Answer:
<h3>473.8 m/s; 473.8 m/s</h3>
Explanation:
Given the initial velocity U = 670m/s
Horizontal velocity Ux = Ucos theta
Vertical component of the cannon velocity Uy = Usin theta
Given
U = 670m/s
theta = 45°
horizontal component of the cannonball’s velocity = 670 cos 45
horizontal component of the cannonball’s velocity = 670(0.7071)
horizontal component of the cannonball’s velocity = 473.757m/s
Vertical component of the cannonball’s velocity = 670 sin 45
Vertical component of the cannonball’s velocity = 670 (0.7071)
Vertical component of the cannonball’s velocity = 473.757m/s
Hence pair of answer is 473.8 m/s; 473.8 m/s
Answer:
8.854 pF
Explanation:
side of plate = 0.1 m ,
d = 1 cm = 0.01 m,
V = 5 kV = 5000 V
V' = 1 kV = 1000 V
Let K be the dielectric constant.
So, V' = V / K
K = V / V' = 5000 / 1000 = 5
C = ε0 A / d = 8.854 x 10^-12 x 0.1 x 0.1 / 0.01 = 8.854 x 10^-12 F
C = 8.854 pF
Answer:
hydrogen
helium
oxygen
Explanation:
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Answer:
C. 0.25J
Explanation:
Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;
L is the inductance
I is the current flowing in the inductor
Given parameters
L = 20mH = 20×10^-3H
I = 5A
Required
Energy stored in the magnetic field.
E = 1/2 × 20×10^-3 × 5²
E = 1/2 × 20×10^-3 × 25
E = 10×10^-3 × 25
E = 0.01 × 25
E = 0.25Joules.
Hence the energy stored in the magnetic field of this inductor is 0.25Joules
