Average velocity = (x( 2.08 ) - x ( 0 )) / ( 2.08 s - 0 s )
x ( 2.08 ) = 1.42 * 2.08² - 0.05 * 2.08³ =
= 1.42 * 4.3264 - 0.443456 = 6.143484 - 0.443456 ≈ 5.7 m
v = ( 5.7 m - 0 m) / (2.08 s - 0 s ) = 5.7 / 2.08 m/s = 27.4 m/s
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Answer:
Explanation:
net force on the skier = mg sin 39 - μ mg cos39
mg ( sin39 - μ cos39 )
= 73 x 9.8 ( .629 - .116)
= 367 N
impulse = net force x time = change in momentum .
= 367 x 5 = 1835 kg m /s
velocity of the skier after 5 s = 1835 / 73
= 25.13 m /s
b )
net force becomes zero
mg ( sin39 - μ cos39 ) = 0
μ = tan39
= .81
c )
net force becomes zero , so he will continue to go ahead with constant speed of 25.13 m /s
so he will have speed of 25.13 m /s after 5 s .
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
Answer:
1)
a) f = 1m × 2 × (5A / √2) × (5A / √2) / 0.003m = 0.00166... (66 is repeating)
b) The currents on two wires on a AC chord are always moving in opposite direction and so they are always replusing.
c) There needs to be a sheath to dampen the replusing, fluctuating force of the wires.
2)
a) v = √( ( (-2)(-1.6 × 10^(-16))(3000V) ) / (2.84 × 10^(-20)kg) ) = 5.81227 × 10^3
b) Any ion transversing a chamber having a magnetic field will deflect.
c) The direction of the electric field is vertical because it's perpendicular to the plates. The electric field magnitude is independent from the magnitude of the magnetic field and charge. So it's not possible to find the magnitude of the electric field, without knowing the voltage on the plates, the distance between the plates, and the dielectric constant.
d) Assuming the mangetic field remained, the path of the negative ions will be deflected vertically given that the magnetic field is horizontally perpendicular to the negative charged ions movement.
Sorry it took so long :) If anything is incorrect please let me know.