Answer: Ok so We already know that velocity is on the x-axis.
Since acceleration = Force / Mass
Here the Force is downward due to the gravitational pull or we can say it is along y-axis.
Since acceleration is directly proportional to force, so acceleration is also along y-axis. This means that velocity & acceleration are perpendicular to each other.
Example:
Let us assume that an aeroplane is flying parallel to the horizontal plane. The aeroplane will experience the acceleration in several directions. One of them here is the gravitational pull which is perpendicular to the the apparent velocity. So the net velocity & its direction will depend upon the vector sum total of all the forces/acceleration acting on it. Also because of this gravitational pull the aeroplane rotates along with the earth, which is a proof that the force/g experienced by it does not go waste.
<h3>Hope this helps have a awesome day/night❤️✨</h3>Explanation:
Answer:
t = 444.125 sec
Explanation:
Given data:
V = 24 volt
I = 0.1 ampere
mass of water mw = 51 gm
cr = 4.18 J/gm degree K^-1
mass of resistor = 8 gm
cr = 3.7 J/gm degree K^-1
we know that power is given as
Power P = VI
But P =E/t
so equating both side we have
solving for t
t = 444.125 sec
Answer:
T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N
Explanation:
This is a balance exercise where we must apply the expressions for translational balance in the two axes
∑ F = 0
Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º
decompose the tension of the two upper cables
cos 41 = T₁ₓ / T1
sin 41 = T₁y / T1
T₁ₓ = T₁ cos 41
T₁y= T₁ sin 41
for cable gold
cos 63 = T₂ / T₂
sin 63 = / T₂
We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.
Let's start by analyzing the point where the traffic light meets the vertical cable
T₃ - W = 0
T₃ = W
T₃ = 200 N
now let's write the equations for the single point of the three wires
X axis
- T₁ₓ + T₂ₓ = 0
T₁ₓ = T₂ₓ
T1 cos 41 = T2 cos 63
T1 = T2 cos 63 / cos 41 (1)
y Axis
+ T_{2y} - T3 = 0
T₁ sin 41 + T₂ sin 63 = T₃ (2)
to solve the system we substitute equation 1 in 2
T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W
T₂ (cos 63 tan 41 + sin 63) = W
T₂ = W / (cos 63 tan 41 + sin 63)
We calculate
T₂ = 200 / (cos 63 tan 41 + sin 63)
T₂ = 200 / 1,2856
T₂ = 155.6 N
we substitute in 1
T₁ = T₂ cos 63 / cos 41
T₁ = 155.6 cos63 / cos 41
T₁ = 93.6 N
therefore the tension in each cable is
T₁ = 93.6 N
T₂ = 155.6 N
T₃ = 200 N