The power applied to move the box anywhere is
(20 n) x (distance moved) / (time to move the distance) .
The rate of acceleration of the crate would be 1 m/s^2 because the equation for force is F=ma and when you plug in your numbers you get 10=10a so a=1
Answer:
(a) m = 33.3 kg
(b) d = 150 m
(c) vf = 30 m/s
Explanation:
Newton's second law to the block:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
Data
F= 100 N
a= 3.0 m/s²
(a) Calculating of the mass of the block:
We replace dta in the formula (1)
F = m*a
100 = m*3
m = 100 / 3
m = 33.3 kg
Kinematic analysis
Because the block moves with uniformly accelerated movement we apply the following formulas:
d= v₀t+ (1/2)*a*t² Formula (2)
vf= v₀+a*t Formula (3)
Where:
d:displacement in meters (m)
t : time interval in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
a= 3.0 m/s²
v₀= 0
t = 10 s
(b) Distance the block will travel if the force is applied for 10 s
We replace dta in the formula (2):
d= v₀t+ (1/2)*a*t²
d = 0+ (1/2)*(3)*(10)²
d =150 m
(c) Calculate the speed of the block after the force has been applied for 10 s
We replace dta in the formula (3):
vf= v₀+a*t
vf= 0+(3*(10)
vf= 30 m/s
To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.
The work done would be defined as
Where,
PE = Potential Energy
KE = Kinetic Energy
Where,
m = Mass
g = Gravitational energy
h = Height
v = Velocity
Considering power as the change of energy as a function of time we will then have to
The rate of mass flow is,
Where,
= Density of water
A = Area of the hose 
The given radius is 0.83cm or 
m, so the Area would be
We have then that,
Final the power of the pump would be,
Therefore the power of the pump is 57.11W
<span>We can answer this using
the rotational version of the kinematic equations:</span><span>
θ = θ₀ + ω₀<span>t + ½αt²
-----> 1</span></span>
ω² = ω₀² + 2αθ
-----> 2
Where:
θ = final angular
displacement = 70.4 rad
θ₀ = initial
angular displacement = 0
ω₀ = initial angular
speed
ω = final angular speed
t = time = 3.80 s
α = angular acceleration
= -5.20 rad/s^2
Substituting the values
into equation 1:<span>
70.4 = 0 + ω₀(3.80)
+ ½(-5.20)(3.80)² </span><span>
ω₀ = (70.4
+ 37.544) / 3.80 </span><span>
ω₀ = 28.406
rad/s </span><span>
Using equation 2:
ω² = (28.406)² + 2(-5.2)70.4
ω = 8.65 rad/s
</span>
