He produced the first orderly arrangement of known elements, he used patterns to predict undiscovered elements
The speed is changing its direction all the time. There
is an acceleration which changes the direction of the speed – that is called
centripetal acceleration. Only uniform linear motions are considered to have no
acceleration.
This is the general formula for acceleration
a = dv/dt
When calculating dv, you should keep in mind the change
in the velocity vector’s direction. You can easily see in a graph that with dt
tending to 0 (so the length of the arc covered is also tending to 0), the difference
between vectors Vf and V0 has a direction which is perpendicular to velocity
(the shorter the arc, the closest the angle is to 90 degrees).
There is a formula (which can be deducted from the
previous formula) which allows you to calculate the acceleration:
a = v^2/r
Let’s talk about the units:
v is in m/s
r is in m
so v^2/r
is in (m/s)^2/m = (m^2/s^2)/m = m/s^2
which is the same unit as dv/dt:
dv/dt = (m/s)/s= m/s^2
Answer:
v = 8.09 m/s
Explanation:
For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.
Let's calculate the energy
starting point. Higher
Em₀ = U = m gh
final point. To go down the slope
Em_f = K = ½ m v²
The work of the friction force is
W = fr L cos 180
to find the friction force let's use Newton's second law
Axis y
N - W_y = 0
N = W_y
X axis
Wₓ - fr = ma
let's use trigonometry
sin θ = y / L
sin θ = 11/110 = 0.1
θ = sin⁻¹ 0.1
θ = 5.74º
sin 5.74 = Wₓ / W
cos 5.74 = W_y / W
Wₓ = W sin 5.74
W_y = W cos 5.74
the formula for the friction force is
fr = μ N
fr = μ W cos θ
Work is friction force is
W_fr = - μ W L cos θ
Let's use the relationship of work with energy
W + ΔU = ΔK
-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²
v² = - 2 μ g L cos 5.74 +2 (gh)
v² = 2gh - 2 μ gL cos 5.74
let's calculate
v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74
v² = 215.6 -150.16
v = √65.44
v = 8.09 m/s
The alpha particle is emitted at 4235 m/s
Explanation:
We can use the law of conservation of momentum to solve the problem: the total momentum of the original nucleus must be equal to the total momentum after the alpha particle has been emitted. Therefore:
where:
is the mass of the original nucleus
is the initial velocity of the nucleus
is the mass of the alpha particle
is the final velocity of the alpha particle
is the mass of the daughter nucleus
is the final velocity of the nucleus
Solving for
, we find the final velocity of the alpha particle:

Learn more about momentum:
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