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4 years ago

8

Starting from rest, approximately how far does an object fall freely in 3 seconds?

1 answer:

Sergio039 [100]4 years ago

8 0

Hi! In 3 seconds the object will fall approximately 44 meters.

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The three basic economic questions societies face are (1) what goods and services to produce, (2) how to produce those goods and

lord [1]

Answer:

For whom are goods and services to be produced? In other words, who gets what?

What should we produce?

For whom should we produce it?

Explanation:

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3 years ago

According to peacemaking criminology, toward which option should a country redirect the money it currently spends on imprisoning

ratelena [41]

Answer:

Explanation:

It is C. Environmental campaigns

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3 years ago

Read 2 more answers

In a building with 10.000 cubic feet where the air changes every two hours, what the rate of air change? A. 167.7 cfm B. 83.3 cf

Naya [18.7K]

Answer:

Flow rate of air is given as 83.33 cubic feet per minute

Explanation:

As we know that total volume of the air flow is given as

$V = 10,000 cubic feet$

also we know that total time is

$t = 2 hours = 120 min$

now we have flow rate given as

$Q = \frac{V}{t}$

$Q = \frac{10000}{120}$

$Q = 83.3 cf/m$

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3 years ago

A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at

pav-90 [236]

Answer:

Explanation:

Given

mass of drop $m=2 kg$

height of fall $h=1 m$

ball leaves the foot with a speed of 18 m/s at an angle of $55^{\circ}$

Velocity of ball just before the collision with the floor

$u^=2gh$

$u=\sqrt{2gh}$

$u=\sqrt{2\times 9.8\times 1}=4.42 m/s$

Impulse delivered in Y direction

$J_y=m(v\sin (55)-(-u))$

$J_y=2(18\sin (55)+4.42)$

$J_y=38.32 kg-m/s$

Impulse in x direction

$J_x=m\times v\cos (55)$

$J_x=2\times 4.42\cos (55)=20.646$

$J_{net}=\sqrt{J_x^2+J_y^2}$

$J_{net}=\sqrt{(38.32)^2+(20.64)^2}$

$J_{net}=43.52 kg-m/s$

at an angle of $\tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}$

$\phi =tan^{-1}(1.856)$

$\phi =61.7^{\circ}$

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3 years ago

Calculate how much force is required to give a 0.15-kg baseball an acceleration of 12m/s2

Nutka1998 [239]

<u>F = M A </u> (Newton's #2)

Force = (0.15 kg) (12 m/s²)

Force = (0.15 · 12) (kg·m/s²)

<em>Force = 1.8 Newtons</em>

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3 years ago