Answer:
Explanation:
A ) When gymnast is motionless , he is in equilibrium
T = mg
= 63 x 9.81
= 618.03 N
B )
When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.
T = mg
= 618.03 N
C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2
Net force on it = T - mg , acting in upward direction
T - mg = m a
T = mg + m a
= m ( g + a )
= 63 ( 9.81 + .6)
= 655.83 N
D ) If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2
Net force acting in downward direction
mg - T = ma
T = m ( g - a )
= 63 x ( 9.81 - .6 )
= 580.23 N
Answer:
450 kJ
Explanation:
Q = mCΔT
where Q is heat (energy),
m is mass,
C is specific heat capacity,
and ΔT is the temperature change.
Q = (1.2 kg) (4180 J/kg/°C) (100°C − 10°C)
Q = 451,440 J
Q ≈ 450 kJ
Answer:
1.47 J/kg
120 m
Explanation:
The maximum height it can jump in a single leap 15 cm, = 0.15 m
to get the kinetic energy per kg of mass, we need to find the takeoff speed. The take off speed can be calculated by using the formula
v =
where h = 0.15
v =
v =
v = 1.715 m/s
energy per kg of mass =
1/2 * (1.715) = 1.47 J/Kg
TJe height a human can jump when compared to the flea is
2 m/2.5 mm * 15 cm =
2/0.0025 *0.15 =
800 * 0.15 = 120 m
Just as the ball begins to touch the first speck of dust on the ground,
its kinetic energy is 49 joules and its potential energy is zero.
Answer:
The compression is .
Explanation:
A Hooke's law spring compressed has a potential energy
where k is the spring constant and the distance to the equilibrium position.
A mass m moving at speed v has a kinetic energy
.
So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity . Knowing that the energy is constant.
If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:
But, in the left side we can use the previous equation to obtain:
And this is the compression we are looking for