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3 years ago

9

What is the rate of heat conduction (in W) through the 3.50 cm thick fur of a large animal having a 1.70 m2 surface area? Assume

that the animal's skin temperature is 33.0°C, that the air temperature is −5.50°C, and that fur has the same thermal conductivity as air. (Assume the thermal conductivity of air is 0.023 J/(s · m · °C).)

1 answer:

IgorC [24]3 years ago

6 0

Answer: Q=42.45W

Explanation:

Q=πA × ∆t/∆X, where;

Q= Rate of heat

π=coefficient of thermal conductivity

A= Area

∆t= change in temperature

∆x= change in thickness

Q= 0.023 × 1.7 × 33-(-5)/0.035

Q= 0.0391×38/0.035

Q = 1.4858/0.035

Q= 42.45w

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17. A 1350 g projectile is launched with a force of 150 N. The length of the firing arm is 1.25 m.

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a) The exit velocity of the projectile is 16.7 m/s

b) The maximum height achieved by the projectile is 14.2 m

c) The total time of flight is 3.40 s

d) The distance covered by the projectile is 46.5 m

Explanation:

a)

We solve this first part of the problem by applying the work-energy theorem, which states that the work done on the projectile is equal to the gain in kinetic energy of the projectile. Mathematically:

$W=Fd = \frac{1}{2}mv^2-\frac{1}{2}mu^2=\Delta K$

where:

F = 150 N is the force applied

d = 1.25 m is the displacement of the projectile (the length of the firing arm)

m = 1350 g = 1.35 kg is the mass of the projectile

u = 0 is the initial velocity of the projectile

v is the exit velocity of the projectile

Solving for v, we find:

$v=\sqrt{\frac{2Fd}{m}}=\sqrt{\frac{2(150)(1.25)}{1.35}}=16.7 m/s$

b)

Assuming the projectile is fired vertically upward, then the initial kinetic energy of the projectile as soon as he leaves the cannon is fully converted into gravitational potential energy as it reaches the top of its trajectory. So we can write:

$K_i = U_f$

$\frac{1}{2}mv^2=mgh$

where:

$K_i$ is the initial kinetic energy

$U_f$ is the final potential energy

m = 1350 g = 1.35 kg is the mass of the projectile

v = 16.7 m/s is the velocity at which the projectile leaves the cannon

$g=9.8 m/s^2$ is the acceleration of gravity

h is the maximum height reached by the projectile

And solving for h, we find

$h=\frac{v^2}{2g}=\frac{(16.7)^2}{2(9.8)}=14.2 m$

c)

Assuming the projectile is launched vertically upward, then the total time of flight is twice the time it takes for reaching the maximum height. This time can be found by using the following suvat equation:

$v=u-gt$

where:

u = 16.7 m/s is the initial velocity

$g=9.8 m/s^2$ is the acceleration of gravity

t is the time

The projectile reaches the maximum height when the vertical velocity becomes zero, so when v = 0. Therefore, substituting,

$0=u-gt\\t=\frac{u}{g}=\frac{16.7}{9.8}=1.70 s$

So, the total time of flight is

$T=2t=2(1.70)=3.40 s$

d)

The motion of a projectile consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

First we have to analyze the vertical motion, to find the time of flight of the apple. We can do it by using the following suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where

s = -27 m is the vertical displacement of the apple

$u_y=u sin \theta = (16.7)(sin 42^{\circ})=11.2 m/s$ is the initial vertical velocity

t is the time if flight

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Substituting we have:

$-27=11.2t - 4.9t^2\\4.9t^2-11.2t+27=0$

which has two solutions:

t = -1.46 s (negative, we discarde)

t = 3.75 s (this is our solution)

Now we can analyze the horizontal motion: the projectile moves horizontally with a constant velocity of

$v_x = u cos \theta = (16.7)(cos 42^{\circ})=12.4 m/s$

So, the distance it covers during its fall is given by

$d=v_x t=(12.4)(3.75)=46.5 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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4 years ago

A small package is attached to a helium-filled balloon rising at 2 m/s. The package drops from the balloon when it is 14 meters

scZoUnD [109]

I would assume air resistance is negligible and so the acceleration of the package would be approximately 9.81 m/s².

Taking downwards as positive, use v²=u²+2as.
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Answer:

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Explanation:

Writing Newton's 2nd Law and Newton's Gravitational Law on the satellite (of mass <em>m,</em> experimenting an acceleration <em>a)</em> orbiting Earth (of mass <em>M</em>) with <em>r</em> as the distance between their centers we have:

$F=ma=\frac{GMm}{r^2}$

Since this acceleration is centripetal, we can write:

$\frac{v^2}{r}=a=\frac{GM}{r^2}$

So we have:

$v^2=\frac{GM}{r}$

Or:

$r=\frac{GM}{v^2}$

This distance <em>r</em> is the sum of Earth's radius <em>R</em> and the satellite's altitude <em>h </em>(<em>r=R+h</em>), so for our values we have (in S.I.):

$h=\frac{GM}{v^2}-R=\frac{(6.67259 \times10^{-11}Nm^2/kg^2)(5.98\times10^{24}kg)}{(5000m/s)^2}-6370000m=9590835m$

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3 years ago