Answer:
B. 2nmv
Explanation:
Pressure is force over area.
P = F / A
Force is mass times acceleration.
F = ma
Acceleration is change in velocity over change in time.
a = Δv / Δt
Therefore:
F = m Δv / Δt
P = m Δv / (A Δt)
The total mass is nm.
The change in velocity is Δv = v − (-v) = 2v.
A = 1 and Δt = 1.
Plugging in:
P = (nm) (2v) / (1 × 1)
P = 2nmv
Answer:
The population size decreases.
Explanation:
If more of a species are dying than being born, the population size will decrease.
Answer: 4100 Mpc
Explanation:
Since H o = 70 km/s/Mpc
Redshift z = 5.82
Recessional velocity vr = 287,000 km/s
Then, the distance to the galaxy in light years will be:
= Recessional velocity / H o
= 287000 / 70
= 4100 Mpc
The temperature of an air parcel and the kinetic energy of an air parcel are directly related. this means that as the temperature of the air parcel increases, the kinetic energy increases.
<h3>
What is temperature?</h3>
Temperature is the measure of degree of hotness or coldness of a body.
Temperature is also the measure of the average kinetic energy of a system.
When the heat is applied to body, its temperature increases as the body gains heat.
Thus, the temperature of an air parcel and the kinetic energy of an air parcel are directly related. this means that as the temperature of the air parcel increases, the kinetic energy increases.
Learn more about temperature here: brainly.com/question/25677592
#SPJ1
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
