2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s

a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)


2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
Answer:
The average acceleration of the ball during the collision with the wall is 
Explanation:
<u>Known Data</u>
We will asume initial speed has a negative direction,
, final speed has a positive direction,
,
and mass
.
<u>Initial momentum</u>

<u>final momentum</u>

<u>Impulse</u>

<u>Average Force</u>

<u>Average acceleration</u>
, so
.
Therefore, 
Answer:
Directly proportional: as one amount increases another amount increases at the ... The "constant of proportionality" is the value that relates the two amounts ... Example: y is directly proportional to x, and when x=3 then y=15. ... Speed and travel time are Inversely Proportional because the faster we go the shorter the time.
Answer:
39.6 m/s
Explanation:
Taking down to be positive:
Given:
v₀ = 0 m/s
a = 9.8 m/s²
Δy = 80 m
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (9.8 m/s²) (80 m)
v = 39.6 m/s
Answer:
Cp = 4756 [J/kg*°C]
Explanation:
In order to calculate the specific heat of water, we must use the equation of energy for heat or heat transfer equation.
Q = m*Cp*(T_f - T_i)/t
where:
Q = heat transfer = 2.6 [kW] = 2600[W]
m = mass of the water = 0.8 [kg]
Cp = specific heat of water [J/kg*°C]
T_f = final temperature of the water = 100 [°C]
T_i = initial temperature of the water = 18 [°C]
t = time = 120 [s]
Now clearing the Cp, we have:
Cp = Q*t/(m*(T_f - T_i))
Now replacing
Cp = (2600*120)/(0.8*(100-18))
Cp = 4756 [J/kg*°C]