Answer:
(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).
(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .
Explanation:
m= 3kg
a= 2 i + 5 j = 5 .38 < 68.19 º
F= m * a
F= 3* ( 5.38 < 68.19º )
F= 16.4 N < 68.19º
Fx= F * cos(68.19º)
Fx= 5.99
Fy= F* sin(68.19º)
Fy= 14.98
Answer:
(i) The angular speed of the small metal object is 25.133 rad/s
(ii) The linear speed of the small metal object is 7.54 m/s.
Explanation:
Given;
radius of the circular path, r = 30 cm = 0.3 m
number of revolutions, n = 20
time of motion, t = 5 s
(i) The angular speed of the small metal object is calculated as;

(ii) The linear speed of the small metal object is calculated as;

Answer:
The diagram assigned B
explanation:
Check the direction of the two vectors, their resultant must be in the same direction.
Answer:
a = 2 [m/s²]
Explanation:
To be able to solve this problem we must make it clear that the starting point when the time is equal to zero, the velocity is 5 [m/s] and when three seconds have passed the velocity is 11 [m/s], this point is the final point or the final velocity.
We can use the following equation.

where:
Vf = final velocity = 11 [m/s]
Vo = initial velocity = 5 [m/s]
a = acceleration [m/s²]
t = time = 3 [s]
![11 = 5 + a*3\\6=3*a\\a= 2[m/s^{2} ]](https://tex.z-dn.net/?f=11%20%3D%205%20%2B%20a%2A3%5C%5C6%3D3%2Aa%5C%5Ca%3D%202%5Bm%2Fs%5E%7B2%7D%20%5D)
Answer:
Both speed and velocity are changing.
Explanation:
They are both going up so both are changing