On half life is 5370 years; 6 half lives have passed. You just multiply,
5370*6 = 32,220 years
<span>This question is based on conservation of energy as the work done would lead to change in kinetice energy of car
change in KE = 1/2 mv(f)^2 - 1/2mv(i)^2 = 1/2m(v(f)^2-v(i)^2)
where v(f) and v(i) are the final and initial speeds
change in KE = 185kJ = 185,000J = 1/2 m((28m/s)^2-(23m/s)^2)
185,000=1/2 m(255m^2/s^2)
solving for m
m=1451kg</span>
I believe the correct gravity on the moon is 1/6 of Earth.
Take note there is a difference between 1 6 and 1/6.
HOWEVER, we should realize that the trick here is that the
question asks about the MASS of the astronaut and not his weight. Mass is an
inherent property of an object, it is unaffected by external factors such as
gravity. What will change as the astronaut moves from Earth to the moon is his
weight, which has the formula: weight = mass times gravity.
<span>Therefore if he has a mass of 50 kg on Earth, then he will
also have a mass of 50 kg on moon.</span>
Answer:
0.71 m/s
Explanation:
We find the time it takes the stone to hit the water.
Using y = ut - 1/2gt² where y = height of bridge, u = initial speed of stone = 0 m/s, g = acceleration due to gravity = -9.8 m/s² (negative since it is directed downwards)and t = time it takes the stone to hit the water surface.
So, substituting the values of the variables into the equation, we have
y = ut - 1/2gt²
82.2 m = (0m/s)t - 1/2( -9.8 m/s²)t²
82.2 m = 0 + (4.9 m/s²)t²
82.2 m = (4.9 m/s²)t²
t² = 82.2 m/4.9 m/s²
t² = 16.78 s²
t = √16.78 s²
t = 4.1 s
This is also the time it takes the raft to move from 5.04 m before the bridge to 2.13 m before the bridge. So, the distance moved by the raft in time t = 4.1 s is 5.04 m - 2.13 m = 2.91 m.
Since speed = distance/time, the raft's speed v = 2.91 m/4.1 s = 0.71 m/s
