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Fiesta28 [93]
3 years ago
10

1. You have a cat who has a mass of 10 kg and is

Physics
1 answer:
disa [49]3 years ago
7 0

Answer:

1) F = 100N

2) a = 2 m/s²

3) m = 25 kg

Explanation:

1) F = ma ( F = ?, m = 10 kg, a = 10 m/s² )

  F = 10×10

  F = 100 N

2) F = ma ( F = 20N, m = 10 kg, a = ? )

   20 = 10×a

   10a = 20

   a = 20/10

   a = 2 m/s²

3)F = ma ( F = 100N, m = ?, a = 4 m/s² )

  100 = m×4

  4m = 100

  m = 100/4

  m = 25 kg

Hope that helps! Good luck!

 

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2 years ago
If you measured 25 g jelly beans and 25 g of gumdrops, the number of jellybeans would be
KengaRu [80]

The number of jellybeans would be B.more than the gumdrops.

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3 years ago
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How is air resistance similar to gravity? give me two ways.
OLga [1]

Answer:

1. they both act on an object in free fall

Explanation:

2. both help determine how fast the object will accelerate

4 0
2 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.

<h3>a)</h3>

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.

<h3>b)</h3>

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
A 4.00-m long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end
Natalka [10]

Answer:

The angular acceleration α = 14.7 rad/s²

Explanation:

The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod  and L = length of rod = 4.00 m. α = angular acceleration of rod

Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.

So Iα = Wr

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Simplifying by dividing through by mL, we have

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multiplying both sides by 12, we have

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αL = 6g

α = 6g/L

α = 6 × 9.8 m/s² ÷ 4.00 m

α = 58.8 m/s² ÷ 4.00 m

α = 14.7 rad/s²

So, the angular acceleration α = 14.7 rad/s²

5 0
3 years ago
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