Answer:
Explanation:
Work is defined as the scalar product of force and distance
W=F•d
Given that
F = 8.5i + -8.5j. +×-=-
F=8.5i-8.5j
d = 2.5i + cj
If the work in the practice is zero, then W=0
therefore,
W=F•ds
0=F•ds
0=(8.5i -8.5j)•(2.5i + cj)
Note that
i.i=j.j=k.k=1
i.j=j.i=k.i=i.k=j.k=k.j=0
So applying this
0=(8.5i -8.5j)•(2.5i + cj)
0= (8.5×2.5i.i + 8.5×ci.j -8.5×2.5j.i-8.5×cj.j)
0=21.25-8.5c
Therefore,
8.5c=21.25
c=21.25/8.5
c=2.5
Answer:
F = Force (Measured in Newtons, N), m = Mass (Measured in kilograms, kg), and a = acceleration (Measured in metres per second squared, 
Explanation:
This is Newton's Second Law!
Hope this helps!
PLS mark as brainliest, hope this helps!
Answer:
7500 Newtons
Explanation:
Mass of the sportscar= 1500 kg
Acceleration of the sportscar= 5m/s^2
Hence, let the Force acting on it be F
Answer:
I. 0 m/s
II. 20 m/s
III. Part BC
Explanation:
I. Determination of the initial velocity.
From the diagram given above,
The motion of the car starts from the origin. This implies that the car start from rest and as such, the initial velocity of the car is 0 m/s
II. Determination of the maximum velocity attained.
From the diagram given above, we can see clearly that the maximum velocity is 20 m/s.
III. Determination of the part of the graph that represents zero acceleration.
It important that we know the meaning of zero acceleration.
Zero acceleration simply means the car is not accelerating. This can only be true when the car is moving with a constant velocity.
From the graph given above, the car has a constant velocity between B and C.
Therefore, part BC illustrates zero acceleration.
An element refers to a collection of atoms having the same number of protons and electrons (an atomic number). In each element there is a different atomic number due to a different amount of protons in the nucleus.
An isotope is a variation of an element that contains a different number of neutrons, therefore adding weight to the atom.
An ion is a charged atom, and its charge shows how many electrons it needs to gain or lose in order to become stable.
