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exis [7]
3 years ago
10

L 2.5.2 Test (CST): The Brain and the Body

Physics
1 answer:
ahrayia [7]3 years ago
8 0

Answer:

I believe it's A frontal. Hope this can help

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A man steps off a diving platform lands in the water below. If he were to miss the water and land on the concrete, his change in
kari74 [83]

Answer:

if he were to land on concrete, he would most likely die and stop all momentum and for the water, he'd most likely live/die (mostly live) and slow momentum to a stop and rise up to the surface

Explanation:

that's the explanation now i gotta execute order 66

7 0
3 years ago
A 0.150-kg cart that is attached to an ideal spring with a force constant (spring constant) of 3.58 N/m undergoes simple harmoni
SVETLANKA909090 [29]

Answer:

E = 0.01 J

Explanation:

Given that,

The mass of the cart, m = 0.15 kg

The force constant of the spring, k = 3.58 N/m

The amplitude of the oscillations, A = 7.5 cm = 0.075 m

We need to find the total mechanical energy of the system. It can be given by the formula as follows :

E=\dfrac{1}{2}kA^2

Put all the values,

E=\dfrac{1}{2}\times 3.58\times (0.075)^2\\\\=0.01\ J

So, the value of total mechanical energy is equal to 0.01 J.

3 0
3 years ago
*Physical Science* *E2020* *Unit Test*
vladimir2022 [97]

i think its b

hope this helps

5 0
4 years ago
Read 2 more answers
Your friend says that for a moving object to continue moving, a force must be continually applied to it. Do you agree with your
Zigmanuir [339]

Answer:

I would disagree with my friend because that according to Newton's first law, an object in motion will continue to be in motion until stopped by another object/force. If the object is already moving, it will stay in motion until something else stops it. There is no need for a force to be <em>continually</em> applied to it while it's already moving.

4 0
2 years ago
A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be
FrozenT [24]

Answer:

\omega=0.31\frac{rad}{s}

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

a_c=\frac{v^2}{r}(1)

Here, r is the radius and v is the tangential speed, which is given by:

v=\omega r(2)

Here \omega is the angular velocity, we replace (2) in (1):

a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r

Recall that r=\frac{d}{2}=\frac{200m}{2}=100m.

Solving for \omega:

\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}

3 0
4 years ago
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