The half-life observed by an observer at rest is 1.25 s
Explanation:
In this problem, the particle is travelling at a significant fraction of the speed of light: therefore, we have to consider the phenomenon of the time dilation. The half-life of the particle as measured by an observer at rest is given by
where
is the proper half-life of the particle
v is the speed of the particle
c is the speed of light
For the particle in this problem,
Substituting into the equation, we find:
#LearnwithBrainly
As we know that in cellular phone all the signals are transmitted by antenna in the form of electromagnetic waves
All these electromagnetic waves are transmitted through the phones and then received as a signal to the receiver of another phone
These signals are then processed and converted into sound which is produced by the phone
So here correct answer will be
<em>C. The cellular telephone transmits information by electromagnetic waves to a receiver which then encodes them and produces sound.</em>
Answer:
6.13 s
219 N
Explanation:
Newton's law in the x direction:
∑F = ma
150 cos 30° N − 50 N = (30 kg) a
a = 2.66 m/s²
Δx = v₀ t + ½ at²
(50 m) = (0 m/s) t + ½ (2.66 m/s²) t²
t = 6.13 s
Newton's law in the y direction:
∑F = ma
Fn + 150 sin 30° N − (30 kg) (9.8 m/s²) = 0
Fn = 219 N
Here, we are required to determine which combination of molecules will produce ammonia with no leftovers.
Option A: 2N2 and 6H2 is the correct combination of molecules that will produce ammonia with no leftovers.
First, it is important to know that both Nitrogen and Hydrogen used in the production of ammonia are diatomic.
Secondly, Nitrogen and Hydrogen are in the ratio 1 : 3.....
As such, the coefficient of hydrogen should be thrice that of Nitrogen to ensure that there are no leftovers.
Therefore, option A which has:
2N2 and 6H2 is the correct combination of molecules that will produce ammonia with no leftovers.
Read more:
brainly.com/question/24396848
That gives you the magnitude of velocity, but it doesn't handle
the directions that are involved.
