The resultant vector is 5.2 cm at a direction of 12⁰ west of north.
<h3>
Resultant of the two vectors</h3>
The resultant of the two vectors is calculated as follows;
R = a² + b² - 2ab cos(θ)
where;
- θ is the angle between the two vectors = 45° + (90 - 57) = 78⁰
- a is the first vector
- b is the second vector
R² = (3.7)² + (4.5)² - (2 x 3.7 x 4.5) cos(78)
R² = 27.02
R = 5.2 cm
<h3>Direction of the vector</h3>
θ = 90 - 78⁰
θ = 12⁰
Thus, the resultant vector is 5.2 cm at a direction of 12⁰ west of north.
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Answer:3.49 m/s
Explanation:
Given
Speed of marble at Bottom 
Diameter of loop 
As Energy is conserved therefore Energy at top is equal to energy at bottom
,where 
is the velocity at bottom
Answer:
All are correct except the age of earth. The age of Earth is approximately 4.543 Billion Years.
Answer:
The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules
Explanation:
The given spring constant of the of the spring, k = 88.0 N/m
The length by which the hose is stretched, x = 4.20 m
For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose
The elastic potential energy, P.E., of a compressed spring is given as follows;
P.E. = 1/2·k·x²
∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²
1/2 × 88.0 N/m × (4.20 m)² = 776.16 J
The work done on the hose = The potential energy given to hose, P.E. = 776.16 J
Answer:
Final temperature is equal to 1291.63°R
Explanation:
given,
p₁ = 100 lb f/in², v₁ = 3.704 ft³/lb, and T₁ = 1000 °R
p₂ = 30 lb f/in² n = 1.4
Δ u = 0.171(T₂ - T₁)
we know for poly tropic process
p vⁿ = constant
p₁ v₁ⁿ = p₂ v₂ⁿ
100 × 3.704¹°⁴ = 30 × v₂¹°⁴
v₂ = 8.753 ft³/lb
work done for poly tropic process
W = 
= 
= 269.525 lbf/in².ft³
W = 
Btu/lb
= 49.87 Btu/lb
in the piston cylinder arrangement air is expanding acrobatically
Δ q = Δu + w
Δ u = - w
0.171(T₂ - T₁) = -49.87
0.171(T₁ - T₂) = -49.87
0.171 T₂ = 0.171 × 1000 + 49.87
T₂ = 1291.63 °R
Final temperature is equal to 1291.63°R
