Answer:
The force of friction that acts on him is

Explanation:
The firefighter with an acceleration of 3m/s^2 take the gravity acceleration as 10m/s^2 isn't necessary to know the coefficient of friction just to know the force of friction:




Sole to Fk



Answer:
420 L
Explanation:
Applying Boyle's Law,
PV = P'V'.................... Equation 1
Where P = Initial pressure, P' = Final pressure, V = Initial volume, V' = Final volume.
make V' the subject of the equation
V' = PV/P'.................... Equation 2
From the question,
Given: P = 720 mmHg, V = 350 L, P' = 600 mmHg
Substitute these values into equation 2
V' = (720×350)/600
V' = 252000/600
V' = 420 L
From laws of motion:

Where S is the distance/displacement (as you would call it) which is unknown
v = final velocity which is 0m/s (this is because the car stops)
u = initial velocity which is 36m/s (from the data given)
t = time taken for the distance to be covered and it is 6s
Substitute the values, hence:


But this is merely the distance he travelled in the 6 seconds he was trying to stop the car.
Therefore, the distance between the car and the cows = 160-108
Distance = 52m
Answer:
0.423m
Explanation:
Conversion to metric unit
d = 4.8 cm = 0.048m
Let water density be 
Let gravitational acceleration g = 9.8 m/s2
Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.
Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:

Where
N is the spring force,
is the buoyancy force, which equals to the weight
of the water displaced by the submerged portion of the cylinder, which is the product of water density
, submerged volume
and gravitational constant g. W = mg is the weight of the metal cylinder.

The submerged volume would be the product of cross-section area and the submerged length x

Plug that into our force equation and we have


