Answer:
V = 331.59m/s
Explanation:
First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.
S = ut + 1/2at²
Given height of the cliff S = 80m
initial velocity u = 0m/s²
a = g = 9.81m/s²
Substitute
80 = 0+1/2(9.81)t²
80 = 4.905t²
t² = 80/4.905
t² = 16.31
t = √16.31
t = 4.04s
Next is to get the vertical velocity
Vy = u + gt
Vy = 0+(9.81)(4.04)
Vy = 39.6324
Also calculate the horizontal velocity
Vx = 1330/4.04
Vx = 329.21m/s
Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.
V² = Vx²+Vy²
V² = 329.21²+39.63²
V² = 329.21²+39.63²
V² = 108,379.2241+1,570.5369
V² = 109,949.761
V = √ 109,949.761
V = 331.59m/s
Hence the speed of the shell as it hits the ground is 331.59m/s
Answer:
a) Vf = 4.95 m/s
b) t = 0.51s
Explanation:
take downwards as positive.
let Vf be the final velocity as the cucumber reach the bottom of the bin and Vi be the initial velocity of the cucumber when they dropped.
a ) from equations of motion:
(Vf)^2 = (vi)^2 +2g×(x - x0) ,
<em>since x0 = 0 m and Vi = 0 m/s.</em>
Vf = \sqrt{2g×(x)} = \sqrt{2(9.8)×(1.25)} = 4.95 m/s, downwards
therefore, the cucumber will reach the bottom of the bin with a speed of 4.95 m/s.
b) from equations of motion:
Vf = Vi + g×t
Vi = 0 then:
t = Vf/g = 4.95/9.8 = 0.51 s
therefore, the cucumber will take 0.51 seconds to reach the bottom of the bin.
The statement that accurately describes properties of valence is A. The smaller the number of electrons an atom has to borrow or to lend, the greater the activity of the atom.
If the temperature is increased the particles gain more kinetic energy or vibrate faster. This means that they move faster and take more space.