PLS HELP ME AS QUICK AS POSSIBLE,
THANKS :)) I'm a bit confused
Can you answer 1 and 2, then confirm 3 :))))
12.5 times 14 and convert to meters its 1.75 meters per second
What is he minumum coating of thickness needed to ensure that lifght of waveelntght 5660 mbnd si
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
Answer:
Ro = 7.8 [g/cm³]
Explanation:
According to the principle of Archimedes, the volume of a body immersed in a liquid is equal to the volume displaced by water. That is, in this problem The displacement volume is equal to the new volume minus the original volume.
![V_{n}=110[cm^{3} ]\\V_{o}=100[cm^{3} ]\\V_{d}=110-100 = 10 [cm^{3} ]](https://tex.z-dn.net/?f=V_%7Bn%7D%3D110%5Bcm%5E%7B3%7D%20%5D%5C%5CV_%7Bo%7D%3D100%5Bcm%5E%7B3%7D%20%5D%5C%5CV_%7Bd%7D%3D110-100%20%3D%2010%20%5Bcm%5E%7B3%7D%20%5D)
We now know that density is defined as the relationship between mass and volume.
where:
Ro = density [g/cm³]
m = mass = 78 [g]
Vd = displacement volume [cm³]
![Ro = 78/10\\Ro = 7.8 [g/cm^{3} ]](https://tex.z-dn.net/?f=Ro%20%3D%2078%2F10%5C%5CRo%20%3D%207.8%20%5Bg%2Fcm%5E%7B3%7D%20%5D)
