What happens when you stick your head in a microwave?

If you run at an average speed of 10mi/h how long will it take for you to run 2.5miles

givi [52]

Answer:

If the avg speed is 10mi/h and you want to know how long it will take to run 2.5mi/h you put that as a ratio 2.5/10 which is 1/4 of an hour so it will take 15 minutes to run 2.5 miles

Explanation:

3 0

2 years ago

??? True or false. An element can only exist as an individual atom.

KengaRu [80]

I think it's false because Oxygen is diatomic so it's found as O2 which means that it has two atoms of itself.

7 0

3 years ago

A typical student listening attentively to a physics lecture has a heat output of 100 w . how much heat energy does a class of 1

nexus9112 [7]

Since each student emits 100 W, so 170 students will emit:

total heat = 100 W * 170 = 17,000 W

Convert minutes to seconds:

time = 50 min * (60 s / min) = 3000 s

The energy is therefore:

E = 17,000 W * 3000 s

E = 51 x 10^6 J = 51 MJ

7 0

2 years ago

A rectangular box has side 10cm x 5cm x 2cm. If it lies on a horizontal surface and has weight 1.00N, calculate the maximum and

Vladimir [108]

Answer:

P max = 1000 pa

P min = 200 pa

Explanation:

P = F/A

pressure will be maximum when surface gets minimum. so to find the maximum amount of pressure we need to calculate the minimum surface. it is 2cm×5cm = 10cm² = 0.001m² . then we have:

P = 1 / 0.001 = 1000 pa

to find minimum pressure the surface that must be chosen is 10cm×5cm = 50cm² = 0.005m² .

P = 1 / 0.005 = 200 pa

7 0

2 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.