Physics, 21.08.2019 04:30 ScardeySpider

During a very quick stop, a car decelerates at 28.4 rad/s? . assume the tires initially rotated in the positive direction and radius of tires is 0.32m a) how many revolutions do the tires make before coming to rest, given their initial angular velocity is 93.0 rad/s? b) how long does the car take to stop completely? c) what was the car's initial velocity (in m/s)? d) how far the car moves forward before it stops?

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Answer from: floresjacqueline

a) 24

b) 3.3 sec

c) 29.8 m/s

d) 48.85 m

Explanation:

α = angular acceleration = - 28.4 rad/s²

r = radius of the tire = 0.32 m

w₀ = initial angular velocity = 93 rad/s

w = final angular velocity = 0 rad/s

θ = angular displacement

Using the equation

w² = w₀² + 2αθ

0² = 93² + 2 (- 28.4) θ

θ = 152.3 rad

n = number of revolutions

Number of revolutions are given as

$n = \frac{\theta }{2\pi }$

$n = \frac{152.3 }{2(3.14) }$

n = 24

t = time taken to stop

using the equation

w = w₀ + αt

0 = 93 + (- 28.4) t

t = 3.3 sec

v₀ = initial velocity of the car

initial velocity of the car is given as

v₀ = r w₀ = (0.32) (93) = 29.8 m/s

v = final velocity = 0 m/s

a = linear acceleration = rα = (0.32) (- 28.4) = - 9.09 m/s²

d = distance traveled by car before stopping

Using the equation

v² = v₀² + 2 a d

0² = 29.8² + 2 (- 9.09) d

d = 48.85 m

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Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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During a very quick stop, a car decelerates at 28.4 rad/s? . assume the tires initially rotated in t...

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