Answer:
c.
Explanation:
We are given that
Acceleration due to gravity on the moon=
Acceleration due to gravity on the earth=

Net force due to am on an object on moon=
There is no friction and no drag force and there is no gravity involved
Then, the force acting on an object on earth=
(given)


Hence, option c is true.
Eisenhower started the NASA project to develop technology for military application.
Answer:
Psm = 30.66 [Psig]
Explanation:
To solve this problem we will use the ideal gas equation, recall that the ideal gas state equation is always worked with absolute values.
P * v = R * T
where:
P = pressure [Pa]
v = specific volume [m^3/kg]
R = gas constant for air = 0.287 [kJ/kg*K]
T = temperature [K]
<u>For the initial state</u>
<u />
P1 = 24 [Psi] + 14.7 = 165.47[kPa] + 101.325 = 266.8 [kPa] (absolute pressure)
T1 = -2.6 [°C] = - 2.6 + 273 = 270.4 [K] (absolute Temperature)
Therefore we can calculate the specific volume:
v1 = R*T1 / P1
v1 = (0.287 * 270.4) / 266.8
v1 = 0.29 [m^3/kg]
As there are no leaks, the mass and volume are conserved, so the volume in the initial state is equal to the volume in the final state.
V2 = 0.29 [m^3/kg], with this volume and the new temperature, we can calculate the new pressure.
T2 = 43 + 273 = 316 [K]
P2 = R*T2 / V2
P2 = (0.287 * 316) / 0.29
P2 = 312.73 [kPa]
Now calculating the manometric pressure
Psm = 312.73 -101.325 = 211.4 [kPa]
And converting this value to Psig
Psm = 30.66 [Psig]
Answer:
c. probablistic view of nature.
Explanation:
According to the problem of particle in a box in one dimension. If the particle energy E is taken less than the height of the barrier V.
Then with the help of classical mechanics it can be prove that the particle can not cross the barrier but according to the quantum mechanics, there is a small but a finite probability to cross the barrier.
Therefore by the above discussion it can be concluded that quantum mechanics can be thought as a probablistic view of nature.
Answer: hello the complete question is attached below
answer :
r2 = 4r1
Explanation:
Electric field strength = F / q
we will assume the rod has an infinite length
For an infinitely charged rod
E ∝ 1/ r
considering two electric fields E1 and E2 at two different locations as described in the question
E1/E2 = r1/r2 ----- ( 2 )
<u>Calculate for r2 when E2 = E1/4 </u>
back to equation 2
E1 / (E1/4) = r1 / r2
∴ r2 = 4r1