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ollegr [7]
3 years ago
8

Calculate how many Kilo-calories of food (CAL) consumed by a 163-lb man due to the force he applies during one repetition of a b

odyweight only (single leg squat) exercise that lifts 87% of his bodyweight through a controlled full-range of motion that measures 31" in both the down and up directions of the movement? Assume that the exercise takes place under comfortable conditions, and that only about 22% of all the potential energy stored in food is available for useful work, and that 1 ft-lb = 0.000324048 Kilocalories.
Physics
1 answer:
Lilit [14]3 years ago
7 0

Answer: the man consumed 0.539598 Kilo-calories of food

 

Explanation:

Given the data in the question;

first we calculate the work required to raise the given weight at the given height;

ΔW = mgh = w'h = (0.87) × h

= ( 0.87 × 163) × (31/12)ft     { 1 foot = 12 inch}

= 366.34 lb-ft

Now since 22% of all the potential energy stored in food is available for useful work;

ΔW = 0.22 × ΔPE

ΔPE = ΔW/0.22

we substitute

ΔPE = 366.34 / 0.22

= 1665.18 lb-ft

Now given that; 1 ft-lb = 0.000324048 Kilocalories

= 1665.18 × 0.000324048

= 0.539598 Kilo-calories

Therefore the man consumed 0.539598 Kilo-calories of food

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Explanation:

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8 0
4 years ago
A particle moves along the x axis. It is initially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration 20.3
mixas84 [53]

Answer:

a) -2.34 m

b) -1.3 m/s

c) 0.056 m

d) 0.320 m/s

Explanation:

part a

Given:

s(0) = 0.27 m

v(0) = 0.14 m/s

a = -0.320 m/s^2

t = 4.50s

Using kinematic equation of motion for constant acceleration:

s (t) = s(0) + v(0)*t + 0.5*a*t^2

s ( 4.5 s ) = 0.27 + 0.14*4.5 + 0.5*-0.32*4.5^2

s(4.5) = -2.34 m

part b

Using kinematic equation of motion for constant acceleration:

v(t) = v(0) + a*t

v(4.5s) = (0.14) + (-0.32)(4.5) = -1.3 m/s

part c

Use equation for simple harmonic motion:

s(t) = A*cos(w*t)

v(t) = -A*w*sin (w*t)

a(t) = -A*(w)^2 * cos (w*t)

0.27 m = A*cos(w*t)   .... Eq 1

0.14 m/s = - A*w*sin (w*t)  .....Eq2

-0.320 = -A*(w)^2 * cos (w*t)   .... Eq3

Solve the three equations above for A, w, and t

Divide 1 and 3:

w^2 = 0.32 / 0.27

w = 1.0887 rad / s

Divide 2 and 1:

w*tan(wt) = 0.14 / 0.27

tan(1.0887*t) = 0.476289

t = 0.4083 s

A = 0.27 / cos (1.0887*0.4083) = 0.3 m

Hence, the SHM is s(t) = 0.3*cos(1.0887*t)

s(4.5) = 0.3*cos(1.0887*4.5) = 0.056 m

part d

v(t) = - 0.32661 * sin (1.0887*t)

v(4.5) = - 0.32661 * sin (1.0887*4.5) = 0.320 m/s

3 0
3 years ago
A triangle has two sides of length 10 cm and 14 cm. what can you say about the length of the third side?
Anton [14]

A side of any triangle is less than the sum of the other two sides so the third side must be less than 24cm.

<h3>What is a triangle?</h3>

A triangle is a 3-sided shape that is occasionally referred to as a triangle. There are three sides and three angles in every triangle, some of which may be the same.

The sum of all three angles inside a triangle will be 180° and the area of a triangle is given as (1/2) × base × height.

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In any triangle, the sum of two sides is always greater than the third side.

If the third side is equal to the sum of two sides then it will be a straight line rather than a triangle.

It means 10 + 14 > third side

24 cm > third side

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3 0
1 year ago
A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA. What amount of vars must be added to bring the pf to 0.85
kvasek [131]

Answer:

\mathtt{Q_{sh} = 600.75 \ vars}

Explanation:

Given that:

A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA

Here:

the initial power factor  i.e cos θ₁ = 0.7 lag

θ₁ = cos⁻¹ (0.7)

θ₁ = 45.573°

Active power P = 1500 watts

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What amount of vars must be added to bring the pf to 0.85

i.e the required power factor here is cos θ₂ = 0.85 lag

θ₂ =  cos⁻¹   (0.85)

θ₂ = 31.788°

However; the initial reactive power Q_1 = P×tanθ₁

the initial reactive power Q_1 = 1500 × tan(45.573)

the initial reactive power Q_1 = 1500 × 1.0202

the initial reactive power Q_1 =  1530.3 vars

The amount of vars that must therefore be added to bring the pf to 0.85

can be calculated as:

Q_{sh} = P( tan \theta_1 - tan \theta_2)

Q_{sh} = 1500( tan \ 45.573 - tan \ 31.788)

Q_{sh} = 1500( 1.0202 - 0.6197)

Q_{sh} = 1500( 0.4005)

\mathtt{Q_{sh} = 600.75 \ vars}

3 0
3 years ago
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