Answer:
q = - 93.334 nC
Explanation:
GIVEN DATA:
Radius of ring 73 cm
charge on ring 610 nC
ELECTRIC FIELD p FROM CENTRE IS AT 70 CM
E = 2000 N/C
Electric field due tor ring is guiven as
![E = \frac{KQx}{[x^2+ R^2]^{3/2}}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7BKQx%7D%7B%5Bx%5E2%2B%20R%5E2%5D%5E%7B3%2F2%7D%7D)
E1 = 3714.672 N/C
electric field due to point charge q
now the eelctric charge at point P is
E = E1 + E2
solving for q
q = - 93.334 nC
At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²
Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s
Point 2 (stagnation):
At the stagnation point, the velocity is zero.
The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
= 314.37 lb/ft²
= 314.37/144 = 2.18 lb/in²
Answer: 2.2 psi
Answer:
X and Z
Explanation:
Conduction occurs through direct physical contact. Heat transferred from the pot to the handle, and from the handle to the hand, are both examples of conduction.
Answer:
6.5 m/s^2
Explanation:
The net force acting on the yo-yo is
F_net = mg-T
ma=mg-T
now T= mg-ma
net torque acting on the yo-yo is
τ_net = Iα
I= moment of inertia (= 0.5 mr^2 )
α = angular acceleration
τ_net = 0.5mr^2(a/r)
Tr= 0.5mr^2(a/r)
(mg-ma)r=0.5mr^2(a/r)
a(1/2+1)=g
a= 2g/3
a= 2×9.8/3 = 6.5 m/s^2
Answer:
D
Explanation:
pressure change have nothing to do with the spontaneity.
Entropy change , enthalpy change , temperature have roles in deciding spontaneity.
