Answer:
Explanation:
The sealed bottle serves as an open system for the water molecules, thus heat energy can be exchanged with its surroundings. When the bottle is placed in a warm environment, there would be a gradual increase in the thermal energy of the molecules of water due to a change in the temperature of its surroundings.
Thus, the molecules of water closer to the boundary of the bottle absorbs heat, becomes lighter and changes position. This gradual process continues until all molecules attains an average temperature which equals that of the surroundings of the bottle.
Therefore, both the thermal energy and temperature of the water increases.
Answer:
a)The electric Field will be zero at the point between the sheets
b)
c)
Explanation:
Let
be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field so the Electric Flux due to it is zero.
Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

The Field will be away from the sheet and perpendicular to it.
a) The Electric Field between them

b)The Electric Field to the right of the sheets

c)The Electric Field to the left of the sheets

Answer:
Answered
Explanation:
a) What is the work done on the oven by the force F?
W = F * x
W = 120 N * (14.0 cos(37))
<<<< (x component)
W = 1341.71
b) 

= 29.4 N


W_f= 328.72 J = 329 J
c) increase in the internal energy
U_2 = mgh
= 12*9.81*14sin(37)
= 991 J
d) the increase in oven's kinetic energy
U_1 + K_1 + W_other = U_2 + K_2
0 + 0 + (W_F - W_f ) = U_2 + K_2
1341.71 J - 329 J - 991 J = K_2
K_2 = 21.71 J
e) F - F_f = ma
(120N - 29.4N ) / 12.0kg = a
a = 7.55m/s^2
vf^2 = v0^2 + 2ax
vf^2 = 2(7.55m/s)(14.0m)
V_f = 14.5396m/s
K = 1/2(mv^2)
K = 1/2(12.0kg)(14.5396m/s)
K = 87.238J
Answer:
Explanation:
Given
Cannon is fired with a velocity of 
Using Equation of motion

where




after time 



So after 3.3 s cannon ball is at a height of 185.89 m