From p1v1/t1 = p2v2/t2
pressure unchanged ... cancelled out
v1=605 , t1=27C = 300K,
t2=-3C = 270K
***remember temperature must be in Kelvin
we got
605/300 = v2/270
v2 = 545
'A' and 'C' both show that behavior.
'D' also shows it, but the object is moving backwards when time begins.
First it moves faster and faster backwards, then it moves slower and slower backwards.
Answer:
No, neutrons have about the same mass as a proton, but both have more mass than electrons.
Hope this helps a bit,
Flips
Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.
Answer:
Explanation:
To solve this problem we use the Hooke's Law:
(1)
F is the Force needed to expand or compress the spring by a distance Δx.
The spring stretches 0.2cm per Newton, in other words:
1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm
The force applied is due to the weight
We replace in (1):
We solve the equation for m:
