There are 10⁹ picoseconds in 1 Ms
1 picosecond= 10¹² s
1 Ms = 10⁻³ s
so the number of picoseconds in one Ms=(10⁻³ s/1 Ms) * (10¹² Ps/ 1 s)=10⁹
Thus there are 10⁹ picoseconds in 1 Ms
Answer:
the spear will end up above the fish relative to the actual position of the fish.
Explanation:
due to refraction of light coming from the fish the fish will appear slightly above from its real position
So due to this refraction the spearfisher will throw the spear directly at the image of the fish due to which it will not reach the position of fish but it will reach the position above the fish.
So here we can say that the spear will end up above the fish relative to the actual position of the fish
Answer:
a) 17.33 V/m
b) 6308 m/s
Explanation:
We start by using equation of motion
s = ut + 1/2at², where
s = 1.2 cm = 0.012 m
u = 0 m/s
t = 3.8*10^-6 s, so that
0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²
0.012 = 0.5 * a * 1.444*10^-11
a = 0.012 / 7.22*10^-12
a = 1.66*10^9 m/s²
If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where
E = electric field
m = mass of proton
a = acceleration
q = charge of proton
E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19
E = 2.77*10^-18 / 1.6*10^-19
E = 17.33 V/m
Final speed of the proton can be gotten by using
v = u + at
v = 0 + 1.66*10^9 * 3.8*10^-6
v = 6308 m/s
Explanation:
The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.
The relation between wave period and frequency is as follows.
T = \frac{1}{f}T=
f
1
where, T = time period
f = frequency
It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.
T = \frac{1}{f}T=
f
1
or, f = \frac{1}{T}f=
T
1
= \frac{1}{18 sec}
18sec
1
= 0.055 per second (1cycle per second = 1 Hertz)
or, f = 5.5 \times 10^{-2} hertz5.5×10 −2 hertz
<h3>Thus, we can conclude that the frequency of the wave is 5.5 \times 10^{-2} hertz5.5×10 −2 hertz .</h3>
Answer:
Heat is transferred by the hot air or water moving to a cooler area. The elements rotate in circular motions, giving the geyser pressure.
