Using the kinematic equation d = V_0 * t + 1/2 * a * t^2, where d is height you can rewrite this to be d = 1/2*g*t^2 or 4.9t^2
g = a because this is a free fall
d = 1/2 * 9.81m/s^2 * 2.5^2
d = 30.65625m
d = 30.7m
MgCl2
Mg = magnesium
Cl = chlorine
Magnesium + chlorine = magnesium chloride.
This is because compounds are always written with the METAL FIRST and the NON METAL SECOND. the non metal ends in - ide when it reacts with a metal.
So ur answer would be magnesium chloride. :)
Answer:
The frog takes 8 jumps to reach top of well
Explanation:
Given data
Frog at bottom=17 foot
Each time frog leaps 3 feet
Frog has not reached the top of the well, then the frog slides back 1 foot
To Find
Total number of leaps the frog needed to escape from well
Solution
in 1 jump distance jumped=3+(-1)
=2 feet
=2×1 feet
The "-1" is because the frog goes back
Now After 2 jumps the distance jumped as:
Distance Jumped=2+2
Distance Jumped=2*2
=4 feet
Similarly after 7 jumps
Distance Jumped=2+2+......+2
Distance Jumped=2*7
=14 feet
Now after 8th jump the frog climbs but doesnot slide back as it is reached to the top of well.
So
Distance Jumped=(Distance Jumped after 7 jumps)+3
=14+3
=17 feet
The frog takes 8 jumps to reach top of well
Explanation:
LD₁ = 10⁵ mm⁻²
LD₂ = 10⁴mm⁻²
V = 1000 mm³
Distance = (LD)(V)
Distance₁ = (10⁵mm⁻²)(1000mm³) = 10×10⁷mm = 10×10⁴m
Distance₂ = (10⁹mm⁻²)(1000mm³) = 1×10¹² mm = 1×10⁹ m
Conversion to miles:
Distance₁ = 10×10⁴ m / 1609m = 62 miles
Distance₂ = 10×10⁹m / 1609 m = 621,504 miles.
Answer:
V₁ = √ (gy / 3)
Explanation:
For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2
Starting point
Em₀ = U₁ + U₂
Em₀ = m₁ g y₁ + m₂ g y₂
Let's place the reference system at the point where the mass m1 is
y₁ = 0
y₂ = y
Em₀ = m₂ g y = 2 m₁ g y
End point, at height yf = y / 2
= K₁ + U₁ + K₂ + U₂
= ½ m₁ v₁² + ½ m₂ v₂² + m₁ g
+ m₂ g ![y_{f}](https://tex.z-dn.net/?f=y_%7Bf%7D)
Since the masses are joined by a rope, they must have the same speed
= ½ (m₁ + m₂) v₁² + (m₁ + m₂) g ![y_{f}](https://tex.z-dn.net/?f=y_%7Bf%7D)
= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g ![y_{f}](https://tex.z-dn.net/?f=y_%7Bf%7D)
How energy is conserved
Em₀ = ![E_{mf}](https://tex.z-dn.net/?f=E_%7Bmf%7D)
2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g ![y_{f}](https://tex.z-dn.net/?f=y_%7Bf%7D)
2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2
3/2 v₁² = 2 g y -3/2 g y
3/2 v₁² = ½ g y
V₁ = √ (gy / 3)