Answer:
The position of my house is a little uphill as compared to the position of my school. The distance I have to travel from my house to school is nearly 2 kilometers. The displacement is in the 2000 m towards the left from my house. The speed of the bus which I usually take is 40 km/ hour.
Answer:
Emechanical=mgh+
mν²
Explanation:
The equation for the total mechanical energy is:
Emechanical=Epotential+Ekinetic
In which,
Epotential=mgh; m: mass of the body, g: gravity; h: height
Ekinetic=
mν²; m: mass of the body, ν: velocity of the body
So,
Emechanical=mgh+
mν²
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
Here when car in front of us applied brakes then it is slowing down due to frictional force on it
So here we can say that friction force on the car front of our car is given as

So the acceleration of car due to friction is given as



now it is given that


so here we have


so the car will accelerate due to brakes by a = - 8.52 m/s^2