Answer:
0.75 m
Explanation:
Let's call the distance between the bulb and the mirror x.
The bulb and the length of the mirror form a triangle. The mirror and the illuminated area on the floor form a trapezoid. If we extend the lines from the mirror edge to the reflected image of the bulb, we turn that trapezoid into a large triangle. This triangle and the small triangle are similar. So we can say:
x / 0.4 = (3 + x) / 2
Solving for x:
2x = 0.4 (3 + x)
2x = 1.2 + 0.4 x
1.6 x = 1.2
x = 0.75
So the bulb should located no more than 0.75 m from the mirror.
Answer:
1. -8.20 m/s²
2. 73.4 m
3. 19.4 m
Explanation:
1. Apply Newton's second law to the car in the y direction.
∑F = ma
N − mg = 0
N = mg
Apply Newton's second law to the car in the x direction.
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
Given μ = 0.837:
a = -(9.8 m/s²) (0.837)
a = -8.20 m/s²
2. Given:
v₀ = 34.7 m/s
v = 0 m/s
a = -8.20 m/s²
Find: Δx
v² = v₀² + 2aΔx
(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx
Δx = 73.4 m
3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.
d = v₀t
d = (34.7 m/s) (0.56 s)
d = 19.4 m
Answer:
See below
Explanation:
See attached diagram
280 km east then 190 km north
Use Pythagorean theorem to find the resultant displacement
d^2 = 280^2 + 190^2
d = 338.4 km
Angle will be arctan ( 190/280) = 34.16 °
Mem me e m even have. Jags. Shah. Shiv side esicjm is n meh dish so do indbbd
Final velocity = 5km/sec = 5000 m/sec
Initial velocity = 0, mass = 5kg and force = 500,000 N
F = m(Vf-Vi)/time
Time = m(Vf-Vi)/force = 5*5000/500000 = 0.05 seconds
So, it takes 0.05 sec to arrive at the speed of 5km/s.
