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MaRussiya [10]
3 years ago
9

The slowest animal ever discovered was a crab found in the Red Sea. It traveled with an average speed of 5.70km/yr. how long wou

ld it take this crab to travel 100km?
Physics
2 answers:
Vaselesa [24]3 years ago
8 0
If it takes 1 year for this crab to travel 5,70km

Then, it will takes approximatly 1000/5,70 = 175 years to travel 1000 km 
svlad2 [7]3 years ago
7 0

Answer:

Time taken by the crab, t = 17.54 years.

Explanation:

It is given that,

Speed of a crab, v = 5.70 km/yr

Distance travelled by the crab, d = 100 km

We need to find the time taken by the crab to cover this much distance. Time taken is given by :

t=\dfrac{d}{v}

t=\dfrac{100\ km}{5.70\ km/yr}

t = 17.54 years

So, the time taken by the crab is 17.54 years. Hence, this is the required solution.

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our lab partner wears a new pair of sneakers to lab and, rather than performing the required experiments, you decide to measure
Dafna1 [17]

Answer:

The coefficient of static friction between your partner and the floor is 0.55

Explanation:

Given:

Mass m = 59 Kg

Frictional force F_{s}  = 318.3 N

From the formula of frictional force,

 F_{s} = \mu_{s} mg

Where \mu _{s} = coefficient of static friction, g = 9.8 \frac{m}{s^{2} }

Put the above values and find the coefficient of static friction.

318.3 = \mu_{s} \times 59 \times 9.8

\mu_{s} = 0.55

Therefore, the coefficient of static friction between your partner and the floor is 0.55

4 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}}

<u>Deriving the half-life formula</u>

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.

<u>Solving for the amount remaining</u>

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
Suppose an air bubble is trapped in the eudiometer before starting the experiment. After the experiment is finished, the resulti
seraphim [82]

Answer:

Explanation:

The resulting valie would be too large

8 0
3 years ago
The___ if a gas can affect the pressure of a gas
sergejj [24]

I think it’s D I am sorry if I am wrong

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3 years ago
Is it True or false a muscle strain occurs when a ligament is torn away from the bone
Anit [1.1K]
I think the answer its false i dont know how to explain that but it is false that what my teacher told me 
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3 years ago
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