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zhannawk [14.2K]
3 years ago
15

A 1.5 kg baseball is pitched at 38 m/s and is hit by the batter. It heads directly back at the pitcher at 27 m/s.

Physics
1 answer:
lesantik [10]3 years ago
5 0

Answer:

The initial velocity of the ball is +38.0 m/s. The final velocity is -27.0 m/s.  

momentum = mass x velocity

(mass x final velocity) – (mass x initial velocity)

(1.5)(-27) – (1.5)(38) = -97.5

The impulse applied to the ball is -97.5 kg m/s

impulse = force x time interval

force = impulse/time interval

-97.5/0.45 s = - 216.6  

The force applied was 216.6 N.

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Answer:

Gallium

Explanation:

Gallium is one such element used as a do/pant in a p-type semiconductor.

A do/pant is an impurity added to a semi-conductor used to alter its properties. Semi-conductors have a wide range of applications. They will conduct heat and electricity only under certain conditions. This property is highly desirable and find a wide application in electronics.

For p-type conductors, they are best do/ped with elements with 3 valence electrons. These are group 3 elements. From the choices, only gallium belongs to this group.

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8 0
4 years ago
What type of radiation does not have to be observed above earths atmosphere? Visible light, X rays, Gamma rays, Ultraviolet radi
juin [17]
The correct answer is visible light.

Hope this helps.
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3 years ago
The bae chart shows the percentage of adults in the U.S. living with certain heart disease risk factors. Which conclusion can be
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3 years ago
You are in a hot-air balloon that, relative to the ground, has a veloc- ity of 6.0 m/s in a direction due east. You see a hawk m
Ann [662]

Answer:

6.32 m/s 18.43° northeast

Explanation:

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8 0
3 years ago
g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9
Elza [17]

Answer:

The answer is "\bold{dosage = 0.031 rem}"

Explanation:

please find the complete question in the attached file.

Given value:

m = 79\  kg  \\\\n = 3.4 \times  10^9 \\\\E = 5.5  \times  10^{-13} \\\\ RBE = 15

\to E = n E\\

        = 3.4  \times  10^9  \times  5.5  \times  10^{-13} \\\\      = 1.87 \times  10^{-3}

\to E(absorbed) = 1.87  \times 10^{-3}  \times  0.87 = 1.63  \times  10^{-3}

calculating the radiation absorbed per kg:

= \frac{1.63  \times  10^{-3}}{79}  \\\\ = 2.06  \times  10^{-5} \\\\ = 0.00206 \  rad

\to Dosage = 0.00206  \times  15 \\

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4 0
3 years ago
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