow experiments a parabolic motion, which is the combination of horizontal motion at constant velocity and vertical uniform accelerated motion. In this case we need to find the horizontal distance between the archer and the tree, calculated by the following kinematic equation:
[tex]x = x_{o} +v_{o}\cdot t \cdot \cos \theta[/tex] (Eq. 1)
Where:
[tex]x_{o}[/tex] - Initial position of the arrow, measured in meters.
[tex]x[/tex] - Final position of the arrow, measured in meters.
[tex]v_{o}[/tex] - Initial speed of the arrow, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]\theta[/tex] - Launch angle, measured in sexagesimal degrees.
If we know that [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 65\,\frac{m}{s}[/tex], [tex]t = 0.85\,s[/tex] and [tex]\theta = 25^{\circ}[/tex], the horizontal distance between the archer and the tree is:
[tex]x = 0\,m + \left(65\,\frac{m}{s}\right)\cdot (0.85\,s)\cdot \cos 25^{\circ}[/tex]
[tex]x = 50.074\,m[/tex]
The distance between the archer and the tree is 50.074 meters.
11) The final speed of the banana ([tex]v[/tex]), measured in meters per second, just before hitting the water is determined by the Pythagorean Theorem:
[tex]v = \sqrt{v_{x}^{2}+v_{y}^{2}}[/tex] (Eq. 2)
Where:
[tex]v_{x}[/tex] - Horizontal speed of the banana, measured in meters per second.
[tex]v_{y}[/tex] - Vertical speed of the banana, measured in meters per second.
Each component of the speed are obtained by using these kinematic equations:
[tex]v_{x} = v_{o}\cdot \cos \theta[/tex] (Eq. 3)
[tex]v_{y} = v_{o}\cdot \sin \theta +g\cdot t[/tex] (Eq. 4)
Where [tex]g[/tex] is the gravitational acceleration, measured in meters per square second.
If we know that [tex]v_{o} = 7.6\,\frac{m}{s}[/tex], [tex]\theta = -40^{\circ}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex] and [tex]t = 0.75\,s[/tex], the components of final speed are, respectively:
[tex]v_{x} = \left(7.6\,\frac{m}{s} \right)\cdot \cos (-40^{\circ})[/tex]
[tex]v_{x} = 5.822\,\frac{m}{s}[/tex]
[tex]v_{y} = \left(7.6\,\frac{m}{s}\right)\cdot \sin (-40^{\circ})+\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (0.75\,s)[/tex]
[tex]v_{y} = -12.240\,\frac{m}{s}[/tex]
And the speed of the banana right before hitting the water is:
[tex]v = \sqrt{\left(5.822\,\frac{m}{s} \right)^{2}+\left(-12.240\,\frac{m}{s} \right)^{2}}[/tex]
[tex]v \approx 13.554\,\frac{m}{s}[/tex]
The speed of the banana when it hits the water is approximately 13.554 meters per second.
find the vector sum of the 3 vectors. A= -3i - 2j + 7k, B= i + 3j + 3k, B= -5k
Answer:
34k+B plus 9 :)
Explanation:
Alejandro made 6.4 liters of punch using half apple juice and half orange juice. How many milliliters of apple juice are in the punch?
Answer:
3.2
Explanation:
I hope that this helps! Have a good day!!
An 80-kg bungee jumper jumps off a bridge. Rubber bungee cords act as a large spring attaching the jumper to the bridge. A bear standing in the river below catches the jumper. If the spring constant of the bungees is 20 N/m and they stretch 50 m. How much force must the bear apply to keep the jumper from moving?
Answer:
The force is [tex]F_b = 216 \ N [/tex]
Explanation:
From the question we are told that
The mass of the bungee jumper is m = 80 kg
The spring constant is [tex]k = 20 \ N/ m[/tex]
The extension of the rubber bungee cords is x = 50 m
Generally the weight of the jumper is
[tex]W = m * g[/tex]
=> [tex]W = 80 * 9.8 [/tex]
=> tex]W = 784 \ N [/tex]
Generally the returning force of the rubber bungee cords is mathematically represented as
[tex]F = k * x[/tex]
=> [tex]F = 20 * 50 [/tex]
=> [tex]F = 1000 \ N [/tex]
The force to be applied by the bear is
[tex]F_b = F - W[/tex]
=> [tex]F_b = 100 - 784[/tex]
=> [tex]F_b = 216 \ N [/tex]
what is the meaning of relative as a noun?
Answer:
noun. a person who is connected with another or others by blood or marriage. something having, or standing in, some relation or connection to something else. something dependent upon external conditions for its specific nature, size, etc. (opposed to absolute).
Velocity which stone gains when falling from height of 80 m is approximately equal to *
A. 0
B. 1 m/s
C. 8 m/s
D. 40 m/s
E. 300 m/s
Answer:
40
Explanation:
1. Name three branches of
science.
Answer:
Biology
Physics
Chemistry
Explanation:
Biology (life like cells, human reproduction, etc.)
Physics (Studies forces, like gravity.)
Chemistry (studies the atoms, the elements, etc.)
Help solve these two problems im having trouble trying to start these problems?
Answer:
25. Approximately 8.1 meters
26. North 1.31 km, and East 2.81 km
Explanation:
25.
Notice that the displacements: 6 meters east and 5.4 south create the legs of a right angle triangle. The hypotenuse of that triangle will be the distance (d) needed to cover in order to get the ball in the hole in one putt. That is:
[tex]d=\sqrt{6^2+5.4^2} =\sqrt{65.16} \approx 8.072\,\,\,m[/tex]
which can be rounded to 8.1 m.
26.
Notice that the 3.1 km at an angle of 25 degrees north of east, is the hypotenuse of a right angle triangle that has for legs the east and north components of that distance.
We can find the leg corresponding to the east displacement using the cosine function (that relates adjacent side with hypotenuse):
[tex]east\,\, comp=3.1 * cos(25^o)\approx 2.809\,\,km[/tex]
and we can calculate the north component using the sine function that relates the opposite side to the angle with the hypotenuse.
[tex]north\,\,component = 3.1 * sin(25^o) \approx 1.31 \,\,km[/tex]
There are two unitless vectors:
F1 = 8.92 i + 17.37 j
F2 = 8.31 i - 10.97 j
A third vector is added to them such that they add up to to the null vector:
F1 + F2 + F3 = 0
What is the angle of the third vector with respect to the +x-axis?
Answer:
[tex]200.38^0[/tex]
Explanation:
Given the forces:
F1 = 8.92 i + 17.37 j
F2 = 8.31 i - 10.97 j
If a third vector is added to them such that they add up to to the null vector as F1 + F2 + F3 = 0
then to get F3:
F3 = -F2-F1
F3 = -(8.31 i - 10.97 j)-(8.92 i + 17.37 j )
F3 = -8.31 i + 10.97 j-8.92 i - 17.37 j
F3 = -8.31i-8.92i+10.97j-17.37j
F3 = -17.23i-6.4j
from the vector:
x = -17.23 and y = -6.4
angle of the third vector with respect to the +x-axis is expressed as:
[tex]\theta = tan^{-1}\frac{y}{x}\\ \theta = tan^{-1}\frac{-6.4}{-17.23}\\\theta = tan^{-1}0.3715\\\theta = 20.38^0[/tex]
Hence the angle the vector makes with the x axis will be [tex]\theta = 180+20.38 = 200.38^0[/tex]
A student studies the effect of an object's speed on its amount of kinetic energy. This graph summarizes the data from the study Kinetic energy Speed Which statement best describes what the graph shows?
A. As speed increases, kinetic energy increases exponentially
B. As speed increases, kinetic energy stays the same
C. As speed decreases, kinetic energy doubles each time.
D. As kinetic energy increases, speed decreases exponentially
The answer is A I Hope this answer helps you (i got the question right)
Answer:
A. As speed increases, kinetic energy increases exponentially.
Explanation:
The amount of kinetic energy an object has depends on the speed. Kinetic energy is also known as "motion energy." This being said, if speed is increasing, decreasing, or staying constant, the kinetic energy of the object will too.
Two cars come to a stop from the same initial speed, one braking gently and the other braking hard. Which car converts more kinetic energy to thermal energy
Answer:
Both cars convert more kinetic energy to thermal energy.
Explanation:
Given that,
Two cars come to a stop from the same initial speed, one braking gently and the other braking hard.
We know that,
The conservation of energy :
The energy of the system is always constant.
The energy can be change one form to another form.
So, we can say that when both cars come to a stop from the same speed, one braking softly and the other braking strongly then both cars convert more kinetic energy to thermal energy.
Hence, Both cars convert more kinetic energy to thermal energy.
Based on the law of consevation of energy, the car that brakes hard will convert more kinetic energy to thermal energy.
Which car will convert more kinetic energy to thermal energy?According to the law of conservation of energy, the total energy is conserved in any closed system.
Since the cars have the same mass and had the same initial speed, their kinetic energies are the same.
However, the car braking hard comes to rest in a shorter time and in a shorter distance.
Using the formula for calculating power:
Power = energy/timeThe car that braked hard will have greater power due to shorter time of action.
This power is seen as greater thermal energy, whereas some of the energy of the other car will still be used in travelling some distance before stopping.
Therefore, the car that brakes hard will convert more kinetic energy to thermal energy.
Learn more about kinetic energy and thermal energy at: https://brainly.com/question/3064483
un avión aterriza en la superficie de un portaaviones a 50 m/s y se detiene por completo en 120 metros, ¿cuál es la aceleración necesaria para detenerlo?
Answer:
La aceleración necesaria para detener el avión es - 10.42 m/s².
Explanation:
Un movimiento uniformemente acelerado (M.U.A) es aquél cuya aceleración es constante y la velocidad de un objeto cambia a medida que el movimiento evoluciona.
Siendo la aceleración "a" el cambio de velocidad al tiempo transcurrido en un punto A a B, la velocidad inicial la velocidad que tiene un cuerpo al iniciar su movimiento en un período de tiempo y la velocidad final la velocidad que tiene un cuerpo al finalizar su movimiento en un período de tiempo, entonces en M.U.A se cumple:
Vf² - Vo² = 2*a*d
donde:
Vf: Velocidad final Vo: Velocidad inicial a: Aceleración d: Distancia recorridaEn este caso:
Vf: 0 m/s, porque el avión se detieneVo: 50 m/sa: ?d: 120 mReemplazando:
(0 m/s)² - (50 m/s)² = 2*a*120 m
Resolviendo:
[tex]a=\frac{(0 m/s)^{2} -(50 m/s)^{2} }{2*120 m}[/tex]
a= - 10.42 m/s²
La aceleración necesaria para detener el avión es - 10.42 m/s².
A man is passing barrels of water to another person from a height of 7 meters. The barrel is attached to a roof that is 10 meters above the ground. The rope holding the barrel is 8 meters long and will break if the tension exceeds 638 N. How many liters of water can the man put in the barrel without the rope breaking (assuming that the barrel is massless)? (The density of water = 1000 kg/m3 and 1 m3 = 1000 L. Density = mass/volume.)
Which of the following is not true about taxes? A. Mandatory sum of money by government so that it can operate B. Due on April 15th C. largely collected to support private businesses D. Collected by the Internal Revenue Service (IRS)
***ECONOMICS***
A government that wants to increase its GDP would most likely take which
action?
A. Increase the money supply to make it easier to borrow money
Ο Ο
B. Decrease the money supply to slow the growth of inflation
C. Increase taxes on businesses that operate outside the country
O O
D. Decrease taxes on citizens who are poor or unemployed
Answer:
The correct answer is A. A government that wants to increase its GDP would most likely increase the money supply to make it easier to borrow money.
Explanation:
If the government wanted to increase its GDP, the most appropriate way to do so would be to increase the money supply both through issuance and through a reduction in bank reserve requirements, thereby increasing the circulating money in the hands of society.
This, in turn, would make citizens reinvest that money, increasing economic production and, therefore, the national GDP.
Answer: A. Increase the money supply to make it easier to borrow money
Explanation: I just took the test on Ap ex
What is the maximum speed with which a 1200-kg car can round a turn of radius 92.0 m on a flat road if the coefficient of static friction between tires and road is 0.65
Answer:
24.22m/sExplanation:
Given data
mass m= 1200kg
radius r=92m
coefficient of friction μ=0.65
We know that
F=mv2/r
F=μmg
mv2/r = μsmg
v^2/r = μsg
vmax = √(rμg)
Substituting our data we have
vmax=√(92*0.65*9.81)
vmax=√586.638
vmax=24.22m/s
PLS ANSWER WILL MARK BRANLIEST!!!!!!!!!!!!
Describe the life cycle of a star before it collapses into a black hole.
Describe the life cycle of a star before it becomes a black dwarf.
What is the likely outcome of our sun? *
The sun will supernova and become a black hole.
The sun will swell, encompassing the inner planets and collapses into a dwarf star.
The sun will become a pulsar.
How Do You Know?
P.S. the how do you know is only for the last question
1) describe the life cycle of a star before it collapses into a black hole.
1) describe the life cycle of a star before it collapses into a black hole.ans: A star's life cycle is determined by its mass. The larger its mass, the shorter its life cycle. A star's mass is determined by the amount of matter that is available in its nebula, the giant cloud of gas and dust from which it was born. Over time, the hydrogen gas in the nebula is pulled together by gravity and it begins to spin. As the gas spins faster, it heats up and becomes as a protostar. Eventually the temperature reaches 15,000,000 degrees and nuclear fusion occurs in the cloud's core. The cloud begins to glow brightly, contracts a little, and becomes stable. It is now a main sequence star and will remain in this stage, shining for millions to billions of years to come. This is the stage our Sun is at right now.
2) describe the life cycle of a star before it becomes a dwarf.
ans: The life cycle of a low mass star (left oval) and a high mass star (right oval). ... As the core collapses, the outer layers of the star are expelled. A planetary nebula is formed by the outer layers. The core remains as a white dwarf and eventually cools to become a black dwarf.
3) what is the likely outcome of our sun?
ans: All stars die, and eventually — in about 5 billion years — our sun will, too. Once its supply of hydrogen is exhausted, the final, dramatic stages of its life will unfold, as our host star expands to become a red giant and then tears its body to pieces to condense into a white dwarf.
Question 1 of 15
All digits shown on the measuring device, plus one estimated digit, are
considered
Answer here
SUBMIT
Answer:
significant
Explanation:
The digits in a measurement that are considered significant are all of those digits that represent marked calibrations on the measuring device plus one additional digit to represent the estimated digit (tenths of the smallest calibration).
61) A bicycle wheel of radius 0.36 m and mass 3.2 kg is set spinning at 4.00 rev/s. A very light bolt is attached to extend the axle in length, and a string is attached to the axle at a distance of 0.10 m from the wheel. Initially the axle of the spinning wheel is horizontal, and the wheel is suspended only from the string. We can ignore the mass of the axle and spokes. At what rate will the wheel process about the vertical
Answer:
The rate the wheel will process about the vertical is 2.86 RPM
Explanation:
Given;
radius of the bicycle wheel, R = 0.36 m
mass of the wheel, m = 3.2 kg
angular velocity, ω = 4 rev/s
The rate at which the wheel will process about the vertical is given by;
Ф = τ/L
Where;
τ is the torque
L is the angular momentum
τ = Fr
τ = mgr = 3.2 x 9.8 x 0.1 = 3.126 N.m
L = Iω = MR²ω
L = 3.2 x (0.36)² x (4 x 2π)
L = 10.4244 kg.m²/s
Ф = τ/L
Ф = (3.126) / (10.4244)
Ф = 0.29987 rad/s
Ф = 0.29987 rad/s x (60 / 2π)
Ф = 2.86 RPM
Therefore, the rate the wheel will process about the vertical is 2.86 RPM