Answer:
3, 43,968.88 m
Explanation:
Let mass of the missile be m
velocity = 3500 m/s
smaller part will have zero horizontal velocity and larger part will have velocity v in horizontal direction
Applying conservation of momentum
m x 3500 = .35m x 0 + .65m x v
v = 5384.61 m / s
Height of explosion
h = 20000 m
time to fall be t
for vertical fall u = 0 , g = 9.8
20000 = 1/2 x 9.8 t²
= 63.88 s
Horizontal range
time of fall x horizontal velocity
= 63.88 x 5384.61
= 3, 43,968.88 m
the 4 resistor wired in series is equal to 0.639 ohms
Answer:
T₂ = 2482.34 N
Explanation:
Equations of balance of forces
Look at the force diagram in the attached graph:
∑Fx=0
T₂cos α -T₁senα = 0 Equation (1)
∑Fy=0
T₁cos α +T₂sinα-W =0 Equation(2)
Data
m=539 kg
g= 9.8 m/s²
W= m*g= 539 kg* 9.8 m/s²= 5282.2 N
T₁ = 1.88T₂
Problem development
in the equation (1)
T₂cos α-(1.88T₂)senα = 0 We divided the equation by ( T₂cos α)
1 - (1.88)tanα = 0
tanα= 1/(1.88)
tanα= 0.5319
α = 28°
T₂=(1.88T₂)tanα
in the equation (2)
(1.88T₂)cos α+ T₂sinα - 5282.2 =0 We divided the equation by cos α
1.88T₂+ T₂tanα - 5282.2/cos α =0
1.88T₂+ T₂tan(28°) - 5282.2/(cos 28°) =0
1.88T₂+ (0.53)T₂- 5982.46 =0
(2.41)T₂ = 5982.46
T₂ = 5982.46/(2.41)
T₂ = 2482.34 N