Answer:
T₂ = 123.9 N, θ = 66.2º
Explanation:
To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.
The tension T1 = 100 N, we create a reference frame centered on the pole
X axis
T₁ₓ - 
= 0
T_{2x}= T₁ₓ
Y axis y
T_{1y} + T_{2y} - 200N = 0
T_{2y} = 200 -T_{1y}
let's use trigonometry to find the component of the stresses
sin 60 = T_{1y} / T₁
cos 60 = t₁ₓ / T₁
T_{1y} = T₁ sin 60
T1x = T₁ cos 60
T_{1y}y = 100 sin 60 = 86.6 N
T₁ₓ = 100 cos 60 = 50 N
for voltage 2 it is done in the same way
T_{2y} = T₂ sin θ
T₂ₓ = T₂ cos θ
we substitute
T₂ sin θ= 200 - 86.6 = 113.4
T₂ cos θ = 50 (1)
to solve the system we divide the two equations
tan θ = 113.4 / 50
θ = tan⁻¹ 2,268
θ = 66.2º
we caption in equation 1
T₂ cos 66.2 = 50
T₂ = 50 / cos 66.2
T₂ = 123.9 N
The light coming out of a concave lens will never meet.
So, the answer is A. will never meet.
Happy Studying! ^^
Complete question is:
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.
Answer:
(V_A) = 31.32 m/s
Explanation:
We are given;
car's mass, m = 1200 kg
h_A = 100 m
h_B = 150 m
v_B = 0 m/s
From law of conservation of energy,
the distance from point A to B is;
h = 150m - 100 m = 50 m
From Newton's equations of motion;
v² = u² + 2gh
Thus;
(V_B)² = (V_A)² + (-2gh)
(negative next to g because it's going against gravity)
Thus;
(V_B)² = (V_A)² - (2gh)
Plugging in the relevant values;
0² = (V_A)² - 2(9.81 × 50)
(V_A) = √981
(V_A) = 31.32 m/s
Mass of the bird(m) = 150 g = 0.15 kg
Speed (v) = 10 m/s
Kinetic Energy = 
= 7.5 J
Altitude (h) = 15 m
Gravitational Potential Energy = (0.15)(9.81)(15) = 22.0725 J
Mechanical Energy = Kinetic Energy + Potential Energy = 7.5 + 22.0725
= 29.5725 J
Answer:
The SI unit of time is second (s) and temperature is Kelvin (K)
Explanation:
hope it is helpful to you
