Answer:
A. Scientists use seismic computer models to measure the atmospheric conditions above the Earth's crust
Explanation:
why would use atmosphere to study the layers of earth? dont think thats possible
Answer
given,
mass of the drop, m = 0.0014 g
speed of the drop, u = 8.1 m/s
a) Change in momentum is equal to impulse
final velocity of the drop, v = 0 m/s
J = m ( v - u )
J = 0.0014 x 10⁻³ x ( 0 - 8.1 )
J = -1.134 x 10⁻⁵ kg.m/s
impulse of the roof = - J = 1.134 x 10⁻⁵ kg.m/s
b) time, t = 0.37 m s
impact of force = ?
we know
J = F x t
1.134 x 10⁻⁵ = F x 0.37 x 10⁻³
F = 0.031 N
the magnitude of the force of the impact is equal to F = 0.031 N
Answer:
a)
Weight in Air = 0.3N
Weight in Water = 0.25N
Weight in Liquid = 0.24N.
Upthrust /Buoyant Force = Weight in Air – Weight in Fluid(Water in this case)
= 0.3 – 0.25
= 0.5N.
b) R.D of Body = Density of Body/Density of Standard Fluid(Water).
There's a Derived Formula for RD.
I'm gonna Apply it here.
Ask me for the derivation in the Comment section if you need it.
RD = α/ρ = (Weight in Air) / (Upthrust Force)
Where
α = density of the Body(or reference substance)
ρ = density of standard fluid (water)
= 0.3/0.05 = 6.
c) RD of Liquid = (Density of Liquid) /(Density of standard Fluid(water)
Or we just go by that formula
RD of Liquid = Weight in Air/Upthrust(In Liquid)
We'll be using the Upthrust in that Liquid now.
= 0.3 – 0.24 = 0.06
RD = 0.3/0.06 = 5.
#1 is 12 u just have to count that one was counting by 5's figure out the pattern on what its counting by and start at the bottom and count your way up till you get to the shaded line
B is the answer you need and i honestly got this question on a middle school test
you must be in different area then me
