Answer: 6.48m/s
Explanation:
First, we know that Impulse = change in momentum
Initial velocity, u = 19.8m/s
Let,
Velocity after first collision = x m/s
Velocity after second collision = y m/s
Also, we know that
Impulse = m(v - u). But then, the question said, the guard rail delivered a "resistive" impulse. Thus, our impulse would be m(u - v).
5700 = 1500(19.8 - x)
5700 = 29700 - 1500x
1500x = 29700 - 5700
1500x = 24000
x = 24000/1500
x = 16m/s
Also, at the second guard rail. impulse = ft, so that
Impulse = 79000 * 0.12
Impulse = 9480
This makes us have
Impulse = m(x - y)
9480 = 1500(16 -y)
9480 = 24000 - 1500y
1500y = 24000 - 9480
1500y = 14520
y = 14520 / 1500
y = 9.68
Then, the velocity decreases by 3.2, so that the final velocity of the car is
9.68 - 3.2 = 6.48m/s
Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3
<span>Two objects move toward each other because of gravitational attraction. As the objects get closer and closer, the force between them increases. </span>
