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jarptica [38.1K]
2 years ago
15

If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2

.5 cm thick and the thermal conductivity may be taken as 0.2 W/m K, what is the temperature difference across the material?
Physics
1 answer:
kaheart [24]2 years ago
5 0

Answer:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

Explanation:

For this case we can use the Fourier Law of heat conduction given by the following equation:

Q = -kA \frac{\Delta T}{\Delta x}   (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 represent the cross sectional area

Q= 3KW represent the rate of heat transfer

\Delta T is the temperature of difference that we want to find

\Delta x=2.5 cm =0.025 m represent the thickness of the material

If we solve \Delta T in absolute value from the equation (1) we got:

\Delta T =\frac{Q \Delta x}{Ak}

First we convert 3KW to W and we got:

Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W

And we have everything to replace and we got:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

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A 70 kg student jumps down to form a 1 m high platform. She forgets to bend her knees and her downward motion stops in 0.02 seco
34kurt

Answer:

15,505 N

Explanation:

Using the principle of conservation of energy, the potential energy loss of the student equals the kinetic energy gain of the student

-ΔU = ΔK

-(U₂ - U₁) = K₂ - K₁ where U₁ = initial potential energy = mgh , U₂ = final potential energy = 0, K₁ = initial kinetic energy = 0 and K₂ = final kinetic energy = 1/2mv²

-(0 - mgh) = 1/2mv² - 0

mgh = 1/2mv² where m = mass of student = 70kg, h = height of platform  = 1 m, g = acceleration due to gravity = 9.8 m/s² and v = final velocity of student as he hits the ground.

mgh = 1/2mv²

gh = 1/2v²

v² = 2gh

v = √(2gh)

v = √(2 × 9.8 m/s² × 1 m)

v = √(19.6 m²/s²)

v = 4.43 m/s

Upon impact on the ground and stopping, impulse I = Ft = m(v' - v) where F = force, t = time = 0.02 s, m =mass of student = 70 kg, v = initial velocity on impact = 4.43 m/s and v'= final velocity at stopping = 0 m/s

So Ft = m(v' - v)

F = m(v' - v)/t

substituting the values of the variables, we have

F = 70 kg(0 m/s - 4.43 m/s)/0.02 s

= 70 kg(- 4.43 m/s)/0.02 s

= -310.1 kgm/s ÷ 0.02 s

= -15,505 N

So, the force transmitted to her bones is 15,505 N

3 0
2 years ago
In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is
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Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

\theta = Angle = 20°

y = -20 cm

Velocity components

u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s

u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s

Acceleration components

a_x=0

a_y=-9.81\ m/s^2

y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0

t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629

Time taken is 0.26088 seconds

x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m

The distance the beetle travels on the ground is 0.3677181864 m

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Which trailer has more downward pressure where it attaches to the car?
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dimensions of the bed is given as

length = 2.55 m

thickness = 2.53 dm = 0.253 m

width = 245 cm = 2.45 m

now the volume of the bed is given as

V = 2.55 * 0.253 * 2.45 m^3

V = 1.581m^3

now the mass of water in it is given as

mass = density * volume

mass = 1000* 1.581

mass = 1581 kg

<em>so it will contain 1581 kg mass in it</em>

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D is the correct answer
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