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Anni [7]
2 years ago
10

At room temperature, a typical person loses energy to the surroundings at the rate of 62 W. If this energy loss has to be made u

p by an equivalent food intake, how many kilocalories (food calories) does this person need to consume every day just to make up this heat loss
Physics
1 answer:
Alex_Xolod [135]2 years ago
4 0

To solve this problem it is necessary to use the given proportions of power and energy, as well as the energy conversion factor in Jules to Calories.

The power is defined as the amount of energy lost per second and whose unit is Watt. Therefore the energy loss rate given in seconds was

P = \frac{E}{t} \rightarrow E= Energy, t = time

P = 62W = 62 \frac{J}{s}

The rate of energy loss per day would then be,

P = 62\frac{J}{s} (\frac{86400s}{1day})

P = 5356800 \frac{J}{day}

That is to say that Energy in Jules per lost day is 5356800J

By definition we know that 1KCal = 4.184*10^{6}J

In this way the energy in Cal is,

E = 5356800J \frac{1KCal}{4.184*10^{6}J}

E = 1279.694 KCal

The number of kilocalories (food calories) must be 1279.694 KCal

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A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra
skelet666 [1.2K]

Answer:

So the acceleration of the child will be 8.05m/sec^2

Explanation:

We have given angular speed of the child \omega =1.25rad/sec

Radius r = 4.65 m

Angular acceleration \alpha =0.745rad/sec^2

We know that linear velocity is given by v=\omega r=1.25\times 4.65=5.815m/sec

We know that radial acceleration is given by a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2

Tangential acceleration is given by

a_t=\alpha r=0.745\times 4.65=3.464m/sec^

So total acceleration will be a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2

7 0
2 years ago
A Micro –Hydro turbine generator is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity
Salsk061 [2.6K]

Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

initial angular speed of the turbine is

\omega_i = 2\pi f_1

\omega_1 = 2\pi(\frac{610}{60})

\omega_1 = 63.88 rad/s

similarly final angular speed is given as

\omega_f = 2\pi f_2

\omega_2 = 2\pi(\frac{837}{60})

\omega_2 = 87.65 rad/s

angular acceleration of the turbine is given as

\alpha = 5.9 rad/s^2

now we have to find the angular displacement is given as

\theta = \omega t + \frac{1}{2}\alpha t^2

\theta = (63.88)(3.2) + (\frac{1}{2})(5.9)(3.2^2)

\theta = 234.62 radian

3 0
2 years ago
A 1500 W radiant heater is constructed to operate at 115 V. (a) What will be the current in the heater? (b) What is the resistan
OlgaM077 [116]

Answer:

a) I = 13.04 A

b)  R = 8.82 ohms

c) 1291.87 kilocalories are generated an hour.

Explanation:

let P be the power of the heater, V be the voltage of the heater, I be the current of the heater, R be the resistance.

a) we know that:

P = I×V

I = P/V

  = (1500)/(115)

  = 13.04 A

Therefore, the current of the heater is 13.04 A

b) we now have voltage and current, according to Ohm's law:

R = V/I

  = (115)/(13.04)

  = 8.82 ohms

Therefore, the resistance of the heating coil is 8.82 ohms.

c) the number of kilocalories generated in one hour by the heater is just the energy the heater produces in one hour which is given by:

E = P×t

  = (1500)(1×60×60)

  = 5400000 J

since 1 calorie = 4.81 J

1 kilocalorie = 0.001 calories

E = 5400000/4.18 ≈ 1291866.029 calories ≈1291.87 kilocalories

Therefore, 1291.87 kilocalories are produced/generated in one hour.

8 0
2 years ago
he drawing shows two perpendicular, long, straight wires, both of which lie in the plane of the paper. The current in each of th
AleksandrR [38]

Answer:

The magnitudes of the net magnetic fields at points A and B is 2.66 x 10^{-6} T

Explanation:

Given information :

The current of each wires, I = 4.7 A

dH = 0.19 m

dV = 0.41 m

The magnetic of straight-current wire :

B= μ_{0}I/2πr

where

B = magnetic field (T)

μ_{0} = 1.26 x 10^{-6} (N/A^{2})

I = Current (A)

r = radius (m)

the magnetic field at points A and B is the same because both of wires have the same distance. Based on the right-hand rule, the net magnetic field of A and B is canceled each other (or substracted). Thus,

BH = μ_{0}I/2πr

     = (1.26 x 10^{-6})(4.7)/(2π)(0.19)

     = 4.96 x 10^{-6} T

BV = μ_{0}I/2πr

     = (1.26 x  10^{-6})(4.7)/(2π)(0.41)

     = 2.3 x 10^{-6} T

hence,

the net magnetic field = BH - BV

                                     = 4.96 x 10^{-6} - 2.3 x 10^{-6}

                                     = 2.66 x 10^{-6} T

4 0
2 years ago
An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion,
Ne4ueva [31]
A)  f = 1.8 rev/s = 2 Hz 
<span>T = 1 / f = 0.55s

B)  not really sure..srry

C)  </span><span>T = 2 pi √ ( L / g ) </span>
<span>0.57 = 2 x 3.14 x √ ( 0.2 / g )
</span><span>
g = 25.5 m/s²
</span>
Hope this helps a little at least.. :)

5 0
2 years ago
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