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Oksana_A [137]
3 years ago
11

A person consumes a snack containing 14 food calories (14kcal). what is the power this food produces if it is to be "burned off"

due to exercise in 6 hours? (1 cal=4.186 J).
A. 9763 W
B. 2.7 W
C. 0.6 W
D. .0027 W
Physics
2 answers:
Inessa05 [86]3 years ago
8 0

Answer:

B) 2.7W

Explanation:

Converting Cal to Joule

        1 cal = 4.186J

        14 kcal = 14 x 1000 x 4.186

                     = 58604 J

Converting hour to seconds

             6 hours = 6 x 60 x 60 seconds

                           = 21600 seconds

Power is the time rate of doing work.

Power = Work/Time

P = (58604) / (21600)

P = 2.7W

max2010maxim [7]3 years ago
7 0

Answer:

D) 0.0027

Explanation:

Power is the rate of work that is power = work/time

we are given the following:

Work = 14kcal this is equal to 58.576 J

time = 6hours this is equal to 21600 seconds

therefore:

Power = 58.576/21600 = 0.0027W

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Say you want to make a sling by swinging a mass M of 1.7 kg in a horizontal circle of radius 0.048 m, using a string of length 0
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Answer:

Tension in the string is equal to 58.33 N ( this will be the strength of the string )

Explanation:

We have given mass m = 1.7 kg

radius of the circle r = 0.48 mF=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N

Kinetic energy is given 14 J

Kinetic energy is equal to KE=\frac{1}{2}mv^2

So \frac{1}{2}\times 1.7\times v^2=14

v^2=16.47

v = 4.05 m/sec

Centripetal force is equal to F=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N

So tension in the string will be equal to 58.33 N ( this will be the strength of the string )

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What is one property of a suspension that is different from that of a solution or a colloid?
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What makes it difficult to see Mercury​
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Mercury is the planet nearest the Sun and is the most difficult to observe, because it is always quite close to the Sun in the sky. Because of its orbital motion, it appears to swing back and forth around the Sun, reaching a maximum angular distance of about 28°.

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I hope that helps you

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A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s 2 starts from rest at t = 0. At the instant
umka21 [38]

Answer:

a = 0.27 rad/s^2

Explanation:

First we will find the time t using the next equation:

θ = \frac{1}{2}at^2

where θ is the angle in radians and a is the angular aceleration. So:

0.4 = \frac{1}{2}(1.6)t^2

Solving for t:

t = 0.707s

Second, with that time, we will find the angular velocity w using the next equation:

w = at

where a is the angular aceleration, so:

w = (1.6)(0.707s)

w = 1.1312 rad/s

Now, the radial aceleration a_r is calcualted as:

a_r = w^2r

a_r = (1.1312)^2(0.13)

a_r = 0.166 rad/s^2

Additionally, the tangential aceleration a_t is calculated as::

a_t = ar

a_t = (1.6)(0.13)

a_t = 0.208 rad/s^2

Finally, by pythagoras theorem, we find the total linear acceleration as:

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a = \sqrt{(0.208)^2+(0.166)^2 }

a = 0.27 rad/s^2

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3 years ago
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