It hardens because you are pressing it against something.
a₀). You know ...
-- the object is dropped from 5 meters
above the pavement;
-- it falls for 0.83 second.
a₁). Without being told, you assume ...
-- there is no air anyplace where the marshmallow travels,
so it free-falls, with no air resistance;
-- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .
b). You need to find how much LESS than 5 meters
the marshmallow falls in 0.83 second.
c). You can use whatever equations you like.
I'm going to use the equation for the distance an object falls in
' T ' seconds, in a place where the acceleration of gravity is ' G '.
d). To see how this all goes together for the solution, keep reading:
The distance that an object falls in ' T ' seconds
when it's dropped from rest is
(1/2 G) x (T²) .
On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall
(9.81 / 2) x (0.83)² = 3.38 meters .
It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was
(5.00 - 3.38) = 1.62 meters
above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.
Answer:
the time needed for her to close the door is 1.36 s.
Explanation:
given information:
Force, F = 220 N
width, r = 1.40 m
weight, W = 790 N
height, h = 3.00 m
angle, θ = 90° = π/2
to find the times needed to close the door we can use the following equation
θ = ω₀t + 1/2 αt²
where
θ = angle
ω = angular velocity
α = angular acceleration
t = time
in this case, the angular velocity is zero. thus,
θ = 1/2 αt²
now, we can find the angular speed by using the torque formula
τ = I α
where
τ = torque
I = Inertia
we know that
τ = F r
and
I = 1/3 mr²
so,
τ = I α
F r = 1/3 mr² α
α = 3 F/mr
= 3 F/(w/g)r
= 3 (220)/(790/9.8) 1.4
= 5.85 rad/s²
θ = 1/2 αt²
π/2 = 1/2 5.85 t²
t = 1.36 s
Answer: The equilibrium constant can help us understand whether the reaction tends to have a higher concentration of products or reactants at equilibrium. We can also use K c K_\text c KcK, start subscript, start text, c, end text, end subscript to determine if the reaction is already at equilibrium.
Explanation:
