Answer:
The new volume is 4.15 L
Explanation:
Given
Initial Volume (,V1) = 4L
Initial Temperature (T1) = 294K
Final Temperature (T2) = 305k
Required
Determine the new volume (V2)
This will be calculated using ideal gas equation where pressure is constant.
We have
V1/T1 = V2/T2
Substitute values for V1, T1 and V2
4/294 = V2/305
Solve for V2
V2 = 305 * 4/294
V2 = 1220/294
V2 = 4.15L (approximated)
<span>Answer:
KE = (11/2)mω²r²,
particle B must have mass of 2m, while A has mass m.
Then the moment of inertia of the system is
I = Σ md² = m*(3r)² + 2m*r² = 11mr²
and then
KE = ½Iω² = ½ * 11mr² * ω² = 11mr²ω² / 2
So I'll proceed under that assumption.
For particle A, translational KEa = ½mv²
but v = ω*d = ω*3r, so KEa = ½m(3ωr)² = (9/2)mω²r²
For particld B, translational KEb = ½(2m)v²
but v = ω*r, so KEb = ½(2m)ω²r²
so total translational KE = (9/2 + 2/2)mω²r² = 11mω²r² / 2
which is equal to our rotational KE.</span>
Bedrock might be the answer
m = mass of the person bouncing = 78 kg
k = spring constant of the bathroom scale = 1.5 x 10⁶ N/m
v = maximum velocity of the person
A = maximum compression of the spring of scale = 0.115 cm = 0.00115 m
h = height gained at maximum speed = A = 0.00115 m
Using conservation of energy
spring potential energy = kinetic energy + gravitational potential energy
(0.5) k A² = (0.5) m v² + m g h
k A² = m v² + 2 m g h
(1.5 x 10⁶) (0.00115)² = (78) v² + 2 (78) (9.8) (0.00115)
v = 0.0538 m/s