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Gemiola [76]
3 years ago
9

A behavior that is developed by observing others or being taught is

Physics
2 answers:
Anit [1.1K]3 years ago
5 0
B. Learned behavior
mamaluj [8]3 years ago
5 0
B. Learned behavior. Your parent didn’t pass down that trait to you , you’re observing someone / being taught that behavior
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A bowling ball has an initial momentum of +30 kg m/s and hits a stationary bowling pin. After the collision, the bowling ball le
dangina [55]

Momentum should be conserved. The momentum of both objects must balance with their initial and final momentum.

Let m1 and v1 be the mass and velocity of the bowling ball

And m2 and v2 be the mass and velocity of the bowling pin

(m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f

30 kg m/s + (1.5 kg)(0 m/s) = 13kg m/s + 1.5v2f

V2f = 11.33 m/s

<span>So the momentum = 1.5 kg(11.33 m/s) = 17 kg m/s</span>

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Differences between concave and convex meniscus​
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Explanation:

A concave meniscus,(normally seen) occurs when the molecules of the liquid are attracted to those of the container. This occurs with water and a glass tube. A convex meniscus occurs when the molecules have a stronger attraction to each other than to the container, as with mercury and glass.

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What are three skills that move objects?
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Gravity, oxygen, mass
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A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What
AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, \alpha=3.3 rad/s^2

Diameter of the wheel, d=21 cm

Radius of wheel, r=\frac{d}{2}=\frac{21}{2} cm

Radius of wheel, r=\frac{21\times 10^{-2}}{2} m

1m=100 cm

Magnitude of total linear acceleration, a=1.7 m/s^2

We have to find the linear speed  of a  at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration,a_t=\alpha r

a_t=3.3\times \frac{21\times 10^{-2}}{2}

a_t=34.65\times 10^{-2}m/s^2

Radial acceleration,a_r=\frac{v^2}{r}

We know that

a=\sqrt{a^2_t+a^2_r}

Using the formula

1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}

Squaring on both sides

we get

2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}

\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}

v^4=r^2\times 2.7699

v^4=(10.5\times 10^{-2})^2\times 2.7699

v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}

v=0.418 m/s

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

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Which of the following is the center of the universe according to the geocentric model?
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Earth is the center according to the geocentric model.
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