the answer is 1a as rearrange gives I = v divided by r
Answer:
10.4 m/s
Explanation:
First, find the time it takes for the projectile to fall 6 m.
Given:
y₀ = 6 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 1.11 s
Now find the horizontal position of the target after that time:
Given:
x₀ = 6 m
v₀ = 5 m/s
a = 0 m/s²
t = 1.11 s
Find: x
x = x₀ + v₀ t + ½ at²
x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²
x = 11.5 m
Finally, find the launch velocity needed to travel that distance in that time.
Given:
x₀ = 0 m
x = 11.5 m
t = 1.11 s
a = 0 m/s²
Find: v₀
(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²
v₀ = 10.4 m/s
Hey there!
In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).
Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:
Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:
Now, at this temperature and pressure, the volume of a saturated vapor is 2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:
This means that the volume of the container will be:
Regards!