<h2>
Answer:</h2>
(a) 10N
<h2>
Explanation:</h2>
The sketch of the two cases has been attached to this response.
<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>
In this case, a frictional force
is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;
∑F = ma -------------------(i)
Where;
∑F = effective force acting on the object (box)
m = mass of the object
a = acceleration of the object
∑F = F - 
m = 50kg
a = 0 [At constant velocity, acceleration is zero]
<em>Substitute these values into equation (i) as follows;</em>
F -
= m x a
F -
= 50 x 0
F -
= 0
F =
-------------------(ii)
<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>
In this case, the same frictional force
is opposing the movement of the box.
∑F = 1.5F - 
m = 50kg
a = 0.1m/s²
<em>Substitute these values into equation (i) as follows;</em>
1.5F -
= m x a
1.5F -
= 50 x 0.1
1.5F -
= 5 ---------------------(iii)
<em>Substitute </em>
<em> = F from equation (ii) into equation (iii) as follows;</em>
1.5F - F = 5
0.5F = 5
F = 5 / 0.5
F = 10N
Therefore, the value of F is 10N
<em />
<span>Earth (and hence the observer) moves.</span>
Answer: A. Her speed is 4.4 m/s, and her velocity is 0 m/s.
Explanation: i took the test on edgenuity
Answer:
155.38424 K
2.2721 kg/m³
Explanation:
= Pressure at reservoir = 10 atm
= Temperature at reservoir = 300 K
= Pressure at exit = 1 atm
= Temperature at exit
= Mass-specific gas constant = 287 J/kgK
= Specific heat ratio = 1.4 for air
For isentropic flow

The temperature of the flow at the exit is 155.38424 K
From the ideal equation density is given by

The density of the flow at the exit is 2.2721 kg/m³