Answer:
h> 2R
Explanation:
For this exercise let's use the conservation of energy relations
starting point. Before releasing the ball
Em₀ = U = m g h
Final point. In the highest part of the loop
Em_f = K + U = ½ m v² + ½ I w² + m g (2R)
where R is the radius of the curl, we are considering the ball as a point body.
I = m R²
v = w R
we substitute
Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R
em_f = m v² + 2 m g R
Energy is conserved
Emo = Em_f
mgh = m v² + 2m g R
h = v² / g + 2R
The lowest velocity that the ball can have at the top of the loop is v> 0
h> 2R
Answer:
Explanation:
From newton's equation of motion of uniform acceleration
v = u + at
where v is final velocity , u is initial velocity , a is acceleration and time is t .
putting the values
v = 0 + .5 x 3 x 60 ( time in second = 3 x 60 s )
= 90 m /s
So , final velocity is 90 m /s .
One of the many random useless factoids that I carry around
in my head is the factoid that 60 miles per hour is equivalent
to exactly 88 feet per second.
So in three seconds at that speed, you would cover exactly
(3 x 88) = 264 feet.
The insulin levels lead to the cause of type 2 diabetes
(a)
KE = m v^2 / 2 = (1200 kg)(20 m/s)^2 / 2 = 240,000 J
(b)
The energy is entirely dissipated by the force of friction in the brake system.
(c)
W = delta KE = KEf - KEi = (0 - 240,000) J = -240,000 J
(d)
Fd = delta KE
F = (delta KE) / d = (-240,000 J) / (50 m) = -4800 N
The magnitude of the friction force is 4800 N.
