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3 years ago

Why is science literacy important?

1 answer:

8 0

Science literacy is important because it provides a context for addressing societal problems, and because a science- literate populace can better cope with many of its prob- lems and make intelligent and informed decisions that will affect the quality of their lives and those of their children

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A train is travelling at a uniform speed of 140 km/h. How long will it take to cover a

Firlakuza [10]

Hope this helps!!!!!!!!!!!!

5 0

3 years ago

The same strength force was exerted in the same direction on both Object A and Object B. Why did Object A go faster than Object

VashaNatasha [74]

Answer:

$m_B > m_A$

the mass of body B must be greater than the mass of body A

Explanation:

Newton's second law establishes a linear relationship between the force, the mass of the body and its acceleration

F = m a

a = F / m

Let's analyze this expression tells us that the force is of equal magnitude for the two bodies, but body A goes faster than body B, this implies that it has more relationships

a_A > a_B

$\frac{F}{m_A} > \frac{F}{m_B}$

$m_B > m_A$

Therefore, for this to happen, the mass of body B must be greater than the mass of body A

6 0

3 years ago

A bus traveling at 24 m/s slows down to 12 m/s in 5.0 seconds. What is the acceleration?

Contact [7]

Answer:

a = -2.4 m/s²

Explanation:

Given,

The initial speed of the bus, u = 24 m/s

The final speed of bus, v = 12 m/s

Time taken to reach final speed is, t = 5.0 s

The acceleration of the body is given by the change in velocity by time

a = (v - u) / t

= (12 - 24) / 5

= -2.4 m/s²

The negative sign in the acceleration indicates that the bus is decelerating.

Therefore, the acceleration of the bus is, a = -2.4 m/s²

8 0

3 years ago

Two coulombs of charge pass through a wire in 5 seconds. Calculate the current in the wire. 1

gogolik [260]

Answer:

0.4A.

Explanation:

Current (A) = Charge (coulomb)/Time (secs)

2 coulombs/5 secs = 0.4A

7 0

3 years ago

Given the isotope 2Fes, which has an actual mass of 55.934939 u: a) b) Determine the mass defect of the nucleus in atomic mass u

SSSSS [86.1K]

Answer:

Mass defect of each iron-56 nuclei:

The binding energy per nucleon of Iron-56 is approximately 8.6 MeV.

Explanation:

According to the physics constants table on Chemistry Libretexts:

Proton rest mass: $\rm 1.0072765\;amu$ ;
Neutron rest mass: $\rm 1.0086649\; amu$ .
Speed of light in vacuum: $\rm 2.99792458\times 10^{8}\;m\cdot s^{-1}$ .
Charge on an electron: $\rm 1.6021765\times 10^{-19}\;C$ .

The mass defect of a nucleus is equal to the sum of the mass of its parts (protons and, in most cases, neutrons) minus the mass of the nucleus.

The atomic number of iron is 26. There are 26 protons in each iron-56 nucleus. The mass number 56 indicates that there are 56 nucleons (neutrons and protons) in each iron-56 nucleus. The other $56 - 26 = 30$ particles are neutrons.

The mass of protons and neutrons in each iron-56 nucleus will be:

$\rm 26 \times 1.0072765 + 30 \times 1.0086649 = 56.464736\;amu$ .

According to this question, the mass of an iron-56 nucleus is equal to 55.934939 amu. The mass defect will be

$\rm 56.464736 - 55.934939 = 0.514197\;amu$ .

By the mass-energy equivalence,

$E = m\cdot c^{2}$ .

Refer to this equation, the speed of light in vacuum $c^{2}$ is the conversion factor between mass $m$ and energy $E$ . The value of $c$ is usually given only in SI units $\rm m\cdot s^{-1}$ . Accordingly, the value of $c^{2}$ will be in the SI unit $\rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}$ .

Convert million electron-volts to joules.

One electron-volt is equal to the electrical work done moving an electron across a potential difference of one volt.

$\begin{aligned}\rm 1 MeV&= \rm 10^{6}\; eV\\ &= \rm (10^{6}\times 1.6021765\times 10^{-19}\;C)\times 1\; V\\&=\rm 1.6021765\times 10^{-19}\;J\end{aligned}$ .

Convert the unit of $c^{2}$ from $\rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}$ to the desired $\rm MeV \cdot amu^{-1}$ :

$\begin{aligned}c^{2} &= \rm {\left(2.99792458\times 10^{8}\;m\cdot s^{-1}\right)}^{2}\\&=\rm 8.987551787\times 10^{16}\; m^{2}\cdot s^{-2}\\ &= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\\&= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\times \frac{1\;MeV}{1.6021765\times 10^{-13}\;J}\times \frac{1\times 10^{-3}\;kg}{6.022142\times 10^{23}\;amu}\\&\approx \rm 931.602164\;MeV\cdot amu^{-1}\end{aligned}$ .

Total binding energy in each iron-56 nucleus:

$\begin{aligned}E &= m\cdot c^{2}\\&= \rm 0.514197\;amu \times 9.31602164\;MeV\cdot amu^{-1} \\&=\rm 479.027038\; MeV \end{aligned}$ .

Again, the mass number 56 indicates that there are 56 nucleons in each iron-56 nucleus. The binding energy per nucleon of iron-56 $\mathrm{^{56}Fe}$ will be:

$\displaystyle \rm \frac{479.027038\; MeV}{56} \approx 8.6\; MeV$ .

6 0

3 years ago