It is a general exercise that we do before practicing sports activities

worm-up or cool down

Answers

Answer 1

It is called a warm-up.

Answer 2

Answer:

We do warm-ups before we exercise. We do cool-downs after we exercise.

Explanation:

Warming up can help you get ready to do the actual exercise, so you are prepared.


Related Questions

what will create static electricity?

Answers

Answer:

Friction between two objects causes a transfer of electrons from one object to the other.

Explanation:


the actual ratio of weight to mass on earth is 9:8 N:1 kg what is the weight on earth of
an object whose mass is 2 kg

Answers

Answer:

The object weighs 19.6 N

Explanation:

Weight

It's the force exerted on a body by gravity.

This is often expressed in the formula

W = mg

Where

W = weight

m = mass of the object

g  = gravitational acceleration

The ratio of weight to mass on Earth is 9.8 N: 1 kg. This gives us the value of  [tex]g=9.8 \ m/s^2[/tex].

An object with a mass of m = 2 kg has a weight of:

[tex]W = 2\ Kg *9.8 \ m/s^2[/tex]

W = 19.6 N

The object weighs 19.6 N

Neon has 3 naturally occurring isotopes, Neon-20, Neon-21, and Neon-22. Neon's average atomic mass on the periodic table is 20.179 amu. Based on this information which isotope is most abundant? Explain why abundance of an isotope matters when calculating the average atomic mass of an isotope. ​

Answers

Neon - 20 is the most abundant. Abundance matters as the average atomic mass has to take into account the amount of one isotope. The more abundance the more mass, the closed the average atomic mass is to the mass of that one isotope

So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, this picture is not quite complete! You are on the surface of the Earth, which is rotating. While answering the following questions you may ignore the Earth’s motion around the sun, galactic center, etc. (though that is another interesting question!) (a) What is the magnitude of the acceleration of a person sitting in a chair on the equator? (b) At the equator, is your mass times the gravitational acceleration of the Earth greater than, less than, or equal to the normal force exerted on you by the chair you are sitting on? Explain. (c) A classmate of yours asks you why we have ignored this acceleration for the whole first term of physics. "Is everything we’ve learned a lie?" they ask. Ease their fears by calculating the percentage difference between the normal force from the chair and your weight while sitting on equator. (d) The latitude of Corvallis is 44.4˚. What is your acceleration while sitting in your chair?

Answers

Answer:

a) [tex]a=33.73mm/s^{2}[/tex]

b) mg>N

c) [tex]\%_{change}=0.343\%[/tex]

d) [tex]a=24.07mm/s^{2}[/tex]

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

[tex]a_{c}=\omega ^{2}r[/tex]

where:

[tex]\omega=\frac{2\pi}{T}[/tex]

we know the period of rotation of the earth is about 24 hours, so:

[tex]T=24hr*\frac{3600s}{1hr}=86400s[/tex]

so we can now find the angular speed:

[tex]\omega=\frac{2\pi}{86400s}[/tex]

[tex]\omega=72.72x10^{-6} rad/s^{2}[/tex]

So the centripetal acceleration will be:

[tex]a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)[/tex]

which yields:

[tex]a_{c}=33.73mm/s^{2}[/tex]

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

[tex]\sum F=0[/tex]

so we get that:

[tex]N-mg+ma_{c} = 0[/tex]

and solve for the normal force:

[tex]N=mg-ma_{c}[/tex]

In this case, we can clearly see that:

[tex]mg>mg-ma_{c}[/tex]

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

[tex]mg=(60kg)(9.81m/s^{2})=588.6N[/tex]

and let's calculate the normal force:

[tex]N=m(g-a_{c})[/tex]

[tex]N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})[/tex]

N=586.58N

so now we can calculate the percentage change:

[tex]\%_{change} = \frac{mg-N}{mg}x100\%[/tex]

so we get:

[tex]\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%[/tex]

[tex]\%_{change}=0.343\%[/tex]

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

[tex]cos \theta = \frac{AS}{h}[/tex]

In this case:

[tex]cos \theta = \frac{r}{R_{E}}[/tex]

so we can solve for r, so we get:

[tex]r= R_{E}cos \theta[/tex]

in this case we'll use the average radius of earch which is 6,371 km, so we get:

[tex]r = (6371x10^{3}m)cos (44.4^{o})[/tex]

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

[tex]a=\omega ^{2}r[/tex]

[tex]a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m[/tex]

[tex]a=24.07 mm/s^{2}[/tex]

In this exercise we have to use the knowledge of mechanics to solve the magnitude of the acceleration and the acceleration of gravity, in this way we find that:

a)[tex]a=33.73 mm/s^2[/tex]

b)[tex]mg>N[/tex]

c) [tex]Change=0.343%[/tex]

d) [tex]a=24.07 mm/s^2[/tex]

Then calculating from the information given in the text;

a)In this case, we know the centripetal acceleration is given by the following formula:

[tex]a_c=w^2r\\w=\frac{2\pi}{T} \\T=24*\frac{3600}{60} = 86400 s[/tex]

so we can now find the angular speed:

[tex]w=\frac{2\pi}{86400} \\w=72.72*10^{-6} rad/s^2[/tex]

So the centripetal acceleration will be:

[tex]a_c=(72.72*10^{-6}rad/s^2)^2(6478*10^3m)\\a_c=33.73mm/s^2[/tex]

b)So we can do a sum of forces in equilibrium:

[tex]\sum F=o\\N-mg+ma_c=0\\N-mg+ma_c=0\\N=mg-ma_c\\mg>mg-ma_c\\mg>N[/tex]

This exist cause the centripetal increasing speed happen pulling united states of america upwards, that will create the size of the normal force tinier than the result or goods created of the bulk times the increasing speed of importance.

c) So let's calculate our weight and normal force:

[tex]mg=(60)(9.81)=588.6N\\N=m(g-a_c)\\N=(60)(9.81-33.73*10^{-3})\\N=586.58 N[/tex]

So now we can calculate the percentage change:

[tex]\%change= \frac{mg-N}{mg}*100\% \\=\frac{588.6-586.58}{588.6}*100\% \\=0.343\%[/tex]

d) There, we can see that the radius can be found by using the cos function:

[tex]cos\theta=\Delta S/h\\cos\theta=r/R_E\\r=R_E cos\theta\\r=(6371*10^3)cos(44.4)\\r=4,551.91 km[/tex]

And now we can calculate the acceleration at that point:

[tex]a=w^2r\\a=(72.72*10^{-6})^2(4,551.91*10^3)\\a=24.07 mm/s^2[/tex]

See more about acceleration at brainly.com/question/2437624

a body is undergoing non uniform circular motion work done by tangential force on body is​

Answers

Answer:

non-zero

Explanation:

The tangential force acts as a circular moment on the body and increases its velocity, and also increases its kinetic energy.

In unequal circular movement work done on the object is non-zero. If the object's velocity increases in the tangential direction, the force is acting in the same direction.

a crane lifts a 200 kilogram weight to a height of 50 meters. what would be the gravitational potential energy ?

Answers

Answer:

98,000 J

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

GPE = 200 × 9.8 × 50

We have the final answer as

98,000 J

Hope this helps you

the orbit of an object is caused by what two factors *
1. The initial force of motion and an inward-acting force
2. 2 unbalanced forces
3, A push or pull on an object

Answers

The initial force of motion and inward acting force

Wind is ________ rushing in to fill an area of low pressure.

Answers

I think is local winds

Wind is the stream of air rushing in to fill an area of low pressure. Winds are  generated by the differential distribution of temperature resulting in a pressure difference.

What are winds?

The natural movement of air or other gases in relation to a planet's surface is referred to as wind. There are many different sizes of winds, ranging from thunderstorm flows that last only a few minutes to local breezes.

Large-scale atmospheric circulation is primarily brought on by the planet's rotation and the difference in temperature between the equator and the poles (Coriolis effect).

Thermal low circulations over topography and high plateaus can cause monsoon circulations in the tropics and subtropics. Local winds can be defined in coastal regions by the sea breeze/land breeze cycle; in regions with varying terrain, mountain and valley.

Find more on winds:

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Consider two object has mass m1=5.0kg m2=0.5kg and the specific heat c1=c2=4200j/kg oc respectively assume that first object has temperature 20oc and second object has temperature 90oc After they place these two object touch each other determine the mixture temperature of the substance

Answers

Answer: [tex]T=26.36^{\circ}C[/tex]

Explanation:

Given

mass of the first object [tex]m_1=5\ kg[/tex]

mass of the second object [tex]m_2=0.5\ kg[/tex]

specific heat of the two is [tex]c_1=c_2=4200\ J/kg/^{\circ}C[/tex]

The temperature of the first [tex]T_1=20^{\circ}[/tex]

The temperature of the second [tex]T_2=90^{\circ}[/tex]

Heat flows from high temperature to low temperature

Suppose T is the common temperature

[tex]m_1c_1(T-T_1)=m_2c_2(T_2-T)\\5\times 4200\times (T-20)=0.5\times 4200\times (90-T)\\5T-100=45-0.5T\\5.5T=145\\T=26.36^{\circ}C[/tex]

Is the force exerted by the scale on the superball greater than, less than, or equal to the force exerted by the scale on the clay

Answers

Answer: A glob of very soft clay is dropped from above onto a digital scale. The clay sticks to the scale on impact. A graph of the clay's velocity vs. time, clay (t), is given, with the upward direction defined as positive.

The experiment is then repeated, but instead of using the clay glob, a superball with identical mass is dropped from the same height onto the scale.

Both the clay and the superball hit the scale 2.9s after they are dropped. Assume that the duration of the collision is the same in both cases and the force exerted by the scale on the clay and the force exerted by the scale on the superball are constant.

Explanation:

A car is traveling to the right with a speed of 2.0 m/s on an icy road when the brakes are applied. The car begins sliding with constant acceleration for 3.0 m until it comes to a stop. How long does it take the car to slide to a stop? Answer using a coordinate system where rightward is positive. Round the answer to two significant digits.

Answers

Answer:

t = 3.0s

Explanation:

U = 2.0m/s , V = 0 (stop) , S = 3m , t =?

From V^2 - U^2 = 2aS

=) a = -4/6 = -0.667m/s^2

Now again by V-U = at

We have t = -U/a = 2/0.667 = 3s

Required time is t = 3.0s

Unit Test
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TIME REMAININO
01:47:35
Which best describes the energy of a sound wave as it travels through a medium?
It increases
It decreases
It remains the same
It depends on the medium

Answers

Answer:

It depends on the medium is answer.

Explanation:

I hope it's helpful!

The energy of sound waves depends on the nature of medium it travels through. As the speed of the wave increases in the medium its energy increases. Hence, option d is correct.

What are sound waves ?

Sound waves are type of mechanical waves passing through a medium. Sound waves are longitudinal waves hence, the oscillation of particles is in the same direction of of the wave propagation.

The energy of a wave is directly proportional to the frequency of the wave and inversely proportional to the wavelength. As the density of the medium increases, the speed of the wave decreases and thereby energy too.

Sound waves travels with higher speed through solids. As their energy increases, the waves moves with higher speed. Hence, the energy of sound waves depends on the nature of the medium.

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Car A is traveling at 18.0 m/s and car B at 25.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with an acceleration of 1.80 m/s2. How long does it take car A to overtake car B

Answers

Answer:

car A reaches and immediately overtakes the car B at 22.56 s.

Explanation:

After car A accelerate at 1.8 m/s2, it travels a distance x(A) and car B will have travels a distance x(B), let's recall that the initial distance between them is 300 m, so we have:

[tex]x_{A}=300+x_{B}[/tex]

Now, we can rewrite this equation in terms of speed and time

[tex]V_{iA}t+\frac{1}{2}at^{2}=300+V_{iB}t[/tex]

Where:

V(iA) is the initial speed of car A

V(iB) is the initial speed of car B

t is the time when car A reaches the car B

a is the acceleration

[tex]18t+\frac{1}{2}1.8t^{2}=300+25t[/tex]

[tex]0.9t^{2}-7t-300=0[/tex]  

Solving this quadratic equation for t, and taking just the positive value, we will have:

t=22.56 s

Therefore, car A reaches and immediately overtakes the car B at 22.56 s.

I hope it helps you!

A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of the piston and attached shaft are 24.5 kg and 0.5 kg, respectively. The piston diameter is 10 cm. The local atmospheric pressure is 1 bar. T

Answers

Answer:

The force acting on the shaft is 1324.75 N

Explanation:

Given that,

Cross sectional area of shaft [tex]A_{sh}=0.8\ cm^{2}[/tex]

Gas pressure [tex]P_{gas}=3\ bar=3\times10^5\ Pa[/tex]

Total mass M=24.5+0.5=25 kg

Diameter of piston, d=10 cm

We need to calculate the cross section area of piston

Using formula of area

[tex]A_{p}=\pi\times\dfrac{d^2}{4}[/tex]

Put the value into the formula

[tex]A_{p}=\pi\times\dfrac{0.1^2}{4}[/tex]

[tex]A_{p}=0.00785\ m^{2}[/tex]

We need to calculate the weight of the piston and shaft

Using formula of weight

[tex]W=mg[/tex]

Put the value into the formula

[tex]W=25\times9.81[/tex]

[tex]W=245.25\ N[/tex]

We need to calculate the force due to gas pressure

Using formula of force

[tex]F_{gas}=P_{gas}\times A_{p}[/tex]

Put the value into the formula

[tex]F_{gas}=3\times10^{5}\times0.00785[/tex]

[tex]F_{gas}=2355\ N[/tex]

We need to calculate the force due to atmospheric pressure,

Using formula of force

[tex]F_{atm}=P_{atm}\times A_{p}[/tex]

Put the value into the formula

[tex]F_{atm}=1\times10^5\times0.00785[/tex]

[tex]F_{atm}=785\ N[/tex]

We need to calculate the force acting on the shaft

from free body diagram

[tex]F_{gas}=F+F_{atm}+W[/tex]

Put the value into the formula

[tex]2355=F+785+245.25[/tex]

[tex]F=2355-785-245.25[/tex]

[tex]F=1324.75\ N[/tex]

Hence, The force acting on the shaft is 1324.75 N

The complete question is :

The figure (attached) shows a gas contained in a vertical piston-cylinder assembly. A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of the piston and attached shaft are 24.5kg and 0.5kg respectively. The piston diameter is 10cm. The local atmospheric pressure is 1 bar. The piston moves smoothly in the cylinder and g=9.81 m/s2.

Solution :

Given

The cross sectional area of the shaft,  [tex]$A_{s} = 0.8 \ cm^2$[/tex]

Gas pressure, [tex]$P_{gas} = 3 \ bar = 3 \times 10^5 \ Pa$[/tex]

The total mass, m = 24.5 + 0.5

                             = 25 kg

Diameter of the piston, d = 10 cm

The cross sectional area of the piston, [tex]$A_{p} = \frac{\pi}{4} \times (0.1)^2$[/tex]

Weight of the piston and shaft, W = mg

                                                        = 25 x 9.81

                                                       = 245.25 N

Force due to the gas pressure,

[tex]$F_{gas} = P_{gas} \times A_{p}$[/tex]

[tex]$F_{gas} = 3 \times 10^5 \times 0.00785 $[/tex]

       = 2355 N

Force due to atmospheric pressure,

[tex]$F_{atm} = P_{atm} \times A_{p}$[/tex]

        [tex]$= 1 \times 10^5 \times 0.00785$[/tex]

        = 785 N

Now [tex]$F_{gas} = F + F_{atm} + W$[/tex]

     ⇒ 2355 = F + 785 + 245.25

     ⇒ F = 1324.75 N

Acid rain is an example of which type of chemical weathering?

Answers

The answer is Carbonic acid
An example of a chemical weathering is Carbonic acid

. A flute is designed so that it plays a frequency of 261.6 Hz, middle C, when all the holes are covered and the temperature is 20.0o C. (a) Consider the flute to be a pipe, open at both ends, and find its length, assuming that the middle-C frequency is the fundamental. (b) A second player, nearby in a colder room, also attempts to play middle C on an identical flute. A beat frequency of 3.00 Hz is heard. What is the temperature of the room

Answers

Answer:

Explanation:

The first case relates to open end pipe with fundamental frequency . Let the length of flute be l . The length vibrating column  is l .

λ = v / n where n is frequency , v is velocity of light and λ is wavelength of sound produced .

λ = 343 / 261.6

= 1.31 m .

For fundamental frequency

λ / 2 = l

l = λ / 2

= 1.31 / 2

= .655 m

= 65.5 cm .

b )

Nearby colder room will create lower frequency because velocity of sound will be smaller .

frequency of note produced = 261.6 - 3 = 258.6 Hz .

velocity of sound v = n λ

= 258.6 x  1.31 m

= 338.76 m /s

decrease of velocity = 343 - 338.76 = 4.24 m /s

1 degree change in temperature produces a change of .61 m/s change in velocity .

decrease in temperature = 4.24 / .61

= 7⁰C

Temperature of colder room = 20 - 7 = 13⁰C .

An object traveling a circular path of radius 5 m at constant speed experiences an acceleration of 3 m/s2. If the radius of its path is increased to 10 m, but its speed remains the same, what is its acceleration?

A. 0.3 m/s2
B. 1.5 m/s2
C. 6 m/s2
D. 12 m/s2

Answers

........The answer is B

Answer:

1.5 m/s2

Explanation:

Gizmo Explanation: The acceleration of an object in uniform circular motion is inversely proportional to the radius of the motion. This means that, if the radius is multiplied by a number, the acceleration is divided by that same number. In this case, the radius of the path doubles, from 5 m to 10 m. From this you can conclude that the acceleration must be half of the original value, or 1.5 m/s2.

An object is released from a rocket moving upwards at a speed of 8.9 m / s.​
What is the downward acceleration of this object?

Answers

Answer:

9.81 m/s²

Explanation:

It is given that the upward speed of the object is 8.9 m/s.

An object is released.

Now, we know that if the object is released then it will move under the control of gravity which means that the initial velocity of the rocket will not affect the acceleration of the object.

Hence, the downward acceleration of the object will be equal to g i.e. 9.81 m/s².

A 1020 kg car is pulling a 365 kg trailer. Together, the car and trailer have an acceleration of 2.21 m/s2 directly forward. (a) Determine the net force on the car. N forward (b) Determine the net force on the trailer. N forward

Answers

Answer:

a. 2936.2 N b. 773.8 N

Explanation:

Let m = mass of car = 1020 kg and m' = mass of trailer and a = acceleration = 2.12 m/s².

a. Determine the net force on the car. N forward

Since the car pulls both itself and the trailer, the combined mass is m + m' and the net force F on the car is F = (m + m')a

= (1020 kg + 365 kg)2.12 m/s²

= 1385 kg × 2.12 m/s²

= 2936.2 N

b. Determine the net force on the trailer. N forward

The net force F' on the trailer is F = m'a

= 365 kg × 2.12 m/s²

= 773.8 N

The age and crisis of the stage trust vs. mistrust

Answers

Trust vs. mistrust is the first stage in Erik Erikson's theory of psychosocial development. This stage begins at birth continues to approximately 18 months of age. ... If the care has been inconsistent, unpredictable and unreliable, then the infant may develop a sense of mistrust, suspicion, and anxiety.
Hope it is helpful

A 10248 kg car is pulled by a low tow truck that has an acceleration of 2.0m/s what is the net force on the car

Answers

Answer: 20496N

Explanation:

The formula to calculate the net force will be given as:

Net force = Acceleration × Mass

Since 10248 kg car is pulled by a low tow truck that has an acceleration of 2.0m/s, the net force would be:

= 10248 × 2

= 20496N

What is the easiest way to add energy to matter.

Answers

Answer:

Explanation:

Heat is probably the easiest energy you can use to change your physical state. The atoms in a liquid have more energy than the atoms in a solid.

Answer:

The easiest way is to heat matter which raises its internal energy andso the mass as well. Another possibility is to accelerate matter as in case of protons or heavy ions in the LHC, Protons ,for example , are accelerated to the energies of 6500 GeV which means increase of the rest mass by a factor of almost 7000.

Question 1(Multiple Choice Worth 2 points)
(06.01 LC)

How can the luster of a mineral be described? by the way this is Science

A Bland
B Dim
C Dull
D Twinkle

Answers

this would be a hard one but if i was you i would go C as in “Dull”
Dull is your answer

how are all the spheres similar

Answers

All the spheres are similar because they have no edges

As an example, a 4.00-kg aluminum ball has an apparent mass of 2.10 kg when submerged in a particular liquid: calculate the density of the liquid.

Answers

Answer:

1.2825 * 10^3 kg/m³

Explanation:

Given that :

Mass of aluminum ball (m1) = 4kg

Apparent mass of ball (m2) = 2.10 kg

Density of aluminum (d1) = 2.7 * 10^3 kg/m³

Density of liquid (d2) =?

Using the relation :

d1 / d2 = m1 / (m2 - m1)

(2.7 * 10^3) / d2 = 4 / (4 - 2.10)

2700 / d2 = 4 / 1.9

4 * d2 = 2700 * 1.9

4 * d2 = 5130

d2 = 5130 / 4

d2 = 1282.5 kg/m³

Hence, density of liquid = 1.2825 * 10^3

Describe how electric potential energy, kinetic energy, and work change when two charges of opposite sign are placed near each other.

Answers

Answer:

As opposites attract, q is pulled to Q. And that force from Q is working on q, force over distance. Which means the potential energy q started with is being converted into kinetic energy. q is accelerating and picking up speed.  

Explanation:

A 75kg encounters 10n of friction and slides down a hall with an acceleration of 3.60 find the coefficients of kinetic between the box and floor

Answers

Answer:

Explanation:

Given,

Mass of block = 75kg

Force of friction=10N

Acceleration of box = 3.60m/s^2

Acceleration due to gravity = 9.8m/s^2

Let tetha represent the angle of inclination.

Wherefore, we have mgsinθ - force of friction = ma........ 1

Substitute the values into equation 1

75×9.8×sinθ -10 = 75×3.60

735sinθ = 10+270

735sinθ = 280

Divide both sides by 735

Sinθ = 280/735

Sinθ = 0.3809

θ = sin^-1 0.3809

θ = 22.389°

We can now solve for the coefficient of friction by considering the formula:

Fs = us×R

Where R = mgcosθ

Fs = 10

10 = Us×75×9.8×cos22.389

10 = Us× 735×0.9246

10 = Us × 679.595

Divide both sides by 679.595

10/679.595 = Us

Us = 0.01471

Hence the coefficient of friction is 0.01471

1. To get to school, a girl walks 1 km North in 15 minutes. She
then walks 200 m South-west in 160 seconds.
What is the girl's average velocity for her walk to school?

Answers

The girl's average velocity is 1.13 m/s

The formula used to determine average velocity is:

Velocity = distance/ timeV = D/t

This means to determine the girl's average velocity is necessary to find out the total distance and the total time.

Distance:

1 km north and 200 m Southwest

As you can see there are two different distances and they are in different units (meters vs kilometers), let's convert the kilometers to meters so both distances are in the same unit.

1  x 1000 = 1000 (1 km = 1000 meters)

Add the two distances:

1000 meters + 200 meters = 1200 meters

Time

15 minutes and 160 seconds

Let's convert the minutes into seconds:

15 x 60 = 900

Add the time:

900 + 160 = 1060

Finally, calculate the average velocity:

1200 meters / 1060 seconds = 1.13 meters per second

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How much work is done by a person who
pushes a cant with a force of 200 newtons if
the cart moves 20 meters in the direction
of the fonce.

Answers

it is 82729cm still

Stoplight parrotfish:

Answers

Is this what you were meaning?
Other Questions
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