Answer:
1)m=89.01 g
2)V(max) = 97.3 cm/s
Explanation:
Given that
K= 15 N/m
The maximum amplitude ,A=7.5 cm = 0.075 m
Given that 31 oscillations in 15 seconds ,this means that frequency f
f=\dfrac{31}{15}
f=2.066 Hz
lets take mass of the ball = m kg
m=0.08901 kg
m=89.01 g
The maximum speed
V(max)= ω x A
ω = 2π f= 2 x π x 2.066 = 12.98 rad/s
V(max) = 12.98 x 0.075 =0.973 m/s
V(max) = 97.3 cm/s
Answer:
Explanation:
1 ha = 10⁴ m²
1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²
In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³
Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m
Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³
Let Q be the withdrawal in m³
Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶
Q = 26.20 x 10⁶ m³
rate of withdrawal per second
= 26.20 x 10⁶ / 30 x 24 x 60 x 60
= 26.20 x 10⁶ / 2.592 x 10⁶
= 10.11 m³ / s
Answer:
A, I think its A but im not sure.
Explanation:
<span>Answer:
Using 1/f = 1/d' + 1/d ...(where d' object distance and d is image distance)
1/4 = 1/7 + 1/d
1/4 - 1/7 = 1/d
3/28 = 1/d
d = 28/3
d = 9.33 cm</span>