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Complete Question
A boat sails 4km on a bearing of 038 degree and then 5km on a bearing of 067 degree.(a)how far is the boat from its starting point.(b) calculate the bearing of the boat from its starting point
Answer:
a)8.717km
b) 54.146°
Step-by-step explanation:
(a)how far is the boat from its starting point.
We solve this question using resultant vectors
= (Rcos θ, Rsinθ + Rcos θ, Rsinθ)
Where
Rcos θ = x
Rsinθ = y
= (4cos38,4sin38) + (5cos67,5sin67)
= (3.152, 2.4626) + (1.9536, 4.6025)
= (5.1056, 7.065)
x = 5.1056
y = 7.065
Distance = √x² + y²
= √(5.1056²+ 7.065²)
= √75.98137636
= √8.7167296826
Approximately = 8.717 km
Therefore, the boat is 8.717km its starting point.
(b)calculate the bearing of the boat from its starting point.
The bearing of the boat is calculated using
tan θ = y/x
tan θ = 7.065/5.1056
θ = arc tan (7.065/5.1056)
= 54.145828196°
θ ≈ 54.146°
Answer: l = 82.5m
Step-by-step explanation:
11/2 x 15 = 82.5
Answer:
Correct arrangement of equation of displacement to find a is as follows;
1- Vt - d = 1/2 a t^2 (^ represents exponent i.e. t square as given in equation)
2- 2(Vt - d ) = a t^2
3- a = 2(Vt - d )/ t^2 (keep in mind, 2(Vt - d) whole divided by t^2)
Step-by-step explanation:
1- In the first equation, Vt is taken to the left side of the equation (keep in mind, original equation of displacement used for reference as given in question) and multiplied by -1 on the both sides of the equation.
2- In the second equation, 2 is multiplied on the both sides.
3- Multiply t^2 on both sides of the equation, We will get a in correct arrangement, which is required to find.
The answer should be 31, if I’m doing this correctly.
6(3)2-5 (multiply 6 and 3, you get 18)
18(2)-5 (multiply 18 and 2, you get 36)
36-5 (subtract 36 and 5)
31
Answer:
THere will be 8 heads and 2 tails
Step-by-step explanation:
I don't know
