Answer:
Explanation:
capacitance of each capacitor
C₀= Q₀ / V₀
V₀ = Q₀ / C₀
New total capacitance = C₀ ( 1 + K )
Common potential
= total charge / total capacitance
= 2 Q₀ / [ C₀ ( 1 + K ) ]
2 V₀ / ( 1 + K )
b )
Common potential = 2 x V₀ / ( 1 + 7.8 )
= .227 V₀
charge on capacitor with dielectric
= .227 V₀ x 7.8 C₀
= 1.77 V₀C₀
= 1.77 Q₀
Ratio required = 1.77
Answer:
a) v = 88.54 m/s
b) vf = 26.4 m/s
Explanation:
Given that;
m = 1400.0 kg
a)
by using the energy conservation
loss in potential energy is equal to gain in kinetic energy
mg × ( 3200-2800) = 1/2 ×m×v²
so
1400 × 9.8 × 400 = 0.5 × 1400 × v²
5488000 = 700v²
v² = 5488000 / 700
v² = 7840
v = √7840
v = 88.54 m/s
b)
Work done by all forces is equal to change in KE
W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)
we substitute
1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf² -0 )
488000 = 700 vf²
vf² = 488000 / 700
vf² = 697.1428
vf = √697.1428
vf = 26.4 m/s
0 to 60 mph in 4.5 seconds
Explanation:
Answer:
30.9 m
Explanation:
x = 129.9 m y = 30.9 m First of all, let's calculate the horizontal and vertical velocities involved h = 50.0cos(30) = 43.30127 m/s v = 50.0sin(30) = 25 m/s The horizontal distance is simply the horizontal velocity multiplied by the time, so 43.30127 m/s * 3 s = 129.9 m So the horizontal distance traveled is 129.9 m, so x = 129.9 m The vertical distance needs to take into account gravity which provides an acceleration of -9.8 m/s^2, so we get d = 25 m/s * 3s - 0.5*9.8 m/s^2 * (3 s)^2 d = 75 m - 4.9 m/s^2 * 9 s^2 d = 75 m - 44.1 m d = 30.9 m So the vertical distance traveled is 30.9 m, so y = 30.9 m