A)
2revs in 0.08s
so in 1s thats 25revs
therefore thats <u>50π radians</u> in one second
b)
well, ω=2π/T
therefore ω=50π = 157.079rads^-1
and v=rω where r is in meters;
v=0.3x157.079
<u>v=47.123ms^-1</u>
c)
f=1/T
f=1/period for one rotation
1 rotation = 0.08/2 = 0.04
f=1/0.04
<u>f=25Hz</u>
Answer:
<u>36 m</u>
Explanation:
We can consider this to be an AP.
Then,
<u>Subtract a₇ from a₉.</u>
- a + 8d - a + 6d = 24 - 20
- 2d = 4
- d = 2
<u>Finding a₁₅</u>
- a₁₅ = a + 14d
- a₁₅ = 8 + 14(2)
- a₁₅ = 8 + 28
- a₁₅ = <u>36 m</u>
Answer:
1. 80,000 Pa
2. 11.3 m/s
3. 12.5 m/s
Explanation:
<u>Question 1</u>
Pressure, 
Where h is the height that water is to reach, g is gravitational constant and
is the density, in this case, we assume
of pure water as 
Assuming 
P=8*10*1000=80000 Pa
<u>Question 2</u>
Pressure can also be found by the formula
where v is the velocity
Equating the new formula of pressure to the formula used in question 1 above

Notice that
is common hence

Making V the subject of the formula


In this case, h=8-1.6=6.4m and taking g as 10 m/s^{2}

Rounding off to 1 decimal place
v=11.3 m/s
<u>Question 3</u>
As already illustrated

Taking g as 9.8 and h now is 8m

v=12.52198067
Rounding off to 1 decimal place
v=12.5 m/s