To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant a
1 answer:
Answer:
Explanation:
Time taken to complete one revolution is called time period.
So, Time period, T = 1 s
Diameter = 1.6 mm
radius, r = 0.8 mm
Let the angular speed is ω.
The relation between angular velocity and the time period is
ω = 2 x 3.14 = 6.28 rad/s
The relation between the linear velocity and the angular velocity is
v = r x ω
v = 0.8 x 10^-3 x 6.28
v = 0.005 m/s
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Answer:
+12 m/s
Explanation:
The velocity of an object moving of accelerated motion is given by:
where
v0 is the initial velocity of the object
a is the acceleration
t is the time
In this problem, the rocket starts from rest, so . The acceleration is
, so the velocity after
will be
Explanation:
It is given that,
Initially the car is at rest and travels for t₁ seconds with a uniform acceleration a₁. The driver then applies the brakes, causing a uniform acceleration a₂, If the brakes are applied for t₂ seconds.
We need to find the speed of the car just before the beginning of the braking period.
Using the formula of acceleration. It is given by :
u = 0
So, just before the beginning of the braking period the speed of the car is . Hence, this is the required solution.
Answer:
λ = 396.7 nm
Explanation:
For this exercise we use the diffraction ratio of a grating
d sin θ = m λ
in general the networks works in the first order m = 1
we can use trigonometry, remembering that in diffraction experiments the angles are small
tan θ = y / L
tan θ = = sin θ
sin θ = y / L
we substitute
= m λ
with the initial data we look for the distance between the lines
d =
d = 1 656 10⁻⁹ 1.00 / 0.600
d = 1.09 10⁻⁶ m
for the unknown lamp we look for the wavelength
λ = d y / L m
λ = 1.09 10⁻⁶ 0.364 / 1.00 1
λ = 3.9676 10⁻⁷ m
λ = 3.967 10⁻⁷ m
we reduce nm
λ = 396.7 nm