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andrew-mc [135]
3 years ago
10

If the activity of a radionuclide decreases from 5,000 disintegrations per minute to 625 disintegrations per minute

Physics
1 answer:
lys-0071 [83]3 years ago
7 0

Answer:

4 hrs

Explanation:

625 = 5000 (1/2)^n

625/5000 = 1/2^n

n = 3 half lives

3 half lives = 12 hours       1 half life = 4 hrs

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6 0
3 years ago
In a simple 2-bulb series circuit, why does the bulb light when you close the switch?
Nataly [62]

Answer:

Because closing the switch completes the circuit.

Explanation:

8 0
3 years ago
Consider f(x) = -4x2 + 24x + 3. Determine whether the function has a maximum or minimum value. Then find the
murzikaleks [220]

Answer:

The function has a maximum in x=3

The maximum is:

f(3) = 39

Explanation:

Find the first derivative of the function for the inflection point, then equal to zero and solve for x

f(x)' = -4*2x + 24=0

-4*2x + 24=0

8x=24

x=3

Now find the second derivative of the function and evaluate at x = 3.

If f (3) ''< 0 the function has a maximum

If f (3) '' >0 the function has a minimum

f(x)''= 8

Note that:

f(3)''= -8

the function has a maximum in x=3

The maximum is:

f(3)=-4(3)^2+24(3) + 3\\\\f(3) = 39

4 0
3 years ago
you place a 0.13kg can of soup and a 0.34kg jar of pickles on the kitchen counter separated by a distance of 0.42m. what is the
Nonamiya [84]

Answer:

1.67×10^-11 N

Explanation:

m1 = 0.13kg

m2 = 0.34kg

d = 0.42m

G = 6.674 × 10 ^−11

F = Gm1m2/r²

F= (6.674 × 10^ −11×0.13×0.34)/(0.42)²

F = 1.67×10^-11 N

4 0
4 years ago
A 1.0 C charged object and a 2.0 C charged object are separated by 100 m. Where should a -1.0x10-3 C charged object be placed on
horsena [70]

Answer:

x = 41.2 m

Explanation:

The electric force is a vector magnitude, so it must be added as vectors, remember that the force for charges of the same sign is repulsive and for charges of different sign it is negative.

In this case the fixed charges (q₁ and q₂) are positive and separated by a distance (d = 100m), the charge (q₃ = -1.0 10⁻³ C)) is negative so the forces are attractive, such as loads q₃ must be placed between the other two forces subtract

             F = F₁₃ - F₂₃

let's write the expression for each force, let's set a reference frame on the charge q1

           F₁₃ = k \frac{q_1 q_3}{x^2}

           F₂₃ = k \frac{q_2 q_3}{(d-x)^2}

they ask us that the net force be zero

           F = 0

           0 = F₁₃ - F₂₃

           F₁₃ = F₂₃

          k \frac{q_1 q_3}{x^2} =k \frac{q_2 q_3}{(d-x)^2}

          \frac{q_1}{x^2} = \frac{q_2}{(d-x)^2 }q1 / x2 = q2 / (d-x) 2

       

           (d-x)² = \frac{q_2}{q_1} x²

we substitute

           (100 - x)² = 2/1  x²

           100- x = √2  x

           100 = 2.41 x

           x = 41.2 m

6 0
3 years ago
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