(a) The plane makes 4.3 revolutions per minute, so it makes a single revolution in
(1 min) / (4.3 rev) ≈ 0.2326 min ≈ 13.95 s ≈ 14 s
(b) The plane completes 1 revolution in about 14 s, so that in this time it travels a distance equal to the circumference of the path:
(2<em>π</em> (23 m)) / (14 s) ≈ 10.3568 m/s ≈ 10 m/s
(c) The plane accelerates toward the center of the path with magnitude
<em>a</em> = (10 m/s)² / (23 m) ≈ 4.6636 m/s² ≈ 4.7 m/s²
(d) By Newton's second law, the tension in the line is
<em>F</em> = (1.3 kg) (4.7 m/s²) ≈ 6.0627 N ≈ 6.1 N
Answer:
anaemia, low blood pressure etc.
Answer:
A u = 0.36c B u = 0.961c
Explanation:
In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains
u ’= (u-v) / (1- uv / c²)
Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory
The data give is u ’= 0.43c and the initial core velocity v = 0.94c
Let's clear the speed with respect to the observer (u)
u’ (1- u v / c²) = u -v
u + u ’uv / c² = v - u’
u (1 + u ’v / c²) = v - u’
u = (v-u ’) / (1+ u’ v / c²)
Let's calculate
u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)
u = 0.51c / (1 + 0.4042)
u = 0.36c
We repeat the calculation for the other piece
In this case u ’= - 0.35c
We calculate
u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)
u = 1.29c / (1- 0.329)
u = 0.961c
Answer:
Momentum of block B after collision =
Explanation:
Given
Before collision:
Momentum of block A = 
= 
Momentum of block B = 
= 
After collision:
Momentum of block A = 
= 
Applying law of conservation of momentum to find momentum of block B after collision 
.
Plugging in the given values and simplifying.
Adding 200 to both sides.
∴ 
Momentum of block B after collision =
