Distance fallen = 1/2 ( V initial + V final ) *t
We know
a = -9.8 m/s2
t=120s
To find distance fallen, we need to find V final
Use the equation
V final = V initial + a*t
Substitute known values
V final = 0 + (-9.8)(120)
V final = -1176 m/s
Then plug known values to distance fallen equation
Distance fallen = 1/2 ( 0 + 1176 )(120)
= 1/2(1776)(120)
=106,560 m
This way plugging into distance equation is actually the long way. A faster way is to plug the values into
Distance fallen = V initial * t + 1/2(a*t)
We won't need to find V final using another equation.
But anyways, good luck!
Answer:
2.12/R mW
Explanation:
The electrical power, P generated by the rod is
P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?
So, the induced emf on the conductor is
E = BLv
= 0.575 T × 0.2 m × 0.4 m/s
= 0.046 V
= 46 mV
The electrical power, P generated by the rod is
P = B²L²v²/R
= B²L²v²/R
So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²
= 0.002116/R W
= 2.12/R mW
Answer:
6.77 m/s
Explanation:
First, in the x direction:
Given:
Δx = 3.17 m
v₀ = v cos 30.8° = 0.859 v
a = 0 m/s²
Δx = v₀ t + ½ at²
(3.17 m) = (0.859 v) t + ½ (0 m/s²) t²
3.17 = 0.859 v t
3.69 = v t
Next, in the y direction:
Given:
Δy = 0.432 m
v₀ = v sin 30.8° = 0.512 v
a = -9.81 m/s²
Δy = v₀ t + ½ at²
(0.432 m) = (0.512 v) t + ½ (-9.81 m/s²) t²
0.432 = 0.512 v t − 4.905 t²
Two equations, two variables. Solve for t in the first equation and substitute into the second equation:
t = 3.69 / v
0.432 = 0.512 v (3.69 / v) − 4.905 (3.69 / v)²
0.432 = 1.89 − 66.8 / v²
66.8 / v² = 1.458
v² = 45.8
v = 6.77
Honestly, giving. I know that receiving may feel very good, and you get that new thing that you wanted but if you give, you feel really good when you see that smile on someone else’s face.
