Answer:
The buoyant force =98 N (OPTION A)
Explanation:
Buoyant force refers to whether an object can float in water or air caused by the upward force exerted on a submerged object, which is equal to the weight of the fluid displaced.
Buoyant force is given by:
Fbuoyant = PA
Where, P = pressure
A = area
While, in a situation where there is volume, area and height involved we have:
Fbuoyant = pgha = gpv
Where, p = density of the fluid
g = gravity
h = height of immersed part
a = area
v = volume of immersed part of the body of the fluid
given data:
density = 2g/cm3
volume = 50 cm3
Fbuoyant = gpv
= (9.8m/s²) x (2 x 10∧
x 10∧
Kg/m³) x (50 x 10∧-
m²) = 98 N
Where, gravity = 9.8m/s²
Explanation:
Given parameters:
Frequency = 500Hz
Unknown:
a. Period of rotation
b. Time required to complete 7 rotation
Solution:
The period of rotation is the time taken to complete a turn.
P =
P is the period
t is the time
n is the number of turns
Period is the inverse of frequency;
P =
= 0.002s
b.
Time required to complete 7 rotation;
P =
t = P x n
t is the time
P is the period
n is the number of rotations
t = 0.002 x 7 = 0.014s
Answer:
a) 6.26 m/s
b) 7.67 m/s
Explanation:
The potential energy at height h0 is initially ...
PE0 = mgh0
At height h1, the potential energy is ...
PE1 = mgh1
The difference in potential energy is converted to kinetic energy:
PE0 -PE1 = KE1 = (1/2)m(v1)^2
Solving for v1, we have ...
mg(h0 -h1) = (1/2)m(v1)^2
2g(h0 -h1) = (v1)^2
v1 = √(2g(h0 -h1))
__
a) When the body is 1 m high, its speed is ...
v = √(2(9.8)(3 -1)) ≈ 6.26 m/s . . . at 1 m high
__
b) When the body is 0 m high, its speed is ...
v = √(2(9.8)(3 -0)) ≈ 7.67 m/s . . . when it reaches the ground
To solve the problem it is necessary to apply the conservation equations for the moment, specifically for collision. In addition to that, the concepts of vector velocity are necessary, in which the components are obtained and if the total magnitude is necessary.
Conservation of Momentum equation is given by,
![m_1v_1+m_2v_2+m_3v_3 = 0](https://tex.z-dn.net/?f=m_1v_1%2Bm_2v_2%2Bm_3v_3%20%3D%200)
We need to find the speed 3, therefore by readjusting the equation we have,
![v_c = \frac{-(m_1v_1+m_2v_2)}{m_3}](https://tex.z-dn.net/?f=v_c%20%3D%20%5Cfrac%7B-%28m_1v_1%2Bm_2v_2%29%7D%7Bm_3%7D)
![v_c = \frac{-(-70*2\hat{j}-85*2.5\hat{j})}{80}](https://tex.z-dn.net/?f=v_c%20%3D%20%5Cfrac%7B-%28-70%2A2%5Chat%7Bj%7D-85%2A2.5%5Chat%7Bj%7D%29%7D%7B80%7D)
![v_c = 2.656\hat{i}+ 1.75\hat{j}](https://tex.z-dn.net/?f=v_c%20%3D%202.656%5Chat%7Bi%7D%2B%201.75%5Chat%7Bj%7D)
Therefore the components of the velocity would be,
![v_{cx} = 2.656m/s](https://tex.z-dn.net/?f=v_%7Bcx%7D%20%3D%202.656m%2Fs)
![v_{cy} = 1.75m/s](https://tex.z-dn.net/?f=v_%7Bcy%7D%20%3D%201.75m%2Fs)
Answer:
103.3 Kg
Explanation:
Use the following formula;
K.E = I/2 mv^2
529*2/(3.2)^2 =m
103.3 Kg=m