Answer:
B
Explanation:
Resolve the 75N force into 2 components; horizontal and vertical. And remember that there is no acceleration in the downward direction, so apply Newton's second law and equate it to 0.
Answer:
54 km/hr
Explanation:
m/s to km/hr => 18/5
15 m/s to km/hr => 15 x 18/5 =>3 x 18 => 54km/hr
Answer:
Proof in explanataion
Explanation:
The basic dimensions are as follows:
MASS = M
LENGTH = L
TIME = T
i)
Given equation is:

where,
H = height (meters)
u = speed (m/s)
g = acceleration due to gravity (m/s²)
Sin Ф = constant (no unit)
So there dimensions will be:
H = [L]
u = [LT⁻¹]
g = [LT⁻²]
Sin Ф = no dimension
Therefore,
![[L] = \frac{[LT^{-1}]^2}{[LT^{-2}]}\\\\\ [L] = [L^{(2-1)}T^{(-2+2)}]](https://tex.z-dn.net/?f=%5BL%5D%20%3D%20%5Cfrac%7B%5BLT%5E%7B-1%7D%5D%5E2%7D%7B%5BLT%5E%7B-2%7D%5D%7D%5C%5C%5C%5C%5C%20%5BL%5D%20%3D%20%5BL%5E%7B%282-1%29%7DT%5E%7B%28-2%2B2%29%7D%5D)
<u>[L] = [L]</u>
Hence, the equation is proven to be homogenous.
ii)

where,
F = Force = Newton = kg.m/s² = [MLT⁻²]
G = Gravitational Constant = N.m²/kg² = (kg.m/s²)m²/kg² = m³/kg.s²
G = [M⁻¹L³T⁻²]
m₁ = m₂ = mass = kg = [M]
r = distance = m = [L]
Therefore,
![[MLT^{-2}] = \frac{[M^{-1}L^{3}T^{-2}][M][M]}{[L]^2}\\\\\ [MLT^{-2}] = [M^{(-1+1+1)}L^{(3-2)}T^{-2}]\\\\](https://tex.z-dn.net/?f=%5BMLT%5E%7B-2%7D%5D%20%3D%20%5Cfrac%7B%5BM%5E%7B-1%7DL%5E%7B3%7DT%5E%7B-2%7D%5D%5BM%5D%5BM%5D%7D%7B%5BL%5D%5E2%7D%5C%5C%5C%5C%5C%20%5BMLT%5E%7B-2%7D%5D%20%3D%20%5BM%5E%7B%28-1%2B1%2B1%29%7DL%5E%7B%283-2%29%7DT%5E%7B-2%7D%5D%5C%5C%5C%5C)
<u>[MLT⁻²] = [MLT⁻²]</u>
Hence, the equation is proven to be homogenous.
<span>The
heavier the body is, the stronger its gravitational pull. Example, the Milky Way
Galaxy has a gravitational pull because of the heavenly bodies such as stars and planets are surrounding it. A strong force is exerted if the mass of another body is bigger than the other body.</span>
Answer:
(1) 42.94 m
(2) 
Explanation:
Let us first draw a figure, for the given question as below:
In the figure, we assume that the person starts walking from point A to travel 11 m exactly
south of west to point B and from there, it walks 21 m exactly
west of north to reach point C.
Let us first write the two displacements in the vector form:

Now, the vector sum of both these vectors will give us displacement vector from point A to point C.

Part (1):
the magnitude of the shortest displacement from the starting point A to point the final position C is given by:

Part (2):
As the vector AC is coordinates lie in the third quadrant of the cartesian vector plane whose angle with the west will be positive in the north direction.
The angle of the shortest line connecting the starting point and the final position measured north of west is given by:
