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11Alexandr11 [23.1K]
3 years ago
5

The element carbon has 3 naturally occurring isotopes. About 99% of carbon isotopes are C-12, about 1% are C-13, and a tiny amou

nt are C-14. What is the average atomic mass of carbon atoms?
A) 11.01 amu

B. 12.01 amu

C. 13.01 amu

D. 14.01 amu
Physics
1 answer:
mezya [45]3 years ago
3 0
Average atomic mass of carbon is 12.01 amu
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vovikov84 [41]

Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity=70 mph\approx 31.29 m/s

velocity after 5 s is 30 mph\approx 13.41 m/s

Therefore acceleration during these 5 s

a=\frac{v-u}{t}

a=\frac{13.41-31.29}{5}=-3.576 m/s^2

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

0=31.29-3.576\times t

t=\frac{31.29}{3.576}=8.75 s

(b)total distance traveled before stoppage

v^2-u^2=2as

0^2-31.29^2=2\times (-3.576)\cdot s

s=136.89 m

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Determine one way you can contribute to water in the atmosphere in your day-to-day activities pleaseeeee helppp
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agricultural production of food

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Valence electrons determine an atom’s __________. a. mass c. atomic number b. chemical properties d. period Please select the be
Lemur [1.5K]
Hello,

Here is your answer:

The proper answer to this question is option B "<span>chemical properties".

Here is how:

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When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius
Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

\tau = I\alpha

where

I = Moment of inertia \rightarrow \frac{1}{2}mr^2 For a disk

\alpha = Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

2 \alpha \theta = \omega_f^2-\omega_i^2

Where

\omega_{f,i} = Final and Initial Angular velocity

\alpha = Angular acceleration

\theta = Angular displacement

Our values are given as

\omega_i = 0 rad/s

\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega_f = 47.12rad/s

\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad

r = 7cm = 7*10^{-2}m

m = 17g = 17*10^{-3}kg

Using the expression of angular acceleration we can find the to then find the torque, that is,

2\alpha\theta=\omega_f^2-\omega_i^2

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}

\alpha = \frac{47.12^2-0^2}{2*6\pi}

\alpha = 58.89rad/s^2

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

\tau = I\alpha

\tau = (\frac{1}{2}mr^2)\alpha

\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)

\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m

Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

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3 years ago
PLSSSS HELPPPP PLSSSS 25 POINTS PLS I REALLY NEED THIS AND GET THEM RIGHT SO PLS DONT ANSWER IF U DONT KNOW
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Answer:

1- C   2-B    3-B   - these are ur best answers

Explanation:

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3 years ago
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