The answer is 570 J. The kinetic energy has the formula of 1/2mV². The total work in this process W= 1/2m(V2²-V1²) = 1/2 * 15.0 * (11.5²-7.50²) = 570 J.
Answer:
earth
sorry I need points but I don’t know the answer again I’m really sorry
Momentum should be conserved. The momentum of both
objects must balance with their initial and final momentum.
Let m1 and v1 be the mass and velocity of the
bowling ball
And m2 and v2 be the mass and velocity of the
bowling pin
(m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f
30 kg m/s + (1.5 kg)(0 m/s) = 13kg m/s + 1.5v2f
V2f = 11.33 m/s
<span>So the momentum = 1.5 kg(11.33 m/s) = 17 kg m/s</span>
Answer:
C = 4,174 10³ V / m^{3/4}
, E = 7.19 10² / ∛x, E = 1.5 10³ N/C
Explanation:
For this exercise we can calculate the value of the constant and the electric field produced,
Let's start by calculating the value of the constant C
V = C
C = V / x^{4/3}
C = 220 / (11 10⁻²)^{4/3}
C = 4,174 10³ V / m^{3/4}
To calculate the electric field we use the expression
V = E dx
E = dx / V
E = ∫ dx / C x^{4/3}
E = 1 / C x^{-1/3} / (- 1/3)
E = 1 / C (-3 / x^{1/3})
We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E
E = 3 / C (0- (-1 / x^{1/3}))
E = 3 / 4,174 10³ (1 / x^{1/3})
E = 7.19 10² / ∛x
for x = 0.110 cm
E = 7.19 10² /∛0.11
E = 1.5 10³ N/C
Answer:
b
Explanation:
they both have a neutral charge so they couldn't be positive or negative since that wouldn't come from anywhere