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3 years ago

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A student evaluates a weight loss program by calculating the number of times he would need to climb a 14.0 m high flight of step

s in order to lose one pound (0.45 kg) of fat. Metabolizing 1.00 kg of fat can release 3.77 x 10' 1 of chemical energy and the body can convert about 22.0% of this into mechanical energy (the rest goes into internal energy.)(a) How much mechanical energy (in 3) can the body produce from 0.450 kg of fat?_______.(b) How many trips up the flight of steps are required for the 68.0 kg student to lose 0.450 kg of fat? Ignore the relatively small amount of energy required to return down the stairs_______trips.

1 answer:

Liula [17]3 years ago

7 0

Answer:

400 trips

Explanation:

Mechanical energy needed to climb 14 m by a man of 68 kg

= mgh

= 68 x 9.8 x 14

= 9330 J

1 Kg of fat releases 3.77 x 10⁷ J of energy

.45 kg of fat releases 1.6965 x 10⁷ J of energy

22% is converted into mechanical energy

so 22% of 1.6965 x 10⁷ J

= 3732.3 x 10³ J of mechanical energy will be available for mechanical work.

one trip of climbing of 14 m requires 9330 J of mechanical energy

no of such trip possible with given mechanical energy

= 3732.3 x 10³ / 9330

= 400 trips

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Very large accelerations can injure the body, especially if they last for a considerable length of time. The severity index (SI)

Ludmilka [50]

Answer:

a) The severity index (SI) is 3047.749, b) The injured travels 0.345 meters during the collision.

Explanation:

a) The g-multiple of the acceleration, that is, a ratio of the person's acceleration to gravitational acceleration, is:

$a' = \frac{35\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }$

$a' = 3.569$

The time taken for the injured to accelerate to final speed is given by this formula under the assumption of constant acceleration:

$v_{f} = v_{o} + a \cdot t$

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v_{f}$ - Final speed, measured in meter per second.

$a$ - Acceleration, measured in $\frac{m}{s^{2}}$ .

$t$ - Time, measured in seconds.

$t = \frac{v_{f}-v_{o}}{a}$

$t = \frac{\left(12\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)}{35\,\frac{m}{s^{2}} }$

$t = 0.095\,s$

Lastly, the severity index is now determined:

$SI = \frac{a'^{5}}{2\cdot t}$

$SI = \frac{3.569^{5}}{2\cdot (0.095\,s)}$

$SI = 3047.749$

b) The initial and final speed of the injured are $1.944\,\frac{m}{s}$ and $5.278\,\frac{m}{s}$ , respectively. The travelled distance can be determined from this equation of motion:

$v_{f}^{2} = v_{o}^{2} + 2\cdot a \cdot \Delta s$

Where $\Delta s$ is the travelled distance, measured in meters.

$\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a}$

$\Delta s = \frac{\left(5.278\,\frac{m}{s} \right)^{2}-\left(1.944\,\frac{m}{s} \right)^{2}}{2\cdot \left(35\,\frac{m}{s^{2}} \right)}$

$\Delta s = 0.345\,m$ .

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2 years ago

A 15.0 mW laser puts out a narrow beam 2.00 mm indiameter.

Murljashka [212]

Answer:

1341.03 V/m

Explanation:

The power output per unit area is the intensity and also the is the magnitude of the Poynting vector.

$S = \frac{P}{A}$ = cε₀ $E^{2} _{rms}$

⇒ $\frac{P}{A}$ = cε₀ $E^{2}_{rms}$

Where;

P is the power output

A is the area of the beam

c is speed of light

ε₀ is permittivity of free space 8.85 × 10⁻¹² F/m

$E_{rms}$ is the average (rms) value of electric field

Making electricfield $E_{rms}$ the subject of the equation

$E^{2}_{rms}$ = P / Acε₀

$E_{rms}$ = √(P / Acε₀)

But area A = πr²

$E_{rms}$ = √(P / πr²cε₀)

Given:

Output power, P = 15 mW = 0. 015 W

Diameter, d = 2 mm = 0.002 m

⇒ Radius, $r = \frac{d}{2} = \frac{0.002}{2} = 0.001 m$

Solving for average (rms) value of electric field;

$E_{rms} = \sqrt{\frac{0.015 W}{\pi * (0.001 m)^2 * (3 * 10^8 m/s) * (8.85 * 10^-12) C^2/Nm^2} }$

$E_{rms}$ = 1341.03 V/m

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3 years ago

A tennis player tosses a tennis ball straight up and then catches it after 1.64 s at the same height as the point of release.

ipn [44]

A. The acceleration of the ball while it is in flight?

magnitude is 0 m/s² (magnitude is zero)

B. The velocity of the ball when it reaches its maximum height is 0 m/s (magnitude is zero)

C. The initial velocity of the ball 8.036 m/s upward

D. The maximum height reached by the ball is 3.29 m

<h3>A. How to determine the acceleration in the flight</h3>

Considering that the ball came to rest after 1.64s, it means the entire acceleration of the flight is zero as the ball was not moving in any form again.

<h3>B. How to determine the velocity at maximum height</h3>

At maximum height, the velocity of the ball is zero as it no longer has magnitude to keep going upwards. Hence the ball begins to ball down.

<h3>C. How to determine the initial velocity</h3>

Acceleration due to gravity (g) = 9.8 m/s²
Final velocity (v) = 0 m/s
Time of flight (T) = 1.64 s
Time to reach maximum height (t) = T / 2 = 1.64 / 2 = 0.82 s
Initial velocity (u) =?

v = u - gt (since the ball is going against gravity)

0 = u - (9.8 × 0.82)

0 = u - 8.036

Collect like terms

u = 0 + 8.036

u = 8.036 m/s upward

<h3>D. How to determine the maximum height reached by the ball</h3>

Time to reach maximum height (t) = T / 2 = 1.64 / 2 = 0.82 s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h)

h = ½gt²

h = ½ × 9.8 × 0.82²

h = 3.29 m

Learn more about motion under gravity:

brainly.com/question/20385439

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1 year ago

Which of the following is same dimension quantity?

hram777 [196]

Answer:

Find the dimension of each and every quantity in all the options to check whether they are the same or not. We can use any one formula of each identity to find its dimension.

Complete step by step solution:

To find the dimension of a quantity, we can use any formula related to that quantity but we will use the easiest ones to save time.

Force-

from Newton’s law of motion,

F=maF=ma

Dimension of force =[M][LT−2]=[MLT−2]=[M][LT−2]=[MLT−2]

Work done-

W=F×sW=F×s

Dimension of work=[MLT−2][L]=[ML2T−2]=[MLT−2][L]=[ML2T−2]

Momentum-

p=mvp=mv

Dimension of momentum=[M][LT1]=[MLT−1]=[M][LT1]=[MLT−1]

Impulse-

I=F×tI=F×t

Dimension of impulse=[MLT−2][T]=[

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2 years ago

How many photons with 10 ev are required to produce 20 joules of energy?

oee [108]

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