A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 34.5 kg.
2 answers:
Answer:
Explanation:
First convert from revolution per minute to rad per second knowing that a revolution is 2π rad and every minute has 60 seconds
According to the law of momentum conservation, the total angular before and after the child jump on must be the same:
(1)
where is the moments of inertia of the merry-go-round, which we are looking for, and
is the moments of inertia of the child, which can be calculated if we treat him as a point mass:
Therefore from eq. (1) we have:
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Answer:
- The difference in length for steel is 2.46 x 10⁻⁴ m
- The difference in length for invar is 1.845 x 10⁻⁵ m
Explanation:
Given;
original length of steel, L₁ = 1.00 m
original length of invar, L₁ = 1.00 m
coefficients of volume expansion for steel, = 3.6 × 10⁻⁵ /°C
coefficients of volume expansion for invar, = 2.7 × 10⁻⁶ /°C
temperature rise in both meter stick, θ = 20.5°C
Difference in length, can be calculated as:
L₂ = L₁ (1 + αθ)
L₂ = L₁ + L₁αθ
L₂ - L₁ = L₁αθ
ΔL = L₁αθ
Where;
ΔL is difference in length
α is linear expansivity =
Difference in length, for steel at 20.5°C:
ΔL = L₁αθ
Given;
L₁ = 1.00 m
θ = 20.5°C
ΔL = 1 x 1.2 x 10⁻⁵ x 20.5 = 2.46 x 10⁻⁴ m
Difference in length, for invar at 20.5°C:
ΔL = L₁αθ
Given;
L₁ = 1.00 m
θ = 20.5°C
ΔL = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m
Answer:
21 m
Explanation:
The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using
where
d is the distance covered
v is the speed
t is the time
The frog in this problem has a speed of
v = 2.1 m/s
and therefore, after t = 10 s, the distance it covered is