Answer:
1750 m
Explanation:
The distance traveled is the 750 meters to the Aunt's house plus the 1000 m from there to Chick-fil-A.
750 +1000 = 1750 . . . meters traveled
To be able to determine the original speed of the car, we use kinematic equations to relate the acceleration, distance and the original speed of the car moving.
First, we manipulate the one of the kinematic equations
v^2 = v0^2 + 2 (a) (x) where v = 0 since the car stopped
Writing the equation in such a way that the initial velocity or v0 is written on one side of the equation,
<span>we get v0 = sqrt (2(a)(x))
Substituting the known values,
v0 = sqrt(2(3.50)(30.0))
v0 = 14.49 m/s
</span>
Therefore, before stopping the car the original speed of the car would be 14.49 m/s
Explanation:
003 (part 1 of 2)
Pressure is force divided by area.
P = F / A
P = (117 kg × 9.8 m/s²) / (2 × (0.05 m)²)
P = 229,320 Pa
003 (part 2 of 2)
There are approximately 6895 Pa in 1 psi.
P = 229,320 Pa × (1 psi / 6895 Pa)
P = 33.3 psi
004 (part 1 of 2)
Since the collisions are elastic, the angle of reflection is the same as the angle of incidence (it bounces off at the same angle).
Impulse = change in momentum
F Δt = m Δv
F (36 s) = (300 × 0.003 kg) (5.2 sin 57° m/s − (-5.2 sin 57° m/s))
F = 0.218 N
004 (part 2 of 2)
Pressure is force over area.
P = F / A
P = 0.218 N / 0.712 m²
P = 0.306 N/m²
<span>Answer:
Therefore, x component: Tcos(24°) - f = 0 y component: N + Tsin(24°) - mg = 0 The two equations I get from this are: f = Tcos(24°) N = mg - Tsin(24°) In order for the crate to move, the friction force has to be greater than the normal force multiplied by the static coefficient, so... Tcos(24°) = 0.47 * (mg - Tsin(24°)) From all that I can get the equation I need for the tension, which, after some algebraic manipulation, yields: T = (mg * static coefficient) / (cos(24°) + sin(24°) * static coefficient) Then plugging in the values... T = 283.52.
Reference https://www.physicsforums.com/threads/difficulty-with-force-problems-involving-friction.111768/</span>