Answer:
(a) 161.57 N
(b) 0.958 m/s^2
Explanation:
Force applied, F = 220 N
mass of crate, m = 61 kg
μ = 0.27
(a) The magnitude of the frictional force,
f = μ N
where, N is the normal reaction
N = m x g = 61 x 9.81 = 598.41 N
So, the frictional force, f = 0.27 x 598.41
f = 161.57 N
(b) Let a be the acceleration of the crate.
Fnet = F - f = 220 - 161.57
Fnet = 58.43 N
According to newton's second law
Fnet = mass x acceleration
58.43 = 61 x a
a = 0.958 m/s^2
Thus, the acceleration of the crate is 0.958 m/s^2.
Answer:
The buoyant force is 3778.8 N in upward.
Explanation:
Given that,
Mass of balloon = 222 Kg
Volume = 328 m³
Density of air = 1.20 kg/m³
Density of helium = 0.179 kg/m³
We need to calculate the buoyant force acting
Using formula of buoyant force

Where,
= density of air
V = Volume of balloon
g = acceleration due to gravity
Put the value into the formula


This buoyant force is in upward direction.
Hence, The buoyant force is 3778.8 N in upward.
Decreasing the distance between two objects having a considerable mass would increase the attraction on gravitation. The reverse is true that if you separate or inrease the objects distance would substantially decrease their gravitational attraction. Most object in our planet is held by its gravitational force towards it's center.
Answer:
Angular velocity is same as frequency of oscillation in this case.
ω =
x ![[\frac{L^{2}}{mK}]^{3/14}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BL%5E%7B2%7D%7D%7BmK%7D%5D%5E%7B3%2F14%7D)
Explanation:
- write the equation F(r) = -K
with angular momentum <em>L</em>
- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.
- Write the energy of the orbit in relative to r = 0, and solve for "E".
- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.
- Solve for effective potential
- ω =
x ![[\frac{L^{2}}{mK}]^{3/14}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BL%5E%7B2%7D%7D%7BmK%7D%5D%5E%7B3%2F14%7D)