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Trava [24]
3 years ago
12

To lift a load of 100 kg a distance of 1 m an effort of 25 kg must be applied over an inclined plane of length 4 m. What must be

done to lift the load 2 m, using the same effort?
A) Use an inclined plane of length 8 m.


B) Use an inclined plane of length 2 m.


C) Use an inclined plane of length 10 m.


D) Use an inclined plane of length 16 m.
Physics
1 answer:
andreyandreev [35.5K]3 years ago
7 0
Work = force * distance.
We must produce twice as much energy as we are lifting the weight twice as high.
But we are not increasing the force so we must increase the length of the ramp ( distance ) instead.
The new length will be twice as great as the previous length.
So 8 metres is required. 


25 kg * 8 m = work = 100 kg * 2 m
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Show the weight of the ladder and draw the missing Frictional force.
madreJ [45]
Since there is no friction between the ladder and the wall, there can be no vertical force component. That's the tricky part ;)

So to find the weight, divide the 100N <em>normal</em> force by earths gravitational acceleration, 9.8m/s^2

W =  \frac{N}{g}  =  \frac{100N}{9.8m/s^{2}} =  \frac{100}{9.8} = 10.2kg


Then;
Draw an arrow at the base of the ladder pointing towards the wall with a value of 30N, to show the frictional force.
5 0
3 years ago
The 2.5-Mg van is traveling with a speed of 100 km&gt;h when the brakes are applied and all four wheels lock. If the speed decre
Andrew [12]

Answer:

0.34

Explanation:

2.5 Mg = 2500 kg

The change in speed from 100 km/h to 40 km/h is

\Delta v = 100 - 40 = 60 km/h = 60 * 1000 / (60 * 60) = 16.67 m/s

The deceleration caused by friction force is the change in speed per unit of time

a = \Delta v / \Delta t = 16.67 / 5 = 3.33 m/s^2

Using Newton 2nd law we can calculate the friction force that caused this deceleration:

F = ma = 2500 * 3.33 = 8333 N

Let g = 9.8m/s2. Friction force is the product of normal (gravity) force and friction coefficient

F = mg\mu

8333 = 2500*9.8\mu

\mu = \frac{8333}{2500 * 9.8} = 0.34

4 0
3 years ago
A 0.12-kg metal rod carrying a current of current 4.1 A glides on two horizontal rails separation 6.3 m apart. If the coefficien
Neporo4naja [7]

Answer:

The magnetic field is B  =  8.20 *10^{-3} \  T

Explanation:

From the question we are told that

   The  mass of the metal rod is  m  = 0.12 \ kg

    The current on the rod is  I  = 4.1 \ A

    The distance of separation(equivalent to length of the rod ) is L   = 6.3 \ m

     The coefficient of kinetic friction is \mu_k  =  0.18

      The kinetic frictional force is  F_k  = 0.212 \ N

     The constant speed is v  = 5.1 \ m/s

Generally the magnetic force on the rod is mathematically represented as  

      F  =  B * I  *   L

For  the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so

        F_ k  =  B*  I  *  L

=>      B  =  \frac{F_k}{L  *  I  }

=>       B  =  \frac{0.212}{ 6.3   *  4.1   }

=>       B  =  8.20 *10^{-3} \  T

7 0
3 years ago
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