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3 years ago

According to the graph, how many atoms would remain after two half-lives?

Physics

1 answer:

Fudgin [204]3 years ago

4 0

Answer:

Let No be initial no of atoms

N = N0 / 2 after 1 half-life

N = N0 / 4 after 2 half-lives

So after 2 half-lives 20 of the 80 atoms remain

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Type the correct answer in each box. Use numerals instead of words.

ser-zykov [4K]

Answer:

6.23 newtons per second?

Explanation:

4 0

3 years ago

0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained a

klio [65]

Answer:

The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

Explanation:

Given the data in the question;

Using the Clapeyron equation

$(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}$

$(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }$

where $h_{fg$ is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu

T is the temperature ( 15 + 460 )R

m is the mass of water ( 0.5 Ibm )

$V_{fg$ is specific volume ( 1.5 ft³ )

we substitute

$(\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5})$ / $( (15+460)\frac{1.5}{0.5})$

$(\frac{dP}{dT} )_{sat } =$ 272.98 Ibf-ft²/R

Now,

$(\frac{dP}{dT} )_{sat } =$ $(\frac{P_2 - P_1}{T_2 - T_1})_{sat$

where P₁ is the initial pressure ( 50 psia )

P₂ is the final pressure ( 60 psia )

T₁ is the initial temperature ( 15 + 460 )R

T₂ is the final temperature = ?

we substitute;

$T_2$ $= ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}$

$T_2 = 475 + 5.2751\\$

$T_2 =$ 480.275 R

Therefore, boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

3 0

3 years ago

a ball was projected in such a way that it attained the maximum horizontal distance with a velocity of 20m/s calculate the verti

Eduardwww [97]

Answer:

80m

Explanation:

u=20,R=?,sin theta=1,g=10

R=u²sin2theta/g

R=20²x2/10

R=400x2=800/10

R=80m

7 0

3 years ago

An object is thrown vertically upward such that it has a speed of 25 m/s when it reaches two thirds of its maximum height above

AlekseyPX

Answer:

v0 = 25 m/s

vf = 0 m/s

a = -9.80 m/s^2

change in x = 31.89m

but that's only 1/3 of the hight, so i time it by 3 to get 96m

4 0

3 years ago

The only force acting on a 3.1 kg body as it moves along the positive x axis has an x component fx = -8x n, where x is in meters

My name is Ann [436]

<span>(a) -9.97 m/s
(b) x = 2.83

This is a simple problem in integral calculus. You've been given part of the 2nd derivative (acceleration), but not quite. You've been given the force instead. So let's setup a function for acceleration.

f''(x) = -8x N / 3.1 kg= -8x kg*m/s^2 / 3.1 kg = -2.580645161x m/s^2

So the acceleration of the body is now expressed as
f''(x) = -2.580645161x m/s^2

Let's calculate the anti-derivative from that.
f''(x) = -2.580645161x m/s^2
f'(x) = -1.290322581x^2 + C m/s

Now let's use the known velocity value at x = 2.0 to calculate C
f'(x) = -1.290322581x^2 + C 1
1 = -1.290322581*2^2 + C
11 = -1.290322581*4 + C
11 = -5.161290323 + C
16.161290323 = C

So the velocity function is
f'(x) = -1.290322581x^2 + 16.161290323

(a) The velocity at x = 4.5
f'(x) = -1.290322581x^2 + 16.161290323
f'(4.5) = -1.290322581*4.5^2 + 16.161290323
f'(4.5) = -1.290322581*20.25 + 16.161290323
f'(4.5) = -26.12903227 + 16.161290323
f'(4.5) = -9.967741942

So the velocity is -9.97 m/s

(b) we want a velocity of 5.8 m/s
5.8 = -1.290322581x^2 + 16.161290323
0 = -1.290322581x^2 + 10.36129032
1.290322581x^2 = 10.36129032
x^2 = 8.029999998
x = 2.833725463</span>

4 0

3 years ago