D) The speed of a wave slows as it travels at different speed in different media.
Answer:
1. 150C.
2. 50sec
3.1.5a
Explanation:
1. I = Q/T
Q= 30x5
=150c
2.applying the formulae, I = Q/T
T= Q/I
=500/10
=50sec.
3. using the formulae i=q/t
i= 120/80
=1.5a.
Scattering occurs when light changes direction after colliding with particles of matter.
Answer:
v = 21.03 m/s
Explanation:
given,
mass of skier = 45 kg
the slope of the snow = 10.0◦
coefficient of friction = 0.114
distance traveled = 300 m
speed = ?
Acceleration = g sin θ - µ g Cos θ
= 9.8 × Sin (10°) - 0.10 × 9.8 × Cos(10°)
= 0.737 m/s²
using equation of motion
v² = u² + 2 a s
v² = 0 + 2 × 0.737 × 300
v = 21.03 m/s
Speed of skier's after travelling 300 m speed is equal to 21.03 m/s
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
