The proton would be 2 and d a part of 1 then calculate that hope this helped
Definitely not the last 2. My bet is on the first option. If it is wrong don't hit me please...
Answer:
Explanation:
The problem is solved using the law of conservation of energy,
So
The power of is series combination is Vn^2 times that of a parallel combination.
For series combination :
Req = R + R + R + ............... n times = nR
I = Δv/nr
Power = (Δv/nr)^2 × nr = Δv^2/nr
For parallel combination
1/req = 1/R + 1/R + 1/R +................(n times) = n/R
Req = R/n
Power = Δv/(R/n) = nΔv^2/R
Ratio = Δv^2/nr/n·Δv^2/R = 1/n^2
Hence, power of is series combination is Vn^2 times that of a parallel.
Learn more about parallel combination here:
brainly.com/question/12400458
#SPJ4
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
