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8_murik_8 [283]
3 years ago
11

The slope of the line tangent to the curve on a position-time graph at a specific time is the

Physics
1 answer:
Rudik [331]3 years ago
7 0

Answer:

I do I make a brinliest can you please can me

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A honey bee flaps its wings 200 times per second. How much time is required for one wingbeat? Give your answer in milliseconds.
kakasveta [241]
A bees wings move so rapidly that studying them, even seeing them, has proved difficult.
The honeybees have a rapid wing beat honeybee flaps its wings 230 times every second.

Therefore, a bee flaps its wings over 200 times in about 1ms
3 0
3 years ago
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A wheel with a tire mounted on it rotates at the constant rate of 2.73 revolutions per second. Find the radial acceleration of a
Lostsunrise [7]

Answer:

110.9 m/s²

Explanation:

Given:

Distance of the tack from the rotational axis (r) = 37.7 cm

Constant rate of rotation (N) = 2.73 revolutions per second

Now, we know that,

1 revolution = 2\pi radians

So, 2.73 revolutions = 2.73\times 2\pi=17.153\ radians

Therefore, the angular velocity of the tack is, \omega=17.153\ rad/s

Now, radial acceleration of the tack is given as:

a_r=\omega^2 r

Plug in the given values and solve for a_r. This gives,

a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]

Therefore, the radial acceleration of the tack is 110.9 m/s².

4 0
3 years ago
The _____ period refers to the first two weeks of development after conception.
Mrac [35]

Answer: Geminal

Explanation:

5 0
3 years ago
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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 \\

                 = 0.031 \ \ rem

4 0
3 years ago
Which of these would not improve accuracy? 
fgiga [73]
I think it’s D, mostly just because all the other ones seem like they couldn’t be the answer.
6 0
3 years ago
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