Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Answer:
11:10 will be the time. reflection causes the object to be flipped when you see its image at the mirror
Answer= 8m/s
Because total Momentum before= total momentum after
Momentum before (p=mu)
p=(4)(12)= 48
p=2(0)=0
So total momentum before=48
Momentum after (p=mu)
Masses combined —2+4=6kg
p=6u
Mb=Ma
48=6u
u=8m/s
Answer:
a) p = 4.96 10⁻¹⁹ kg m / s
, b) p = 35 .18 10⁻¹⁹ kg m / s
,
c) p_correst / p_approximate = 7.09
Explanation:
a) The moment is defined in classical mechanics as
p = m v
Let's calculate its value
p = 1.67 10⁻²⁷ 0.99 3. 10⁸
p = 4.96 10⁻¹⁹ kg m / s
b) in special relativity the moment is defined as
p = m v / √(1 –v² / c²)
Let's calculate
p = 1.67 10⁻²⁷ 0.99 10⁸/ √(1- 0.99²)
p = 4.96 10⁻¹⁹ / 0.141
p = 35 .18 10⁻¹⁹ kg m / s
c) the relationship between the two values is
p_correst / p_approximate = 35.18 / 4.96
p_correst / p_approximate = 7.09
Answer:
B. moving faster than car B, but not necessarily accelerating
Explanation:
Velocity is the speed of something. So car A's velocity is greater than car B but does not mean car A is accelerating.
