¿Por qué los mensajes llegan en forma inmediata de un celular a otro?

xeze [42]

Answer:

The air in the soccer ball in cold weather will decrease slightly in size and it becomes flat. The air in the soccer ball in hot weather will seem flat because the low preasure leads to lower bounce in the ball.

The metal door frame in cold weather contracts and the wood contracts more in the winter. The metal door frame in hot weather thermal blowing can occur on the outer surface of the metal door frame. Hopefully that is what you were looking for have a good day.

4 0

2 years ago

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Which of the

kompoz [17]

C. Amount of oxygen

The others either change but don’t decrease or they increase.

8 0

2 years ago

An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m

Aneli [31]

Answer:

$\displaystyle y_m=3.65m$

Explanation:

<u>Motion in The Plane</u>

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of $v_o$ and $\theta\\$ as the initial speed and angle, then we have

$\displaystyle v_x=v_o\ cos\theta$

$\displaystyle v_y=v_o\ sin\theta-gt$

$\displaystyle x=v_o\ cos\theta\ t$

$\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}$

If we want to know the maximum height reached by the object, we find the value of t when $v_y$ becomes zero, because the object stops going up and starts going down

$\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt$

Solving for t

$\displaystyle t=\frac{v_o\ sin\theta }{g}$

Then we replace that value into y, to find the maximum height

$\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2$

Operating and simplifying

$\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}$

We have

$\displaystyle v_o=20\ m/s,\ \theta=25^o$

The maximum height is

$\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}$

$\displaystyle y_m=3.65m$

7 0

2 years ago

A chair of weight 95.0N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 3

AlladinOne [14]

Answer:

118.4 N

Explanation:

weight of chair, mg = 95 N

Push, F = 39 N

Ф = 37 ° below x axis

Let n be the normal force.

So, by using the diagram and resolve the components of Force F.

n = mg + F SinФ

n = 95 + 39 Sin 37°

n = 95 + 39 x 0.6

n = 118.4 N

Explanation:

7 0

3 years ago

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A shot-putter projects the shot at 42.00˚ to the horizontal from a height of 2.100 m. It lands 17.00 m away horizontally. Next,

Tamiku [17]

Answer:

Explanation:

Let 100 m/s be the velocity of projection.

So horizontal component

= 100 cos42

= 74.31 m /s

Vertical component = - 100 sin 42 . in upward direction

66.91 m/s

Net displacement = 2.1 downwards ( + ve )

Using s = ut + 1/2 gt²

2.1 = - 66.91 t + .5 x 9.8 x t²

4.9 t² - 66.91 t - 2.1 = 0

t = 13.685 s

Horizontal distance covered

= 13.685 x 74.31

= 1016.93 m

If angle of projction is 40°

So horizontal component

= 100 cos40

= 76.60 m /s

Vertical component = - 100 sin 42 . in upward direction

64.27 m/s

Net displacement = 2.1 downwards ( + ve )

Using s = ut + 1/2 gt²

2.1 = -76.60 t + .5 x 9.8 x t²

4.9 t² - 76.60 t - 2.1 = 0

t = 15.659 s

Horizontal distance covered

= 15.659 x 76.60

= 1199.49 m

So horizontal range is increased , if angle of projection is increased .

8 0

2 years ago