I think it’s 10 kg since 20/2=10
It is Potential energy it's at rest
Answer:
12.0 m/s²
Explanation:
Newton's second law:
F = ma
Solving for acceleration:
a = F/m
If force is doubled and mass is tripled:
a' = (2F)/(3m)
a' = ⅔ (F/m)
a' = ⅔ (18.0 m/s²)
a' = 12.0 m/s²
First, let us derive our working equation. We all know that pressure is the force exerted on an area of space. In equation, that would be: P = F/A. From Newton's Law of Second Motion, force is equal to the product of mass and gravity: F = mg. So, we can substitute F to the first equation so that it becomes, P = mg/A. Now, pressure can also be determined as the force exerted by a fluid on an area. This fluid can be measure in terms of volume. Relating volume and mass, we use the parameter of density: ρ = m/V. Simplifying further in terms of height, Volume is the product of the cross-sectional area and the height. So, V = A*h. The working equation will then be derived to be:
P = ρgh
This type of pressure is called the hydrostatic pressure, the pressure exerted by the fluid over a known height. Next, we find the literature data of the density of seawater. From studies, seawater has a density ranging from 1,020 to 1,030 kg/m³. Let's just use 1,020 kg/m³. Substituting the values and making sure that the units are consistent:
P = (1,020 kg/m³)(9.81 m/s²)(11 km)*(1,000 m/1km)
P = 110,068,200 Pa or 110.07 MPa
Answer: 5,640 s (94 minutes)
Explanation:
the tangential speed of the HST is given by
(1)
where
is the length of the orbit
r is the radius of the orbit
T is the orbital period
In our problem, we know the tangential speed: 
. The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:
So, we can re-arrange equation (1) to find the orbital period:
Dividing by 60, we get that this time corresponds to 94 minutes.
