Which simple machine belongs to the inclined plane family

The chart below shows the average surface temperature or temperature range for each of the eight planets.

Vaselesa [24]

Answer:

A) Earth and the other inner planets have higher average surface temperatures than the outer planets.

Explanation:

the earth and the other inner planets have higher average surface temperatures than the outer planets.

The reason for this response is due to the distance between the sun and the respective planet, the source of energy generation is the sun and the only way in which the temperature increase of each planet is guaranteed is by radiation, the further away a planet is from its star, its temperature will decrease. Although it is also important to highlight the atmospheric composition of the planet if this planet in its stratosphere has high density clouds that do not allow the entry of solar radiation, the temperature of the planet's surface will not increase, independent of the distance from the sun, but these are more complex cases where specialists in that area enter to perform a study in detail.

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If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th

mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

$x=ut$

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

$t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s$

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

$y=\frac{1}{2} gt^2$

Substitute 9.8 m/s² for g and 0.461 s for t.

$y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m$

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by, $(2.0 m)-(1.04 m)=0.96 m$