Answer:
50 watts
Explanation:
Applying,
Power (P) = Workdone (W)/Time(t)
But,
Work done (W) = Force (F)×distance(d)
Therefore,
P = Fd/t..................... Equation 1
Where P = power of the weightlifter, F = Force applied, d = distance, t = time.
From the question,
Given: F = 200 N, d = 0.5 m, t = 2 s
Substitute these values into equation 1
P = (200×0.5)/2
P = 100/2
P = 50 watts
Jet streams are tied to global warming so I’m guessing hotter
Answer:
Explanation:
From the given information:
The initial PE 
= m×g×h
= 5 kg × 9.81 m/s² × 10 m
= 490.5 J
The change in Potential energy P.E of the box is:
ΔP.E = 
ΔP.E = 0 -
ΔP.E = 
If we take a look at conservation of total energy for determining the change in the internal energy of the box;
this can be re-written as:
Here, K.E = 0
Also, 70% goes into raising the internal energy for the box;
Thus,
ΔU = 343.35 J
Thus, the magnitude of the increase is = 343.35 J
<h2>Answer:</h2>
<u>Acceleration is </u><u>-10.57 rad/s² </u>
<u>Time is </u><u>6 seconds</u>
<h2>Explanation:</h2><h3>a) </h3>
u=900rpm= 94.248 rad/s
v =300rpm= 31.416 rad/s
s=60 revolutions= 377 rad
v²= u²+ 2as
31.416² = 94.248²+ 2 * a * 377
a = v²-u² / 2s
a= -10.57 rad/s²
<h3>b) </h3>
Using 1st equation of motion
v-u/a = t
Putting the values
t = (31.4 - 94.2)/-10.57
t = 6 seconds
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
