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suter [353]
3 years ago
11

If a moving car speeds up until it is going twice as fast, how much kinetic energy doe s it have compared with its initial kinet

ic enegy
Physics
1 answer:
dezoksy [38]3 years ago
6 0
The kinetic energy of an object of mass m and velocity v is given by
K= \frac{1}{2} mv^2

Let's call v_i the initial speed of the car, so that its initial kinetic energy is
K_i =  \frac{1}{2} mv_i^2
where m is the mass of the car. 

The problem says that the car speeds up until its velocity is twice the original one, so 
v_f = 2 v_i
and by using the new velocity we can calculate the final kinetic energy of the car
K_f =  \frac{1}{2} mv_f^2 =  \frac{1}{2}m (2 v_i)^2 = 4 ( \frac{1}{2} mv_i^2)=4 K_i
so, if the velocity of the car is doubled, the new kinetic energy is 4 times the initial kinetic energy.
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Tracy stands on a skateboard and tosses her backpack to her friend who is standing in front of her. Which best describes the acc
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In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the p
Bond [772]

Answer:

B_0 = 1.69 \times 10^{-4}\ T

Explanation:

given,

total deflection = 4.12 cm

Electric field = 1.1 ×10³ V/m

plate length = 6 cm

distance between them = 12 cm

using formula

v_0 = \sqrt{\dfrac{q\epsilon_0d}{ym}(\dfrac{d}{2}+L)}

q = 1.6 × 10⁻¹⁹ C

m = 9.11 x 10⁻³¹ kg

d = 0.06 m

L = 0.12 m

v_0 = \sqrt{\dfrac{1.6 \times 10^{-19}\times 1.1 \times 10^{3}\times 0.06}{0.0412\times 9.11 \times 10^{-31} }(\dfrac{0.06}{2}+0.12)}

v_0 = 6496355.63 m/s

v_0 = \dfrac{E}{B_0}

B_0 = \dfrac{E}{v_0}

B_0 = \dfrac{1.1\times 10^{3}}{6496355.63}

B_0 = 1.69 \times 10^{-4}\ T

5 0
3 years ago
What are other countries public reaction to climate change?
tangare [24]
Other countries have reacted the same way as the United States has.
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3 years ago
The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it
MrRissso [65]

Answer:

Explanation:

From the question we are told that

   The moment of inertia is  I = 2.50 \ kg \cdot m^2

    The final  angular speed is w_f =  400 rev/min  =  \frac{400 * 2\pi}{60}  = 41.89 \ rad/s

     The time taken is  t =  8.0 s

      The initial angular speed is  w_i  =  0\ rad/s

Generally the average angular acceleration is mathematically represented as

        \alpha  =  \frac{w_f - w_i }{t}

=>     \alpha  =  \frac{41.89}{8}

=>      \alpha  = 5.24 \ rad/s^2

Generally the torque is mathematically represented as

   \tau  =  I  *  \alpha

=>    \tau   =  5.24 *  2.50

=>     \tau   =  13.09 \  N \cdot m

5 0
3 years ago
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