Answer:
Hip and knee extensors
Explanation:
These are gluteus muscles and hamstring muscles the there are the major movers for your body and are very important in pelvic alignment and lower back support during weight lifting
Convert 252 cL into uL
Answer:
Explanation:
252,000
Question 8 of 10
It takes a person 22 seconds to swim in a straight line from the south end of
a pool to the north end of the pool, a distance of 28 meters. What is the
swimmer's velocity?
A. 1.3 m/s south
B. 1.3 m/s north
C. 0.8 m/s south
D. 0.8 m/s north
The correct answer is B. 1.3 m/s north
Explanation:
Velocity is a factor that describes how fast or slow the motion of a body occurs and its direction. Moreover, this can be calculated by dividing the total displacement into the time of movement, and the final result is expressed in units such as meters per second followed by the direction, for example, 152 m/s south. The process to calculate the velocity of the swimmer is shown below.
[tex]v = \frac{d}{t}[/tex]
[tex]v = \frac{28 meters}{22 seconds}[/tex]
[tex]v = 1.27 m/s[/tex]
This means the velocity of the swimmer is 1.27 m/s, which can be rounded as 1.3 m/s. Additionally, if the direction is considered it would be 1.3 m/s north because the swimmer went from the south of the pool to its north.
Answer:
the answer is B
Explanation:
confirmed
An object moving with uniform acceleration has a velocity of 14.0 cm/s in the positive x-direction when its x-coordinate is 2.94 cm. If its x-coordinate 1.85 s later is −5.00 cm, what is its acceleration?
Answer:
-19.9m/s²
Explanation:
Given that
vi = 14cm/s
xi = 2.94 cm
xf=-5.00 cm
t= 1.85s
So
Using xf-xi= vi t + 1/2at²
Which is = -5-2.94= 14*1.85+1/2a1.85²
= -7.94= 25.9+1.7a
-33.9= 1.7a
a= - 19.9m/s²
Determine the discharge of a river that is 10 m wide, has a semicircular cross-sectional area, and has a velocity of 3 m/s.
Answer:
117.8 m³/s
Explanation:
given that
the diameter of the river, d = 10 m wide
Velocity of flow of water, v, = 3 m/s
To get the discharge of a river, we have to find the area of the said river first.
Area of the river, a = area of semi circle = 1/2 * area of a circle
Area = 1/2 * πr²
Area = 1/2 * 3.142 * 5²
Area = 39.275 m²
Discharge of the river therefore is
Discharge, q = A * V, where
A is area of the river, and
V is the velocity of flow of the river.
Discharge, q = 39.275 * 3
Discharge, q = 117.825
Why do we perceive the sky as blue instead of red?
The blue sky we observe depends upon two factors: how sunlight interacts with Earth's atmosphere, and how our eyes perceive that light.
If this helps please mark me the brainiest
how sunlight interacts with Earth's atmosphere, and how our eyes perceive that light.nation:
We will use the same coordinate system for a number of the questions on this exam.
Place a sheet of paper flat on a table in front of you as you sit facing it.
In the horizontal plane toward the right side of the paper is the positive x direction.
In the horizontal plane toward the top of the paper ( away from you in the direction you are facing ) is the positive y direction.
In the vertical direction toward the ceiling is the positive z direction
A spatially uniform magnetic field is steadily changing its size. When a clock reads 2.00s the field is 1.75T in the positive z direction. In the x y plane there is a coil of wire with 55 loops and an area of 0.12m%. The EMF ( voltage change) from one end of the coil to the other is a steady 12.7 Volts. One of these is the magnetic field when the clock reads 2.37s. Which is it?
Answer:
Explanation:
Let the magnetic field after .37 s is B
magnetic flux associated with coil = n BA , where B is magnetic field and A is area of coil and n is no of turns
= 55 x .12 x 1.75
= 11.55 Tm²
magnetic flux after .37 s
= 55 x B x .12
= 6.6 B
rate of change of flux = (66B - 11.55) / .37 = EMF induced
(6.6B - 11.55) / .37 = 12.7
6.6 B - 11.55 = 4.7
6.6 B = 16.25
B = 2.46 T .
The prediction that alcohol slows reaction time is:_________.
a) nondirectional
b) directional
c) semi directional
d) partially directional
Answer:
b) directional
Explanation:
The prediction that alcohol slows reaction time is Directional . This is because for alcohol to be known as one which slows reaction time means there have been various hypothesis conducted which supports this.
In this case, formulation of a null hypothesis is usually necessary which means that alcohol does not slow reaction time and another alternative hypothesis that suggests that alcohol slows reaction time.
What are the definitions of momentum, angular momentum, and kinetic energy for a single particle? What are the dimensions of these quantities?
Answer:
Explanation:
Momentum is defined as a physical quantity that takes into cognisance the product of mass of a body and its velocity. It's unit is kgm/s
Momentum = mv
Angular Momentum is defined as the quantity of rotation of a body that measures the product of its moment of inertia with respect to its angular velocity. Because it deals with rotation is what differs it from ordinary momentum. It's unit is kgm²/s
Angular Momentum = mvr
Kinetic Energy is defined as the energy a body posses as a result of it being in motion. The unit of Kinetic Energy is J
Kinetic Energy = ½mv²
The dimensions of these quantities are
Momentum => MLT^−1
Angular Momentum => ML²T^−1
Kinetic Energy => ML²T^-2
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What does F equal???.......ANSWER FAST A. ma B. m/a C. m+a D. m-a
Answer:
A. maExplanation:
Force = mass × acceleration
F = ma
At what speed, in m/s, would a moving clock lose 4.5 ns in 1.0 day according to experimenters on the ground
Answer:
the speed at which a moving clock lose 4,5ns in 1.0 day is 98.58 m/s
Explanation:
Let's first find the relation for the time dilation by using the formula:
[tex]\Delta t' = \Delta t \gamma[/tex]
Making [tex]\gamma[/tex] the subject of the formula, we have:
[tex]\gamma= \dfrac{\Delta t'}{\Delta t}[/tex]
Number of day(s) = 1
1 day in seconds = 24 × 60 × 60
= 84600 seconds
[tex]\gamma= \dfrac{86400 \ s}{86400 \ s - 4.5 \ ns}[/tex]
[tex]\sqrt{1- \dfrac{v^2}{c^2} }= \dfrac{86400 \ s - 4.5 \ ns}{86400 \ s}[/tex]
[tex]\sqrt{1- \dfrac{v^2}{c^2} }= 1 - 5.4 \times 10^{-14}[/tex]
From the above equation, if we apply binomial expansion to it, we have the following,
[tex]\sqrt{1- \dfrac{v^2}{c^2} }=(1-\dfrac{v^2}{c^2})^{\frac{1}{2}}[/tex]
⇒[tex]1 -\dfrac{1}{2}(\dfrac{v^2}{c^2})[/tex]
So;
[tex]1 -\dfrac{1}{2}(\dfrac{v^2}{c^2}) = 1 - 5.4 \times 10^{-14}[/tex]
[tex]\dfrac{1}{2}(\dfrac{v^2}{c^2}) = 5.4 \times 10^{-14}[/tex]
[tex]\dfrac{v^2}{c^2} = 2 \times 5.4 \times 10^{-14}[/tex]
[tex]\dfrac{v^2}{c^2} = 1.08 \times 10^{-13}[/tex]
[tex]v^2 = 1.08 \times 10^{-13} \times c^2[/tex]
where ;
[tex]c = 3 \times 10^8 \ m/s[/tex]
[tex]v^2 = 1.08 \times 10^{-13} \times (3 \times 10^8 \ m/s)^2[/tex]
[tex]v = \sqrt{ 1.08 \times 10^{-13} \times (3 \times 10^8 \ m/s)^2}[/tex]
[tex]v = \sqrt{ 1.08 \times 10^{-13} } \times (3 \times 10^8 \ m/s)[/tex]
[tex]v =3.286 \times 10^{-7} \times (3 \times 10^8 \ m/s)[/tex]
[tex]\mathbf{v =98.58 \ m/s}[/tex]
Therefore, the speed at which a moving clock lose 4,5ns in 1.0 day is 98.58 m/s
which surface has more friction
A. an ice rink
B. a grassy field
C. a paved road
Answer:
The answer is C. a paved roadExplanation:
Generally, roads are meant to have very rough surfaces to allow for friction between car tyres hence the motion of vehicles.
What is a paved surface?
Paved roads are roads that are covered with a firm surface suitable for travel, like paving stones or concrete or asphalts.
Another good example of a paved road is a small paved courtyard, covered with a hard layer of concrete/cement.
The other two examples sited in this problem are glossy materials with very low surface friction, that is their surfaces are very smooth.
(a) A standard sheet of paper measures 8 1/2 by 13 inches. Find the area of one such sheet of paper in m2.8.5(!meter/39.37in) — 0.2159 13 (!meter/39.37in ) — 0.3302 0.2159 X 0.3302 = 0.0713
= 0.0713 m2 (b) A second sheet of paper is three-quarters as long and three-quarters as wide as the one described in part (a). By what factor is its area less than the area found in part(a)?= _____________ times less
Answer:
a) A = 0.07129 m²
b) A / A ’= 1.77
Explanation:
In this exercise we are asked to find the area in SI units, so let's start by reducing the dimensions to SI units.
width a = 8.5 inch (2.54 cm 1 inch) (1 m / 100 cm)
a = 0.2159 m
length l = 13 inch (2.54 cm / 1 inch) (1 m / 100 cm)
l = 0.3302 m
The area of a rectangle is
A = l a
A = 0.3302 0.2159
A = 0.07129 m²
b) we have a second sheet with reduced dimensions
a ’= 3/4 a
l ’= ¾ l
Let's find the area of this glossy sheet
A ’= l’ a ’
A ’= ¾ l ¾ a
A ’= 9/16 l a
To find the factor we divide the two quantities
A / A ’= l a 16 / (9 l a
A / A ’= 1.77
Which of the following statements is/are correct?
A. The instantaneous speed is always equal to the magnitude of instantaneous velocity.
B. The average speed is always equal to the magnitude of average velocity.
C. Neither of these statements is correct.
D. Both of these statements are correct.
Answer:
The correct answer is:
The instantaneous speed is always equal to the magnitude of instantaneous velocity. (A)
Explanation:
Speed is the ratio of distance travelled to time. It is a scalar quantity.
Velocity is the rate of change of displacement with time. It is a scalar quantity, having both magnitude and direction
Instantaneous velocity is the rate of change of position for a very small time interval (at a particular point in time), while the instantaneous speed at any given time, is the magnitude of instantaneous velocity. Note that the velocity has both magnitude and direction, so at a particular point in time, the magnitude part of velocity is the same as the speed.
The formula of instantaneous velocity is: [tex]V_{i}=\lim_{\Delta t\rightarrow 0}\frac{ds}{dt}[/tex]
The formula of instantaneous speed is: [tex]Speed_{(i)}=\frac{ds}{dt}[/tex]
On the other hand, average speed does not equal the magnitude of average velocity, because velocity depends on displacement while speed depends on distance, and if an object in motion changes direction at any point in the motion, the velocity reduces because the angle of change in the direction will be considered, hence the speed will be greater than the velocity, therefore, the average speed is not the same as the magnitude of average velocity.
The force of gravity pulls down on your school with a total force of 400,000 newtons. The force of gravity pulling down on your school would be exactly twice as much if your school: a Had twice as much mass b Was twice as tall c Had twice as much volume d Covered twice as much area
Answer: a Had twice as much mass
Explanation:
The data that we have is:
"The force of gravity pulls down on your school with a total force of 400,000 newtons. "
First, remember that, by the second Newton's law that:
F = a*m
F = force
a = acceleration
m = mass
In the case of the gravitational force, the gravitational acceleration is a constant: a = 9.8m/s^2
Then, if we want to have twice as much force the only thing that we can change in the equation is the mass:
Then if the initial force is written as:
F = a*m
twice as much that force is written as:
2*F = a*x
x is a variable that represents the new mass.
We know that F = a*m
2*F = 2*a*m
2*a*m = a*x
2*m = x
Then, if we want to have twice as much force, we should have twice as much mass.
If you walk at a constant speed of 4.5 m/s, how many km can you walk in half an hour? how do you solve this?
What are two examples of healthcare fraud ?
What is the binding energy (in MeV per nucleon) for the ⁶₃Li nucleus?
Answer:
Explanation:
⁶₃Li will have 3 protons and 3 neutrons .
mass of proton in amu = 1.00727 amu
mass of neutron in amu = 1.00866 amu
mass of lithium nucleus in amu = 6.01512 amu
mass defect = 3 ( 1.00727 + 1.00866 ) - 6.01512 amu
= .03267 amu
Binding energy = mass defect in amu x 931 Mev
= 30.41 MeV
binding energy per nucleon
no of nucleon = 3 + 3 = 6
binding energy per nucleon = 30.41 / 6 Mev
= 5.068 MeV .
The binding energy of ⁶₃Li nucleus in MeV per nucleon is 5.26 MeV per nucleon.
We can see that Li has three protons and three neutrons. The total mass of the Li nucleus is obtained from;
3(1.007277) + 3(1.008665) = 3.023 + 3.026 = 6.049 amu
Actual mass of Li - 6 = 6.0151 amu
Mass defect = 6.049 amu - 6.0151 amu = 0.0339 amu
We can obtain the binding energy in MeV as follows;
Binding energy = 0.0339 amu × 931 = 31.56 MeV
The binding energy per nucleon in MeV per nucleon = 31.56 MeV/6
= 5.26 MeV per nucleon
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A 11,000-kg train car moving due east at 21.0 m/s collides with and couples to a 23,000-kg train car that is initially at rest. What is the common velocty of the two-car train after the collisions?
Answer:
v = 6.79 m/s
Explanation:
It is given that,
Mass of a train car, m₁ = 11000 kg
Speed of train car, u₁ = 21 m/s
Mass of other train car, m₂ = 23000 kg
Initially, the other train car is at rest, u₂ = 0
It is a case based on inelastic collision as both car couples each other after the collision. The law of conservation of momentum satisied here. So,
[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]
V is the common velocity after the collisions
[tex]V=\dfrac{m_1u_1}{(m_1+m_2)}\\\\V=\dfrac{11000\times 21}{(11000+23000)}\\\\V=6.79\ m/s[/tex]
So, the two car train will move with a common velocity of 6.79 m/s.
If the position versus time graph of an object is a horizontal line, the object is
In a position-time graph, the velocity of the moving object is represented by the slope, or steepness, of the graph line. If the graph line is horizontal, like the line after time = 5 seconds in Graph 2 in the Figure below, then the slope is zero and so is the velocity. The position of the object is not changing.
Explanation:
The slope of a position time graph gives the velocity of an object. It can be calculated as follows:
[tex]v=\dfrac{dx}{dt}[/tex]
Where
[tex]dx[/tex] is the change in position
[tex]dt[/tex] is the change in time
If the position vs time graph of an object is a horizontal line, it shows that the object is at rest. At this point the object is not moving.
Answer:
If the position versus time graph of an object is a horizontal line, the object is at rest.
Reference:
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scott and jafar are ready to work a problem. in this simulation, assume the coordinates of the points are as follows. a (0 s, 30 m); b (10 s, 50 m); c (20 s, 38 m); d (30 s, 0 m); e (40 s, −38 m); and f (50 s, −50 m) recall the definitions of average speed, vavg ≡ d δt , and average velocity in the x direction, vx ,avg ≡ δx δt . find the average velocity from circled a to circled b.
Answer:
Explanation:
point a represents time 0 and position coordinate 30
point b represents time 10 s , and position coordinate 50 m .
time elapsed = 10 - 0 = 10 s .
displacement = 50 m - 30 m
= 20 m
average velocity = displacement / time elapsed
= 20 / 10
= 2 m /s .
you travel 4.0km east, 4.0km north, then 5.0km at 53.1 degrees north of west in a total of 5 hours. What is the magnitude and direction of your average velocity
Answer:
Explanation:
We shall convert all the displacement in vector form .
i and j represents east and north respectively .
D₁ = 4 i
D₂ = 4 j
D₃ = - 5 cos 53.1 i + 5 sin 53.1 j
= -3i + 4 j
Total displacement = D₁ + D₂ + D₃
= 4i + 4 j - 3i + 4 j
= i + 8j
magnitude of displacement = √( 1² + 8² )
= 8.06 km
velocity = 8.06 / 5
= 1.61 km / h
Direction from x axis in anticlockwise direction .
Tanθ = 8 / 1 = 8
θ = 83° north of east .
A merry-go-round is rotating at constant angular speed. Two children are riding the merry-go-round: Ana is riding at point A and Bobby is riding at point B.
Part A Which child moves with greater magnitude of velocity? Ana has the greater magnitude of velocity. Bobby has the greater magnitude of velocity. Both Ana and Bobby have the same magnitude of velocity.
Part B Who moves with greater magnitude of angular velocity? Ana has the greater magnitude of angular velocity. Bobby has the greater magnitude of angular velocity. Both Ana and Bobby have the same magnitude of angular velocity.
Part C Who moves with greater magnitude of tangential acceleration? Ana has the greater magnitude of tangential acceleration. Bobby has the greater magnitude of tangential acceleration. Both Ana and Bobby have the same magnitude of tangential acceleration.
Part D Who has the greater magnitude of centripetal acceleration? Ana has the greater magnitude of centripetal acceleration. Bobby has the greater magnitude of centripetal acceleration. Both Ana and Bobby have the same magnitude of centripetal acceleration.
Part E Who moves with greater magnitude of angular acceleration? Ana has the greater magnitude of angular acceleration. Bobby has the greater magnitude of angular acceleration. Both Ana and Bobby have the same magnitude of angular acceleration.
Answer:
bobby has a greater magnitude of velocity because because when angular speed is constant linear velocity is proportional to radius of the circular path
B. They both have same magnitude of angular velocity since the angular speed of the merrygoround is constant
C. Also they both have the same tangential acceleration because the angular speed is constant and tangential is zero for both of them
D. Centripetal acceleration of Bobby is greater
E.they both have the same angular acceleration because angular Speed I constant so angular acceleration is zero for both
What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2 A
Complete Question
A very long, straight solenoid with a cross-sectional area of 2.39cm2 is wound with 85.7 turns of wire per centimeter. Starting at t= 0, the current in the solenoid is increasing according to i(t)=( 0.162A/s2) t2. A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A ?
Answer:
The value is [tex]\epsilon = 1.83 *10^{-5} \ V[/tex]
Explanation:
From the question we are told that
The cross-sectional area is [tex]A = 2.39 \ cm^2 = \frac{2.39}{10000} = 0.000239 \ m^2[/tex]
The number of turns is [tex]N = 85.7 \ turns/cm = 8570 \ turns / m[/tex]
The initial time is t = 0s
The current on the solenoid is [tex]I(t) = (0.162 \ A/s^2) t^2[/tex]
The number of turns of the secondary winding is [tex]n = 5 \ turns[/tex]
Generally At I = 3.2 A
[tex]3.2 = (0.162 )t^2[/tex]
=> [tex]t^2 = 19.8[/tex]
=> [tex]t = 4.4 \ s[/tex]
Generally induced emf is mathematically represented as
[tex]\epsilon = A * \mu_o * n * N \frac{d(I)}{dt}[/tex]
[tex]\epsilon = 0.000239 * 4\pi * 10^{-7} * 8570 * 5 * (0.162) * 2t[/tex]
[tex]\epsilon = 0.000239 * 4\pi * 10^{-7} * 8570 * 5 * (0.162) * 2 * 4.4[/tex]
[tex]\epsilon = 1.83 *10^{-5} \ V[/tex]
The following three questions refer to a situation in which a driver is in a car that crashes into a solid wall. The car comes to a complete stop without bouncing back. The car has a mass of 1500 kg and has a speed of 30 m/s before the crash (this is about 65 mph). The driver has a mass of 60 kg.
1) What is the car's initial momentum? What is the person's initial momentum? What is the change in the momentum of the car? What is the change in the person's momentum?
2) What is the net impulse that acts on the car to bring it to a stop? What is the net impulse that acts on the person to bring them to a stop? Do these numbers depend on the details of the crash? Why or why not?
3) What object delivers the impulse that brings the car to a stop? What object delivers the impulse that brings the person to a stop? Think about what happens during such a crash, and discuss the role each of the factors below plays in the ultimate fate of the driver. Be sure to include what objects exert forces on what other objects, and use the model of impulse & momentum to estimate the strength of these forces.
a) "crumple zone" of the front of the car.
b) use or non-use of seat belt.
c) deployment or non-deployment of airbag.
Answer:
1) p₀ = 45000 N / s , p₀ '= 1800 , b) I = -45000 N s , I = 1800 Ns
Explanation:
Impulse equals the change in momentum
I = Δp
1) the initial moment of the car
p₀ = M v
p₀ = 1500 30
p₀ = 45000 N / s
the change at the moment is
Δp = 45000
because the end the car is stopped
moment of the person
P₀ ’= m v
p₀ '= 60 30
p₀ '= 1800
D₀ '= 1800
2) of the momentum change impulse ratio
car
I = Δp
I = -45000 N s
person
I = Δpo '
I = 1800 Ns
3) the object that give the momentum to stop the wall motoring
The person is stopped by the impulse given by the car
a) This area is the one that absorbs most of the vehicle impulse
be) If using a safety painter, the time during which the greater force will act, therefore the lessons decrease
c) The air bag helps reduction in the speed of the person relatively quickly.
A runner traveling with an initial velocity of 1.1 m/s accelerates at a constant rate of 0.8 m/s2fora time of 2.0 s.(a).What is his velocity at the end of this time
Answer:
The final velocity of the runner at the end of the given time is 2.7 m/s.
Explanation:
Given;
initial velocity of the runner, u = 1.1 m/s
constant acceleration, a = 0.8 m/s²
time of motion, t = 2.0 s
The velocity of the runner at the end of the given time is calculate as;
[tex]v = u + at[/tex]
where;
v is the final velocity of the runner at the end of the given time;
v = 1.1 + (0.8)(2)
v = 2.7 m/s
Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.
(a)Suppose during 2020due to relief in Taxation Marginal Propensity to Consume (MPC) will increase from 0.65 to 0.75then calculate Tax multiplier (KT). (b)By using the Tax multiplier value calculated in part (a), find the change in national income if Tax amount is increased from 4500billion rupees to 5100billion rupees. (c)How Subsidies and Ehsas program amount will affect the Aggregate Demand (AD)?
Answer:
abcd
Explanation:
Answer:
where is the answer pls send me answer on my what.s up number
Explanation:
When sunlight interacts with a raindrop to produce a rainbow, how many refractive events occur for each drop
Answer:
The correct answer will be "2".
Explanation:
The mechanism of rainbow formation starts whenever sunlight is shone on a raindrop. When the sun's light waves hit and go through a droplet of water, their pace changes a little. This induces bending or "refracting" of the direction of light.Light moves in such a single direction, unless someone represents something, refracts, or diffracts everything that. When some of these things have happened, the balanced information of color is isolated, in which each could still be seen.Physics help please..............
Answer:
3.68 m/s
Explanation:
Full answer in the attached picture
Which of the following surface environments is the most likely site for deposits in this photograph?
A. steep mountain front.
B. glacier.
C. sand dunes.
D. beach.
E. lake
Answer:
d. beach is the correct answer
Beach is the most likely site for deposits in this photograph. So, the correct option is (D).
What is Beach?A beach is defined as a narrow strip of land separating a body of water from inland areas usually made up of small grains of sand, rocks and minerals that have been eroded by frequent pounding by wind and waves. become The beach has both sandy and rocky features.
Beaches provide protection to residents living near the sea by acting as a buffer against high winds and powerful storms or waves of rough seas which also play an important role in the economy. This is the most likely site for deposits in this photograph,
Thus, Beach is the most likely site for deposits in this photograph. So, the correct option is (D).
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1) A plane lands on a runway with a speed of 125 m/s, moving east, and it slows to a stop in 13.0 s. What is the magnitude (in m/s2) and direction of the plane's average acceleration during this time interval?2) Influenced by the gravitational pull of a distant star, the velocity of an asteroid changes from from +20.8 km/s to −17.1 km/s over a period of 2.01 years.(a) What is the total change in the asteroid's velocity? (Indicate the direction with the sign of your answer.)(b) What is the asteroid's average acceleration during this interval? (Indicate the direction with the sign of your answer.)
Answer: 1) a = 9.61m/s² pointing to west.
2) (a) Δv = - 37.9km/s
(b) a = - 6.10⁷km/years
Explanation: Aceleration is the change in velocity over change in time.
1) For the plane:
[tex]a=\frac{\Delta v}{\Delta t}[/tex]
[tex]a=\frac{125}{13}[/tex]
a = 9.61m/s²
The plane is moving east, so velocity points in that direction. However, it is stopping at the time of 13s, so acceleration's direction is in the opposite direction. Therefore, acceleration points towards west.
2) Total change of velocity:
[tex]\Delta v = v_{f}-v_{i}[/tex]
[tex]\Delta v = -17.1-(+20.8)[/tex]
[tex]\Delta v= -37.9[/tex]km/s
The interval is in years, so transforming seconds in years:
v = [tex]\frac{-37.9}{3.15.10^{-7}}[/tex]
[tex]v=-12.03.10^{7}[/tex]km/years
Calculating acceleration:
[tex]a=\frac{-12.03.10^{-7}}{2.01}[/tex]
[tex]a=-6.10^{7}[/tex]
Acceleration of an asteroid is a = -6.10⁷km/years .