The electric potential at point A in the electric field= 0.099 x 10 ⁻¹v
<u>Explanation</u>:
Given data,
charge = 5.5 x 10¹² C
k =9.00 x 10⁹
The electric potential V of a point charge can found by,
V= kQ / r
Assuming, r=5.00×10⁻² m
V= 5.5 x 10⁻¹²C x 9.00 x 10⁹ / 5.00×10⁻² m
V= 49.5 x 10⁻³/ 5.00×10⁻²
Electric potential V= 0.099 x 10⁻¹v
Answer:
#Where 
is in meters and 
in seconds.
Explanation:
Given that :
From 
we have:
From 
we have that:
Now,given that the initial value problem is given by:
Hence,the position of u at time t is given by
, in meters, in seconds.
Explanation:
(a) Given:
Δx = 150 m
v₀ = 27 m/s
v = 54 m/s
Find: a
v² = v₀² + 2aΔx
(54 m/s)² = (27 m/s)² + 2a (150 m)
a = 7.29 m/s²
(b) Given:
Δx = 150 m
v₀ = 0 m/s
a = 7.29 m/s²
Find: t
Δx = v₀ t + ½ at²
150 m = (0 m/s) t + ½ (7.29 m/s²) t²
t = 6.42 s
(c) Given:
v₀ = 0 m/s
v = 27 m/s
a = 7.29 m/s²
Find: t
v = at + v₀
27 m/s = (7.29 m/s²) t + 0 m/s
t = 3.70 s
(d) Given:
v₀ = 0 m/s
v = 27 m/s
a = 7.29 m/s²
Find: Δx
v² = v₀² + 2aΔx
(27 m/s)² = (0 m/s)² + 2 (7.29 m/s²) Δx
Δx = 50 m
Answer:
true
Explanation:
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