Answer:
0 deg C = 4.81E4 pressure at 0 deg
100 deg C = 6.48E4 pressure at steam point
100 deg C - 50 deg C = (6.48 - 4.81) * 10^4 = 1.67E4 Pa
50 deg C = 50 / 100 * 1.67E4 + 4.81E4 = 5.65E4 Pa Just the halfway point between the two given pressures
The kinetics formula that applies to this problem is
1. To find height, we use 1/2mv12−mg∗μ∗h sin θ = mgh where μ is equal to 0.20, v is 11 m/s and <span>θ is 40 degrees. we cancel mass, h is equal to 5.47 meters.
2. The final speed is from 2(ug</span>) * h sin <span>θ </span><span> = vf2
</span>2(0.2*9.8) * <span>sin 40 * 5.47= vf2 ; vf is equal to 3.71 m/s.
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Answer:
a)
b)
Explanation:
Given:
- weight of the stone,
- initial velocity of vertical projection,
- air drag acting opposite to the motion of the stone,
The mass of the stone:
Now the acceleration of the stone opposite of the motion:
where:
d = deceleration
<u>In course of going up the net acceleration on the stone will be:</u>
a)
Now using the equation of motion:
where:
final velocity when the stone reaches at the top of the projectile = 0
h = height attained by the stone before starting to fall down
b)
during the course of descend from the top height of the projectile:
initial velocity,
The acceleration will be:
here the gravity still acts downwards but the drag acceleration acts in the direction opposite to the motion of the stone, now the stone is falling down hence the drag acts upwards.
Using equation of motion:
(+ve acceleration because it acts in the direction of motion)
A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at rest (at x = 900 m). Through the first 1/4 of that distance, its acceleration is +6.25 m/s2. Through the next 3/4 of that distance, its acceleration is -2.08 m/s2. What are (a) its travel time through the 900 m and (b) its maximum speed?
<span>Solve for the time at the 1/4 mark. That's 225 m. How? d = (1/2)at^2 ( initial velocity zero). Thus 225 = (1/2) 6.25 t^2. t^2 = ( 225 * 2 ) / 6.25. t = 8.5 sec. </span>
<span>At the other end t^2 = (675 * 2) / 2.08 -- we reversed the sign and ran time backwards. t = 25.5 sec. </span>
<span>So total time is 8.5 + 25.5 or 34 sec. </span>
<span>Since zero initial velocity: v^2 = 2 a d. Here, v^2 = 2 * 6.25 * 225. v = 53 m/s. That's the fastest speed since braking then occurs.</span>