Because,
In left image pin is not touch to the wire.
In right image pin is touch to the wire.
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Answer:
V = 576 V
Explanation:
Given:
- The area of the two plates A = 0.070 m^2
- The space between the two plates d = 6.3 mm
- Te energy density u = 0.037 J /m^3
Find:
- What must the potential difference between the plates V?
Solution:
- The energy density of the capacitor with capacitance C and potential difference V is given as:
u = 0.5*ε*E^2
- Where the Electric field strength E between capacitor plates is given by:
E = V / d
Hence,
u = 0.5*ε*(V/d)^2
Where, ε = 8.854 * 10^-12
V^2 = 2*u*d^2 / ε
V = d*sqrt ( 2*u / ε )
Plug in values:
V = 0.0063*sqrt ( 2 * 0.037 / (8.854 * 10^-12) )
V = 576 V
Answer:
a) 51.8 m, b) 27.4 m/s, c) 142 m
Explanation:
Given:
v₀ = 42.0 m/s
θ = 60.0°
t = 5.50 s
Find:
h, v, and H
a) y = y₀ + v₀ᵧ t + ½ gt²
0 = h + (42.0 sin 60.0) (5.50) + ½ (-9.8) (5.50)²
h = 51.8 m
b) vᵧ = gt + v₀ᵧ
vᵧ = (-9.8)(5.5) + (42.0 sin 60.0)
vᵧ = -17.5 m/s
vₓ = 42.0 cos 60.0
vₓ = 21.0 m/s
v² = vₓ² + vᵧ²
v = 27.4 m/s
c) vᵧ² = v₀ᵧ² + 2g(y - y₀)
0² = (42.0)² + 2(-9.8)(H - 51.8)
H = 142 m
Answer:
i cant even see that no one can i will answer it if u can make it bigger
Explanation: eyes
Explanation:
Given that,
Mass of the particle, m = 4 kg
Speed of the particle, v = 2.5 m/s
The radius of the circle, r = 2 m
We need to find the angular momentum about the center of the circle. The formula for the angular momentum is given by :
Substitute all the values,
So, the angular momentum of the particle is 20 kg-m² s.
