Answer:
5.3 m/s
Explanation:
First, find the time it takes for him to fall 7m.
y = y₀ + v₀ t + ½ at²
0 = 7 + (0) t + ½ (-9.8) t²
0 = 7 − 4.9 t²
t ≈ 1.20 s
Now find the velocity he needs to travel 6.3m in that time.
x = x₀ + v₀ t + ½ at²
6.3 = 0 + v₀ (1.20) + ½ (0) (1.20)²
v₀ ≈ 5.27 m/s
Rounded to two significant figures, the man must run with a speed of 5.3 m/s.
The distance you free-fall from rest is D = (1/2) (g) (T²) <== memorize this
Height of the platform = (1/2) (9.8 m/s²) (2.4 sec)²
Height = (4.9 m/s²) (5.76 s²)
Height = (4.9/5.76) meters
Height = 28.2 meters (a VERY high platform ... about 93 ft off the water !)
Without air-resistance, your horizontal speed doesn't change. It's constant. Traveling 3.1 m/s for 2.4 sec, you cover (3.1 m/s x 2/4 s) = 7.4 m horizontally.
The added weight of the sand puts more downward pressure on the wheels contacting the rails, which would cause the trains speed to decrease.
Answer:

Explanation:
The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:

The maximum speed is:



Answer:
increase
decrease
Explanation:
using formula
Vt=mg/6πηr
so if m increases V increases
r is the denominator so if r increases V decreases