Answer:
Points downward, and its magnitude is 9.8 m/s^2
Explanation:
The motion of a projectile consists of two independent motions:
- A uniform horizontal motion, with constant velocity and zero acceleration. In fact, there are no forces acting on the projectile along the horizontal direction (if we neglect air resistance), so the acceleration along this direction is zero.
- A vertical motion, with constant acceleration g = 9.8 m/s^2 towards the ground (downward), due to the presence of gravity wich "pulls" the projectile downward.
The total acceleration of the projectile is given by the resultant of the horizontal and vertical components of the acceleration. But we said that the horizontal component is zero, therefore the total acceleration corresponds just to its vertical component, therefore it is a vector with magnitude 9.8 m/s^2 which points downward.
Answer:
calcium carbonate is transformed into calcium sulfate
Explanation:
Limestone stones are formed by calcium carbonate (Ca Co3), when acid rain falls with a little H2SO2, a chemical reaction of the type
Ca Co3 + H2SO4 → CO2 + H2O + CaSo4
Calcium sulfate is much weaker than alatium carbonate, which is why the stone tends to break under its weigh
Earthquakes can also occur far from the edges of plates, along faults. Faults<span> are cracks in the earth where sections of a plate (or two plates) are moving in different directions. Faults are caused by all that bumping and sliding the plates do. They are more common near the edges of the plates.</span>
Answer:
Explanation:
a ) It is given that bomb was at rest initially , so , its momentum before the explosion was zero.
b ) We shall apply law of conservation of momentum along x and y direction separately because no external force acts on the bomb.
If v be the velocity of the third part along a direction making angle θ
with x axis ,
x component of v = vcosθ
So momentum along x axis after explosion of third part = mv cosθ
= 10 v cosθ
Momentum along x of first part = - 5 x 42 m/s
momentum of second part along x direction =0
total momentum along x direction before explosion = total momentum along x direction after explosion
0 = - 5 x 42 + 10 v cosθ
v cosθ = 21
Similarly
total momentum along y direction before explosion = total momentum along y direction after explosion
0 = - 5 x 38 + 10 v sinθ
v sinθ= 21
squaring and and then adding the above equation
v² cos²θ +v² sin²θ = 21² +19²
v² = 441 + 361
v = 28.31 m/s
Tanθ = 21 / 19
θ = 48°
