Answer:
Δ P = 13.24 Pa
Explanation:
Given that
Density of oil ,ρ₁ = 9 x 10² kg/m³
We know that density for mercury ,ρ₂ = 13.6 x 10³ kg/m³
The change in the height of column ,Δh = 1.5 mm
The pressure given as
P = ρ g h
Change in the pressure
Δ P = ρ₁ g Δh
Now by putting the values
Δ P = 9 x 10² x 9.81 x 1.5 x 10⁻³ Pa
Δ P = 13.24 Pa
Therefor the change in the pressure will be 13.24 Pa.
This question was apprently selected from the "Sneaky Questions" category.
The store is 3 km from his home, and he walks there with a speed of 6 km/hr. So it takes him (3 km) / (6 km/hr) = 1/2 hour to get to the store.
That's 30 minutes. So the whole part-(a.) of the question refers to only that part of the trip, and we don't care what happens once he reaches the store.
a). Over the first 30 minutes of his travel, Greg walks 3.0 km on a straight road, and he ends up 3.0 km away from where he started.
Average speed = (distance/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>
Average velocity = (displacement/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>
There's probably some more questions in part-(b.) where you'd need to use Greg's return trip to find the answers, but johnaddy210 is only asking us for part-(a.).
Answer:

Explanation:
given data:
flow Q = 9 m^{3}/s
velocity = 8 m/s
density of air = 1.18 kg/m^{3}
minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as

here
is mass flow rate and given as


Putting all value to get minimum power


Answer: Resistance = 
The approximate diameter of a penny is, <em>d</em> = 20 mm
thickness of penny is, <em>L = </em> 1.5×
mm
The area of penny along circular face is,
= 3.14×
m²
The resistivity of copper is <em>ρ</em> = 1.72 x 10-8 Ωm.
Resistance,

Answer:
F = 69.3 N
Explanation:
For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by
fr = μ N
We define a reference system parallel to the floor
block B ( lower)
Y axis
N - W₁-W₂ = 0
N = W₂ + W₂
N = (M + m) g
X axis
F -fr = M a
for block A (upper)
X axis
fr = m a (2)
so that the blocks do not slide, the acceleration in both must be the same.
Let's solve the system by adding the two equations
F = (M + m) a (3)
a =
the friction force has the formula
fr = μ N
fr = μ (M + m) g
let's calculate
fr = 0.34 (2.0 + 0.250) 9.8
fr = 7.7 N
we substitute in equation 2
fr = m a
a = fr / m
a = 7.7 / 0.250
a = 30.8 m / s²
we substitute in equation 3
F = (2.0 + 0.250) 30.8
F = 69.3 N