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2 years ago

A person is standing outdoors in the shade where the temperature is 35 °C.

Physics

1 answer:

Free_Kalibri [48]2 years ago

6 0

Answer:

(a) Eₐ = 6.36 J/s

(b) Eₐ = 4.64 J/s

Explanation:

Stefan-Boltzmann law: States that the total energy per second radiated or absorbed by a black body is directly proportional to the absolute temperature.

Using, Stefan-Boltzmann equation

Eₐ =eσAT⁴ ................ Equation 1

where Eₐ = Radiant energy absorbed per seconds, e = emissivity, σ = stefan - boltzman constant, A = Surface area. and T = temperature in kelvin

(a) Where e = 0.89, σ = 5.67 ×10⁻⁸ watt/m²/K⁴, A = 140 cm² = 140 cm²(m²/10000cm²) = 0.014 m², T = 35 °C = (35 + 273) K = 308 K.

Applying these values in equation 1 above,

Eₐ = 0.89 × 5.67 ×10⁻⁸ × 0.014 × (308)⁴

Eₐ =6.36 J/s

(b) when e = 0.65,

∴ Eₐ = 0.65 × 5.67 × 10⁻⁸ × 0.014 × (308)⁴

Eₐ = 4.64 J/s

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2 years ago

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2 years ago

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A golfer hits a golf ball at an angle of 25.0Â° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is

kvasek [131]

<u>Answer:</u>

Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

0 = u sin θ - gt

t = u sin θ/g

Total time for vertical motion is two times time taken for upward vertical motion of projectile.

So total travel time of projectile = 2u sin θ/g

Horizontal motion:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

Vertical motion (Maximum height reached, H) :

We have equation of motion, $v^2=u^2+2as$ , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

$0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}$

In the give problem we have R = 301.5 m, θ = 25° we need to find H.

So $\frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g$

Now we have $H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m$

So maximum height reached = 35.15 meter.

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2 years ago

Because the soles of your shoes have cleats, you can exert a forward force of 100 N even on slippery ice. A 10-kg picnic cooler

Brilliant_brown [7]

Answer:

you must throw 3 snowballs

Explanation:

We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment

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p₀ = n m v₀

Where n is the snowball number

Final. When the refrigerator moves

pf = (n m + M) v

The moment is preserved because the forces during the crash are internal

n m v₀ = (n m + M) v

n m (v₀ - v) = M v

n = M/m v/(vo-v)

Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton

F = m a

a = F / m

The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm

v₁² = v₀² + 2 a x

v₁² = 0+ 2 (100/1) 1

v₁ = 14.14 m / s

This is the initial speed for the crash

v₀ = v = 14.14 m / s

Let's calculate

n = M/m v/ (v₀-v)

n = 10/1 3 / (14.14 -3)

n = 2.7 balls

you must throw 3 snowballs

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2 years ago

A person drops a stone down a well and hears the echo 8.9 s later. if it takes 0.9 s for the echo to travel up the well, approxi

Temka [501]

Total time in between the dropping of the stone and hearing of the echo = 8.9 s

Time taken by the sound to reach the person = 0.9 s

Time taken by the stone to reach the bottom of the well = 8.9 - 0.9 = 8 seconds

Initial speed (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s^2

Time taken (t) = 8 seconds

Let the depth of the well be h.

Using the second equation of motion:

$h = ut + \frac{1}{2}\times a \times t^2$

$h = 0 \times 8 + \frac{1}{2} \times 9.8 \times 8^2$

h = 313.6 m

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2 years ago