<h3><em>physical</em><em> </em><em>science</em><em> </em><em>deals</em><em> </em><em>with</em><em> </em><em>the</em><em> </em><em>study</em><em> </em><em>of</em><em> </em><em>physics</em><em> </em><em>chemistry</em></h3>
Explanation:
yan lng po Alam ko
Answer :
I think it is A. Distance
Answer:
38.47 m
Explanation:
To find the height of the building, we will use the following equation

Where yf is the final height, yi is the initial height, viy is the initial vertical velocity, t is the time, and a is the acceleration due to gravity.
If the brick is in flight for 3.1 s, we can say that when t = 3.1s, yf = 0 m. So, replacing
viy = (16 m/s)sin(10) = 2.78 m/s
a = -9.8 m/s²
we get

Solving for yi

Therefore, the height of the building is 38.48 m
Answer:
It can go back to it's original shape
Explanation:
Answer:
2.77 * 10^5 m/s
Explanation:
Let us recall that kinetic energy is given by 1/2 mv^2
Where;
m = mass of the body
v = velocity of the body
In this case,
m = 3.38 * 10^31 kg
KE= 1.30 * 10^42 J
KE = 1/2 mv^2
v = √2KE/m
v = √2 * 1.30 * 10^42/3.38 * 10^31
v = √7.69 * 10^10
v = 2.77 * 10^5 m/s