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3 years ago

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Describe the motion for each segment below. Include start position, relative speed, and direction. Then calculate the speed of e

ach section. Finally, write the equation of the line segment for segment C.

1 answer:

ratelena [41]3 years ago

3 0

Where is the picture?

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How many grams of CO-60 result in 1 Millicuire of activity? How many years until the activity decays to 1 microcuire tl/2 =5.3 Y

vredina [299]

Answer:

m =8.81*10^{-6}grams

time t = 52.8 year

Explanation:

GIVEN DATA:

the half life of the CO-60 is, T_1/2 = 5.27 years = 1.663 e+8 s

activity dN/dt = 1 mCi = 3.7 X 10^7 decay/s

activity , $dN/dt = N* \lambda$

$\frac{dN}{dt} = N* \frac{ln2}{T_1/2}$

$N = \frac{(dN/dt )(T1/2)}{ln2}$

= ( 3.7 X 10^7 )(1.663*10^8 ) / ln2

= 8.877*10^{16}

Number of moles:

n = N/NA = 8.877*10^{16} / 6.022X10^23 = 1.474*10^{-7} mol

mass of the CO-60 is,

m = n*M = [1.474*10^{-7} mol]*[59.93 grams /mol] = 8.81*10^{-6}grams

-----------------------------------------------------------------------------------------

time t = -[T1/2 / ln2]*ln[N/N0]

= - [5.3 years / ln2]*ln[1x10-6/1x10-3]

= 52.8 year

8 0

3 years ago

Can any one answer these two

dalvyx [7]

) (2.68 x 10¯5) x (4.40 x 10¯8)

The calculator display gives 1.1792 x 10¯12. Rounded off to three significant figures gives 1.18 x 10¯12 as the answer.

2) (2.95 x 107) ÷ (6.28 x 1015)

The calculator display gives 4.6975 x 10¯9. Rounded off to three significant figures gives 4.70 x 10¯9 as the answer.

3) (8.41 x 106) x (5.02 x 1012)

The calculator display gives 4.2218 x 1019. Rounded off to three significant figures gives 4.22 x 1019 as the answer.

When done as a division, the answer to this problem is 1.68 x 10¯6.

4) (9.21 x 10¯4) ÷ (7.60 x 105)

The calculator display gives 1.2118 x 10¯9. Rounded off to three significant figures gives 1.21 x 10¯9 as the answer.

1) (2.68 x 10¯5) x (4.40 x 10¯8)

The calculator display gives 1.1792 x 10¯12. Rounded off to three significant figures gives 1.18 x 10¯12 as the answer.

2) (2.95 x 107) ÷ (6.28 x 1015)

The calculator display gives 4.6975 x 10¯9. Rounded off to three significant figures gives 4.70 x 10¯9 as the answer.

3) (8.41 x 106) x (5.02 x 1012)

The calculator display gives 4.2218 x 1019. Rounded off to three significant figures gives 4.22 x 1019 as the answer.

When done as a division, the answer to this problem is 1.68 x 10¯6.

4) (9.21 x 10¯4) ÷ (7.60 x 105)

The calculator display gives 1.2118 x 10¯9. Rounded off to three significant figures gives 1.21 x 10¯9 as the answer

7 0

3 years ago

An arrow of mass 20 g is shot horizontally into a bale of hay, striking the hay with a velocity of 60 m/s. It penetrates a depth

goblinko [34]

From the question, The kinetic energy of the fired arrow is equal to the work done by the bale of hale in stopping the arrow.

We make use of the following formula

mv²/2 = F'd................... Equation 1

Where

m = mass of the arrow
v = velocity of the arrow
F' = average stopping force acting on the arrow
d = distance of penetration

Make F' the subject of the equation

F' = mv²/2d.................. Equation 2

From the question,

Given:

m = 20 g = 0.02 kg
v = 60 m/s
d = 40 cm = 0.4 m

Substitute these values into equation 2

F' = 0.02(60²)/(0.4×2)

F' = 72/0.8

F' = 90 N

Hence, The average stopping force acting on the arrow is 90 N

Learn more about average stooping force here: brainly.com/question/13370981

5 0

2 years ago

As the building collapses, the volume of air inside the building decreases, while the mass of the air stays the same. This means

Ymorist [56]

Answer:

Density

Explanation:

4 0

3 years ago

A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 6.3 m/s2. For what value of the coe

Levart [38]

Answer:

Coefficient of static friction will be equal to 0.642

Explanation:

We have given acceleration $a=6.3m/sec^2$

Acceleration due to gravity $g=9.8m/sec^2$

We have to find the coefficient of static friction between truck and a cabinet will

We know that acceleration is equal to $a=\mu g$ , here $\mu$ is coefficient of static friction and g is acceleration due to gravity

So $\mu =\frac{a}{g}=\frac{6.3}{9.8}=0.642$

So coefficient of static friction will be equal to 0.642

3 0

3 years ago