I think only if they were too overpowered maybe, but the modern world doesn't except this kind of dictatorship. Most armies are much more powerful than in the past.
P=IV
V=IR
P=I(IR)
P=I²R
375=5²R
R=375/25
R=15
Answer:
hellooooo :) ur ans is 33.5 m/s
At time t, the displacement is h/2:
Δy = v₀ t + ½ at²
h/2 = 0 + ½ gt²
h = gt²
At time t+1, the displacement is h.
Δy = v₀ t + ½ at²
h = 0 + ½ g (t + 1)²
h = ½ g (t + 1)²
Set equal and solve for t:
gt² = ½ g (t + 1)²
2t² = (t + 1)²
2t² = t² + 2t + 1
t² − 2t = 1
t² − 2t + 1 = 2
(t − 1)² = 2
t − 1 = ±√2
t = 1 ± √2
Since t > 0, t = 1 + √2. So t+1 = 2 + √2.
At that time, the speed is:
v = at + v₀
v = g (2 + √2) + 0
v = g (2 + √2)
If g = 9.8 m/s², v = 33.5 m/s.
Answer:
(a) 7.72×10⁵ J
(b) 4000 J
(c) 1.82×10⁻¹⁶ J
Explanation:
Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,
Ek = 1/2mv²................... Equation 1
Where Ek = Kinetic energy, m = mass, v = velocity
(a)
For a moving automobile,
Ek = 1/2mv².
Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s
Substitute into equation 1
Ek = 1/2(2.0×10³)(27.78²)
Ek = 7.72×10⁵ J
(b)
For a sprinting runner,
Given: m = 80 kg, v = 10 m/s
Substitute into equation 1 above,
Ek = 1/2(80)(10²)
Ek = 40(100)
Ek = 4000 J
(c)
For a moving electron,
Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s
Substitute into equation 1 above,
Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²
Ek = 1.82×10⁻¹⁶ J
For a constant-velocity object, the average and instantaneous are the same. So the answer is no. It's like taking a running average of a string of numbers that are all the same number. The average is always the sum of the numbers divided by how many have accumulated, which will always equate to the repeated number.
