Answer:
The amount of each gas that can dissolve in the ocean depends on the solubility and saturation of the gas in water. Solubility refers to the amount of a dissolved gas that the water can hold under a particular set of conditions, which are usually defined as 0o C and 1 atmosphere of pressure.
Explanation:
hope this helps
Let's use Newton's 2nd law of motion:
Force = (mass) x (acceleration)
Force = (68 kg) x (1.2 m/s²) = 81.6 newtons .
Answer:
Final Speed of Dwayne 'The Rock' Johnson = 15.812 m/s
Explanation:
Let's start out with finding the force acting downwards because of the mass of 'The Rock':
Dwayne 'The Rock' Johnson: 118kg x 9.81m/s = 1157.58 N
Now the problem also states that the kinetic friction of the desk in this problem is 370 N
Since the pulley is smooth, the weight of Dwayne Johnson being transferred fully, and pulls the desk with a force of 1157.58 N. The frictional force of the desk is resisting this motion by a force of 370 N. Subtracting both forces we get the resultant force on the desk to be: 1157.58 - 370 = 787.58 N
Now lets use F = ma to calculate for the acceleration of the desk:
787.58 = 63 x acceleration
acceleration = 12.501 m/s
Finally, we can use the motion equation:

here u = 0 m/s (since initial speed of the desk is 0)
a = 12.501 m/s
and s = 10 m
Solving this we get:


Since the desk and Mr. Dwayne Johnson are connected by a taught rope, they are travelling at the same speed. Thus, Dwayne also travels at 15.812 m/s when the desk reaches the window.
Answer:
Buoyancy causes things to float in water or liquid which appears to defeat the force of gravity. Hope this helps.
Answer:
-4*10⁴ units.
Explanation:
As the metal rod was initially neutral (which means that it has the same quantity of positive and negative charges), after being close to the charged sphere, as charge must be conserved, the total charge of the metal rod must still remain to be zero.
So, if due to the influence of the negative charge in the sphere, the half of the road closer to the sphere has a surplus charge of +4*10⁴ units, the charge on the half of the rod farther from the sphere must be the same in magnitude but of the opposite sign, i.e., -4*10⁴ units.