Answer:
Explanation:
given,
constant rate for the reaction(k) = 5.64 × 10⁻⁴ s⁻¹
to calculate the half life = ?
on solving the above equation
hence, the half-life of cyclopropane at this temperature =
If everything is in meters and kilograms, then the gravitational force
between 2 masses is
F = G M₁ M₂ / R²
G = 6.67 x 10⁻¹¹
M₁ and M₂ = the masses of the two masses
R = the distance between their centers.
When we plug in 3 x 10⁻¹⁰ for the force, and 0.45 for the distance,
we get the product of the 2 objects M₁M₂ = about 0.911 , and
that's as far as we can go without some more information.
The question says that the two masses add up to 4.0 kg.
So now we know that M₁M₂ = about 0.911 and (M₁ + M₂) = 4.0 kg.
With this, we immediately step into a big squishy quadratic equation,
whose solutions are:
one mass = <em>3.758 kg</em>
the other mass = <em>0.242 kg</em> .
Answer:
volume of the bubble just before it reaches the surface is 5.71 cm³
Explanation:
given data
depth h = 36 m
volume v2 = 1.22 cm³ = 1.22 × m³
temperature bottom t2 = 5.9°C = 278.9 K
temperature top t1 = 16.0°C = 289 K
to find out
what is the volume of the bubble just before it reaches the surface
solution
we know at top atmospheric pressure is about P1 = Pa
so pressure at bottom P2 = pressure at top + ρ×g×h
here ρ is density and h is height and g is 9.8 m/s²
so
pressure at bottom P2 = + 1000 × 9.8 ×36
pressure at bottom P2 =4.52 × Pa
so from gas law
here p is pressure and v is volume and t is temperature
so put here value and find v1
V1 = 5.71 cm³
volume of the bubble just before it reaches the surface is 5.71 cm³
Answer:
I think the Answer is B. Longitudinal waves
The answer is Of frequencies