Explanation:
760 mmHg (millimeters of mercury) is a measure of atmospheric pressure. It represents the height of a column of mercury at which the static pressure at the bottom is equal to the atmospheric pressure.
1 atm = 760 mmHg = 101,325 Pa = 14.7 psi
From conservation of momentum, the ram force can be calculated similarly to rocket thrust:
F = d(mv)/dt = vdm/dt.
<span>In other words, the force needed to decelerate the wind equals the force that would be needed to produce it.
</span><span> v = 120/3.6 = 33.33 m/s
</span><span> dm/dt = v*area*density
</span> dm/dt = (33.33)*((45)*(75))*(1.3)
dm/dt = <span>
146235.375 </span><span>kg/s
</span><span> F = v^2*area*density
</span> F = (33.33)^2*((45)*(75))*(1.3) = <span>
<span>4874025 </span></span><span>N
</span> This differs by a factor of 2 from Bernoulli's equation, which relates velocity and pressure difference in reference not to a head-on collision of the fluid with a surface but to a fluid moving tangentially to the surface. Also, a typical mass-based drag equation, like Bernoulli's equation, has a coefficient of 1/2; however, it refers to a body moving through a fluid, where the fluid encountered by the body is not stopped relative to the body (i.e., brought up to its speed) like is the case in this problem.
The car travels at a speed of 25m/s.
<u>Explanation:</u>
Given-
Mass, m = 1500kg
Coefficient of friction, μk = 0.47
Distance, x = 68m
Speed, s = ?
We know,

and
F = μ X m X g
Therefore,
μ * m * g = m * a
μ * g = a
Let, g = 9.8m/s²
So,


We know,

where, v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
If the car comes to rest, the final velocity, v becomes 0.
So,

The car travels at a speed of 25m/s.
Because the effective charge of the nucleus increase from left to eight due to the increasing number of protons.
The greater charge pulls the electrons closer to the nucleus, decreasing the radius.
<span>A transverse wave is characterized by peaks and dips.</span>