C. is correct : ) it is applied when you push and that causes the desk to move
Answer:
The answer to your question is a = 0.25 m/s²
Explanation:
Data
mass = m = 400 kg
Force = F = 100 N
acceleration = a = ? m/s²
Process
To solve this problem use Newton's second law that states that the force applied to an object is directly proportional to the mass of the body times its acceleration.
Formula
F = ma
solve for a
a = 
Substitution
Simplification and result
a = 0.25 m/s²
Answer:
this vehichle has 896,000 jules of energy
Explanation: KE=1/2mv squared, KE is Kinetic energy. m is mass and v is velocity
1) Force = m*a = 1.00 g * (1kg / 1000 g) * 225 m/s^2 = 0.225 N
2) Charge
Force = K (charge)^2 /(distance)^2 => charge = √ [Force * distance^2 / k]
k = 9.00 * 10^9 N*m^2 / C^2
charge = √ [0.225 N * (0.02 m)^2 / 9.00* 10^9 N*m^2 / C^2 ]
charge = 0.0000001 C = 0.0001 mili C
This question is based on the fundamental assumption of vector direction.
A vector is a physical quantity which has magnitude as well direction for its complete specification.
The magnitude of a physical quantity is simply a numerical number .Hence it can not be negative.
A negative vector is a vector which comes into existence when it is opposite to our assumed direction with respect to any other vector. For instance, the vector is taken positive if it is along + X axis and negative if it is along - X axis.
As per the first option it is given that a vector is negative if its magnitude is greater than 1. It is not correct as magnitude play no role in it.
The second option tells that the magnitude of the vector is less than 1. Magnitude can not be negative. So this is also wrong.
Third one tells that a vector is negative if its displacement is along north. It does not give any detail information about the negativity of a vector.
In a general sense we assume that vertically downward motion is negative and vertically upward is positive. In case of a falling object the motion is vertically downward. So the velocity of that object is negative .
So last option is partially correct as the vector can be negative depending on our choice of co-ordinate system.
