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3 years ago

6

A balloon filled with helium gas at 1.00 atm occupies 15.6 L. Will the volume of the balloon increase or decrease in the upper a

tmosphere, at a pressure of 0.150 atm?

1 answer:

Mariana [72]3 years ago

6 0

Answer:

The volume of the balloon increases in the upper atmosphere.

Explanation:

p1= 1 atm

p2= 0.15 atm

V1= 15.6 L

V2= ?

p1*V1= p2 * V2

V2= (p1/p2)*V1

V2= 104 L

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In the 25 ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of int

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Answer:

a) <em>8.33 x 10^-6 Pa</em>

b) <em>8.23 x 10^-11 atm</em>

c) <em>1.67 x 10^-5 Pa</em>

d) <em>1.65 x 10^-10 atm</em>

<em></em>

Explanation:

Intensity of the light $I$ = 2500 W/m^2

speed of light $c$ <u> </u>= 3 x 10^8 m/s

a) we know that the pressure for for a totally absorbing surface is given as

$P_{abs}$ = $I/c$ = 2500/(3 x 10^8) = <em>8.33 x 10^-6 Pa</em>

b) 1 atm = 101325 Pa

$P_{abs}$ = (8.33 x 10^-6)/101325 = <em>8.23 x 10^-11 atm</em>

c) for a totally reflecting surface

$P_{ref}$ = $2I/c$ = twice the value for totally absorbing

$P_{ref}$ = 2 x 8.33 x 10^-6 = <em>1.67 x 10^-5 Pa</em>

d) 1 atm = 101325 Pa

$P_{ref}$ = 2 x 8.23 x 10^-11 = <em>1.65 x 10^-10 atm</em>

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How do you do this problem?

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Explanation:

First, find the velocity of the projectile needed to reach a height h when fired straight up.

Given:

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v₀ = √(2gh) sin 45° = √(2gh) / √2 = √(gh)

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Read 2 more answers

A vertical spring (spring constant =160 N/m) is mounted on the floor. A 0.340-kg block is placed on top of the spring and pushed

AleksandrR [38]

(a) 3.5 Hz

The angular frequency in a spring-mass system is given by

$\omega=\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

Here in this problem we have

k = 160 N/m

m = 0.340 kg

So the angular frequency is

$\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s$

And the frequency of the motion instead is given by:

$f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz$

(b) 0.021 m

The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at

x = A

where A is the amplitude of the motion.

The maximum displacement is given by Hook's law:

$F=kA$

where

F is the force applied initially to the spring, so it is equal to the weight of the block:

$F=mg=(0.340 kg)(9.81 m/s^2)=3.34 N$

k = 160 N/m is the spring constant

Solving for A, we find

$A=\frac{F}{k}=\frac{3.34 N}{160 N/m}=0.021 m$

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3 years ago