If the mass of 293nv is 293.15 amu and that of 295nv is 295.30 amu , what is the atomic weight of nv?

1 answer:

5 0

Honestly, I am quite confused with what Nv stands for because there is no element with that symbol. However, I still get the concept of finding the average molecular mass of an element. Let's just assume that nv stands for a specific type of element and it has two isotopes: nv-293 and nv-295. Isotopes have the same number of protons but differ in mass number (protons+neutrons).

To find the average atomic weight, just multiply the individual weights with the respective composition of the isotope. Since there are only two isotopes, they constitute 50% each. So, the average atomic weight is

(50%)(293.15 amu) + (50%)(295.30 amu) = 294.225 amu

Hence, the atomic weight of nv is 294.225 atomic mass units.

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Stells [14]

<h3>Answer;</h3>

= 64 N/m

<h3>Explanation;</h3>

According to Hooke's Law for a helical spring or an elastic material, extensional force is directly proportional to the distance the material has extended.

F = ke; where F is the extension force, k is the spring constant, and e is the distance extended.

Thus;

k = F/e

= 44N/0.69 m

= 63.768 N/m

= 64 N/m

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3 years ago

think of two school rules and briefly describe how these rules help protect health and why its important to follow them.

Softa [21]

Answer:

Wash your hand when you are done using the restroom because you could spread germs if not.

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4 years ago

Read 2 more answers

At time t=0 a grinding wheel has an angular velocity of 30.0 rad/s . It has a constant angular acceleration of 35.0 rad/s2 until

Rama09 [41]

Answer:

(A) 570 rad

(B) 10 s

Explanation:

The equations of motion for circular motions are used.

Initial angular velocity, $\omega_0 = 30.0 \text{ rad/s}$

Angular acceleration, $\alpha =35.0 \text{ rad/s}^2$

(A)

At t = 2.00 s, the angular displacement, θ, is given by

$\theta = \omega_0t+\frac{1}{2}\alpha t^2 = (30\times 2) + \frac{1}{2}\times35\times2^2=60+70 = 130\text{ rad}$

After this time, it decelerates through an angular displacement of 440 rad.

Total angular displacement = 130 + 440 rad = 570 rad

(B)

At the time the circuit breaker tips, the angular velocity is given by

$\omega = \omega_0+\alpha t = 30.0+(35.0\times 2) = 30.0+70.0 =100.0\ \text{rad/s}$

This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:

$\theta_2 = \left(\dfrac{\omega_i+\omega_f}{2}\right)t$