Una pelota de golf se lanza con una rapidez de 50 m/seg y un ángulo de elevación de 40º. Calcular:

What did Charles Darwin’s conclude on the Galapagos Island? Plz answer fast

kvasek [131]

He discovered several species of finches that varied from island to island and it helped him make his theory of natural selection.

hope this helps !

3 0

4 years ago

Galileo Galilei was the first scientist to perform experiments in order to test his ideas. He was also the first astronomer to s

arsen [322]

Explanation:

Geocentric model said that the Earth is at the center of the universe and everything revolves around it. It was considered to be stationary. Galileo proved this model incorrect with the help of his astronomical observations. Some of the key observation that he used to support the heliocentric model were:

1. He proposed the theory that the tides on the Earth occur because of its motion.

2. He observed the phases of the Venus which meant that the Venus revolved around the Sun and not the Earth.

3. He observed other planets and thus noted that they also move around the Sun and not Earth.

4. He discovered the Moons of other planets.

6 0

3 years ago

When you drink cold water, your body must expend metabolic energy in order to maintain normal body temperature (37Â° C) by warmi

Gelneren [198K]

Answer:

7.72 Liters

Explanation:

normal body temperature = T_body =37° C

temperature of ice water = T_ice =0°c

specfic heat of water = c_{water} =4186J/kg.°C

if the person drink 1 liter of cold water mass of water is = m = 1.0kg

heat lost by body is Qwater =mc_{water} ΔT

= mc{water} ( T_ice - T_body)

= 1.0×4186× (0 -37)

= -154.882 ×10^3 J

here negative sign indicates the energy lost by body in metabolic process energy expended due to brisk - hour long walk is Q_{walk} = 286 kilocalories

= 286×4186J

so number of liters of ice water have to drink is

n×Q_{water} =Q_{walk} n= Q_{walk}/ Q_{water}

= 286×4186J/154.882×10^3 J

= 7.72 Liters

7 0

3 years ago

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A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turn

kumpel [21]

Answer:

a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

Explanation:

Statement is incomplete. The complete description is now described below:

A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off.

(a) What is the magnitude of the satellite's velocity when the thruster turns off

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis

Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite ( $\vec{v}_{S}$ ), measured in meters per second, is:

$\vec{v}_{S} = (v_{o,x}+a_{x}\cdot t)\,\hat{i}+v_{y}\,\hat{j}$

Where:

$v_{o,x}$ - Initial velocity in +x direction, measured in meters per second.

$a_{x}$ - Acceleration in +x direction, measured in meter per square second.

$t$ - Time, measured in seconds.

$v_{y}$ - Velocity in +y direction, measured in meters per second.

If we know that $v_{o,x} = 0\,\frac{m}{s}$ , $a_{x} = 0.250\,\frac{m}{s^{2}}$ , $t = 45\,s$ and $v_{y} = 21.4\,\frac{m}{s}$ , the final velocity of the satellite is:

$\vec{v}_{S} = \left[0\,\frac{m}{s}+\left(0.250\,\frac{m}{s^{2}} \right)\cdot (45\,s) \right]\,\hat{i}+\left(21.4\,\frac{m}{s} \right)\,\hat{j}$

$\vec{v_{S}} = 11.25\,\hat{i}+21.4\,\hat{j}\,\,\left[\frac{m}{s} \right]$

a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:

$\|\vec {v}_{S}\| = \sqrt{\left(11.25\,\frac{m}{s} \right)^{2}+\left(21.4\,\frac{m}{s} \right)^{2}}$

$\|\vec{v}_{S}\| \approx 24.177\,\frac{m}{s}$

The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:

$\tan \alpha = \frac{v_{y}}{v_{x}} = \frac{21.4\,\frac{m}{s} }{11.25\,\frac{m}{s} }$

$\tan \alpha = 1.902$

$\alpha = \tan^{-1}1.902$

$\alpha \approx 62.266^{\circ}$

The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

4 0

3 years ago

HEY CAN ANYONE ANSWER DIS PLS!!!

SOVA2 [1]

Answer:

Multiplied.

Explanation:

I have a positive feeling about my answer. If I'm wrong, I am terribly sorry!!!

8 0

3 years ago

Read 2 more answers