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marissa [1.9K]
3 years ago
10

A batter hits a fly ball into the outfield. The

Physics
2 answers:
Anon25 [30]3 years ago
6 0
Answer: gravitational force
Vaselesa [24]3 years ago
4 0
The gravitational force :))))
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Which of the following best describes what alveolar are
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Alveoli are tiny balloon shaped structures and are the smallest passageway in the respiratory system. The alveoli are only one cell thick, allowing the relatively easy passage of oxygen and carbon dioxide (CO2) between the alveoli and blood vessels called capillaries.
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Temperatures of gases inside the combustion chamber of a four‑stroke automobile engine can reach up to 1000°C. To remove this en
Alona [7]

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thats a lot, which one u want me to do?

Explanation:

8 0
3 years ago
Altitude means the what?
andrew-mc [135]

Answer:

the height of a object or point relation to sea level or ground level

7 0
3 years ago
When AC Electric is producing 120 volts with 20 amps, What controls the the 120 volts at 60cycles a second?
lubasha [3.4K]

Answer:

Cycles per second is dependent on the construction of the alternator and the 120 volts is dependent upon the current and resistance in the circuit according to the ohms law.

Explanation:

We are given with AC of 120 volts, 20 amperes and 60 hertz frequency.

<u>According to the Ohm's law, we find its resistance:</u>

R=\frac{V}{I}

R=\frac{120}{20}

R=6\ \Omega

So, this 6 ohm resistance controls the current controls the magnitude of the AC current, while the frequency of the current remains constant and depends upon the construction and rotational speed of the armature of the alternator producing the current.

Here the value of frequency is the number of times the current changes its direction or the polarity in one second.

6 0
4 years ago
In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, f=5\times 10^{14}\ Hz

Separation between slits, d=2.2\times 10^{-5}\ m

(a) The condition for maxima is given by :

d\ sin\theta=n\lambda

For third maxima,

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{nc}{fd})  

\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})  

\theta=4.69^{\circ}

(b) For second dark fringe, n = 2

d\ sin\theta=(n+1/2)\lambda

\theta=sin^{-1}(\dfrac{5\lambda}{2d})

\theta=sin^{-1}(\dfrac{5c}{2df})

\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})

\theta=3.90^{\circ}

Hence, this is the required solution.

8 0
3 years ago
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