To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:
Where,
P = Pressure
V = Volume
T = Temperature
n = amount of substance
R = Ideal gas constant
We start by calculating the volume of inhaled O_2 for it:
Our values are given as
P = 1atm
T=293K 
Using the equation to find n, we have:
Number of molecules would be found through Avogadro number, then
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
Answer:
Explanation:
λ = wave length = 632 x 10⁻⁹
slit width a = 2 x 10⁻³ m
angular separation of central maxima
= 2 x λ /a
= 2 x 632 x 10⁻⁹ / 2 x 10⁻³
= 632 x 10⁻⁶ rad
width in m of light spot.
= 632 x 10⁻⁶ x 376000 km
= 237.632 km
When a relationship between two different things is shown in a fraction it is a ratio.
hope this helps :)
Answer:
a) Yes
b) 7 rad/s
c) 0.01034 J
Explanation:
a)
Yes the angular momentum of the block is conserved since the net torque on the block is zero.
b)
m = mass of the block = 0.0250 kg
w₀ = initial angular speed before puling the cord = 1.75 rad/s
r₀ = initial radius before puling the cord = 0.3 m
w = final angular speed after puling the cord = ?
r = final radius after puling the cord = 0.15 m
Using conservation of angular momentum
m r₀² w₀ = m r² w
r₀² w₀ = r² w
(0.3)² (1.75) = (0.15)² w
w = 7 rad/s
c)
Change in kinetic energy is given as
ΔKE = (0.5) (m r² w² - m r₀² w₀²)
ΔKE = (0.5) ((0.025) (0.15)² (7)² - (0.025) (0.3)² (1.75)²)
ΔKE = 0.01034 J
