- momentum
- Yes, if the elephant is standing still.
- Fullback
- impulse acting on it.
- 2.25 N∙s
- A cannon firing.
- Inelastic
- it stays the same
- When the cue ball contacts the other balls, momentum is transferred causing them to gain momentum and speed.
- less than 3 m/s
<h3><u><em>
these are all correct i got an 100%</em></u><em><u> </u></em></h3>
If they both are moving with the same speed and direction
i.e. covering the same distance in the same time interval in the same direction
Answer:
Angular velocity is same as frequency of oscillation in this case.
ω = 
x ![[\frac{L^{2}}{mK}]^{3/14}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BL%5E%7B2%7D%7D%7BmK%7D%5D%5E%7B3%2F14%7D)
Explanation:
- write the equation F(r) = -K
with angular momentum <em>L</em>
- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.
- Write the energy of the orbit in relative to r = 0, and solve for "E".
- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.
- Solve for effective potential
- ω = x
A=(vf-vi)/t
a=(50-25)/10
a=2.5m/s^2
The magnitude of the induced emf is given by:
ℰ = |Δφ/Δt|
ℰ = emf, Δφ = change in magnetic flux, Δt = elapsed time
The magnetic field is perpendicular to the loop, so the magnetic flux φ is given by:
φ = BA
B = magnetic field strength, A = loop area
The area of the loop A is given by:
A = πr²
r = loop radius
Make a substitution:
φ = B2πr²
Since the strength of the magnetic field is changing while the radius of the loop isn't changing, the change in magnetic flux Δφ is given by:
Δφ = ΔB2πr²
ΔB = change in magnetic field strength
Make another substitution:
ℰ = |ΔB2πr²/Δt|
Given values:
ΔB = 0.20T - 0.40T = -0.20T, r = 0.50m, Δt = 2.5s
Plug in and solve for ℰ:
ℰ = |(-0.20)(2π)(0.50)²/2.5|
ℰ = 0.13V
