I really wish I could be helping you. I don't know.
Answer:
v_f = 17.4 m / s
Explanation:
For this exercise we can use conservation of energy
starting point. On the hill when running out of gas
Em₀ = K + U = ½ m v₀² + m g y₁
final point. Arriving at the gas station
Em_f = K + U = ½ m v_f ² + m g y₂
energy is conserved
Em₀ = Em_f
½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂
v_f ² = v₀² + 2g (y₁ -y₂)
we calculate
v_f ² = 20² + 2 9.8 (10 -15)
v_f = √302
v_f = 17.4 m / s
Answer:

Explanation:
The electric field produced by a single point charge is given by:

where
k is the Coulomb's constant
q is the charge
r is the distance from the charge
In this problem, we have
E = 1.0 N/C (magnitude of the electric field)
r = 1.0 m (distance from the charge)
Solving the equation for q, we find the charge:

Answer:
a) V = 1.866 10² V
, b) V = 3.424 10⁵ V
, c) v = 8.1 10⁶ m / s
Explanation:
a) the potential difference is requested to accelerate the electrons up to 2.7% of the speed of light
v = 0.027 c
v = 0.027 3 10⁸
v = 8.1 10⁶ m / s
for this part we can use the conservation of mechanical energy
starting point. When electrons are at rest
Em₀ = U = q V
final point. Electrons with maximum speed
Em_f = K = ½ m v2
Em₀ = Em_{f}
e V = ½ m v²
V = ½ m v² / e
let's calculate
V = ½ 9.1 10⁻³¹ (8.1 10⁶)² / 1.6 10⁻¹⁹
V = 1.866 10² V
V = 1866 V
b) if this acceleration protons is the mass of the proton is m_{p} = 1.67 10-27
V = ½ 1.67 10⁻²⁷ (8.1 10⁶)² / 1.6 10⁻¹⁹
V = 3.424 10⁵ V
V = 342402 V
c)
this potential difference should give the protons the same speed as the electrons
v = 8.1 10⁶ m / s
Gravity changes as the altitude change.<span> The gravitational force is proportional to 1/R2, where R is the distance from the center of the Earth the radius of earth where gravity is 9.8 m/s^2 is 6400 km this will serve as the zero mark.
g1/(g2) = R2^2/(R1)^2
so we set the constant values to R1 and the unknown distance as x
(9.8)/(8.80) = (6400-x)2/(6400)^2
solving for x we will get
x = 345.85 km above the earths surface
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<span>Hope my answer would be a great help for you.
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