Answer:
Explanation:
25 mm diameter
r₁ = 12.5 x 10⁻³ m radius.
cross sectional area = a₁
Pressure P₁ = 100 x 10⁻³ x 13.6 x 9.8 Pa
a )
velocity of blood v₁ = .6 m /s
Cross sectional area at blockade = 3/4 a₁
Velocity at blockade area = v₂
As liquid is in-compressible
a₁v₁ = a₂v₂
a₁ x .6 m /s = 3/4 a₁ v₂
v₂ = .8m/s
b )
Applying Bernauli's theorem formula
P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²
100 x 10⁻³ x 13.6 x10³x 9.8 + 1/2 X 1060 x .6² = P₂ + 1/2x 1060 x .8²
13328 +190.8 = P₂ + 339.2
P₂ = 13179.6 Pa
= 13179 / 13.6 x 10³ x 9.8 m of Hg
P₂ = .09888 m of Hg
98.88 mm of Hg
Answer:
Explanation:
Let fuel is released at the rate of dm / dt where m is mass of the fuel
thrust created on rocket
= d ( mv ) / dt
= v dm / dt
this is equal to force created on the rocket
= 220 dv / dt
so applying newton's law
v dm / dt = 220 dv / dt
v dm = 220 dv
dv / v = dm / 220
integrating on both sides
∫ dv / v = ∫ dm / 220
lnv = ( m₂ - m₁ ) / 220
ln4000 - ln 2500 = ( m₂ - m₁ ) / 220
( m₂ - m₁ ) = 220 x ( ln4000 - ln 2500 )
( m₂ - m₁ ) = 220 x ( 8.29 - 7.82 )
= 103.4 kg .
Answer:
The ability of our bodies to adapt to different levels of gravity. You would become weaker and your heart is use to zero gravity. Boredom because there isn't much to in space. When intelligent people get bored, it's not pretty all the time...
Explanation:
It is given that,
Frequency of vibration, f = 215 Hz
Amplitude, A = 0.832 mm
(a) Let T is the period of this motion. It is given by the following relation as :
(b) Speed of sound in air, v = 343 m/s
It can be given by :
Hence, this is the required solution.
Answer:
E1 = 10.15 * 10^4 N/C
E2 = 0
E3 = 10.15 *10^4 N/C
Explanation:
Given data:
Two 13 cm-long thin glass rods ( L ) = 0.13 m
charge (Q) = +11nC
distance between thin glass rods = 4 cm .
<u>Calculate the electric field strengths </u>
electric charge due to a single glass rod in the question ( E ) = 
equation 1 can be used to determine E1, E2 and E3 because the points lie within the two rods hence the net electric field produced will be equal to the difference in electric fields produced
applying equation 1 to determine E1
E1 = 
( distance from 1 rod is 0.01 m and from the other rod is 0.03 )
= 
= 10.15 * 10^4 N/C
applying equation 1 to determine E2
E2 = 
therefore E2 = 0
E1 = E3
hence E3 = 10.15*10^4 N/C
