Answer:
A) 6N
Explanation:
the weight of the astronaut on earth can be calculated with the formula
Weight = mass * gravity of earth
Since the gravitational force is one-hundredth of Earth, the formula should be
Weight = mass * (gravity of earth / 100)
Weight = (mass * gravity of earth) / 100
Since you already know the weight, you only need to divide by 100
Weight = (mass * gravity of earth) / 100
Weight = 600N / 100
Weight = 6N
Answer:
(a) Force must be grater than 283.87 N
(B) Force will be equal to 193.945 N
Explanation:
We have given mass of the crate m = 49.6 kg
Acceleration due to gravity 
Coefficient of static friction 
Coefficient of kinetic friction 
(a) Static friction force is given by 
So to just start the crate moving we have to apply more force than 283.87 N
(B) This force will be equal to kinetic friction force
We know that kinetic friction force is given by 
<span>3.2 grams
The first thing to do is calculate how many half lives have expired. So take the time of 72 seconds and divide by the length of a half life which is 38 seconds. So
72 / 38 = 1.894736842
So we're over 1 half life, but not quite 2 half lives. So you'll have something less than 12/2 = 6 grams, but more than 12/4 = 3 grams.
The exact answer is done by dividing 12 by 2 raised to the power of 1.8947. So let's calculate 2^1.8947 power
= 12 g / (e ^ ln(2)*1.8947)
= 12 g / (e ^ 0.693147181 * 1.8947)
= 12 g / (e ^ 1.313305964)
= 12 g / 3.718446464
= 3.227154167 g
So rounded to 2 significant figures gives 3.2 grams.</span>
Answer:
(D) parasite........................
Answer:
Surely Achilles will catch the Tortoise, in 400 seconds
Explanation:
The problem itself reduces the interval of time many times, almost reaching zero. However, if we assume the interval constant, then it is clear that in two hours Achilles already has surpassed the Tortoise (20 miles while the Tortoise only 3).
To calculate the time, we use kinematic expression for constant speed:

The moment that Achilles catch the tortoise is found by setting the same final position for both (and same time as well, since both start at the same time):
