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den301095 [7]
3 years ago
8

Plz help me and thx if you do!

Physics
1 answer:
Rina8888 [55]3 years ago
3 0
I think it’s b... not sure tho sorry
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Manganese-52 has a half-life of 6 days. How many days would a scientist have to wait for the radioactivity to be 12.5% the start
sineoko [7]

Answer:

Let N = N0 where N0 is the number of atoms originally present.

In 6 days    N = N0 / 2

In 12 days   N = N0 / 4

In 18 days   N = N0 / 8    = .125 N0

So it would take 18 days.

4 0
3 years ago
Read 2 more answers
Which describes an object in projectile motion? Check all that apply.A.Gravity acts to pull the object downB.The object moves in
irina1246 [14]

Answer:

A.Gravity acts to pull the object down

D.The object’s inertia carries it forward.

E.The path of the object is curved.

Explanation:

The motion of a projectile consists of two independent motions:

- A uniform motion along the horizontal direction, with constant horizontal speed

- A vertical motion with constant acceleration of g = 9.8 m/s^2 downward (acceleration due to gravity), due to the presence of the force of gravity, so the vertical velocity changes (increases in the downward direction)

As a result, the combined motion of the projectile has a curved trajectory (parabolic, more specifically). So the following options are correct:

A.Gravity acts to pull the object down --> gravity acts along the vertical direction

D.The object’s inertia carries it forward. --> there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal motion continues with constant speed

E.The path of the object is curved

3 0
3 years ago
What and where is the asteroid belt?<br><br> Please ANSWER THIS
Rudiy27

Answer:

The asteroid belt is a region of our solar system, between the orbits of Mars and Jupiter, in which many small bodies orbit our sun.

Explanation:

Hope this helps!

3 0
3 years ago
Read 2 more answers
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
Jack and jill are married. this year jack earned $72,000 and jill earned $80,000 and they received $4,000 of interest income fro
MaRussiya [10]
<span>If Jack is filing married-filing-separate he would report $76,000 gross income as head of household.</span>
4 0
3 years ago
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