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Pavlova-9 [17]
3 years ago
8

a skinny girl devours food that is equivalent to 600 kcal, as she is anorexic she repents and wants to lose twice as many calori

es by lifting 20 kg weights to a height of 1.9 m. If she measures 1.7m ... she calculates the number of times she must lift the weights to lose the same amount of energy that she acquired with food and the time she must be exercising if it takes 2 seconds for repetition.
Physics
1 answer:
blondinia [14]3 years ago
4 0

Answer:

Explanation:

Work done in lifting the weight once = mgh

= 20 x 9.8 x (1.9+1.7)

= 705.6 J  

= 705.6 / 4.2 calorie

= 168 cals

Total energy to be spent = 600 x 10³ cals

No of times weight is required to be lifted

= 600 x 10³ / 168

= 3.57  x 10³ times

Total time to be taken = 2 x 3.57 x 10³

= 7.14 x 10³ s

=119 minutes .

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Answer:

θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

Explanation:

This is an angular kinematic exercise the equation for the angular position

the particle A

       θ = θ₀ + ω₀ t + ½ α t²

They say for the particle B

     w₀B = ½ w₀

     αB = 2 α

In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial

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3 years ago
The uniform rods AB and BC weigh 24 ky and kg, respectively,and the small wheel at C is of negligible weight. If the wheel ismov
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The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

T_{1}  + V_{1} = T_{2}  + V_{2}

h_{ab}  = h_{bc}  = 0.18m

potential energy at position 1,

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kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} +  \frac{1}{2} m_{bc} v^{2}_{G} +  \frac{1}{2} lw^{2}_{bc}

l_{ab} =\frac{m_{ab} l^{2}_{ab}  }{3}

= 1/3 *4 * (0.36)²

=0.10368kg m²

l =\frac{m_{bc} l^{2}_{bc}  }{12}

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

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