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3 years ago

High levels of pollution will MOST likely

Physics

2 answers:

galben [10]3 years ago

4 0

Answer:

Explanation:

A) increase the rate of chemical weathering

Nataly [62]3 years ago

4 0

Answer:

Explanation:

increase the rate of chemical weathering is the right choice

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Bill is learning to play tennis. He does pretty well hitting the ball back to his opponent but, many times he misses the ball wh

KiRa [710]

Answer: Place his feet parallel to the baseline prior to tossing the ball

5 0

3 years ago

Students are asked to create roller coasters for marbles. Their goal is to design a coaster with the tallest possible hill that

ivann1987 [24]

Answer:

Kinetic Energy.

Explanation:

The movement of a roller coaster is accomplished by the conversion of potential energy to kinetic energy. The roller coaster cars gain potential energy as they are pulled to the top of the first hill. As the cars descend the potential energy is converted to kinetic energy.

3 0

3 years ago

5. Put these in order from least to greatest resistance:

zloy xaker [14]

superconductors, conductors, semiconductors, insulators

8 0

3 years ago

The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the

MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations. The half-life formula is $P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}$ where $P_o$ is the initial amount, $k$ is the length of the half-life, $t$ is the amount of time that has elapsed since the initial measurement was taken, and $P$ is the amount that remains at time $t$ .

$P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}$

Deriving the half-life formula

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, $y=a b^{t}$ , where $t$ is still the amount of time, and $y$ is the amount remaining at time $t$ . The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

$y=a b^{t}$

$(14.2)=ab^{(0)}$

$14.2=a *1$

$14.2=a$

So, $a=14.2$ , which represents out initial amount of the substance, and our equation becomes: $y=14.2 b^{t}$

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

$\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}$

$\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}$

$0.5=b^{8.0252}$

$\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}$

$\sqrt[8.0252]{\frac{1}{2}}=b$

Recalling exponent properties, one could find that $\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b$ , which will give the equation identical to the half-life formula. However, recalling this trivia about exponent properties is not necessary to solve this problem. One can just evaluate the radical in a calculator:

$b=0.9172535661...$

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

$y=14.2* (0.9172535661)^t$ or $y=14.2*(0.5)^{\frac{t}{8.0252}}$ where y represents the amount remaining at time t.

Solving for the amount remaining

With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:

$y=14.2* (0.9172535661)^{(31.8)}$ $y=14.2*(0.5)^{\frac{(31.8)}{8.0252}}$

$y=14.2* 0.0641450581$ $y=14.2*(0.5)^{3.962518068}$

$y=0.9108598257$ $y=14.2* 0.0641450581$

$y=0.9108598257$

Since both the initial amount of Iodine, and the amount of time were given to 3 significant figures, the amount remaining after 31.8days is 0.911g.

8 0

2 years ago

A 18.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.9-T magnetic field. The loop is rotated so that its p

Alchen [17]

Answer:

0.3405V

Explanation:

#Given a magnetic field of $1.9T$ , diameter= 18.5cm(r=9.25cm or 0.0925m), we find the magnetic flux of the loop as:

$\phi=B.(\pi r^2)cos 0\textdegree\\=1.9\pi\times 0.0925^2 \times cos 0\textdegree\\=5.107\times10^-^2 \ Tm^2$

we can now calculate the induced emf, $\frac{\phi}{\bigtriangleup t}$ :

$\frac{\phi}{\bigtriangleup t}=\frac{5.107\times 10^-^2}{0.15}\\=3.405\times 10^-^1V$

Hence, the induced emf of the loop is 0.3405V

8 0

4 years ago