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navik [9.2K]
4 years ago
5

When density increases, what happens to the number of molecules in a volume of air?

Physics
1 answer:
solniwko [45]4 years ago
5 0

When density increases the number of molecules stay the same, im not entirely sure but i hope its right.

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A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 330 N/m. The object is droppe
pentagon [3]

Answer:

(A) l = 0.39 m      

(B)  l =0.38 m  

(C) l = 0.14 m

Explanation:

Answer:

Explanation:

Answer:

Explanation:

from the question we are given the following values:

mass (m) = 1.6 kg

height (h) = 1.05 m

compression of spring (l) = ?

spring constant (k) = 330 N/m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) initial potential energy of the object = final potential energy of the spring

         potential energy of the object = mg(1.05 + l)  

         potential energy of the spring = 0.5 x k x l^{2}  (k= spring constant)

 therefore we now have

              mg(1.05 + l)  = 0.5 x k x l^{2}

              1.6 x 9.8 x (1.05 + l)  = 0.5 x 300 x l^{2}

               15.68 (1.05 + l) = 150 x l^{2}

                   16.5 + 15.68l = 150l^{2}

l = 0.39 m        

(B)   with constant air resistance the equation applied in part A above becomes

initial P.E of the object - air resistance = final P.E of the spring

mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}        

     1.6 x 9.8 x (1.05 + l) - 0.750(1.05 + l)  = 0.5 x 300 x l^{2}

         (16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}        

          16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

            15.71 + 14.93l = 150^{2}

            l =0.38 m  

(C)   where g = 1.63 m/s^{2} and neglecting air resistance

      the equation mg(1.05 + l)  = 0.5 x k x l^{2} now becomes

        1.6 x 1.63 x (1.05 + l)  = 0.5 x 300 x l^{2}

        2.608 (1.05 +l) = 0.5 x 300 x l^{2}

        2.74 + 2.608l = 150 x l^{2}

l = 0.14 m

6 0
3 years ago
An increase in the atomic number____the atomic radius moving from
olganol [36]

Answer:

to

Explanation:

7 0
3 years ago
A golf club rotates 215 degrees and has a length (radius) equal to 29 inches. The time it took to swing the club was 0.8 seconds
vichka [17]

Answer:

The average linear velocity (inches/second) of the golf club is 136.01 inches/second

Explanation:

Given;

length of the club, L = 29 inches

rotation angle, θ = 215⁰

time of motion, t = 0.8 s

The angular speed of the club is calculated as follows;

\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega =  (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s

The average linear velocity (inches/second) of the golf club is calculated as;

v = ωr

v = 4.69 rad/s  x  29 inches

v = 136.01 inches/second

Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second

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3 years ago
Assume the following vehicles are all moving at the Sam speed .it would be harder to change the velocity of which vehicle . What
AVprozaik [17]

Answer:what the choices

Explanation:

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What is the difference between a neap and a spring tide in terms of size?
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Spring tides occur when the moon is either new or full, and the sun, the moon, and the Earth are aligned. ... neap tide- A tide in which the difference between high and low tide is the least. Neap tides occur twice a month when the sun and moon are at right angles to the Earth.
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