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Mandarinka [93]
3 years ago
6

a) A bus of mass 760 kg requires 120 m to reach certain velocity value Vf. Ignore friction and drag forces and assume the bus en

gine exerts a constant forward force F. When the bus is towing a 330-kg small car, how long distance needed to reach same Vf? b) If the Vf of the bus is 28 m/s, what is the tension in the tow cable between bus and small car?
Physics
1 answer:
zmey [24]3 years ago
7 0

Answer:

Given :  A bus of mass 760 kg requires 120 m to reach certain velocity value Vf.

the bus engine exerts a constant forward force F.

To Find : When the bus is towing a 330-kg small car, how long distance needed to reach same Vf?

Solution:

V²  - U² = 2aS

V = Vf

U = 0

S = 120 m

=> Vf² - 0 = 2a(120)

=> Vf² = 240a

m = 760  kg

Force = F

F = ma

=> F =760 a

=> a = F/760

Vf² = 240F/760

Case 2  :When the bus is towing a 330-kg small car,

m = 760 + 330 = 1090 kg

a = F/1090

Vf²  = 2aS

=> 240F/760  = 2 (F/1090) S

=> S = 120 x 1090 /760

=> S = 172.1   m  

172.1   m   distance needed to reach same Vf

Explanation:

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7 0
3 years ago
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

8 0
3 years ago
If you are pushing on a box with a force of 20 N and there is a 7 N force on the box due to sliding friction, what is the net fo
vovangra [49]

Answer:

13 N

Explanation:

The Net Force of an object should be the difference between the forces applied to the object if the object is not in equilibrium. This object is not in equilibrium so therefore by finding the difference between the forces, you will find your answer. 20 N - 7 N = 13 N.

6 0
3 years ago
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