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DENIUS [597]
3 years ago
11

A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.90 x 105 P

a. Assuming that the top and bottom surfaces of the cap each have an area of 3.90 x 10-4 m2, obtain the magnitude of the force that the screw thread exerts on the cap in order to keep it on the bottle. The air pressure outside the bottle is one atmosphere.
Physics
1 answer:
worty [1.4K]3 years ago
6 0

Answer:

The magnitude of the force is 34.59 N.

Explanation:

Given that,

Inside pressure P_{in}= 1.90\times10^{5}\ Pa

Area A=3.90\times10^{-4}\ m^2

Outside pressure = 1 atm

We need to calculate the magnitude of the force

Using formula of force

F_{net}=F_{in}-F_{out}

F_{net}=(P_{in}-P_{out})A

Where, P_{in} =inside Pressure

P_{out} =outside Pressure

A = area

Put the value into the formula

F_{net}=(1.90\times10^{5}-1.013\times10^{5})3.90\times10^{-4}

F_{net}=34.59\ N

Hence, The magnitude of the force is 34.59 N.

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A nucleus with a mass number of 64 has a mean radius of about: _________.
dangina [55]

Answer:

The correct option is A

Explanation:

From the question we are told that

The mass number is A= 64

Generally the mean radius is mathematically evaluated as

R  =  R_o A^{\frac{1}{3} }

Here R_o is a constant with a value  R_o =1.2*10^{-15}

So  

     R  =  1.2*10^{-15} *  64^{\frac{1}{3} }

      R  =   4.8 *10^{-15}

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To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

a_g = \frac{GM}{R^2}

Here

M = \text{Mass inside the Orbit of the star}

R = \text{Orbital radius}

G = \text{Universal Gravitational Constant}

Mass inside the orbit in terms of Volume and Density is

M =V \rho

Where,

V = Volume

\rho =Density

Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

a_c = \frac{4}{3} G\pi R\rho

At the same time the general expression for the centripetal acceleration is

a_c = \frac{\Theta^2}{R}

Where \Theta is the orbital velocity

Using this expression in the left hand side of the equation we have that

\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2

\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}

\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

\Theta \propto R

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

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The histogram below shows the number of downloads of a song over time.
Allisa [31]

Given data:

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  • A diagram consists rectangles, whose area is proportional to frequency of a variable and whose width is equal to the class interval.
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<em>From Figure:</em>  

        Each box in the graph (small rectangle box) is assumed to be one download. So, in the graph the time between 8 p.m to 9 p.m, the number of downloads are 8.75 approximately (because the last box is incomplete, therefore 8 complete boxes and 9th is more than half).

<em>So, We conclude that the total number of downloads are approximately 9 in the time span of 8 p.m. to 9 p.m.</em>

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