Answer: a) 127 eV; b) there is no change of kinetic energy.
Explanation: In order to explain this problem we have to use the change of potentail energy ( conservative field) is equal to changes in kinetic energy. So for the proton ther move to lower potential then they gain kinetic energy from the electric field. This means the electric force do work in this trayectory and then the protons increased changes its speed.
If we replace the proton by a electron we have a very different situaction, the electrons are located in a lower potental then they can not move to higher potential if any external force does work on the system.
In resumem, the electrons do not move from a point with V=87 to other point with V=-40 V. The electric force point to high potential so the electrons can not move to lower potential region (V=-40V).
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:
9m^3
Explanation:
Given data
volume v1= 3m^3
volume v2= ???
Temperature T1= 20.0°C.
Temperature T2= 60.0°C.
Applying the relation for temperature and volume
V1/T1= V2/T2
substitute
3/20= V2/60
3*60= V2*20
180= 20*V2
180/20= V2
V2= 9m^3
Hence the final volume is 9m^3
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
