Ask question

Ask question

All categories

4 years ago

10

A weightlifter works out at the gym each day. Part of her routine is to lie on her back and lift a 41kg barbell straight up from

chest height to full arm extension, a distance of 0.45m . Part A: How much work does the weightlifter do to lift the barbell one time? Part B: If the weightlifter does 23 repetitions a day, what total energy does she expend on lifting, assuming a typical efficiency for energy use by the body? Part C: How many 500 Calorie donuts can she eat a day to supply that energy?

1 answer:

larisa86 [58]4 years ago

6 0

Answer:

A) 180.8 J B) 16,632 J C) 8

Explanation:

A) The work done by the weightlifter, is just the product of the weight of

the barbell (which is exactly equilibrated by the force applied by her,

assuming the lifting happens at constant speed) times the distance

travelled by the barbell:

W = F*d = 41 kg * 9.8 m/s * 0.45 m = 180. 8 J

B) If the weightlifter makes 23 repetitions a day, the total work done will

be 23 times the one needed for one try, as follows:

W = 23* 180.8 J = 4,158 J

Assuming a typical body efficiency of 25%, the energy expended by

the body, must be 4 times the used for useful work, as follows:

Ebody = W / 0.25 = 4,158 J / 0.25 = 16,632 J

C) At first, we need to convert J to cal, as follows:

1 cal = 4,186 J

If a donut has 500 cal, this means that it will have 2,093 J.

Dividing the total energy (in J) between the energy added by one

donut, and approximating to the nearest integer, we have:

Ndonuts= 16,632 J / 2,093 J = 7.95 = 8 Donuts

You might be interested in

11. Which compound is the most soluble at 10°C?<br> a. Ce2(S04)3<br> b. NaCl<br> c. KCI<br> d. KCIO3

Rasek [7]

KCIO3 is the answer

6 0

3 years ago

A time-dependent but otherwise uniform magnetic field of magnitude B0(t) is confined in a cylindrical region of radius 6.5 cm. I

Papessa [141]

Answer:

The acceleration is $a = 3.45*10^{3} m/s^2$

Explanation:

From the question we are told that

The radius is $d = 6.5 cm = \frac{6.5}{100} = 0.065 m$

The magnitude of the magnetic field is $B = 5.5 T$

The rate at which it decreases is $\frac{dB}{dt} = 24.5G/s = 24.5*10^{-4} T/s$

The distance from the center of field is $r = 1.5 cm = \frac{1.5}{100} = 0.015m$

According to Faraday's law

$\epsilon = - \frac{d \o}{dt}$

and $\epsilon = \int\limits {E} \, dl$

Where the magnetic flux $\o = B* A$

E is the electric field

dl is a unit length

So

$\int\limits {E} \, dl = - \frac{d}{dt} (B*A)$

${E} l = - \frac{d}{dt} (B*A)$

Now $l$ is the circumference of the circular loop formed by the magnetic field and it mathematically represented as $l = 2\pi r$

A is the area of the circular loop formed by the magnetic field and it mathematically represented as $A= \pi r^2$

So

${E} (2 \pi r)= - \pi r^2 \frac{dB}{dt}$

$E = \frac{r}{2} [ - \frac{db}{dt} ]$

Substituting values

$E = \frac{0.015}{2} (24*10^{-4})$

$E = 3.6*10^{-5} V/m$

The negative signify the negative which is counterclockwise

The force acting on the proton is mathematically represented as

$F_p = ma$

Also $F_p = q E$

So

$ma = qE$

Where m is the mass of the the proton which has a value of $m = 1.67 *10^{-27} kg$

$q = 1.602 *10^{-19} C$

So

$a =\frac{1.60 *10^{-19} *(3.6 *10^{-5}) }{1.67 *10^{-27}}$

$a = 3.45*10^{3} m/s^2$

8 0

3 years ago

What test were used to find out information about the iceman

Blababa [14]

To find information, the would take DNA tests.

6 0

3 years ago

Why neon is in the same group as helium?

guajiro [1.7K]

Neon and helium are both gases that fall under the Noble Gases section in the periodic table of elements.

Hope this helped to answer your question. :D

5 0

3 years ago

After fracture, the total length was 47.42 mm and the diameter was 18.35 mm. Plot the engineering stress strain curve and calcul

Fittoniya [83]

Answer:

Part a: <em>The value of yield strength at 0.2% offset is obtained from the engineering stress-strain curve which is given as 274 MPa.</em>

Part b: <em>The value of tensile strength is obtained from the engineering stress-strain curve which is given as 417 MPa</em>

Part c: <em>The value of Young's modulus at given point is 172 GPa.</em>

Part d: <em>The percentage elongation is 18.55%.</em>

Part e: The percentage reduction in area is 15.81%

Explanation:

From the complete question the data is provided for various Loads in ductile testing machine for a sample of d0=20 mm and l0=40mm. The plot is drawn between stress and strain whose values are calculated using following formulae. The corresponding values are attached with the solution.

The engineering-stress is given as

$\sigma=\frac{F}{A}\\\sigma=\frac{F}{\pi \frac{d_0^2}{4}}\\\sigma=\frac{F}{\pi \frac{(20 \times 10^-3)^2}{4}}\\\sigma=\frac{F}{3.14 \times 10^{-4}}$

Here F are different values of the load

Now Strain is given as

$\epsilon=\frac{l-l_0}{l_0}\\\epsilon=\frac{\Delta l}{40}\\$

So the curve is plotted and is attached.

<h2>Part a</h2>

<em>The value of yield strength at 0.2% offset is obtained from the engineering stress-strain curve which is given as 274 MPa.</em>

<h2>Part b</h2>

<em>The value of tensile strength is obtained from the engineering stress-strain curve which is given as 417 MPa</em>

<h2>Part c</h2>

Young's Modulus is given as

$E=\frac{\sigma}{\epsilon}\\E=\frac{238 /times 10^6}{0.00138}\\E=172,000 MPa\\E=172 GPa$

<em>The value of Young's modulus at given point is 172 GPa.</em>

<h2>Part d</h2>

The percentage elongation is given as

$Elongation=\frac{l_f-l_0}{l_0} \times 100\\Elongation=\frac{47.42-40}{40}\times 100\\Elongation=18.55 \%\\$

So the percentage elongation is 18.55%

<h2>Part e</h2>

The reduction in area is given as

$Reduction=\frac{A_0-A_n}{A_0} \times 100\\Reduction=\frac{\pi \frac{d_0^2}{4}-\pi \frac{d_n^2}{4}}{\pi \frac{d_0^2}{4}}\times 100\\Reduction=\frac{{d_0^2}-{d_n^2}}{{d_0^2}} \times 100\\Reduction=\frac{{20^2}-{18.35^2}}{{20^2}} \times 100\\Reduction=15.81\%$

So the reduction in area is 15.81%

6 0

3 years ago