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rjkz [21]
3 years ago
15

One ounce of a well-known breakfast cereal contains 103 Calories (1 food Calorie = 4186 J). If 1.6% of this energy could be conv

erted by a weight lifter's body into work done in lifting a barbell, what is the heaviest barbell that could be lifted a distance of 1.8 m?
Physics
1 answer:
faust18 [17]3 years ago
5 0

Answer:

the heaviest barbell that could be lifted  is 390.6kg

Explanation:

Hello!

To solve this question you must follow the following steps

1. Find the amount of energy that can transform the body into motion, this is achieved by multiplying the breakfast consumed by the percentage of energy conversion.

W=(103Calories)(0.016)\frac{4186J}{1calorie} =6898.5J

2. We use the equation that defines the work done by a body that has weight when it is lifted, this is defined by the product of mass by gravity by height.

W=mgh

where

W=work=6898.5J

m=mass

g=gravity=9.81m/s^2

h=height=1.8m

now we solve for mass, and use the values.

m=\frac{W}{hg} =\frac{6898.5}{(1.8)(9.81)} =390.6kg

the heaviest barbell that could be lifted  is 390.6kg

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A sinusoidal, transverse wave that propagates in the positive x-direction can be described with the wave function y of xt is equ
myrzilka [38]

Answer:

\omega =\frac{2 \pi }{T} rad/s

Explanation:

The wave function is:

y(xt) = Acos(kx- \omega t)

where :

k = wave number

x = position of a point on the wave

\omega = angular frequency

t = time

What is another way to express the angular frequency (omega)

Angular frequency (omega) can be express as :

\omega =\frac{2 \pi }{T} rad/s ( i.e one repetition that it takes to repeats itself)

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3 years ago
Find the noise level of a sound having an intensity of 1.5x10^-14W/m^2 given I0=10^12 W/m2
Llana [10]

Answer:

Noise level will be -18.2 watt

So option (b) will be correct answer

Explanation:

We have given sound intensity I=1.5\times 10^{-14}w/m^2

And threshold intensity I_0=\times 10^{-12}w/m^2 ( in question it is given as 10^{12}w/m^2 but its standard value is 10^{-12}w/m^2 )

Now noise level  =10log\frac{I}{I_0}=10log\frac{1.5\times 10^{-14}}{10^{-12}}=10log0.015-18.23

So the noise level will be -18.2

So option (b) will be correct answer

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2 years ago
John drives a distance of 90 Km to the right at a steady speed of V1=25.00 m/s. Then he stops at the gas station for 10 minutes.
aleksley [76]

a. Speed is defined as rate of change of distance per unit time whereas velocity is defined as rate of change of displacement per unit time.

b. t=6000\ s is the total time taken in the trip

c. d=126000\ m is the total distance

d. s=54000\ m towards right from the starting point.

e. v_a=21\ m.s^{-1}

f. \vec v_a=9\ m.s^{-1} towards right.

Explanation:

a.

Speed is a scalar quantity while velocity is a vector quantity.

Speed is defined as rate of change of distance per unit time whereas velocity is defined as rate of change of displacement per unit time.

Speed is a directionless quantity while velocity constitutes direction.

b.

<em>Total time of round trip when we're given:</em>

  • distance travelled to the right, d_r=90000\ m
  • speed while travelling to the right, v_r=25\ m.s^{-1}
  • time spent at gas station, t_g=600\ s
  • time spent while travelling back towards the left, t_l=30\times 60=1800\ s
  • speed while travelling to the left, v_{_l}=20\ m.s^{-1}

<em>Now time taken for travelling towards right:</em>

t_r=\frac{d_r}{v_r}

t_r=\frac{90000}{25}

t_r=3600\ s

<u>Therefore total time taken in the round trip:</u>

t=t_r+t_l+t_g

t=3600+600+1800

t=6000\ s

c.

<em>Now, distance travelled towards left:</em>

d_l=v_{_l}\times t_l

d_l=20\times1800

d_l=36000\ m

<u>Therefore total distance:</u>

d=d_l+d_r

d=36000+90000

d=126000\ m

d.

Now, total displacement:

s=d_r-d_l

s=90000-36000

s=54000\ m towards right from the starting point.

e.

<u>Average speed:</u>

v_a=\frac{d}{t}

v_a=\frac{126000}{6000}

v_a=21\ m.s^{-1}

f.

<u>Average velocity:</u>

\vec v_a=\frac{s}{t}

\vec v_a=\frac{54000}{6000}

\vec v_a=9\ m.s^{-1} towards right.

4 0
3 years ago
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