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11Alexandr11 [23.1K]
2 years ago
6

The basal metabolic rate is the rate at which energy is produced in the body when a person is at rest. A 157 lb (71.0 kg ) perso

n of height 5.91 ft(1.80 m ) would have a body surface area of approximately 1.90 m2 .
Reqiuired:
a. What is the net amount of heat this person could radiate per second into a room at 19.0 ∘C (about 66.2∘F) if his skin's surface temperature is 31.0 ∘C? (At such temperatures, nearly all the heat is infrared radiation, for which the body's emissivity is 1.00, regardless of the amount of pigment.)

b. Normally, 80.0 % of the energy produced by metabolism goes into heat, while the rest goes into things like pumping blood and repairing cells. Also normally, a person at rest can get rid of this excess heat just through radiation. Use your answer to part A to find this person's basal metabolic rate(BMR).
Physics
1 answer:
GalinKa [24]2 years ago
6 0

Answer:

A. Net amount of heat radiated = 109.2W

B. Person's basal energy = 136.5

Explanation:

Part A:

Area of person, A = 1.90 m^2

Temperature of person , T = 31 C

T = 304 K

Temperature of surroundings , To = 19 C

To = 282 K

Now, net amount of heat radiated = e*A*sigma *(T^4 - To^4)

Net amount of heat radiated = 1 * 1.8 * 5.6703 *10^-8 *(304^4 - 294^4)

Net amount of heat radiated = 109.2 W

The net amount of heat radiated is 109.2 W

Part B:

Person's basal energy = net amount of heat radiated /(0.80)

Person's basal energy = 109.2/0.80

Person's basal energy = 136.5 W

Person's basal energy is 136.5 W

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an object traveling 200 feet per second slows to 50 feet per second in 5 seconds. Calculate the acceleration of the object
emmainna [20.7K]

Answer:

The acceleration of the object is -30\ m/s^2

Explanation:

Given:

Initial velocity of object v_i = 200 feet/second

Final velocity of object v_f = 50 feet/second

Time of travel = 5 seconds

To calculate acceleration of the object we will find the rate of change of velocity with respect to time.

So, acceleration a is given by:

a=\frac{v_f-v_i}{t}

where v_f represents final velocity, v_i represents initial velocity and t is time of travel.

Plugging in values to evaluate acceleration.

a=\frac{50-200}{5}

a=\frac{-150}{5}

a=-30\ m/s^2

The acceleration of the object is -30\ m/s^2 (Answer). The negative sign shows the object is slowing down.

4 0
3 years ago
A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli
Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

   centripetal force acting on satellite = \dfrac{mv^2}{r}

     gravitational force = \dfrac{GMm}{r^2}

    equating both the above equation

    \dfrac{mv^2}{r} = \dfrac{GMm}{r^2}

      v = \sqrt{\dfrac{GM}{r}}

      v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}

          v = 5578.5 m/s

b) T= \dfrac{2\pi\ r}{v}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

          T = 14416.92 s

          T = \dfrac{14416.92}{3600}\ hr

          T = 4 hr

c) gravitational force acting

  F = \dfrac{GMm}{r^2}

  F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}

     F = 5202 N

4 0
3 years ago
When you apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed ().When you apply a f
elena-s [515]

Answer:

Explanation:

When we apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed. So net force on the block is zero .

It implies that a force ( frictional ) acts on it which is equal to 76 N in opposite direction ( friction )

When we apply  a greater force on it it starts moving with acceleration .

This time kinetic friction acts on it due to rough ground equal to 76 N .This is limiting friction ( maximum friction )

Net force on the body in later case

= 89 - 76

= 13 N

Force by ground on the block in horizontal direction = 76 N ( FRICTIONAL FORCE )

=

3 0
3 years ago
A 5.0-kg box has an acceleration of 2.0 m/s2 when it is pulled by a horizontal force across a surface with μK = 0.50. Find the w
Maslowich

Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J

Explanation: In order to solve this problem we have to used the second Newton law given by:

∑F= m*a

F-f=m*a where f is the friction force (uk*Normal), from this we have

F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N

then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N

the net Force = (34.5-24.5)N= 10 N

Finally the work done by the net force is equal to kinetic energy change so

W=∫Force net*dr= 10 N* 0.1 m= 1J

7 0
3 years ago
Read 2 more answers
When a voltage difference is applied to a piece of metal wire, a 5.0 mA current flows through it. If this metal wire is now repl
zheka24 [161]

Answer:

I = 21.13 mA ≈ 21 mA

Explanation:

If

I₁ = 5 mA

L₁ = L₂ = L

V₁ = V₂ = V

ρ₁ = 1.68*10⁻⁸ Ohm-m

ρ₂ = 1.59*10⁻⁸ Ohm-m

D₁ = D

D₂ = 2D

S₁ = 0.25*π*D²

S₂ = 0.25*π*(2*D)² = π*D²

If we apply the equation

R = ρ*L / S

where (using Ohm's Law):

R = V / I

we have

V / I = ρ*L / S

If V and L are the same

V / L =  ρ*I / S

then

(V / L)₁ = (V / L)₂  ⇒     ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂

If

S₁ = 0.25*π*D²   and

S₂ = 0.25*π*(2*D)² = π*D²

we have

ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)

⇒    I₂ = 4*ρ₁*I₁ / ρ₂

⇒     I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m

⇒     I₂ = 21.13 mA

5 0
3 years ago
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