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andrezito [222]
3 years ago
5

Question 23 of 32

Physics
1 answer:
-BARSIC- [3]3 years ago
4 0

Answer:

2m/s

     

Explanation:

Given parameters:

Mass of truck  = 9000kg

Kinetic energy = 18000J

Unknown:

Speed  = ?

Solution:

Kinetic energy is the energy due to the motion of a body;

  K.E  = \frac{1}{2} mv²  

 So;

 m is the mass

  v is the speed

   2 KE = mv²

  2 x 18000 = 9000 x v²

     v²  = 4

     v = 2m/s

     

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What is the approximate terminal velocity of a sky diver before the parachute opens
anzhelika [568]

Answer:

The approximate terminal velocity of a sky diver before the parachute opens is 320 km/h.

Explanation:

  • The terminal velocity is the maximum magnitude of velocity that is attained by the diver when he or she falls in the air.
  • The terminal velocity of the person diving in air before opening parachute is 320 km/h that means the velocity when the person is experiencing free fall is 320 km/h.
  • During terminal velocity, we can represent mathematical equation as;

                           Buoyancy force + drag force = Gravity

6 0
3 years ago
The flagpole is held vertical by two ropes. The first of these ropes has a tension in it of 100 N and is at an angle of 60° to t
KatRina [158]

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 - T_{2x} = 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

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