Answer: Vector quantity is a measurement that has both magnitude and direction.
Explanation:
Vector quantity is a measurement that has both magnitude and direction. Examples are force, acceleration, displacement and velocity.
Answer:
Acceleration of the object is
.
Explanation:
It is given that, the position of the object is given by :
![r=[2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j](https://tex.z-dn.net/?f=r%3D%5B2%5C%20m%2B%285%5C%20m%2Fs%29t%5Di%2B%5B3%5C%20m-%282%5C%20m%2Fs%5E2%29t%5E2%5Dj)
Velocity of the object, 
Acceleration of the object is given by :

![a=\dfrac{d^2}{dt^2}([2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j)](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bd%5E2%7D%7Bdt%5E2%7D%28%5B2%5C%20m%2B%285%5C%20m%2Fs%29t%5Di%2B%5B3%5C%20m-%282%5C%20m%2Fs%5E2%29t%5E2%5Dj%29)
Using the property of differentiation, we get :

So, the magnitude of the acceleration of the object at time t = 2.00 s is
. Hence, this is the required solution.
Answer:
0.0327 m
Explanation:
m = 2 kg
ω = 24 rad/s
A = 0.040 m
Let at position y, the potential energy is twice the kinetic energy.
The potential energy is given by
U = 1/2 m x ω² x y²
The kinetic energy is given by
K = 1/2 m x ω² x (A² - y²)
Equate both the energies as according to the question
1/2 m x ω² x y² = 2 x 1/2 m x ω² x (A² - y²)
y² = 2 A² - 2 y²
3y² = 2A²
y² = 2/3 A²
y = 0.82 A = 0.82 x 0.040 = 0.0327 m
Answer:


The motion of the block is downwards with acceleration 1.7 m/s^2.
Explanation:
First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.

Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.

Now, let’s investigate the free-body diagram of the block.
Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,

As for the static friction, we will consider the angle 31.8, but just before the block starts the move.
