The speed that a tsunami can travel is modeled by the equation , where s is the speed in kilometers per hour and d is the averag
1 answer:
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<h2>Answer: 10.52m</h2><h2 />
First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).
According to this, the initial velocity has two components, because the brick was thrown at an angle
:
(1)
(2)
(3)
(4)
As this is a projectile motion, we have two principal equations related:
<h2>In the x-axis: </h2> (5)
Where:
is the distance where the brick landed
is the time in seconds
If we already know and
, we have to find the time (we will need it for the following equation):
(6)
(7)
(8)
Where:
is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>
is the acceleration due gravity
Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:
(9)
(10)
Multiplying by -1 each side of the equation:
>>>>This is the height of the building
Answer:
c) At a distance greater than r
Explanation:
If G= Gravitational constant
M= Mass of earth
r= distance from earth center
then orbital speed is ;
v =
==> v²=GM/r
If speed of first satellite = V₁
==> V₁² = GM/r
==> r = GM/V₁²
If speed of second satellite say V₂ is less than V₁ then square of V₂ will be less than square of V₁ , and hence GM will be divided by less number in case of second satellite, and hence will give greater value of r as compared to first satellite.
So our answer is c