I would make a flying car
Answer:
By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.
here weight of the child =21kgx9.8m/s2 = 205.8N
the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.
torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.
net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.
b) Ta = 2.5x151 = 377.5N-m
Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.
c)Ta = 2x151 = 302N-m
Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.
Answer:
Impedance = 93.75 ohms
Current = 1.81 A
Explanation:
Resistance = R = 80 ohms
Inductance = L = 0.2 H
Inductive reactance = XL =
= ωL = (2πf) L
= 2 (3.14) (60)(0.2) = 75.398 Ohms
Capacitive reactance = 1 / ωC = 1/(2πf)C = 1 / [(2π)(60)(0.1 × 10⁻3)]
= 26.526 Ohms
Impedance = Z =
=
= 93.747 ohms
Voltage =
× 120 = 169.7056 V
Current = I = V ÷ R = (169.7056) ÷ 93,747 = 1.81 A
Answer:
v=0.60 m/s
Explanation:
Given that
m ₁= 390 kg ,u ₁= 0.5 m/s
m₂ = 250 kg ,u₂ = 0.76 m/s
As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.
Pi = Pf
m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v
Now putting the values in the above equation
390 x 0.5 + 250 x 0.76 = (390 + 250 ) v

v=0.60 m/s
Therefore the velocity of the system will be 0.6 m/s.
Answer:
Hence the answer is E inside
.
Explanation:
E inside
so if r1 will be the same then
E
proportional to 1/R3
so if R become 2R
E becomes 1/8 of the initial electric field.