The second one because you don't get shocked by plugging in something you can get electricted by putting something thin in the outlet then it will send a shock to your hand
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:
it should be 8,000
Explanation:
because if you multiply the mass value by 1000, you will get 8,000
:):):):)
It depends on the length of the pendulum and the strength of gravitational pull acting upon the pendulum.
Hope this helps!
Answer:
Explanation:
Given
initial speed of Launch=53.7 m/s
Range of Projectile =Maximum height of Projectile
Range is given by R
Maximum height is given by
as u cannot be zero therefore