Answer:
A
Explanation:
Initial gravitational energy = final kinetic energy + heat
mgh = KE + Q
(50 kg) (9.81 m/s²) h = 78400 J + 884000 J
h = 1960 meters
Given:
V1 = 4m3
T1 = 290k
P1 = 475 kpa = 475000 Pa
V2 = 6.5m3
T2 = 277K
Required:
P
Solution:
n = PV/RT
n = (475000 Pa)(4m3) / (8.314 Pa-m3/mol-K)(290k)
n = 788 moles
P = nRT/V
P = (788 moles)(8.314
Pa-m3/mol-K)(277K)/(6.5m3)
P = 279,204 Pa or 279 kPa
Answer
3
Explanation:
x=
<em>t + 1/2 </em>
<em />
As we know that block of steel is continue to be in moving state
so here the friction must be kinetic friction between two surface
so we know that formula of kinetic friction must be

now we have


now from the above equation we have


So here we need atleast 14.25 N force to continue sliding the box now as per given options all forces which are less than 14.25 N is not correct
Hence correct answer must be 18.0 N
<u>D) 18.0 N</u>
Answer:
c
Explanation:
if numbers of protons and electrons are equal, the atom is neutral