a ball rolls off the edge of a 2m high shelf at a speed of 5 m/s and hits the ground the taken to hit the ground is
1 answer:
Answer:
Explanation:
Use the one-dimensional equation
Δx = where delta x is the displacement of the object, v0 is the velocity of the object, a is the pull of gravity, and t is the time in seconds. That's our unknown.
Δx = -2 (negative because where it ends up is lower than the point at which it started),
, and
a = -9.8
Filling in:
and simplified a bit:
this should look hauntingly familiar (a quadratic, which is parabolic motion...very important in physics!!). We begin by getting everything on one side of the equals sign and solving for t by factoring:
(the 0 is also indicative of the object landing on the ground! Isn't this a beautiful thing, how it all just works so perfectly together?)
When you factor this however your math/physics teacher has you factoring you will get that
t = 1.3 sec and t = -.31 sec
Since we all know that time can NEVER be negative, it takes the ball 1.3 sec to hit the ground from a height of 2 m if it is rolling off the shelf at 5 m/s.
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Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
The potential at a distance can be represented as:
V(r) - V(0) =
V(r) - V(0) = ₀
V(r) = ₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) =
V(r) - V(0) =
V(r) =
V(r) =
V(r)
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V