Answer:
A 1.0 min
Explanation:
The half-life of a radioisotope is defined as the time it takes for the mass of the isotope to halve compared to the initial value.
From the graph in the problem, we see that the initial mass of the isotope at time t=0 is
The half-life of the isotope is the time it takes for half the mass of the sample to decay, so it is the time t at which the mass will be halved:
We see that this occurs at t = 1.0 min, so the half-life of the isotope is exactly 1.0 min.
Answer:
v = 2.45 m/s
Explanation:
first we find the time taken during this motion by considering the vertical motion only and applying second equation of motion:
h = Vi t + (1/2)gt²
where,
h = height of cliff = 15 m
Vi = Initial Vertical Velocity = 0 m/s
t = time taken = ?
g = 9.8 m/s²
Therefore,
15 m = (0 m/s) t + (1/2)(9.8 m/s²)t²
t² = (15 m)/(4.9 m/s²)
t = √3.06 s²
t = 1.75 s
Now, we consider the horizontal motion. Since, we neglect air friction effects. Therefore, the horizontal motion has uniform velocity. Therefore,
s = vt
where,
s = horizontal distance covered = 4.3 m
v = original horizontal velocity = ?
Therefore,
4.3 m = v(1.75 s)
v = 4.3 m/1.75 s
<u>v = 2.45 m/s</u>
Answer:
2.89 hours
Explanation:
given :
Vo = 72 km/h
Vt = 90 km/h
S = 234 km
find : the time taken (t) = ?
solution :
2.a.s = Vt² - Vo²
2.a.(234) = 90²- 72²
468.a = 8100 - 5184
= 2916
a = 2916/468 = 6.23 km/h²
so,
t = (Vt-Vo) /a
= (90-72)/ 6.23
= 18/ 6.23
= 2.89 hours
