Equations of motion (EoM) use EoM <span>v2=u2+2ax</span> to establish velocities at positions shown in blue in drawing from EoM v=u+at for final 1 second of flight time, we can say v=u+g(1) <span><span>2gH−−−−√</span>=<span><span>2g1625H</span>−−−−−−√</span>+g</span><span> then, solve for H [in terms of g]
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Answer: 96N
Explanation:
To calculate the velocity of the impact On the persons head, we have
h = gt²/2
14 = 9.81t²/2
t² = 28/9.8
t² = 2.86
t = 1.69s
V = u + at
V = 0 + 9.81*1.69
V = 16.58m/s
a(average) = (v1² + v2²) /2Δy
a(average) = 16.58² + 0)/2 * 0.005
a(average) = 274.8964/0.01
a(average) = 27489.64m/s²
Using newton's second law of motion,
F(average) = m * a(average)
F(average) = 0.0035 * 27489.64
F(average) = 96.21N
Therefore the force needed by the acorn to do much damage starts from 96N
Answer:
A) 140 k
b ) 5.22 *10^3 J
c) 2910 Pa
Explanation:
Volume of Monatomic ideal gas = 1.20 m^3
heat added ( Q ) = 5.22*10^3 J
number of moles (n) = 3
A ) calculate the change in temp of the gas
since the volume of gas is constant no work is said to be done
heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT
make ΔT subject of the equation
ΔT = Q / n.(3/2).R
= (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )
= 140 K
B) Calculate the change in its internal energy
ΔU = Q this is because no work is done
therefore the change in internal energy = 5.22 * 10^3 J
C ) calculate the change in pressure
applying ideal gas equation
P = nRT/V
therefore ; Δ P = ( n*R*ΔT/V )
= ( 3 * 8.3144 * 140 ) / 1.20
= 2910 Pa
Answer:
A. To find the mass flow rate.
We use= 220 x 0.355/ 60
= 1.3kg/s
B. Volume flowrate is = mass flowrate / density
But density is 1000kg/m³
= 1.3kg/s/ 1000kg/m³
= 0.0013m³/s
C. Flow speead at 1
= 0.0013m³/s / (2 x 10-2m)²
= 6.5m/s
D.flow speed at 2
0.0013m³/s / (8x 10-2m)²
=1.63m/s
E. Gauge pressure at point 1
= 152+ 1/1000 ( 1.63)²- 6.5² + 1000( 9.8) ( 0-1.35)
= 119kpa
It is <span>C. Low to moderate level of exertion can be sustained over long periods of time </span>
