a long jumper jumps at a 20 degree angle and attains a maximum altitude of 0.6 m

Answers

Answer 1
Refer to the diagram below.

Assume g = 9.8 m/s² and ignore air resistance.

Let V =  the launch velocity.
The horizontal component of the launch velocity is
Vx = V cos(20°) = 0.9397V m/s
The vertical component of the launch velocity is
Vy = V sin(20°) = 0.342V m/s

Let t =  the time to attain the maximum height of 0.6 m.
At maximum height, the vertical velocity is zero, therefore
Vy - 9.8 t = 0
Vy = 9.8t                   (1)

Also,
Vy*t - (1/2)*9.8*t² = 0.6               (2)

Substitute (1) into (2).
(9.8t)*t - 0.5*9.8*t² = 0.6
4.9t² = 0.6
t = 0.35 s

Therefore,
Vy = 9.8*0.35 = 3.43 m/s
V = Vy/0.342 = 10.0292 m/s
Vx = 0.9397V = 9.4244 m/s

The total time of travel is 2t = 2*0.35 = 0.7 s.
The horizontal travel is
d = 9.4244 *0.7 = 6.597 m

Answers:
The time for the jump is 0.7 s.
The length of the jump is 6.6 m (nearest tenth)

A Long Jumper Jumps At A 20 Degree Angle And Attains A Maximum Altitude Of 0.6 M

Related Questions

Choose the 200 kg refrigerator. Set the applied force to 400 N (to the right). Be sure friction is turned off.What is the net force acting on the refrigerator?

Answers

So, there should be two forces acting on the refrigerator: the applied force and the friction force.

The question mentioned that the friction force was set to zero, so the only effective force now would be the applied force.

We have an applied force of 400 N to the right, this means that:
The magnitude of the net force is 400, directed to the right.

The net force acting on the refrigerator is 400 N to the right

[tex]\texttt{ }[/tex]

Further explanation

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

[tex]\boxed {F = ma }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

mass of refrigerator = m = 200 kg

magnitude of applied force = F = 400 N

magnitude of frictional force = f = 0 N

Asked:

net force = ΣF = ?

Solution:

We will use Newton's Law of Motion to solve this problem as follows:

[tex]\Sigma F = F - f[/tex]

[tex]\Sigma F = 400 - 0[/tex]

[tex]\boxed{\Sigma F = 400 \texttt{ N} }[/tex]

[tex]\texttt{ }[/tex]

We could also calculate the acceleration of the refrigerator as follows:

[tex]\Sigma F = ma[/tex]

[tex]a = \Sigma F \div m[/tex]

[tex]a = 400 \div 200[/tex]

[tex]\boxed{a = 2 \texttt{ m/s}^2 }[/tex]

[tex]\texttt{ }[/tex]

Learn moreImpacts of Gravity : https://brainly.com/question/5330244Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454The Acceleration Due To Gravity : https://brainly.com/question/4189441Newton's Law of Motion: https://brainly.com/question/10431582Example of Newton's Law: https://brainly.com/question/498822

[tex]\texttt{ }[/tex]

Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

Ismarelda has enough money to purchase 27 bottles of root beer for her party at her home she is expecting 4 guests.
What is the largest numbers of bottles of root beer she needs to purchase if she want every one ( including herself ) to have equal number of root beers

Answers

25. That way, (since there are 5 people including herself) everyone will get 5 root beers.

A gymnast practices two dismounts from the high bar on the uneven parallel bars. during one dismount, she swings up off the bar with an initial upward velocity of + 4.0 m/s. in the second, she releases from the same height but with an initial downward velocity of −3.0 m/s. what is her acceleration in each case? how do the final velocities of the gymnast as she reaches the ground differ?

Answers

The working formula to analyze this problem is Vf^2 - Vo^2 = 2gs where Vf = velocity at which the gymnast hits the ground Vo = initial velocity of the gymnast (given as 4 m/s and -3 m/s) g = acceleration due to gravity = 9.8 m/sec^2 (constant) s = height at which gymnast starts her dismount For the first dismount, Vf^2 - (4)^2 = 2(9.8)(s) Vf^2 = 16 + 19.6s and for the second dismount, Vf^2 - (-3)^2 = 2(9.8)(s) Vf^2 = 9 + 1.6s Since "s" is the same for both dismounts, then her final velocity in the first dismount is higher than that of her second dismount.

Find the equivalent capacitance of a 4.20 uf capacitor and an 8.50 uf in parallel and in series

Answers

We are given that there are two capacitors:

4.20 uf

8.50 uf

 

A. In parallel

The equivalent capacitance of capacitors is similar to calculating that of a current, they are added when in parallel. Therefore the equivalent capacitance is:

equivalent capacitance = C1 + C2 + C3 + ...

equivalent capacitance = 4.20 uf + 8.50 uf

equivalent capacitance = 12.70 uf

 

B. In series

When the capacitors are placed in series, the formula for the equivalent capacitance is:

equivalent capacitance = 1 / (1/C1 + 1/C2 + 1/C3 + ...)

equivalent capacitance = 1 / (1/4.20 + 1/8.50)

equivalent capacitance = 2.81 uf

Final answer:

For capacitors in parallel, the equivalent capacitance is the sum of individual capacitances, so we get 12.70 µF. For capacitors in series, we calculate using the reciprocal formula and get approximately 2.79 µF.

Explanation:

To solve this problem, we need to know the formulas to calculate capacitance in a series and parallel circuit. For capacitors in parallel, the total equivalent capacitance (Ceq) is the sum of the capacitors, so Ceq = C1 + C2, where C1 and C2 are the capacitances of your individual capacitors. Here, C1 is 4.20 µF and C2 is 8.50 µF. So, Ceq = 4.20 µF + 8.50 µF which equals 12.70 µF.

For capacitors in a series connection, the total equivalent capacitance is calculated using the reciprocal formula: 1/Ceq = 1/C1 + 1/C2. So, 1/Ceq = 1/4.20 µF + 1/8.50 µF. Solving for Ceq gives us an equivalent capacitance of about 2.79 µF.

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A car changes velocity at a constant acceleration of 2.5m/s to reach 43.7m/s in 2.7 s how fast was the car moving when it began to accelerate?

Answers

The formula we can use in this case is:

v = v0 + a t

where v is final velocity, v0 is initial velocity, a is acceleration and t is time

So finding for v0:

v0 = v – a t

v0 = 43.7 – (2.5) 2.7

v0 =  36.95 m/s

Final answer:

The car was moving at a speed of 36.95 m/s when it began to accelerate.

Explanation:

To find the initial velocity of the car when it began to accelerate, we can use the equation for average velocity:

V = (U + V)/2

Where:

V is the final velocity (43.7 m/s)U is the initial velocity (unknown)

Given that the car changes velocity at a constant acceleration, we can use the kinematic equation:

V = U + at

Where:

a is the acceleration (2.5 m/s²)t is the time (2.7 seconds)

Substituting the known values, we have:

43.7 m/s = U + (2.5 m/s²)(2.7 s)

Simplifying the equation, we find:

U = 43.7 m/s - (2.5 m/s²)(2.7 s)

U = 43.7 m/s - 6.75 m/s

U = 36.95 m/s

Therefore, the car was moving at a speed of 36.95 m/s when it began to accelerate.

The vestibulo-ocular reflex ensures stable vision by moving the eyes in the opposite direction of the head. Please select the best answer from the choices provided T F

Answers

Hi Andijwiltbank

The vestibule-ocular reflex ensures stable vision by moving the eyes in the opposite direction of the head.

Answer: A) True or the first option.

Answer:

The statement is true. The vestibulo-ocular reflex ensures stable vision by moving the eyes in the opposite direction of the head.

Explanation:

The vestibulo-ocular reflex is a reflection of ocular movement that stabilizes the image in the retina during the movement of the head, producing an eye movement in the opposite direction to the movement of the head, conserving the image in the center of the visual field. For example, when the head moves to the right, the eyes move to the left, and vice versa. As there are slight movements of the head at all times, the RVO is very important to stabilize the vision: patients who have damaged RVO find it difficult to read printed media, because they can not stabilize the eyes during small tremors of the head.

The blade on a typical table saw rotates at 3300 revolutions per minute. calculate the linear velocity in miles per hour of one of the teeth at the edge of the 14 inch diameter blade.

Answers

Final answer:

The linear velocity of a tooth at the edge of the 14 inch diameter blade on a typical table saw can be calculated by converting the blade's rotational speed from revolutions per minute to radians per second and using the formula for linear velocity. The calculated value will give the speed of the tooth in miles per hour.

Explanation:

To calculate the linear velocity of a tooth at the edge of the blade, we need to first convert the blade's rotational speed from revolutions per minute to radians per second. We can do this by multiplying the revolutions per minute (rpm) by 2π (since there are 2π radians in one revolution) and dividing by 60 (to convert minutes to seconds). So, the angular velocity of the blade is:

ω = (3300 rpm) x (2π rad/rev) / (60 s/min)

Next, we can use the formula for linear velocity to find the speed of a point on the edge of the blade. The formula is:

v = ω x r

where v is the linear velocity, ω is the angular velocity, and r is the radius. In this case, the blade has a diameter of 14 inches, so the radius is half of that, which is 7 inches or 0.5833 feet. Converting this to miles, we get:

r = 0.5833 ft x (1 mile/5280 ft)

Finally, we can substitute the values into the formula:

v = (3300 rpm) x (2π rad/rev) / (60 s/min) x 0.5833 ft x (1 mile/5280 ft)

Calculating this equation gives the linear velocity of a tooth at the edge of the blade.

What s the density of a gasoline sample that has a volume of 13.3 ml and a mass of 8.86g?

Answers

In this question, you are given the volume( 13.3ml) and the mass(8.86) of the gasoline. You are asked to determine the density. Density is calculated by dividing mass by volume. Then the calculation would be:

Density= mass/ volume
Density= 8.86g/13.3ml= 0.666g/ml

How do constellations differ from other groupings of stars such as galaxies?

Answers

I think a constellation is an area of sky that contains up a picture, object, animal, ect. - a galaxy is a grouping of several million to billion stars, gas and dust, that is gravitationally bound. 

Facial recognition and emotional perception are all tasks that are primarily associated with the __________.
A.
left hemisphere
B.
right hemisphere
C.
cerebral cortex
D.
corpus callosum

Answers

Facial recognition and emotional perception are all tasks that are primarily associated with the right hemisphere.

Answer: B) or the second option.

A car starts from rest and after 5 seconds it is moving at 35 m/s. what is the car's average acceleration?

Answers

Acceleration =velocity /time
=35/5
7

The main reason for positioning many radio telescopes across a large area and combining the signals is

Answers

I think is to produce a higher resolution image. 

a man stands on a flat surface and shoots an arrow vertically into the sky at avelocity of 60 meters per second. calculate the maximum height the arrow reached. what was the velocity of the car when it hit the ground?

Answers

a) Vi= 60 m/s
a= -9.81 m/s^2
Vf= 0m/s^2
d=?

Vf^2=Vi^2+2ad
(0m/s)^2=(60m/s)^2+2(-9.81m/s^2)d
0=3600+(-19.62)d
-3600=(-19.62)d
(-3600)/(-19.62)=(-19.62)d/(-19.62)
d=183.5m

c) Vi= 60 m/s
a= -9.81 m/s^2
Vf= 0m/s^2
t=10s
d=?

d=Vi*t+(1/2)a*t^2
d=60m/s(10s)+(1/2)(-9.81m/s^2)(10s)^2
d=600+(-4.905)(100)
d=600+(-490.5)
d=108.5m
Final answer:

The maximum height reached by the arrow shot vertically at 60 meters per second is 183.67 meters. Upon returning to the ground, the arrow's velocity will be -60 meters per second.

Explanation:

The student is dealing with a problem that involves projectile motion, a topic within physics. To calculate the maximum height the arrow reached when shot vertically into the sky at a velocity of 60 meters per second, we use the kinematic equation that ignores air resistance:

Maximum height (h) = (v^2) / (2g)

where v is the initial velocity and g is the acceleration due to gravity (9.8 m/s^2). Substituting the given values:

h = (60^2) / (2*9.8) = 3600 / 19.6 = 183.67 meters

The maximum height reached by the arrow is 183.67 meters. When the arrow hits the ground, the velocity will be the same as the initial velocity in magnitude but in the opposite direction, so the velocity of the arrow when it hits the ground is -60 meters per second (the negative sign indicates the direction is downward).

Which parameter of a projectile depends on the horizontal as well as the vertical component of velocity of projection?

hang time



range



height



angle of launch

Answers

its the angle of launch, its one of many variables that will effect the over all out come of this test, angle at the ground and launch and well you get the point.

Answer:

range and angle of launch

Explanation:

Let the horizontal component of velocity be VCosθ = Vx and the vertical component of velocity is VSinθ = Vy.

Here , θ be the angle of projection.

So, if we divide the VSinθ by VCosθ, we get the value of Tanθ, i.e., we get the value of angle of launch.

Now we know that the formula fr the range is given by

[tex]R = \frac{V^{2}\times Sin2\Theta }{g}[/tex]

We can write it as

[tex]R = \frac{V^{2}\times 2\times Sin\Theta \times Cos\Theta }{g}[/tex]

Again we rewrite as

[tex]R = \frac{2\times Vx\times Vy}{g}[/tex]

It shows that the range also depends on the horizontal and vertical component of velocity.

1. Some wooden rulers do not start with 0 at the edge, but have it set on a few millimeters. How could this improve the accuracy of the ruler?
2. You find a micrometer that has been badly bent. How would it compare to a new, high-quality meter stick in terms of precision? Its accuracy?
3. Does parallax affect the precision of a measurement that you make? Explain.
4. A box has a length of 18.1 cm and a width of 19.2 cm and it is 20.3 cm tall.
a. what is the volume?
b. how precise is the measure of length? Of volume?
c. How tall is a stack of 12 of these boxes?
d. how precise is the measure of the height of one box? Of 12 boxes?

Answers

1. Some wooden rulers do not start with 0 at the edge, but have it set on a few millimeters because the cut of the woof alters the length of the first measures. Moreover, some objects are difficult to be aligned on the ruler, thus, in order to remove this difficulty, some wooden rulers do not start with 0 at the edge.

2. The micrometer has been badly bent which means that it will not measure properly and will not be perfectly parallel with the surface to be measured. This distortion will  change its precision. The device should be parallel with the surface to be measured in order to maintain its accuracy.

3. No, the parallax will not affect the precision of a measurement that we take as precision  is the closeness of the readings and the readings are already precise, thus taking the reading with the same parallax won't impair the precision but decreases the accuracy.

4. Length of box=18.1 cm, width= 19.2 cm and height=20.3 cm, then the a.volume=length×width×height

=[tex]18.1{\times}19.2{\times}20.3[/tex]

=[tex]7050cm^{3}[/tex]

b. The reading of the length is precise to 0.1 centimeters, while the volume's reading is precise to 10 cm³.

c. The stack of 12 of the boxes has height=[tex]12{\times}19.3[/tex]

=[tex]232cm[/tex]

d.The measure of one box is precise to 0.1 cm, while the measurement of the combined boxes is precise to 1 cm.

an object travels 10 meters in the first second of travel, 10 meters again during the second second of travel, and 10 meters again during the third second of travel. what is the objects accelaration of the object?

A) 0 m/s2
B) 20m/s2
C) 10m/s2
D) 30m/s2

Answers

Final answer:

The object's velocity remains constant as it covers the same distance each second, implying that the acceleration is zero (0 m/s^2). So the correct option is A.

Explanation:

The question relates to constant velocity and acceleration and to identifying the acceleration of an object moving 10 meters each second for three seconds. Since the object travels the same distance in each second, it means that its velocity remains constant. Therefore, the rate of change of velocity, which is acceleration, is zero. The correct answer to the question is A) 0 m/s2. To understand this, consider the definition of acceleration: a change in velocity over some time. Since there is no change in velocity here (the object keeps moving at a constant rate), there is no acceleration.

What are the units of ppb mass mass/mass?

Answers

Mass concentration is typically represented as mass units of chemical per million units total mass or parts per million (ppmm), where the subscript 'm' indicates mass). The conversion factors are 109 and 1,012 for parts per billion (ppbm) and parts per trillion (pptm), respectively. hope this helps 

The __________ houses the nucleus and other cell parts.
A.
neural receptor
B.
myelin sheath
C.
cell body
D.
axon terminal

Answers

C.  cell body encapsulate the nucleus and the other parts of a cell

The part of the neuron that houses the nucleus and other cell parts is the cell body that is in Option c, as the cell body, also known as the soma, is a central part of a eukaryotic cell.

What is the cell body?

The cell body includes the nucleus and other organelles like the mitochondria, endoplasmic reticulum, Golgi apparatus, and cytoplasm. The cell body is the main site for the metabolic and biochemical processes of the cell, and it is responsible for maintaining the integrity of the cell and regulating its activities. In the case of neurons, the cell body also contains the genetic information of the neuron in the form of DNA, which is used to synthesize the proteins and other molecules needed for the growth, repair, and maintenance of the neuron.

Hence, the part of the neuron that houses the nucleus and other cell parts is the cell body that is in Option c.

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which of the following descriptions could apply to an accelerating car

Answers

There are many descriptions that can describe an accelerating vehicle, some include fast, quickly, Speedster, amount others. Cars accelerate fast for many reasons including but not limited to racing, running from the police, and racing to the scene of a crime.

Which action occurs during a nuclear change?

A.No energy is involved in the reaction.

B.The nucleus disappears.

C.Matter changes from one state to another.

D.Elements are created that differ from the reactants.

Answers

the answer is d bro f  you mean

henry was given this cool toy for his birthday;he can use it to tell how fast someone is going at that moment in time. the toy must calculate the persons....
average velocity
average speed
acceleration
instantaneous speed

Answers

Instantaneous Speed because its happening a that moment in time or instantly. Average speed is wrong because that's the total amount of time you went and the averages of the different speeds you went. So its the last one! 

Answer:

Instantaneous speed

Explanation:

It is given that, Henry was given this cool toy for his birthday. He can use it to tell how fast someone is going at that moment in time. The toy must calculate the person's instantaneous speed.

Instantaneous speed of an object is defined as the speed at a particular instant. Mathematically, it is given by :

[tex]v= \lim_{\Delta t \to 0}(\dfrac{\Delta x}{\Delta t})[/tex]

Instantaneous speed is a scalar quantity. It does not involve any direction.

So, Henry must calculate the person's instantaneous speed. Hence, the correct option is (d).

Ingrid kicks a football with an initial velocity of 12 m/s at an angle of 45 degrees relative to the ground. What is the horizontal component of the initial velocity?

Answers

Vx=v*cos()
=12*cos(45)
=12*0.71
Vx=8.52ms^-1

Answer:

8.5 m/s

Explanation:

Two planes leave wichita at noon. one plane flies east 30 mi/h faster than the other plane, which is flying west. at what time will they be 1200 mi apart?

Answers

The time they will be 1200 mi apart is [tex]\(12:00 PM + \frac{1200}{2S + 30}\)[/tex] hours.


Let's denote the speed of the slower plane as [tex]\(S\) (in mi/h)[/tex]. The faster plane, which is flying east, will have a speed of [tex]\(S + 30\) mi/h[/tex].

The relative speed between the two planes is the sum of their individual speeds:

[tex]\[ \text{Relative speed} = S + (S + 30) = 2S + 30 \, \text{mi/h} \][/tex]

Now, we know that the planes leave at noon, and we want to find the time it takes for them to be 1200 mi apart. Let t be the time in hours.

The distance traveled by the slower plane in t hours is [tex]\(S \cdot t\)[/tex], and the distance traveled by the faster plane is [tex]\((S + 30) \cdot t\)[/tex]. The sum of these distances is the total distance between the planes:

[tex]\[ S \cdot t + (S + 30) \cdot t = 1200 \][/tex]

Combine like terms:

[tex]\[ 2S \cdot t + 30 \cdot t = 1200 \][/tex]

Factor out t:

[tex]\[ t(2S + 30) = 1200 \][/tex]

Now, solve for t:

[tex]\[ t = \frac{1200}{2S + 30} \][/tex]

We know that when t is the time, it should be in hours, and we want to find at what time they will be 1200 mi apart. If they leave at noon, the time will be [tex]\(12:00 PM\) + \(t\) hours[/tex].

So, the time they will be 1200 mi apart is [tex]\(12:00 PM + \frac{1200}{2S + 30}\)[/tex] hours.

The outer planets are mostly formed of the lighter elements, hydrogen and helium, because these elements condense at a __________ temperature, further from the Sun.
1. Lower
2.Equal
3.Higher

Answers

lower is the correct answer

the correct answer is A  *LOWER*

What is the primary reason that we launch x-ray telescopes into space rather than building them on the ground?

Answers

The primary reason is that on earth, x-rays are partly blocked by the atmosphere because x rays do not penetrate into the atmosphere of earth, so that's why there is a need to bring that telescope to space.
 And also on ground there is a blurring on telescope that is caused by atmospheric turbulence and on space it reduces that effect. The wavelength regions of X-rays and ultraviolet rays cannot be studied with telescopes that are on the ground.

Answer:

X-Rays from space do not reach the ground.

Explanation:

This map shows the Mid-Atlantic Ridge, which spreads approximately 2.5 cm/year. What is the rate of the spread in m/month?

A) .002
B) .3
C) 216
D) 777600

Answers

  2.5 cm                 1 meter                  1 Year
_______       x     ________      x     ________        = 0.002 m/month
  Year                     100 cm                12 mo.s

The answer is A 00.2

A 590-kg rocket is at rest on the launch pad. what upward thrust force is needed to accelerate the rocket uniformly to an upward speed of 28 m/s in 3.3 s?

Answers

Final answer:

To find the upward thrust force needed to accelerate the rocket, we use Newton's second law of motion. The upward thrust force needed is approximately 4989.2 N.

Explanation:

To find the upward thrust force needed to accelerate the rocket, we can use Newton's second law of motion, which states that force is equal to mass times acceleration. In this case, the mass of the rocket is 590 kg and the acceleration is the change in velocity divided by the time taken.

Using the formula acceleration = (final velocity - initial velocity) / time, we can calculate the acceleration:

acceleration = (28 m/s - 0 m/s) / 3.3 s =  8.48 m/s²

Now that we have the acceleration, we can calculate the upward thrust force:

force = mass x acceleration = 590 kg x 8.48 m/s² = 4989.2 N

Therefore, the upward thrust force needed to accelerate the rocket uniformly to an upward speed of 28 m/s in 3.3 s is approximately 4989.2 N.

To accelerate the 590-kg rocket uniformly to an upward speed of 28 m/s in 3.3 s, a thrust force of approximately 5010.2 N is required, calculated using Newton's second law.

To determine the upward thrust force needed to accelerate the rocket uniformly, we can use Newton's second law, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma).

First, calculate the acceleration (a) of the rocket using the kinematic equation:

v = u + at

where:

v is the final velocity (28 m/s),

u is the initial velocity (0 m/s, as the rocket is at rest),

a is the acceleration, and

t is the time (3.3 s).

Rearrange the equation to solve for acceleration:

a = (v - u) / t

Substitute the values:

a = (28 m/s - 0) / 3.3 s ≈ 8.48 m/s^2

Now, use Newton's second law to find the force (F):

F = ma

F = 590 kg × 8.48 m/s^2 ≈ 5010.2 N

The density of water is 1.00 g/cm3. What is its density in kg/m3?

Answers

1 g = 1 ÷ 1000 kg
= 0.001 kg

1 cm³ = 1 ÷ 100 ÷ 100 ÷ 100 m³
= 0.000001 m³

1 g/cm³ = 1 g / 1 cm³
= 0.001 kg / 0.000001 m³
= 1000 kg/m³

The density is 1000 kg/m³.

The density of water is 1.00 g/cm3, then in kilograms per centimeter cube unit, its density would be 1000 kg/m³.

What is density?

It can be defined as the mass of any object or body per unit volume of the particular object or body. Generally, it is expressed as in gram per cm³ or kilogram per meter³.

As given in the problem we have to calculate the density of water in kilogram per centimeter cube,

1 cm³ = 1×10⁻⁶ m³

1 gram = 1×10⁻³ kg

The density of the water= mass of the water /volume of the water

                                 =1.00 g/cm3

                                 =1×10⁻³ kg /1×10⁻³ kg

                                 =1000 kg/m³

Thus, if the density of water is 1.00 g/cm3, then in kilograms per centimeter cube unit its density would be 1000 kg/m³.

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A car travels 20 km north. It then turns around and travels 30 km south. What is the car's displacement?

Answers

10Km South. Displacement is the total distance the subject is from the starting point.

Viteza de aterizare a unui avion este de 45 m/s iar timpul deplasarii pe pista aerodromului este egal cu 30 s

Answers

Final answer:

The landing speed of the airplane is 1.5 m/s.

Explanation:

The question is asking for the landing speed of an airplane given its landing time and the distance traveled on the airstrip. To find the landing speed, we can use the formula:



Speed = Distance / Time



Plugging in the values given:



Speed = 45 m/sTime = 30 s



Using the formula:



Landing Speed = 45 m/s / 30 s = 1.5 m/s



Therefore, the landing speed of the airplane is 1.5 m/s.

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Final answer:

The question pertains to the deceleration of an airplane after it lands, specifically calculating the final velocity given an initial velocity and a constant deceleration over a certain time span.

Explanation:

The student's question is related to the deceleration of an airplane after landing. An airplane that lands with an initial velocity of 70.0 m/s, and then decelerates at 1.50 m/s2 for 40.0 s, will have a final velocity that can be calculated using the following formula: final velocity = initial velocity + (acceleration × time). Here, the acceleration is negative because it is opposite the direction of motion (deceleration).

In the example given, if an airplane lands with a velocity of 70.0 m/s and decelerates at 1.50 m/s2, the final velocity after 40 seconds can be found by:

Final Velocity = 70.0 m/s - (1.50 m/s2 × 40.0 s) = 70.0 m/s - 60.0 m/s = 10.0 m/s. The plane would slow down to 10.0 m/s before heading to the terminal.

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