A) it gets more negative
B) it becomes neutral
C) it gets more positive
D) it becomes reactive
(C) ;)
Answer:
Explanation:
The heat lost by the water in the cooling process is transferred to the muscles. Therefore, we must calculate this water lost heat, which is defined as:
Where m is the water's mass, c is the specific heat capacity of the water and 
is the change in temperature. Replacing the given values:
Answer:
a) Tb = m (g + vb² / R)
, b) Tt = m (g - vt² /R), c) Tb = Ty + (vb² -vt²) /R
Explanation:
a) For this exercise we will use Newton's second law, at the bottom
Tb - W = m a
Where the acceleration is centripetal
a = v² / R
Tb = W + m vb² / r
Tb = m (g + vb² / R)
b) the tension at the top
-Tt -w = m a
Tt = W - m vt² / R
Tt = m (g - vt² / R)
c) write the two equations and solve system
Tb = m (g + vb² / R)
Tt = m (g - vt² / R) (-1)
We multiply by (1) and add
Tb -Ty = vb² /R -vt² /R
Tb = Ty + (vb² -vt²) / R
Answer:
3.536*10^-6 C
Explanation:
The magnitude of the charge is expresses as Q = CV
C is the capacitance of the capacitor
V is the voltage across the capacitor
Get the capacitance
C = ε0A/d
ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m
A is the area = 0.2m²
d is the plate separation = 0.1mm = 0.0001m
Substitute
C = 8.84 x 10-12 * 0.2/0.0001
C = 1.768 x 10-8 F
Get the potential difference V
Using the formula for Electric field intensity
E = V/d
2.0 × 10^6 = V/0.0001
V = 2.0 × 10^6 * 0.0001
V = 2.0 × 10^2V
Get the charge on each plate.
Q = CV
Q = 1.768 x 10-8 * 2.0 × 10^2
Q = 3.536*10^-6 C
Hence the magnitude of the charge on each plate should be 3.536*10^-6 C
