Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is
, the final velocity is
, and the total time of the motion is
, so the acceleration is given by
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
So, the bullet penetrates the sandbag 1.8 meters.
Is there more information ?
Answer:
a) y₂ = 49.1 m
, t = 1.02 s
, b) y = 49.1 m
, t= 1.02 s
Explanation:
a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero
² = ² - 2 g (y –yo)
The origin of the coordinate system is on the floor and the ball is thrown from a height
y-yo = = - g t
t = / g
t = 10 / 9.8
t = 1.02 s
b) the maximum height
y- 44.0 = ² / 2 g
y - 44.0 = 5.1
y = 5.1 +44.0
y = 49.1 m
The time is the same because it does not depend on the initial height
t = 1.02 s
Answer:
As an object’s temperature increases, the Rate at which it radiates energy increases.
A. 10 rations = 1 deca-ration.
b. 2000 mockingbirds = 2 x 10³ = 2 kilo-mockingbirds.
c. 10⁻⁵ phones = 1 micro-phones.
d. 10⁻⁹ goats = 1 nano-goats.
e. 1018 miners = 1.018 x 10³ = 1.018 kilo-miners.