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icang [17]
3 years ago
9

Suppose you have a rock that, when it solidifies, contains 1 microgram of a radioactive isotope. How much of this isotope remain

s after five half-lives? a) none b) 1/8 microgram c) 1/16 microgram d) 1/32 microgram e) Cannot be determined without knowing the daughter isotope.
Physics
1 answer:
algol133 years ago
7 0

Answer:

d) 1/32 microgram

Explanation:

First half life is the time at which the concentration of the reactant reduced to half.

Second half reaction is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/4.

Third half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/8.

Forth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/16.

Fifth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/32.

The initial mass of the sample = 1 microgram

After 5 half-lives, the mass should reduce to 1/32 of the original.

So the concentration left = 1/32 of 1 microgram = 1/32 microgram

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A(n) 636 kg elevator starts from rest. It moves upward for 4.5 s with a constant acceleration until it reaches its cruising spee
zmey [24]

Answer:

The average power delivered by the elevator motor during this period is 6.686 kW.

Explanation:

Given;

mass of the elevator, m = 636 kg

initial speed of the elevator, u = 0

time of motion, t = 4.5 s

final speed of the elevator, v = 2.05 m/s

The upward force of the elevator is calculated as;

F = m(a + g)

where;

m is mass of the elevator

a is the constant acceleration of the elevator

g is acceleration due to gravity = 9.8 m/s²

a = \frac{v-u}{t} \\\\a = \frac{2.05 -0}{4.5} \\\\a = 0.456 \ m/s^2

F = (636)(0.456 + 9.8)

F = (636)(10.256)

F = 6522.816 N

The average power delivered by the elevator is calculated as;

P_{avg} = \frac{1}{2} (Fv)\\\\P_{avg} = \frac{1}{2} (6522.816 \ \times \ 2.05)\\\\P_{avg} = 6685.89 \ W\\\\P_{avg} =  6.68589 \ kW\\\\P_{avg} =  6.686 \ k W

Therefore, the average power delivered by the elevator motor during this period is 6.686 kW.

3 0
3 years ago
I’ll give brainliest and 10 points
leonid [27]

Answer:

what's these its D okkkk

7 0
3 years ago
How long must a current of 100mA flow so as to transfer a charge of 80 C? ​
defon

Answer:

800s

Explanation:

Q=It

where,

Q=80c

I =100MA=0.1A

t=?

Q=It

800=0.1×t

t=80×0.1

t=800s

6 0
3 years ago
Read 2 more answers
An arch carries the thrust of weight to its _____(1)______. With a _____(2)______, the horizontal part of the structure supports
finlep [7]

Option D is correct. An arch carries the thrust of weight to its <u>sides </u>with a <u>post-and-lintel.</u>

<u></u>

<h3>What is an arch?</h3>

An arch is indeed a vertical curving construction that covers an elevated space that may or may not sustain the load above it or the pressure gradient against it

In the case of a horizontally arched, such as an embankment dam. While arches and vaults are often confused, A vault is defined as an ongoing arch forming a roof.

Option D satisfies the fill-in blanks option.

Hence option D is correct. An arch carries the thrust of weight to its <u>sides </u>with a <u>post-and-lintel.</u>

<u></u>

To learn more about the arch refer to the link;

brainly.com/question/18162421

6 0
2 years ago
A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.
denpristay [2]

Answer:

Approximately 5.5\; \rm N, assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let q_1 and q_2 denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, q_1 = 20 \times 10^{-6}\; \rm C  and q_2 = 15 \times 10^{-6}\; \rm C.

Let r denote the distance between these two point charges. In this question, r = 0.7\; \rm m.

Let k denote the Coulomb constant. In standard units, k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}.

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}.

Substitute in the values and evaluate:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}.

8 0
3 years ago
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