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3 years ago

Explicar los avances que dieron los estudioas de robert hooke y antonie van leeuwnhook

Physics

1 answer:

Dafna11 [192]3 years ago

5 0

Answer:

El microscopio y el descubrimiento de microorganismos. Antonie van Leeuwenhoek (1632-1723) fue una de las primeras personas en observar microorganismos, utilizando un microscopio de su propio diseño, e hizo una de las contribuciones más importantes a la biología. Robert Hooke fue el primero en usar un microscopio para observar seres vivos.

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A helicopter flies 25 km north, 5 km east, then 5 km S, then 15 km W. What is the resultant displacement and direction of the he

ch4aika [34]

Answer:

Explanation:

Plotting the original location of the helicopter before it flies 25 km north, it would be at the origin, (0,0) then after it flies north, the y vertex gains 25 points, so it would be (0,25)

After it flies east, the x coordinate gains 5 points, so it would now be (5,25)

After it flies south, the y coordinate loses or is subtracted by 5 points. so it would now be (5,20)

After flying west, the x coordinate loses 15 points. So the final vertex would be at (-10,20)

East = Right

West = Left

South= Down

North = Up

I used mainly mathematical methods by adding and subtracting the x and y coordinate values, but this could be graphed easily since I gave the coordinates just incase!

Hope this helps!

7 0

2 years ago

Each of the four expansion models (recollapsing, critical, coasting, and accelerating) predict different ages for the universe,

cupoosta [38]

Answer:

This is because the age of the universe is determined by the pace of expansion in the past, and each model forecasts a different pace.

Explanation:

The age of the universe is determined by the pace of expansion in the past, and each model forecasts a different pace.

This is due to the fact that the expansion rate in the coasting model is constant and never changes. Because the cosmos is growing faster now than during the old days, recollapsing and critical models give shorter ages. According to the accelerating model, the universe is growing at a slower rate currently than in the past, implying an older age.

4 0

3 years ago

An elevator filled with passengers has a mass of 1603 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 f

Galina-37 [17]

Answer:

T = 17649.03 N = 17.65 KN

Explanation:

The tension in the cable must be equal to the apparent weight of the passenger. For upward acceleration:

$T = W_A = m(g+a)\\$

where,

T = Tension in cable = ?

$W_A$ = Apparent weight

m = mass = 1603 kg

g = acceleration due to gravity = 9.81 m/s²

a = acceleration of elevator = 1.2 m/s²

Therefore,

$T = (1603\ kg)(9.81\ m/s^2+1.2\ m/s^2)\\\\$

8 0

2 years ago

A cyclist rides at 6.20 m/s through a intersection. A stationary car begins to

Xelga [282]

Answer:

The width of the intersection is 20 meters

Explanation:

The speed with which the cyclist is riding, v₁ = 6.20 m/s

The rate at which the car starts to accelerate, a = 3.844 m/s²

The initial velocity of the car = The car is stationary at the start = 0 m/s

The time at which the cyclist and the car reach the other side of the intersection = The same time;

Let 't' represent the time at which the cyclist and the car both reach the other side of the intersection, we have;

The distance travelled by the cyclist = The distance traveled by the car

∴ v₁ × t = 1/2 × a × t²

Plugging in the values for 'v₁', and 'a' in the above equation, we get;

6.20 × t = 1/2 × 3.844 × t²

∴ 1.922·t² - 6.20·t = 0

∴ t·(1.922·t - 6.20) = 0

t = 0, or t = 6.20/1.922 = 100/31

The time at which the cyclist and the car both reach the other side of the intersection, t = 100/31 seconds

The with of the intersection, w = v₁ × t

∴ w = 6.20 × 100/31 = 100/5 = 20

The width of the intersection, w = 20 meters.

8 0

3 years ago

Note that the simulation allows you to also display the force of the smaller moon

Lelu [443]

the force that the planet exerts on the moon is equal to the force that the moon exerts on the planet

Explanation:

In this problem we are analzying the gravitational force acting between a planet and its moon.

The magnitude of the gravitational attraction between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, we are considering a planet and its moon. According to Newton's third law of motion,

"When an object A exerts a force (action force) on an object B, then object B exerts an equal and opposite force (reaction force) on object A"

If we apply this law to this situation, this means that the force that the planet exerts on the moon is equal to the force that the moon exerts on the planet.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago