Kinetic energy<span> increases with the square of the velocity (KE=1/2*m*v^2). If the velocity is doubled, the KE quadruples. Therefore, the </span>stopping distance<span> should increase by a factor of four, assuming that the driver is </span>can<span> apply the brakes with sufficient precision to almost lock the brakes.</span>
Answer:
v = 2.029 m/s
Explanation:
Given
L = 84.0 cm ⇒ R = 0.5*L = 0.5*84 cm = 42 cm = 0.42 m
m₁ = 0.600 kg
m₂ = 0.200 kg
g = 9.8 m/s²
u₁ = u₂ = 0 m/s
v₁ = ?
v₂ = ?
Due to gravity, the bar oscillates and becomes vertical. The mass that occupies the lower position is the one with the highest torque. The one that reduces the potential energy (the system tends to the position of minimum energy). This is achieved if the mass that goes down is 0.6kg (that goes down 42cm) and the one that goes up is 0.2kg (goes up 42cm).
In this system mechanical energy is conserved, so we can match its value in the horizontal position with the one in the vertical.
then
Ei = Ki + Ui = 0.5*(m₁+m₂)*(0)² + (m₁+m₂)*9.8*(0) = 0 J
Ef = Kf + Uf
⇒ Kf = 0.5*(m₁+m₂)*v² = 0.5*(0.6+0.2)*v² = 0.4*v²
⇒ Uf = m₁*g*h₁ + m₂*g*h₂ = 0.6*9.8*(-0.42) + 0.2*9.8*0.42 = - 1.6464
⇒ Ef = Kf + Uf = 0.4*v² - 1.6464
Since
0 = 0.4*v² - 1.6464 ⇒ v = 2.029 m/s
v is the same value due to the wooden rod is pivoted about a horizontal axis through its center and the masses are on opposite ends.
v₁ = v₂ = v ⇒ ω₁*R₁ = ω₂*R₂ ⇒ ω₁*R = ω₂*R ⇒ ω₁ = ω₂ = ω
⇒ v = ω*R
Answer:
my mind cant comprehend a shing word u just said
Explanation:
Substract two consecutive terms of the sequence to see if there is a common difference:
As we can see, there is a common difference of -6.
Then, if a number of the sequence is given, the next one can be found by adding -6 (which is the same as subtracting 6).
Notice that the first term of the sequence is 3.
Then, the rule for the sequence is to start with 3 and add -6 repeatedly.
Therefore, the correct choice is option A) Start with 3 and add -6 repeatedly.
Given that the block have two applied masses 250 g at East and 100 g at South. In order to make a situation in which block moves towards point A, we have to apply minimum number of masses to the blocks. In order to prevent block moving toward East, we have to apply a mass at West, equal to the magnitude of mass at East but opposite in direction. Therefore, mass of 250 g at West is the required additional mass that has to be added. There is already 100 g of mass acting at South, that will attract block towards South or point A. No need to add further mass in North-South direction.
