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2 years ago

11

Q: What happens when cold air approaches a body of warm air? Optional Answers: A. The Warm air rises. B. The warm a

nd cold air both disappear. C. The warm and cold air mix immediately. D. The cold air rises.

2 answers:

artcher [175]2 years ago

8 0

A. The warm air rises! Hope this helps

Dafna11 [192]2 years ago

8 0

Answer a is correct.
Warm air rises.
EXAMPLE: when your cooking the hot air or warm air rises or moves upward.
Hope I helped.

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En el País Vasco los deportistas rurales levantan enormes piedras hasta su hombro. En un concurso , Jose levanta una piedra de 2

Mama L [17]

Answer:

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

Explanation:

The sportsman that lifts the stone with a greater mass needs a higher force (El deportista que levanta la piedra con mayor masa necesita una mayor fuerza):

José

$F = (200\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 1961.4\,N$

Txomin

$F = (220\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 2157.54\,N$

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

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2 years ago

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Degger [83]

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3 years ago

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A volcano erupts and launches a chunk of hot magma horizontally with a speed of 252 m/s. The magma travels a horizontal distance

ArbitrLikvidat [17]

Answer:

The value is $v_y = -48.61 \ m/s$

Explanation:

From the question we are told that

The horizontal speed is $u_x = 252 \ m/s$

The horizontal distance is $d = 1250 \ m$

Generally the time taken by the hot magma in air before landing is mathematically represented as

$t = \frac{d}{u_x}$

=> $t = \frac{ 1250 }{252}$

=> $t = 4.96 \ s$

Generally the initial vertical velocity of the magma when it was lunched is

$u_y = 0 \ m/ s$

Then the final velocity of the magma when it hits the ground is mathematically represented s

$- v_y = u_y + gt$

Here the negative sign mean that the direction of the velocity is towards the negative y -axis

So

$- v_y = 48.61 \ m/s$

=> $v_y = -48.61 \ m/s$

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2 years ago

How Negative Points Works<br><br>btw help me with this<br><br>How much does the sky weigh?<br>

kipiarov [429]

Answer:

it weighs 237469812734t7162341873498273417234321476281736481273648123764812736481723648273648137468127364872364 million pounds :)

Explanation:

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2 years ago

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If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by

fgiga [73]

Answer:

a) $v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) $v=65-32(2)=1ft/s$

Explanation:

From the exercise we got the ball's equation of position:

$y=65t-16t^{2}$

a) To find the average velocity at the given time we need to use the following formula:

$v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }$

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

$y_{t=2}=65(2)-16(2)^{2} =66ft$

$y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft$

$v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

--

$y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

--

$y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

--

$y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) To find the instantaneous velocity we need to derivate the equation

$v=\frac{df}{dt}=65-32t$

$v=65-32(2)=1ft/s$

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3 years ago