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Genrish500 [490]
3 years ago
15

What are the factors that affect the weather?​

Physics
1 answer:
jarptica [38.1K]3 years ago
6 0

Answer:

temperature, atmosphere pressure,wind, topography, humidity etc

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Which of the following allows the human body to maintain stability in response to a viral infection
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The spine is the one that keeps a huminss body mantaned
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Which of the following is the best example of a primary circular reaction?
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Primary Circular Reactions (1-4 months): This substage involves coordinating sensation and new schemas. For example, a child may suck his or her thumb by accident and then later intentionally repeat the action. These actions are repeated because the infant finds them pleasurable.

4 0
3 years ago
A person swings a 0.546-kg tether ball tied to a 4.56-m rope in an approximately horizontal circle. If the maximum tension the r
Murrr4er [49]

Answer:

2.1 rad/s

Explanation:

Given that,

Mass of a tether ball, m = 0.546 kg

Length of a rope, l =  4.56 m

The maximum tension the rope can withstand before breaking is 11.0 N

We need to find the maximum angular speed of the ball. Let v is the linear velocity. The maximum tension is balanced by the centripetal force acting on it. It can be given by :

F=\dfrac{mv^2}{r}\\\\v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{11\times 4.56}{0.546}} \\\\=9.584\ m/s

Let \omega is the angular speed of the ball. The relation between the angular speed and angular velocity is given by :

v=r\omega\\\\\omega=\dfrac{v}{r}\\\\=\dfrac{9.584}{4.56}\\\\=2.1\ rad/s

So, the maximum angular speed of the ball is 2.1 rad/s.

4 0
3 years ago
In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward
Vladimir79 [104]

Answer:

The answer is "1.01 \times 10^{-13}"

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

KE=U=\frac{kqq'}{r}\\\\

Calculating the closest distance:

\to r=\frac{kqq'}{KE}\\\\

=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\

=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}

5 0
2 years ago
A copper wire 20 m long and 4mm in diameter is attached to the Ceiling and a 400 N
Fiesta28 [93]

Answer:

A

Explanation:

5 0
2 years ago
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