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attashe74 [19]
3 years ago
10

What is the difference between an empirical issue and ethical issue? (sociology question)

Physics
1 answer:
zavuch27 [327]3 years ago
3 0

Answer:

When it comes to research questions, sociologists are best equipped to answer empirical questions. —those that can be answered by real experience in the real world—as opposed to ethical questions. —questions about which people have moral opinions and that may not be answerable in reference to the real world.

Explanation:

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A circus act involves a trapeze artist and a baby elephant. They are going to balance on a teeter-totter that is 10 meters long
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Answer:

Explanation:

Balance point will be achieved as soon as the weight of the baby elephant creates torque equal to torque created by weight of woman about the pivot. torque by weight of woman

weight x distance from pivot

= 500x 5

= 2500 Nm

torque by weight of baby woman , d be distance of baby elephant from pivot at the time of balance

= 2500x d

for equilibrium

2500 d = 2500

d = 1 m

So elephant will have to walk up to 1 m close to pivot or middle point.

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The velocity v(t) in the table below is increasing for 0_t_12.
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When n=2 subdivisions distance traveled = 30 X 6 = 180 

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How can spectroscopy and infrared technology be useful in space?
olchik [2.2K]

Answer:

B

Explanation:

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3 years ago
In which step of the scientific method would a scientist make use of graphs
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6 0
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1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
3 years ago
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