Answer: The temperature of the water falls by 3.3°C
Explanation:
The heat change is related to the change in temperature by the equation
dH = m Cp dT
In this example, -2665 J = 193 g x 4.184 J/g°C x dT
so dT = -3.3 °C
Answer:
The minimum wall thickness required for the spherical tank is 0.0189 m
Explanation:
Given data:
d = inside diameter = 8.1 m
P = internal pressure = 1.26 MPa
σ = 270 MPa
factor of safety = 2
Question: Determine the minimum wall thickness required for the spherical tank, tmin = ?
The allow factor of safety:

The minimun wall thickness:

The motion of the ball on the vertical axis is an accelerated motion, with acceleration

The following relationship holds for an uniformly accelerated motion:

where S is the distance covered, vf the final velocity and vi the initial velocity.
If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:

So we can rewrite the equation as

from which we can isolate h

(1)
Now let's assume that

is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball:

. So the maximum height of the second ball is

(2)
Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
Answer is 6 tires.
This is a projectile question.
First make sure units are consistent - express speed in m/s.
20 km/h = 20000m / 3600 s = 5.56 m/s
Assume the takeoff point of the ramp is at ground level (height, h, = 0m). We need to determine how long Joe is in the air, and use that time to calculate the horizontal distance he traveled.
Joe is traveling 5.56 m/s on a ramp angled at 20 degrees. There are vertical and horizontal components to his speed:
Vertical speed = 5.56sin20 = 1.90 m/s
Horizontal speed = 5.56cos20 = 5.22 m/s
An easy way to proceed is to calculate the time it takes for Joe’s vertical speed to reach 0m/s - this represents the time when Joe is at his maximum height and is therefore halfway through the trip. Double whatever time this is to find the total time of the trip. Remember he is decelerating due to gravity:
Time to peak:
a = Δv / Δt
-9.8 = -1.9 / Δt
Δt = 0.19s
Total trip time:
0.19 x 2 = 0.38s
Now that we have the total tome Joe is in the air, we can find the horizontal distance he traveled:
v = d / t
5.22 = d / 0.38
d = 1.98m
Now divide this total distance by the length of an individual tire to find the number of tires he will clear:
1.98 / 0.3 = 6.6 tires
Therefore he can jump 6 tires safely (he will land in the middle of the 7th tire).
Lots of steps I know but just try to think of the situation and keep track of the vertical and horizontal things!