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3 years ago

11

During light activity, a 70-kg person may generate 200 kcal

1 answer:

Lerok [7]3 years ago

3 0

Answer:

$2.75387\ ^{\circ}$

Explanation:

m = Mass of person = 70 kg

c = Specific heat of human body = $0.83\ J/kg^{\circ}C$

$\Delta T$ = Change in temperature

Time taken is 1 hour

Heat is given by

$Q=mc\Delta T\\\Rightarrow \Delta T=\dfrac{Q}{mc}\\\Rightarrow \Delta T=\dfrac{0.8\times 200\times 1}{70\times 0.83}\\\Rightarrow \Delta T=2.75387\ ^{\circ}C$

The rise in temperature is $2.75387\ ^{\circ}C$

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In which changes of state do atoms lose energy

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3 years ago

If the sprinter accelerates at that rate for a distance of 15 m, and then maintains the velocity he has at that point for the re

ahrayia [7]

Answer:

The time for the entire race is 11.39 sec.

Explanation:

Given that,

Distance = 15 m

Remainder distance = 100 m

Suppose A sprinter begins a race with an acceleration of 3.4 m/s².

We need to calculate the time

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value in the equation

$15=0+\dfrac{1}{2}\times3.4\times t^2$

$t^2=\dfrac{30}{3.4}$

$t=2.97\ sec$

We need to calculate the final velocity of sprinter

Using equation of motion again

$v=u+at$

Put the value into the formula

$v=0+3.4\times2.97$

$v=10.09\ m/s$

We need to calculate the distance covers by sprinter

$d=100-15=85\ m$

The sprinter need to covers only 85 m.

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t'=\dfrac{85}{10.09}$

$t'=8.42\ sec$

We need to calculate the time for the entire race

$t''=t+t'$

Put the value into the formula

$t''=2.97+8.42$

$t''=11.39\ sec$

Hence, The time for the entire race is 11.39 sec.

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3 years ago

Describe what happens to the moving boat when the oars are out of the water and the forward thrust is zero

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Answer:

The boat won't be able to move if the oars were out and there was no thruster. If there was a flow of the water then yes there would be a moving boat.

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2 years ago

The force of gravity on a 2-kg rock is twice as great as that on a 1-kg rock. why then doesn't the heavier rock fall faster?

katrin2010 [14]

It takes twice the force to produce the same acceleration in the 2kg rock.

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3 years ago

Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) .

jek_recluse [69]

Explanation:

Given that,

Force, $F=((-8i)+6j)\ N$

Position of the particle, $r=(3i+4j)\ m$

(a) The toque on a particle about the origin is given by :

$\tau=F\times r$

$\tau=((-8i)+6j) \times (3i+4j)$

Taking the cross product of above two vectors, we get the value of torque as :

$\tau=(0+0-50k)\ N-m$

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$cos\theta=\dfrac{r.F}{|r|.|F|}$

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$cos\theta=\dfrac{0}{50}$

$\theta=90^{\circ}$

6 0

3 years ago