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RSB [31]
3 years ago
10

A car of mass 500kg travelling at 60m/s has it speed reduced to 40m/s by a constant breaking force over a distance of 200m. find

car initial kinetic energy. the final kinetic energy​
Physics
1 answer:
uranmaximum [27]3 years ago
6 0

Answer:

Ek1 = 900000 [J]

Ek1 = 400000 [J]

Explanation:

In order to solve this problem we must remember that kinetic energy is defined as the product of mass by velocity squared by a medium. Therefore using the following equation we have:

E_{k1}=\frac{1}{2}*m*v1^{2}

where:

m = mass = 500 [kg]

v1 = 60 [m/s]

So we have:

Ek1 = 0.5*500*(60^2)

Ek1 = 900000 [J]

and:

Ek2 = 0.5*500*(40^2)

Ek2 = 400000 [J]

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If 61.4 cm of copper wire (diameter = 1.08 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendic
harkovskaia [24]

Answer:

the thermal energy generated in the loop = 6.64*10^{-4}  \ W

Explanation:

Given that;

The length of the copper wire L = 0.614 m

Radius of the loop  r = \frac{L}{2 \pi}

r = \frac{0.614}{2 \pi}

r = 0.0977 m

However , the area of the loop is :

A_L = \pi r^2

A_L = \pi (0.0977)^2

A_L = 0.02999 \ m^2

Change in the magnetic field is \frac{dB}{dt}= 0.0914 \ T/s

Then the induced emf e = A_L \frac{dB}{dt}

e = 0.02999 * 0.0914

e = 2.74 × 10⁻³ V

resistivity of the copper wire \rho = 1.69* 10^{-8} Ω m

diameter of the wire = 1.08 mm

radius of the wire = 0.54 mm = 0.54 × 10⁻³  m

Thus, the resistance of the wire  R = \frac {\rho L}{\pi r^2}

R =  \frac{(1.69*10^{-8})(0.614)}{   \pi (0.54*10^{-3})^2}

R = 1.13× 10⁻² Ω

Finally,  the thermal energy generated in the loop (i.e the power) = \frac{e^2}{R}

= \frac{(2.74*10^{-3})^2}{1.13*10^{-2}}

= 6.64*10^{-4}  \ W

8 0
3 years ago
A 69 g particle is moving to the left at 25 m/s . How much net work must be done on the particle to cause it to move to the righ
LekaFEV [45]

Answer:11.59 J

Explanation:

Given

mass of Particle m=69 gm

Initially Particle moves towards left v_1=25 m/s

Final velocity of Particle is towards Right v_2=31 m/s

According to Work Energy theorem

Work done by all the Forces=change in Kinetic Energy

Work done by Force=\frac{mv_2^2}{2}-\frac{mv_1^2}{2}

W=\frac{69\times 10^{-3}}{2}\left [ 31^-25^2\right ]

W=\frac{69\times 10^{-3}}{2}\left [ 961-625\right ]

W=11.59 J

5 0
3 years ago
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