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RUDIKE [14]
3 years ago
11

A green ball has a mass of 0.525 kg and a blue ball has a mass of 0.482 kg. A croquet player strikes the green ball and it gains

an initial velocity of 2.26 m/s. It then strikes the blue ball, which is initially at rest. After the collision, the green ball has a velocity of 1.14 m/s in the same direction. If the balls roll on a frictionless surface and the collision is head-on, what is the final velocity of the blue ball? (Round your answer to the nearest hundredths place.)
Physics
1 answer:
Gelneren [198K]3 years ago
4 0

Answer:

v' = 1.21 m/s

Explanation:

Mass of a green ball, m = 0.525 kg

Mass of a blue ball, m' = 0.482 kg

Initial velocity of green ball, u = 2.26 m/s

Initial velocity of blue ball, u' = 0 (at rest)

After the collision,

The final velocity of the green ball, v = 1.14 m/s

We need to find the final velocity of the blue ball after the collision if the collision is head on. Let v' is the final velcity of the blue ball. Using the conservation of momentum to find it :

mu+m'u'=mv+m'v'\\\\0.525 (2.26)+0=0.525 (1.14)+0.482v'\\\\0.588=0.482v'\\\\v'=\dfrac{0.588}{0.482}\\\\v'=1.21\ m/s

So, the final velocity of the blue ball is 1.21 m/s.

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The end or finish of an event or process.
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A 75 N box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each
shusha [124]

Answer:

μk = 0.26885

Explanation:

Conceptual analysis

We apply Newton's second law:

∑Fx = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

a= -0.9  m/s²,  

g = 9.81 m/s² : acceleration due to gravity

W= 75 N :  Block weight

W= m*g  

m =  W/g = 75/9.8= 7.65 kg :  Block mass

Friction force : Ff

Ff= μk*N

μk: coefficient of kinetic friction

N : Normal force (N)  

Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W-25 = 0

N = 75 +25

N= 100N

∑Fx = m*ax    

20-Ff= m*ax    

20-μk*100 = 7.65*(-0.90 )

20+7.65*(0.90) = μk*100

μk = ( 20+7.65*(0.90)) / (100)

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4 0
3 years ago
You want to find out how many atoms of the isotope 65Cu are in a small sample of material. You bombard the sample with neutrons
serious [3.7K]

Answer:

a) number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶ atoms

b) m_total Cu = 1.585 10⁹ u = 2.632 10⁻¹⁸ kg

Explanation:

a) For this exercise let's start by using the radioactive decay ratio

           N = N₀  e^{- \lambda t}o e - lambda t

The half-life time is defined as the time it takes for half of the radioactive (activated) atoms to decay, therefore after two half-lives there are

            N = ½ (½ N₀) = ¼ N₀

            N₀ = 4 N

in each decay a photon is emitted so we can use a direct rule of proportions. If an atom emits a photon it has Eo = 1,04 Mev, how many photons it has energy E = 10,000 MeV

          # _atoms = 1 atom (photon) (E / Eo)

          # _atoms = 1 10000 / 1.04

          # _atoms = 9615,4 atoms

          N₀ = 4 #_atoms

          N₀ = 4 9615,4

          N₀=  38461.6  atoms

in the exercise indicates that half of the atoms decay in this way and the other half decays directly to the base state of Zinc, so the total number of activated atoms

          N_activated = 2 # _atoms

          N_activated = 2 38461.6

          N_activated = 76923.2

also indicates that 1% = 0.01 of the nuclei is activated by neutron bombardment

          N_activated = 0.01 N_total

          N_total = N_activated / 0.01

          N_total = 76923.2 / 100

          N_total = 7.692 10⁶ atoms

so the number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶

b) the natural abundance of copper is

  ⁶³Cu     69.17%

  ⁶⁵Cu    30.83%

Let's use a direct proportion rule. If there are 7.692 10⁶  ⁶⁵Cu that represents 30.83, how much ⁶³Cu is there that represents 69.17%

                # _63Cu = 69.17%  (7.692 10⁶    / 30.83%)

                # _63Cu = 17.258 10⁶  atom  ⁶³Cu

the total amount of comatose is

              #_total Cu = #_ 65Cu + # _63Cu

              #_total Cu = (7.692 + 17.258) 10⁶

              #_total Cu = 24.95 10⁶

the atomic mass of copper is m_Cu = 63.546 u

          m_total = #_totalCu m_Cu

          m_total = 24.95 10⁶ 63,546 u

          m_total = 1.585 10⁹ u

let's reduce to kg

           m_total Cu = 1.585 10⁹ u (1,66054 10⁻²⁷ kg / 1 u)

           m_total Cu = 2.632 10⁻¹⁸ kg

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3 years ago
A student was conducting an experiment in which a heater was placed on one side of a large
erastovalidia [21]

Answer:

the transfer of heat through convection

Explanation:

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3 years ago
While accelerating at 5.22 m/s/s an object changes its velocity from 6.73 m/s to 29.88 m/s. Over what
yan [13]

Answer:

Explanation:

v² = u² + 2as

s = (v² - u²) / 2a

s = (29.88² - 6.73²) / (2(5.22))

s = 81.1802203065...

s = 81.18 m

4 0
3 years ago
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