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2 years ago

A cannonball is shot horizontally off a

Physics

1 answer:

inn [45]2 years ago

5 0

Answer:it's 132.2m

Explanation:

Hello !

Because first recall parabolic peak height equation and

given that v=47.4m/s , h=10.5m

v²sin²(θ⁰)=2ℎ₀
(47.4)²²(θ⁰)=2(9.81)(10.5)
θ=17.63

secondly recall parabolic total distance equation

d₀=²(2θ⁰)/g
d₀=(47.4)²(2(17.63)⁰)/9.81
d₀=132.2m

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an alligator crawls 25m to the left with an average velocity of -1.2m/s. how many seconds did the alligator crawl?

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It will take 21s for yhe alligator to crawl that distance. Reported answer contains 2 Significant Figures.

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3 years ago

NBAのバスケットボールコートのサイズは

Marysya12 [62]

28.65m x 15.24m

Explanation:

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3 years ago

What is occurring in the umbra?

matrenka [14]

Answer:

A. a total solar eclipse

Explanation:

During a solar eclipse,the moon comes between the sun and Earth, casting its shadow over the Earth

3 0

3 years ago

A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration

JulsSmile [24]

Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

$v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}$

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

$v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m$

So, the maximum height reached by the rocket is:

$h=y_1+y_2\\h=160m+154.28m=314.28m$

$v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s$

And in the second part:

$t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s$

So, the time it takes to reach the maximum height is:

$t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s$

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

$v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}$

$t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s$

So, the total time is:

$t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s$

7 0

4 years ago

Warm air aloft is normally higher pressure relative to colder air aloft, which is normally of lower pressure (at equivalent alti

Allisa [31]

Answer:

the statement given is False.

Explanation:

The pressure is defined by

P = F / A

The force of the air is its weight,

P = W_air / A

The weight of an air column on us depends on the energy of the molecules that is related to the temperature

E = 3/2 K T

When the air is warmer it has more energy so the molecules can move more distance and therefore the average density of the gas decreases, as the density decreases, the weight of the column on us also decreases, therefore the pressure low.

The correct statement is: When the air is hotter it has less pressure than when the air is cold.

It is summary the statement given is False

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4 years ago