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7nadin3 [17]
3 years ago
7

A compact car has a maximum acceleration of 4.0 m/s2 when it carries only the driver and has a total mass of 1200 kg . you may w

ant to review ( pages 108 - 109) . part a what is its maximum acceleration after picking up four passengers and their luggage, adding an additional 400 kg of mass?
Physics
2 answers:
deff fn [24]3 years ago
4 0

Answer : The maximum acceleration will be 3.0 m/sec^{2}.

Explanation : We can calculate the force using the formula given below;

F = m . a ;

(F - Force; m - mass and a - acceleration);

given in the problem; mass (1) - 1200 kg; acceleration (1) - 4 m/sec^{2}

So, on substituting we get,

F = m_{1} a_{1} = 1200 X 4 = 4800 N; we get F = 4800 N

Now, when mass 400 kg is added then mass (2) will be 1600 kg and acceleration has to be found;

Here, the force remains the same; so it can be equated;

m_{1} a_{1} = m_{2} a_{2};

So, 1200 X 4 = (1200 + 400) x a_{2}

Therefore, the answer will be a_{2} = 3.0 m/sec^{2}

juin [17]3 years ago
3 0
<span>Let F be the maximum thrust of the car's engine. Then F = ma = 1300 x 3.0 = 3900N. With the added load, F remains the same, but we now have F = 1700a giving a = F/1700 = 3900/1700 = 2.3 m/s².</span>
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3 years ago
At the instant the traffic light turns green, an automobile that has been waiting at an intersection starts ahead with a constan
liubo4ka [24]

Answer:

a)  t = 11.407 s, b)   x = 175.66 m,  v = c)         v = 30.80 m / s

Explanation:

a) This is a kinematics exercise, let's write the equation of each vehicle

car

          x = x₀ + v₀ t + ½ a t²

Let's fix our reference system at the point where the car is, indicate that the car stops from rest vo = 0

          x₀ = v₀ = 0

we substitute

          x = ½ a t²

truck

         x₂= v₀ t

         v₀ = 15.4 m / s

at the point where they are, their positions are equal

       ½ a t² = vo t

        t = 2 vo / a

calculate us

        t = 2 15.4 / 2.70

        t = 11.407 s

b) the distance to reach it

        x = ½ to t²

        x = ½ 2.70 11.407²

        x = 175.66 m

c)  the speed of the car is

       v = vo + a t

       vo = 0

       v = at

       v = 2.70 11.407

       v = 30.80 m / s

4 0
3 years ago
A plane cruising at 233 m/s accelerates at 17 m/s 2 for 4.8 s. What is its final velocity? Answer in units of m/s. 013 (part 2 o
Volgvan

Answer:

Final velocity will be 314.6 m/sec

Distance traveled = 1314.24 m

Explanation:

We have given initial velocity u = 233 m/sec

Acceleration a=17m/sec^2

Time t = 4.8 sec

From first equation of motion v=u+at, here v is final velocity, u is initial velocity and t is time

So v=233+17\times 4.8=314.6m/sec

Now we have to find distance traveled

From second equation of motion

S=ut+\frac{1}{2}at^2=233\times 4.8+\frac{1}{2}\times 17\times 4.8^2=1314.24m

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