Answer:
b) Vectors A and B are in the same direction.
Explanation:
To understand this problem we will say that vector A has a magnitude of 5 units and vector B a magnitude of 3 units. In the subtraction of vectors the initial parts of vectors always bind together. And the vector resulting from the subtraction is traced from the end of the second vector (B) to the end of the first vector (A).
The length of the resultant vector will be 5 - 3 = 2
In the attached image, we analyze case a), b), and d)
For a)
As we can see in the attached image the resultant vector has a length of 8 units.
For d)
As we can see in the attached image the resultant vector has a length of 5.83 units.
For b)
The resultant vector has a length of 2 units.
Therefore the case given in b) is true
Based on Newton's second law of motion, the net force applied to an object is equal to the product of the mass of the object and the acceleration it experiences. That is,
F = ma
If we are to assume that the net force is constant and that the mass is increased, the acceleration should therefore decrease in order to make constant the value at the right-hand side of the equation.
Answer:
53
Explanation:
Because there are 1609.34 meters in a mile. 1609.34÷30=53.64 but because you put one at the beginning of the mile it will stay 53 and not round up to 54
Gravitational potential energy is energy.
The unit of energy is the Joule.
1 Joule = 1 kilogram-meter² / sec²
The speed of light is: c
= 3x10^8 m/s <span>
or
c = 186,000,000 miles/sec = 1.86x10^8 mi/s
1 furlong = 0.125 mile
1 fortnight = 2 weeks(7d/wk)(24h/d)(3600s/h)
= 1209600s = 1.2096x10^6 s
Therefore,
c =1.86x10^8 mi/s(1furl/0.125mi)(1.2096x10^6s/fort)
<span>c = 18x10^14 furlong/fortnight = 18x10^8 Mfurlong/fortnight</span></span>
