Where??? complete your question please
Answer:
a. Δx = 2.59 cm
Explanation:
mb = 0.454 kg , mp = 5.9 x 10 ⁻² kg , vp = 8.97 m / s , k = 21.0 N / m
Using momentum conserved
mb * (0) + mp * vp = ( mb + mp ) * vf
vf = ( mp / mp + mb) * vp
¹/₂ * ( mp + mb) * (mp / mp +mb) ² * vp ² = ¹/₂ * k * Δx²
Solve to Δx '
Δx = √ ( mp² * vp² ) / ( k * ( mp + mb )
Δx = √ ( ( 5.9 x 10⁻² kg ) ² * (8.97 m /s) ² / [ 21.0 N / m * ( 5.9 x10 ⁻² kg + 0.454 kg ) ]
Δx = 0.02599 m ⇒ 2.59 cm
Answer:
1.5 times
Explanation:
= depth of the diver initially = 5 m
= density of seawater = 1030 kg m⁻³
= Initial pressure at the depth
= final pressure after rising = 101325 Pa
Initial pressure at the depth is given as
= Initial volume at the depth
= Final volume after rising
Since the temperature remains constant, we have
Answer:
c. V = k Q1 * Q2 / R1 potential energy of Q1 and Q2 separated by R
V2 / V1 = (R1 / R2) = 1/4
V2 = V1 / 4
The answer is B frequency. When frequency increases more wave crests pass a fixed point each second. That means the wavelength shortens. So, as frequency increases, wavelength decreases
