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3 years ago

Research has shown ____ often depends on one's current status, circumstances, and expectations.

Physics

1 answer:

larisa86 [58]3 years ago

6 0

Answer:

A. subjective well-being

Explanation:

A subjective well being does indeed often on the status, expectations and circumstances according to the theory of emotion. Also, there are flashcards on quizlet about this subject so if you want to learn more go there!

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Elza [17]

1) push down on the end of the lever, and 2) 3/4 of the way from the fulcrum

7 0

3 years ago

A 0.05kg dart is thrown at and sticks into a 0.4 kg block hanging on a string. After the collision the block and dart swing in a

zloy xaker [14]

Answer:

v = 1.4 m /s

Explanation:

We shall apply law of conservation of mechanical energy

The kinetic energy of dart and block is converted into potential energy of both dart and block .

1 /2 (m+M) v² = ( m +M) gH

.5 x v² = 9.8 x .1

= v² = 1.96

v = 1.4

v = 1.4 m /s

6 0

3 years ago

What the unit of work?

ra1l [238]

Answer:

yes

Explanation:

5 0

2 years ago

Did i get any of these answers right if not what do i have to change

gregori [183]

Answer:

You got them right

8 0

3 years ago

A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,

alexgriva [62]

The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
$S= \frac{1}{2} a t^2$
where a is the acceleration. Using the data of the problem, we can find a:
$a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2$
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
$v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J$

and so the power output is
$P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W$

b) power output at t=4.0s
The velocity at t=4.0 s is
$v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J$

and so the power output is
$P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W$

c) Power output at t=6.0 s
The velocity at t=2.0 s is
$v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J$

and so the power output is
$P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W$

8 0

3 years ago