Answer:
Explanation:
At the time of a body achieving terminal velocity, the drag force becomes equal to the weight of the body less the buoyant force by the surrounding medium which can be represented by the following equation
Where r is radius of the body , d is density of the material of the body σ is density of the medium and n is coefficient of viscosity of the medium and v is terminal velocity.
Simplifying
v = 
Assuming the value of density of air as 1.225 kg/m³ and putting other given values in the formula we get
v = ![[tex]\frac{2\times (1.2\times10^{-5})^2(2182-1.225)}{9\times 1.8\times10^{-5}}](https://tex.z-dn.net/?f=%5Btex%5D%5Cfrac%7B2%5Ctimes%20%281.2%5Ctimes10%5E%7B-5%7D%29%5E2%282182-1.225%29%7D%7B9%5Ctimes%201.8%5Ctimes10%5E%7B-5%7D%7D)
[/tex]
v = 387 x 10⁻⁵ m/s
Terminal velocity = 387 x 10⁻⁵ m/s
Time taken to fall a distance of 100 m
= 
= 2.6 x 10⁴ s.
Answer:
0.025 m
Explanation:
From the question,
Applying Hook's law
F = ke................... Equation 1
Where F = Force, k = spring constant of the scale, e = maximum distance at which the spring will compress.
make e the subject of the equation
e = F/k....................... Equation 2
Given: F = 10 N, e = 395 N/m
Substitute these values into equation 2
e = 10/395
e = 0.025 m
The answer is neutral charge. An atom element will always and has to be stable, in order for this state to happen. The charge of an electron has to be neutral. For atom with neutral charge, the proton will always equal the number of electron.
Construct a vector diagram. It will be a right-angled triangle. One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I'll call x. Their magnitudes are 20, 3 and x. Let the required angle = theta. We have:
<span>theta = arcsin(3/20) = approx. 8.63° </span>
<span>The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line. </span>
<span>x = sqrt(20^2 - 3^2) </span>
<span>= sqrt(400 - 9) </span>
<span>= sqrt 391 </span>
<span>The boat's crossing time = </span>
<span>0.5 km/(sqrt 391 km/hr) </span>
<span>= (0.5/sqrt 391) hr </span>
<span>= approx. 0.025 hr </span>
<span>= approx. 91 seconds</span>
Force , F = ma
F = m(v - u)/t
Where m = mass in kg, v= final velocity in m/s, u = initial velocity in m/s
t = time, Force is in Newton.
m= 1.2*10³ kg, u = 10 m/s, v = 20 m/s, t = 5s
F = 1.2*10³(20 - 10)/5
F = 2.4*10³ N = 2400 N
