2 years ago

In a catapult, energy is transferred into useful kinetic energy stores. What store is the energy in when it is the catapult?

Physics

1 answer:

yawa3891 [41]2 years ago

6 0

Answer:

actually I was just wondering what you are thinking

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Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal ener

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Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

= 1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

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2 years ago

A vehicle moving with a uniform a acceleration of 2m/s and has a velocity of 4m/s at a certain time .

Evgesh-ka [11]

Answer:

1 second later the vehicle's velocity will be:

$v(1)= 6\,\,\frac{m}{s} \\$

5 seconds later the vehicle's velocity will be:

$v(5)=14\,\,\frac{m}{s}$

Explanation:

Recall the formula for the velocity of an object under constant accelerated motion (with acceleration " $a$ "):

$v(t)=v_0+a\,t$

Therefore, in this case $v_0=4\,\,\frac{m}{s}$ and $a=2\,\,\frac{m}{s^2}$

so we can estimate the velocity of the vehicle at different times just by replacing the requested "t" in the expression:

$v(t)=v_0+a\,t\\v(t)=4+2\,\,t\\v(1)=4+2\,(1) = 6\,\,\frac{m}{s} \\v(5)=4+2\,(5)=14\,\,\frac{m}{s}$

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