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
I believe it’s false, I think I had this before
The least value varies depends on ammeter range. In the given question, ammeter range is not mentioned. So, the least value of an ammeter is 0.1 or 0.5 (depends on ammeter range).
<u>Explanation:</u>
The least value of an ammeter is the measure of the smallest number which can be observed in ammeter. So the least value will be varying depending upon its range. If we consider the range of ammeter is 30 A and the scale readings are 10 numbers, then the least value will be 30/10 = 3 A per scale.
So the least value is determined as 
So among the given options 0.1 is most suitable for an ammeter with range of 3 A with 30 divisions.
Answer:
-210.25
Explanation:
you should start at the braces, and start with multipe the numbers and then have the answer
Sequence of operations:
1.Brackets
2.braces
3.multiply
4. plus
5.negetive