Answer:
4,524,660 N
Explanation:
Assuming the submarine's density is uniform, 1/9th of the submarine's mass is equal to the mass of the displaced water.
m/9 = (1026 kg/m³) (50 m³)
m = 461,700 kg
mg = 4,524,660 N
Vf = Vo + at
Vf = 20 m/s
Vo = 50 m/s
a = ?
t = 15
Therefore
20 = 50 + 15a
20 - 50 = 15a
-30 = 15a
a = -30 / 15
a = -2 m/s²
Answer:
Explanation:
From the question we are told that:
Mass 
Charge 
Velocity 
Length of Wire 
Current 
Generally the equation for Magnetic Field of Wire B is mathematically given by
Generally the equation for Force on the plane F is mathematically given by
Therefore
Therefore in Terms of g's
Answer:
Explanation:
a) Magnification = image height / object height = -9 / 18 = -0.5
b) Magnification = - image distance / object distance = -0.5
so image distance = 0.5 object distance
1/focal length = 1/image distance + 1/object distance
1/6 = 1/(0.5 object distance) + 1/object distance
object distance = 18.0 cm
c) Image appears behind the lens.
Answer:
17280 J or 17.28 kJ
Explanation:
Given that the voltage drop,
U = U2 - U1
U = 9 - 6
U = 3V
Also, we're told that the current, I is equal to 20 mA with the discharge time, t being 80 hrs.
Converting the time from h oi urs to seconds, we have
t = 80 * 3600
t = 288000
Now, to find the energy needed, we're going to use the formula
w = pt, where p = U * I
p = 3 * 20*10^-3
p = 60*10^-3
w = 60*10^-3 * 288000
w = 17280 J or 17.28 kJ
Therefore, the total energy the battery delivers in the 80 hrs is 17.28 kJ
