Answer:
T = 273 + (-50) = 273 – 50 = 223 K
R = 188.82 J / kg K for CO2
Density (Martian Atmosphere) = P / RT = 900 / 188.92 x 223 = 900 / 42129.16 = 0.0213 kg / 
T = 273 +18 = 291 K, R = 287 J / kg k (for air) P = 101.6 k Pa = 101600 Pa
Density (Earth Atmosphere) = P / RT = 101600 / 287 x 291 = 1.216 kg /
To help you I need to assume a wavelength and then calculate the momentum.
The momentum equation for photons is:
p = h / λ , this is the division of Plank's constant by the wavelength.
Assuming λ = 656 nm = 656 * 10 ^ - 9 m, which is the wavelength calcuated in a previous problem, you get:
p = (6.63 * 10 ^-34 ) / (656 * 10 ^ -9) kg * m/s
p = 1.01067 * 10^ - 27 kg*m/s which must be rounded to three significant figures.
With that, p = 1.01 * 10 ^ -27 kg*m/s
The answers are rounded to only 2 significan figures, so our number rounded to 2 significan figures is 1.0 * 10 ^ - 27 kg*m/s
So, assuming the wavelength λ = 656 nm, the answer is the first option: 1.0*10^-27 kg*m/s.
Answer:
a = 4.72 m/s²
Explanation:
given,
mass of the box (m)= 6 Kg
angle of inclination (θ) = 39°
coefficient of kinetic friction (μ) = 0.19
magnitude of acceleration = ?
box is sliding downward so,
F - f = m a
f is the friction force
m g sinθ - μ N = ma
m g sinθ - μ m g cos θ = ma
a = g sinθ - μ g cos θ
a = 9.8 x sin 39° - 0.19 x 9.8 x cos 39°
a = 4.72 m/s²
the magnitude of acceleration of the box down the slope is a = 4.72 m/s²
We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space).
<span>The potential energy of charge q is proportional to </span>
<span>∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, </span>
<span>so if r₂ = 3r₁ and q₂ = q₁/4, then </span>
<span>U₁/U₂ = q₁Qr₂/(r₁q₂Q) = (q₁/q₂)(r₂/r₁) </span>
<span>= 4•3 = 12.</span>
Answer:
We know that the speed of sound is 343 m/s in air
we are also given the distance of the boat from the shore
From the provided data, we can easily find the time taken by the sound to reach the shore using the second equation of motion
s = ut + 1/2 at²
since the acceleration of sound is 0:
s = ut + 1/2 (0)t²
s = ut <em>(here, u is the speed of sound , s is the distance travelled and t is the time taken)</em>
Replacing the variables in the equation with the values we know
1200 = 343 * t
t = 1200 / 343
t = 3.5 seconds (approx)
Therefore, the sound of the gun will be heard at the shore, 3.5 seconds after being fired
