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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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small plastic container, called the coolant reservoir, catches the radiator fluid that overflowswhen the automobile engine becom
Ilia_Sergeevich [38]

Answer:

0.53 quart

Explanation:

The volume expansion of the coolant is gotten from ΔV = VγΔθ where ΔV   = change in volume of the coolant, V = initial volume of coolant = 15 quart, γ = coefficient of volume expansion of coolant = 410 × 10⁻⁶ /°C and Δθ = temperature change = θ₂ - θ₁ where θ₁ = initial temperature of coolant = 6 °C and θ₂ = final temperature of coolant = 92 °C. So, Δθ = θ₂ - θ₁ = 92 °C - 6 °C = 86 °C

Since, ΔV = VγΔθ

substituting the values of the variables into the equation, we have

ΔV = VγΔθ

ΔV = 15 × 410 × 10⁻⁶ /°C × 86 °C

ΔV = 528900 × 10⁻⁶ quart

ΔV = 0.528900 quart

ΔV ≅ 0.53 quart

Since the change in volume of the coolant equals the spill over volume, thus the overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 °C is 0.53 quart.

4 0
3 years ago
When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the
Taya2010 [7]

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

\theta = 54.6^{\circ}

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = \sqrt{\frac{gx}{sin2\theta}}

v = \sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}

v = \sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}

v = \sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s

Now,

The angular velocity can be calculated as:

v = \omega R

Thus

\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s

8 0
3 years ago
A merry-go-round rotates from rest with an angular acceleration of 1.45 rad/s2. How long does it take to rotate through (a) the
elena55 [62]

Answer:

(a) t = 5.66 s

(b) t = 8 s

Explanation:

(a)

Here we will use 2nd equation of motion for angular motion:

θ = ωi t + (1/2)∝t²

where,

θ = Angular Displacement = (3.7 rev)(2π rad/1 rev) = 23.25 rad

ωi = initial angular speed = 0 rad/s

t = time = ?

∝ = angular acceleration = 1.45 rad/s²

Therefore,

23.25 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²

t² = (23.25 rad)(2)/(1.45 rad/s²)

t = √(32.06 s²)

<u>t = 5.66 s</u>

<u></u>

(b)For next 3.7 rev

θ = ωi t + (1/2)∝t²

where,

θ = Angular Displacement = (3.7 rev + 3.7 rev)(2π rad/1 rev) = 46.5 rad

ωi = initial angular speed = 0 rad/s

t = time = ?

∝ = angular acceleration = 1.45 rad/s²

Therefore,

46.5 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²

t² = (46.5 rad)(2)/(1.45 rad/s²)

t = √(64.13 s²)

<u>t = 8 s</u>

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Jake calculates that the frequency of a wave is 500 hertz and that the wave is moving at 1,250 m/s. What is the wavelength of th
Neko [114]
Frequency (f) = 500 hz (SI)
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Wavelength (Lambda) = ? meters

v =  \lambda \times f
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3) Earth is about 150 million km from the Sun, and the apparent brightness of the Sun in our sky is about 1,300 watts per square meter. Determine the apparent brightness we would measure for the Sun if we were located five times Earth's distance from the Sun. Answer: The Sun would appear 1/25 times as bright.
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