Answer:
horizontal direction force move wagon at 18.79 N
Explanation:
given data
force F = 20 N
angle = 20 degree
to find out
What part of the force moves the wagon
solution
we know here as per attach figure
boy pull a wagon at force 20 N at angle 20 degree
so there are 2 component
x in horizontal direction i.e F cos20
and y in vertical direction i.e F sin20
so we can say
horizontal direction force is move the wagon that is
horizontal direction force = F cos 20
horizontal direction force = 20× cos20
horizontal direction force = 18.79 N
so horizontal direction force move wagon at 18.79 N
To solve the problem, we
must know the heat capacity of ice and water.
For Cp = 2090 J/kg C
H = mCpT
H = (10 kg) ( 2090 J/ Kg C)
( -23 C)
H = - 480700 J
For water Cp = 4180 j/kg C
H = (100 kg) ( 4180 J/kg C)
( 60 C)
<span>H = 2508000 J</span>
The <u>speed</u> of the ride is
Speed = (distance covered) / (time to cover the distance)
Speed = (4,000 m) / (45 min)
Speed = <em>88.89 m/s</em>
Speed = <em>1.481 m/s</em>
Speed = <em>5.333 km/hr</em>
There's not enough information given in the question to calculate the velocity of the ride. For example ...
-- If the ride was completely in a straight line, then the velocity would be exactly equal to the speed.
-- If the rider went 2km from his house and then 2km back home again, his velocity for the whole ride would be zero.
We just don't know from the information given.
Since the ladder is standing, we know that the coefficient
of friction is at least something. This [gotta be at least this] friction
coefficient can be calculated. As the man begins to climb the ladder, the
friction can even be less than the free-standing friction coefficient. However,
as the man climbs the ladder, more and more friction is required. Since he
eventually slips, we know that friction is less than what's required at the top
of the ladder.
The only "answer" to this problem is putting lower
and upper bounds on the coefficient. For the lower one, find how much friction
the ladder needs to stand by itself. For the most that friction could be, find
what friction is when the man reaches the top of the ladder.
Ff = uN1
Fx = 0 = Ff + N2
Fy = 0 = N1 – 400 – 864
N1 = 1264 N
Torque balance
T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)
N2 = 439 N
Ff = 439= u N1
U = 440 / 1264 = 0.3481
