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Svetradugi [14.3K]
3 years ago
9

How fast does a 500kg car need to drive to have 100,000 J of kinetic energy?

Physics
1 answer:
olchik [2.2K]3 years ago
8 0

Answer:

The car must move at 2 m/s to have a Ke of 2,000 Joules.

Explanation:

Mark me pls

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false

Explanation:

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A car accelerates uniformly at 2m/s2 for 3minutes. what is the velocity of the car​
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4 0
4 years ago
In wet climates, granite weathers more slowly than ____
RideAnS [48]
Marble. This is because granite is of a different chemistry that is not nearly as vulnerable to acidic rain as marble.
6 0
3 years ago
In a perfectly elastic collision between two perfectly rigid objects
ipn [44]

Both the total momentum and the total kinetic energy are conserved

Explanation:

- In a collision between two or more objects, if there are no external forces acting on the system (isolated system), the total momentum of the objects is always conserved. This is called principle of conservation of momentum, and can be written as follows:

mu+MU = mv+MV

where

m, M are the masses of the two objects

u, U are the initial velocities of the two objects

v, V are the final velocities of the two objects

- The total kinetic energy, however, is not always conserved. In fact, we have two types of collision:

1) In a perfectly elastic collision, the total kinetic energy of the objects is conserved. This means that we can write the following equation:

\frac{1}{2}mu^2 + \frac{1}{2}MU^2 = \frac{1}{2}mv^2+\frac{1}{2}MV^2

2) In an inelastic collision, the total kinetic energy of the object is NOT conserved. This means that part of the total kinetic energy is "lost", converted into other forms of energy (mainly thermal energy, due to the presence of frictional forces within the system). The most extreme case is called perfectly inelastic collision, in which the two objects stick together after the collision, and there is the maximum loss of kinetic energy.

Learn more about collisions:

brainly.com/question/13966693#

brainly.com/question/6439920

LearnwithBrainly

7 0
4 years ago
The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant
uysha [10]

Answer:

The amount of charge on the plates is now equal to half its original  value

Explanation:

From the question we are told that  

  The charge on the plate is  Q

   The  initial  separation is d_1

    The new separation is d_2 = 2 d_1

Generally the capacitance of the capacitor is mathematically represented as

      C =  \frac{\epsilon * A }{d}

Generally the charge of the parallel plate capacitor is mathematically represented as

     Q =CV

=>  Q =\frac{\epsilon *  A  *  V   }{ d}

Here  \epsilon ,  A and  V are constant , so looking at the question we see that Q varies inversely with d

  So  

       Q = \frac{K}{d}

Here K  is  a constant

so

     Qd =  K

=>  Q_1 d_1 =  Q_2 * d_2

=>   Q_1 d_1 =  Q_2 * [2 d_1]

=> Q_2 = \frac{Q_1}{2}

So we see from the mathematically relation that the charge of the parallel plate capacitor will reduce by half of it original  value

     

   

3 0
3 years ago
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