How is it possible for man made things to move?

Answers

Answer 1

Explanation:

Natural things are all the elements present in the world. Whatever is there because of nature, like petrolium and coal which is developed over thousands and millions of year process. There are many more, like even living things do come inside the circle of natural things, because man can not make life we can alter it examples can be found on your dining table, fruits and vegetables. Over the period of time humans have altered the genes of fruits and vegetables to grow the way they want example seedless grapes, seedless banana etc.


Related Questions

Potential energy can be converted into kinetic energy, a good example of this is when a pole-vaulter bends the pole during a leap. When the pole is bent the most, does it store elastic or gravitational potential energy? Question 7 options: Only gravitational potential energy because that is more powerful than elastic potential energy. It has neither elastic or gravitational potential energy. The pole stores elastic potential energy when the pole is bent because its shape is change from its natural shape and it will want to go back to its original form, just like a spring or stretched elastic material. The pole stores gravitational potential energy because it is bent from its natural shape when off the ground.

Answers

Answer:

The pole stores elastic potential energy when the pole is bent because its shape is change from its natural shape and it will want to go back to its original form, just like a spring or stretched elastic material.

Question:

Why did you lie about being in college?

Answer:

The pole stores gravitational potential energy because it is bent from its natural shape when off the ground.

Explanation: i took the test

True or False: The negative oxygen end of one water molecule is attracted
to the positive hydrogen end of another water molecule to form a
hydrogen bond.

Answers

Answer: I believe that it is true

Explanation:

The two general types of air pollution are _____ and _____.

Answers

Answer:

The two general types of air pollution are _____ and _____.

answer: gases and particles

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A(n) 131 g ball is dropped from a height
of 61.1 cm above a spring of negligible mass.
The ball compresses the spring to a maximum
displacement of 4.82755 cm.
The acceleration of gravity is 9.8 m/s2.
h
х
illele
dellee
Calculate the spring force constant k.

Answers

Answer:

26.59 N/m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 131 g

Extention (e) = 4.82755 cm

Acceleration due to gravity (g) = 9.8 m/s²

Spring constant (K) =?

Next, we shall convert 131 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

131 g = 131 g × 1 Kg / 1000 g

131 g = 0.131 Kg

Thus, 131 g is equivalent to 0.131 Kg.

Next, we shall the force exerted by the ball on the spring. This can be obtained as follow:

Mass (m) = 0.131 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = ma

F = 0.131 × 9.8

F = 1.2838 N

Next, we shall convert 4.82755 cm to metre (m)

This can be obtained as follow:

100 cm = 1 m

Therefore,

4.82755 cm = 4.82755 cm × 1 m / 100 cm

4.82755 cm = 0.0482755 m

Thus, 4.82755 cm is equivalent to 0.0482755 m

Finally, we shall determine the spring constant as follow:

Force (F) = 1.2838 N

Extention (e) = 0.0482755 m

Spring constant (K) =?

F = Ke

1.2838 = K × 0.0482755

Divide both side by 0.0482755

K = 1.2838 / 0.0482755

K = 26.59 N/m

Thus the spring constant is 26.59 N/m

3. Your friend says your body is made up of more than 99.9999% empty space. What do you think?

Answers

Answer:

I would agree with the statement. it's not just the body, but everything that we see is almost 99.9999% empty space

A frog leaps straight up from the ground and is caught when he reaches his maximum height 2\,\text s2s2, start text, s, end text later by a little girl. We want to find the vertical velocity of the frog at the moment that it left the ground. Which kinematic formula would be most useful to solve for the target unknown

Answers

Correct question is;

A frog leaps straight up from the ground and is caught when he reaches his maximum height 2 seconds later by a little girl. We want to find the vertical velocity of the frog at the moment that it left the ground. Which kinematic formula would be most useful to solve for the target unknown? Assume the positive direction is upward and air resistance is negligible.

Answer:

v = u + at

Explanation:

We are given time taken to reach maximum height to be 2 seconds.

We also know that since it's going upwards against gravity, our acceleration due to gravity will be g = -9.8 m/s²

Now, since we have values of g and t and we know that at maximum height vertical component of velocity is zero.

Thus, we can use Newton's first equation of motion to solve for the vertical velocity at which it left the ground.

Thus;

v = u + at

Where v is velocity at maximum height and u is velocity at which it left the ground.

Towards the end of a 400m race, Faisal and Edward are leading and are both running at 6m/s. While Faisal is 72m from the finish line Edward is 100m from the finish line. Realising this and to beat Faisal, Edward decides to accelerate uniformly at 0.2 m/s2 until the end of the race while Faisal keeps on the same constant speed. Does Edward succeed in beating Faisal

Answers

Answer:

NO

Explanation:

To determine this, we have to firstly determine the time it took Faisal to complete the race with 72 meters left and then the speed Edward used for the final 100 meters (with both distance occurring at the same point in time).

To determine the time it took Faisal from that onward, we use the formula

Speed = distance/time

Since Faisal maintained his speed at the end of the race, the speed will be 6 m/s while distance at the end to be covered his 72 m. Thus

6 = 72/time

time = 72/6

time = 12 seconds; It will take Faisal 12 seconds to complete the race from that point.

For Edward, he accelerated uniformly (for 0.2 m/s²) after running at a constant speed of 6 m/s. Thus, the time it would have taken him to complete (at a speed of 6 m/s) the final 100 m would be

speed = distance/time

6 = 100/time

time = 100/6

time = 16.67 s.

However, he accelerated uniformly at 0.2m/s², hence to get the speed, the formula; speed = acceleration × time will be used

speed = 0.2 × 16.67

speed = 3.334 m/s

To determine the time it took Edward from 100 meters away,

speed = distance/time

3.334 = 100/time

time = 100/3.334

time = 29.94 seconds

It would have taken Edward 29.94 seconds to complete the race, hence he would not have succeeded in defeating Faisal

A perfect gas undergoes a constant 100 kPa pressure process with a 20 m3 change in volume. What is the work required to cause this volume change?

Answers

Answer:

The work required to cause this volume change is 2 x 10⁶ J

Explanation:

Given;

constant pressure of the gas, P = 100 kPa = 100,000 Pa

change in volume of the gas, ΔV = 20 m³

The work required to cause this volume change is calculated as;

W = PΔV

Substitute the given values and solve for the required work (W).

W = (100,000)(20)

W = 2 x 10⁶ J

Therefore, the work required to cause this volume change is 2 x 10⁶ J

QUICK!! What class of leer is shown below?

first class lever

second class lever

third class lever

fourth class lever

Answers

Answer:

3rd class lever

HOPE IT HELPS YOU OUT PLEASE MARK IT AS BRAINLIEST AND FOLLOW ME PROMISE YOU TO FOLLOW BACK ON BRAINLY.IN

The class of lever that should be given in the attached image is to be considered the third-class lever.

What is a third-class lever?

In a Class Three Lever, the Force should be lies between the Load and the Fulcrum. In the case when the Force is closer to the Load, it should be easier to lift and there is a mechanical advantage.

So based on this, we can say that The class of leer that should be given in the attached image is to be considered as the third-class lever.

Learn more about lever here: https://brainly.com/question/21535944

What is the velocity of a sound wave that has a frequency of 300 Hz and a wavelength of 10
m?

Answers

Answer:

3000 m/s

Explanation:

The velocity of a sound wave is found by the wavelength times the frequency.  Therefore, the velocity = wavelength*frequency = (10 m)(300 Hz) = 3000 m/s

I hope this helps! :)

For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0 cmcm long) after reflecting from the first mirror

Answers

This question is incomplete, the complete question is;

Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point 11.5 cm from their point of intersection, as shown in the figure.

For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0 cmcm long) after reflecting from the first mirror

Answer: angle of incidence is 39.4°

Explanation:

Given that;

two plain mirrors intersect at right angle (90°)

distance d = 11.5 cm

S = 28.0 cm

Now the angle that the reflection ray males with first the mirror equal theta  (∅)

so

tan∅ = (S/2) / d

tan∅ = (28/2) / 11.5

tan∅ = 14 / 11.5

tan∅ = 1.2173

∅ = tan⁻¹ (1.2173)

∅ = 50.6°

so angle of incidence = 90° - ∅

= 90° - 50.6°

= 39.4°

Therefore angle of incidence is 39.4°

Fix this problem please.

Answers

Explanation:

Original Kinetic energy = 0.5mv²

= 0.5(5.0kg)(4.0m/s)² = 40J.

New Kinetic energy = 0.5(5.0kg)(10.0m/s)² = 250J.

Hence net work done = 250J - 40J = 210J.

Answer:

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List down the procedures for each swimming stroke 1.Crawl 2.Breaststoke 3.Butterflystroke 4. Backstoke​

Answers

Answer: Swimming strokes are techniques that includes arm and leg movements to help push the swimmer against water and propel the swimmer forward.

Explanation:

There are different types of swimming strokes these includes:

--> FRONT CRAWL: This is the fastest of all the techniques. The procedure includes:

• the body is kept flat, facing down and in line with the water surface,

• As the swimmer proceeds with movements, the arms are alternately moved in a PULL (with your palms facing down pull in line with the body) and RECOVERY (with the hand closed to the upper thigh, lift one arm out of the water with a bent elbow) actions.

• As you finish the recovery phase, turn quickly side ways to take in some air.

• With ankles relaxed and flexible, point your toes behind you and kick up-and-down in a continuous motion from your thighs.

BUTTERFLY STROKE: The procedures for this technique includes:

• the body is kept flat, facing down and in line with the water surface.

• the arms are moved in three ways, the catch, pull and recovery movements. The Catch involves the arms being straight, shoulder width apart and palms facing down wards, press down and out against the water with both hands at the same time. The pull involves the hands being pulled towards the body in a semicircular motion. The recovery starts at the end of each pull, the arms are moved out and over the water simultaneously and is thrown forward into the starting position.

• the chin is being raised up at the recovery stage to draw in a breath while looking straight.

• With both legs together and toes pointed, kick downwards at the same time.

• the body is moved in a wave-like manner.

BREASTSTROKE: The procedure for this technique includes;

• the body is kept flat, facing down and in line with the water surface

• the arms are also moved in three ways. In the catch movements, with arms out straight and palms facing downwards, press down and out at the same time. With elevated elbows above the arms, pull hard towards the chest. Then while recovering, to reduce drag when pushing against water, the both palms are joined together Infront of the chest and pushed out until the arms are straight again.

• the head is lifted above water at the end of pulling movement to breath in air.

• bend your knees to bring your heel towards your bottom and make a circular motion outwards with your feet until they return to the starting position.

BACKSTROKE: The procedure for this technique includes

• the body kept flat while backing the water surface. But following the arm movement, it rows from side to side.

• the arms performs alternating and opposite movements. As one arm pulls backwards in the water the other arm recovers above the water.

• taking in air should be alternated with the arm movements.

• the legs are moved up and down in a quick succession to enhance movements.

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