When we look up at the stars, we are seeing the future", is this astronomer correct ? Why or why not ?

A student pushes against a wall with a force of 30N. The wall does not move. What amount of force does the wall exert on the stu

True [87]

Answer:

C

Explanation:

they both have to be the same for both to not move

8 0

2 years ago

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What are the three parts of Kinetic Theory

anyanavicka [17]

1. No energy is gained or lost when molecules collide.
2. The molecules in a gas take up a negligible (able to be ignored) amount of space in relation to the container they occupy.
3. The molecules are in constant, linear motion.

8 0

3 years ago

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Directions: Using the T-chart below, compare balanced forces and unbalanced forces.

Salsk061 [2.6K]

Explanation:

unbalanced: a turning vehicle, apple falling on the ground, kicking a ball

balanced: floating on water, fruit hanging from tree, tug of war equally balanced teams

8 0

2 years ago

Problem set 3 » (8) river swimming a swimmer heads directly across a river, swimming at her maximum speed of 1.30 m/s relative t

jasenka [17]

Velocity of swimmer across river = 1.30 m/s

Distance arrived downstream = 48 m

Width of river = 64 m

Time taken to cross river = $\frac{Width of river}{Velocity across river}$

= $\frac{64}{1.30} =49.23 s$

Speed of river current = $\frac{Distance arrived downstream}{Time taken to cross river}$

= $\frac{48}{49.23} = 0.975 m/s$

So, the river is flowing at a speed 0.975 m/s.

5 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.