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3 years ago

Which chart provides the correct number of neutrons for each isotope of tin (Sn)?

Physics

1 answer:

Trava [24]3 years ago

6 0

We know, Atomic mass = Atomic number + number of neutrons
Here, Atomic mass - 50 = number of neutrons [ as A.N. = 50 ]

So, According to our info. only option B can be correct, as the difference between atomic mass & number of neutrons is 50, which is atomic number of Tin (Sb)

In short, Your Answer would be Option B

Hope this helps!

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10. when creating a cardiovascular fitness routine which program would be appropriate?

sasho [114]

Answer:

10. B

13. It's actually 30 seconds but 20 comes closest so B

14. B, any type of stretching is good before you perform any type of physical activity because it loosens up the muscles

15. D, all of the above

Explanation:

I have done many sports in my life and am still very active, I grew to know most of these things through the years.

5 0

3 years ago

A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o

blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

$P=\rho \times g \times h$

Here

P is the pressure which is given as 68 kPa.
ρ is the density of the oil whose SG is 0.92. It is calculated as

$\rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\$

g is the gravitational constant whose value is 9.8 m/s^2
h is the height which is to be calculated

$P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m$

So the height of column is 7.54m

a-3

By the relation of volume and density

$M=\rho \times V$

Here

ρ is the density of the oil which is 920 kg/m^3
V is the volume of cylinder with diameter 16m calculated as follows

$V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3$

Mass is given as

$M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg$

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

$\Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\$

Now corrected pressure is as

$P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa$

Finding the value of height for this corrected pressure as

$P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m$

The original height of column is 5.98m

4 0

3 years ago

How does the gravity which pulls the moon and earth toward each other also affect the ocean?

slamgirl [31]

Gravity is what holds the planets in orbit around the sun and what keeps the moon in orbit around Earth. The gravitational pull of the moon pulls the seas towards it, causing the ocean tides. Gravity creates stars and planets by pulling together the material from which they are made.

5 0

3 years ago

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People in the future may well live inside a rotating space structure that is more than 2 km in diameter. Inside the structure, p

Sergeeva-Olga [200]

Answer:

option B

Explanation:

given,

diameter of the rotating space = 2 Km

Force exerted at the edge of the space = 1 g

force experienced at the half way = ?

As the object is rotating in the circular part

Force is equal to centripetal acceleration.

at the edge

g = ω² r

ω is the angular velocity of the particle

r is the radius.

now, acceleration at the half way

g' = ω² r'

$g' = \omega^2 (\dfrac{r}{2})$

$g' =\dfrac{1}{2}(\omega^2 r)$

$g'=\dfrac{g}{2}$

People at the halfway experience g/2

hence, the correct answer is option B

7 0

3 years ago

1.What type of sediment forms from minerals that crystallize from seawater?

nikdorinn [45]

The correct answers are as follows:
<span>1) hydrogenous sediment

2)sand and gravel

3) They rapidly break down at surface temperatures and pressures.</span>

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3 years ago

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