Answer:
D) 25 m/s
Explanation:
In order to solve this problem we must use the following kinematics equation.
where:
Vf = final speed [m/s]
Vi = initial speed = 0
a = acceleration = 5[m/s^2]
t = time = 5[s]
After 5 seconds the acceleration is equal to 5 [m/s^2]
Now replacing the values in the equation:
Vf = 0 + (5*5)
Vf = 25[m/s]
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:
The jug drowns because the density of the jug is more than that of the density of water.
Answer:
Because the Earth has so much gravity, it can hold water, land, and life in it's atmosphere.
(Not sure what beaker you are talking about, so sorry) But I don't think the moon's gravity would have an effect on a beaker of water because the Earth's gravity is much more than the moon's.
I think you would be able to feel a little bit of Earth's gravity on the moon because the Earth's gravity pulled the moon into orbit, therefore, gravity on Earth my have some effect on the moon.
hope this helps!
Answer:
<em> The distance required = 16.97 cm</em>
Explanation:
Hook's Law
From Hook's law, the potential energy stored in a stretched spring
E = 1/2ke² ......................... Equation 1
making e the subject of the equation,
e = √(2E/k)........................ Equation 2
Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.
Given: k = 450 N/m, e = 12 cm = 0.12 m.
E = 1/2(450)(0.12)²
E = 225(0.12)²
E = 3.24 J.
When the potential energy is doubled,
I.e E = 2×3.24
E = 6.48 J.
Substituting into equation 2,
e = √(2×6.48/450)
e = √0.0288
e = 0.1697 m
<em>e = 16.97 cm</em>
<em>Thus the distance required = 16.97 cm</em>
