Answer:
= 3289.8 m / s
Explanation:
This exercise can be solved using the definition of momentum
I = ∫ F dt
Let's replace and calculate
I = ∫ (at - bt²) dt
We integrate
I = a t² / 2 - b t³ / 3
We evaluate between the lower limits I=0 for t = 0 s and higher I=I for t = 2.74 ms
I = a (2,74² / 2- 0) - b (2,74³ / 3 -0)
I = a 3,754 - b 6,857
We substitute the values of a and b
I = 1500 3,754 - 20 6,857
I = 5,631 - 137.14
I = 5493.9 N s
Now let's use the relationship between momentum and momentum
I = Δp = m - m v₀o
I = m - 0
= I / m
= 5493.9 /1.67
= 3289.8 m / s
Answer:
980 kJ
Explanation:
Work = change in energy
W = mgh
W = (1000 kg/m³ × 5.0 m³) (9.8 m/s²) (20 m)
W = 980,000 J
W = 980 kJ
The pump does 980 kJ of work.
Answer:
assume nitrogen is an ideal gas with cv=5R/2
assume argon is an ideal gas with cv=3R/2
n1=4moles
n2=2.5 moles
t1=75°C <em>in kelvin</em> t1=75+273
t1=348K
T2=130°C <em>in kelvin</em> t2=130+273
t2=403K
u=пCVΔT
U(N₂)+U(Argon)=0
<em>putting values:</em>
=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)
<em>by simplifying:</em>
Tfinal=363K
Answer:
a) V_a = -5.7536 10⁺⁷ V
, b) Vb = -1.92 10⁻⁷ V c) the sign of the potential change
Explanation:
The electrical potential for a point charge
V = k q / r
Where k is the Coulomb constant that you are worth 8.99 10⁹ N m² / C²
a) potential At point x = 0.250 cm = 0.250 10-2m
V_a = -8.99 10⁹ 1.6 10⁻¹⁹ /0.250 10⁻²
V_a = -5.7536 10⁺⁷ V
b) point x = 0.750 cm = 0.750 10-2
Vb = 8.99 10⁹ (-1.6 10⁻¹⁹) /0.750 10⁻²
Vb = -1.92 10⁻⁷ V
potemcial difference
ΔV = Vb- Va
V_ba = (-5.7536 + 1.92) 10⁻⁷
V_ba = -3.83 10⁻⁷ V
c) To know what would happen to a particle, let's use the relationship between the potential and the electric field
ΔV = E d
The force on the particle is
F = q₀ E
F = q₀ ΔV / d
We see that the force on the particle depends on the sign of the burden of proof. Now the burden of proof is negative to pass between the two points you have to reverse the sign of the potential, bone that the value should be reversed
V_ba = 0.83 10⁻⁷ V
