Answer:
the friction force on this box is closest to 45.9 N.
Explanation:
given data
Weight of the box W = 150 N
accelerating uniformly = 3.00 m/s²
coefficient of kinetic friction = 0.400
coefficient of static friction = 0.600
solution
we know box does not move relative to the wagon
we get here friction force that is express as
friction force = mass × acceleration ..............1
here mass = weight ÷ g
mass =
= 15.3 kg
put value in equation 1 we get
friction force = 15.3 kg × 3
friction force = 45.9 N
So, the friction force on this box is closest to 45.9 N.
<h2>
Answer:</h2>
(c) 5m/s²
<h2>
Explanation:</h2>
Total acceleration (a) of a particle in a circular motion is the vector sum of the radial or centripetal acceleration (
) of the particle and the tangential acceleration (
) of the particle and its magnitude can be calculated as follows;
a =
---------------------(i)
<em>But;</em>
=
------------------------------(ii)
Where;
v = instantaneous velocity
r = radius of the circular path of motion
<em>From the question;</em>
v = 30m/s
r = 300m
(i) First let's calculate the centripetal acceleration by substituting the values above into equation (ii) as follows;
= 
= 
= 3m/s²
(ii) From the question, the velocity of the particle is increasing at a constant rate of 4m/s² and that is the tangential acceleration
, of the particle. i.e;
= 4m/s²
(iii) Now substitute the values of
and
into equation (i) as follows;
a = 
a = 
a = 
a = 5m/s²
Therefore, the magnitude of its total acceleration a, is 5m/s²
Answer:
a) xf = 5.1 m
b) u = 0.304
c) x = 10.3 m
Explanation:
we will use the following formula:
u = 0.1 + A*x
Si x = 12.5 m, u = 0.6
Clearing A:
A = 0.5/12.5 = 0.04 m^-1
a) we have to:
W = Ekf - Eki
where Ekf = final kinetic energy
Eki = initial kinetic energy
9.8*(0.1xf + ((0.04*xf^2)/(2))) = (4.5^2)/(2)
Clearing xf, we have:
xf = 5.1 m
b) Replacing values for u:
u = 0.1 + (0.04*5.1) = 0.304
c) Wf = Ekf - Eki
-u*m*x*g = 0 - (m*v^2)/2
Clearing x:
x = v^2/(2*u*g) = (4.5^2)/(2*0.1*9.8) = 10.3 m
Answer:
d = 380 feet
Explanation:
Height of man = perpendicular= 130 feet
Angle of depression = ∅ = 70 °
distance to bus stop from man = hypotenuse = d = 130 sec∅
As sec ∅ = 1 / cos∅
so d = 130 sec∅ or d = 130 / cos∅
d = 130 / cos(70°)
d = 380 feet
As per given conditions there are two directions along which forces are acting
1. Net force along left direction is given as

2. Net force towards right direction is given as

now since the two forces here in opposite direction so here we will have net force given as



so here net forces must be 440 N towards right