When people aboard a plane...the amount of baggage you take has to vary because the plane has a certain carrying capacity.
Answer:
Potential energy of book = 7.5 J
Explanation:
Given:
Weight of book = 5 N
Height of shelf = 1.5 meter
Find:
Potential energy of book
Computation:
Weight = Mass x Acceleration of gravity
Mass x Acceleration of gravity = 5 N
Potential energy = Mass x Acceleration of gravity x Height
Potential energy of book = Mass x Acceleration of gravity x Height
We know that;
Mass x Acceleration of gravity = 5 N
So,
Potential energy of book = 5 x 1.5
Potential energy of book = 7.5 J
Answer:
Explanation:
Balance point will be achieved as soon as the weight of the baby elephant creates torque equal to torque created by weight of woman about the pivot. torque by weight of woman
weight x distance from pivot
= 500x 5
= 2500 Nm
torque by weight of baby woman , d be distance of baby elephant from pivot at the time of balance
= 2500x d
for equilibrium
2500 d = 2500
d = 1 m
So elephant will have to walk up to 1 m close to pivot or middle point.
Answer:
Vi = 0.055 m³ = 55 L
Explanation:
From first Law of Thermodynamics, we know that:
ΔQ = ΔU + W
where,
ΔQ = Heat absorbed by the system = 52.5 J
ΔU = Change in Internal Energy = -102.5 J (negative sign shows decrease in internal energy of the system)
W = Work Done in Expansion by the system = ?
Therefore,
52.5 J = - 102.5 J + W
W = 52.5 J + 102.5 J
W = 155 J
Now, the work done in a constant pressure condition is given by:
W = PΔV
W = P(Vf - Vi)
where,
P = Constant Pressure = (0.5 atm)(101325 Pa/1 atm) = 50662.5 Pa
Vf = Final Volume of System = (58 L)(0.001 m³/1 L) = 0.058 m³
Vi = Initial Volume of System = ?
Therefore,
155 J = (50662.5 Pa)(0.058 m³ - Vi)
Vi = 0.058 m³ - 155 J/50662.5 Pa
Vi = 0.058 m³ - 0.003 m³
<u>Vi = 0.055 m³ = 55 L</u>
Answer:
Explanation:
GIVEN
diameter = 15 fm =m
we use here energy conservation
there will be some initial kinetic energy but after collision kinetic energy will zero
on solving these equations we get kinetic energy initial
J ..............(i)
That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e
and gains kinetic energy K =e∆V ..........(ii)
by accelerating through a potential difference ∆V
Thus the alpha particle will
just reach the nucleus after being accelerated through a potential difference ∆V
equating (i) and second equation we get
e∆V = 35.33 Me V