Answer:
331.7m/s
Explanation:
Given parameters:
Initial velocity = 100m/s
Acceleration = 50m/s²
Distance = 1km = 1000m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we have to apply the right motion equation shown below;
v² = u² + 2aS
v is the final velocity
u is the initial velocity
a is the acceleration
S is the distance
Now insert the parameters and solve;
v² = 100² + (2 x 50 x 1000)
v² = 110000
v = √110000 = 331.7m/s
Answer:
Cumulonimbus
Hail development. Hail is a type of strong precipitation, which is shaped in rainstorms mists (Cumulonimbus). Tempests mists comprises of beads of fluid water (at temperatures lower than 0°с, the beads can be in a thermodynamically instable supercooled condition) and ice gems
Explanation:
Answer:
1 = 5.4 J
2 = 0.1979 C
3 = 5
Explanation:
Energy in a capacitor, E is
E = 1/2 * C * V²
E = 1/2 * 3000*10^-6 * 60²
E = 1/2 * 3000*10^-6 * 3600
E = 1/2 * 10.8
E = 5.4 J
E = Q²/2C = 6.53 J
E * 2C = Q²
Q² = 6.53 * 2 * 3000*10^-6
Q² = 13.06 * 3000*10^-6
Q² = 0.03918
Q = √0.03918
Q = 0.1979 C
The Capacitor, C is inversely proportional to the distance of separation, D. Thus, if D is increased by 5 to be 5D, then C would be C/5. And therefore, our energy stored in the capacitor is increased by a factor of 5.
Answer:
Explanation:
given
m= 17.5kg
F= 75N
d= 5.7m
∪=0.150
θ= 21°
a. W = Fcos θ × d
75cos21° ×5.7
=399.106J
b. normal force is zero. 0 Joules
cos 90°=0
Answer:
C) Both the charge on the plates of the capacitor and its capacitance would change.
Explanation:
The capacitance of parallel plate capacitor without dielectric material is given as;
A parallel plate capacitor with a dielectric between its plates has a capacitance given by;
where;
C is the capacitance
K is the dielectric constant
ε₀ is permittivity of free space
A is the area of the plates
d is the distance of separation of the two plates
- first point to note, is that the capacitance increases when dielectric material is inserted by a factor 'k'
Again, Q = CV (without dielectric material)
(with dielectric material)
- second point to note, is that charge stored in the plates increases due to presence of dielectric material.
Finally, we can conclude that both the charge on the plates of the capacitor and its capacitance would change.