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user100 [1]
2 years ago
5

______________ are chemicals in the body affecting how you feel.

Physics
2 answers:
weeeeeb [17]2 years ago
8 0

Answer:

emotions affect your body and how you feel

7nadin3 [17]2 years ago
3 0

Answer:

emotions

Explanation:

emotions are how you feel and can happen any time

Hope it helps <333

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Please select the word from the list that best fits the definition
lina2011 [118]

Answer:

A subculture.

Explanation:

  • Aspects of a group that is based on the shared values and belief systems and the norms and behaviors aspects that are only a part of the subsection of society are called as subcultures. Like the goths, hippies, etc as its nor accepted by the entire population.
3 0
3 years ago
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Gravity can be described as (2 points)
ladessa [460]

Answer:

"the force of attraction between two objects"

Explanation:

According to Newton's Universal Law of Gravitation, gravity is a force of attraction acting between objects that possess mass. The fact that we only observe gravitational attraction (as opposed to repulsion) makes gravity unique among the known forces.

6 0
3 years ago
How many leg of elephant?<br> I guest, it have four legs.
lawyer [7]

Answer:

I believe it has 4 legs as well

Explanation:

It looks like 4 legs, it never gets on 2 legs so it should be 4.

4 0
3 years ago
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Why does the speed of an object change when a force acts on it?
Crazy boy [7]
It changes because force is somewhat like pressure. Force is continuously against the object so as a result, speed changes .
7 0
3 years ago
The frictionless system shown is released from rest. After the right-hand mass has risen 75 cm, the object of mass 0.50m falls l
Andreas93 [3]

Let a be the acceleration of the masses. By Newton's second law, we have

• for the masses on the left,

1.3mg - T = 1.3ma

where T is the magnitude of tension in the pulley cord, and

• for the mass on the right,

T - mg = ma

Eliminate T to get

(1.3mg - T) + (T - mg) = 1.3ma + ma

0.3mg = 2.3ma

\implies a = \dfrac{0.3}{2.3}g \approx 0.13g \approx 1.3 \dfrac{\rm m}{\mathrm s^2}


Starting from rest and accelerating uniformly, the right-hand mass moves up 75 cm = 0.75 m and attains an upward velocity v such that

v^2 = 2a(0.75\,\mathrm m) \\\\ \implies v \approx \sqrt{2\left(1.3\frac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx 1.4\dfrac{\rm m}{\rm s}

When the 0.5m mass is released, the new net force equations change to

• for the mass on the right,

mg - T' = ma'

where T' and a' are still tension and acceleration, but not having the same magnitude as before the mass was removed; and

• for the mass on the left,

T' - 0.8mg = 0.8ma'

Eliminate T'.

(mg - T') + (T' - 0.8mg) = ma' + 0.8ma'

0.2mg = 1.8 ma'

\implies a' = \dfrac{0.2}{1.8}g = \dfrac19 g \approx 1.1\dfrac{\rm m}{\mathrm s^2}

Now, the right-hand mass has an initial upward velocity of v, but we're now treating down as the positive direction. As it returns to its starting position, its speed v' at that point is such that

{v'}^2 - v^2 = 2a'(0.75\,\mathrm m) \\\\ \implies v' \approx \sqrt{\left(1.4\dfrac{\rm m}{\rm s}\right)^2 + 2\left(1.1\dfrac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx \boxed{1.9\dfrac{\rm m}{\rm s}}

3 0
2 years ago
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