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Genrish500 [490]
2 years ago
6

Consider an acrylic sheet of thickness L = 5 mm that is used to coat a hot, isothermal metal substrate at Th = 300°C. The proper

ties of the acrylic are rho = 1990 kg/m3, c = 1470 J/kg · K, and k = 0.21 W/m · K. Neglecting the thermal contact resistance between the acrylic and the metal substrate, determine how long it will take for the insulated back side of the acrylic to reach its softening temperature, Tsoft = 90°C. The initial acrylic temperature is Ti = 20°C.

Physics
1 answer:
Ad libitum [116K]2 years ago
7 0

Answer:

74.52s

Explanation:

The solution is shown in the picture below

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For metalloids on the periodic table, how do the group number and the period number relate?
lapo4ka [179]
Im guessing it's (a) since the numbers go in chronological order and you read the periodic table left to right
3 0
2 years ago
Read 2 more answers
Among the largest passenger ships currently in use, the Norway has been in service the longest. The Norway is more than 300 m lo
LenaWriter [7]

Answer:

6.33\times 10^8\ kg\cdot m/s

Explanation:

Mass of the ship (m) = 6.9 × 10⁷ kg

Speed of the ship (v) = 33 km/h

First, let us convert the speed from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

So, 33 km/h = 33\times \frac{5}{18}=9.17\ m/s

Now, we know, the momentum of an object only depends on its mass and speed. Momentum is independent of the length of the object.

So, here, length of the ship doesn't play any role in the determination of the momentum.

Magnitude of momentum of the ship = Mass × Speed

                                                             = (6.9\times 10^7\ kg)(9.17\ m/s)

                                                             = 6.33\times 10^8\ kg\cdot m/s

Therefore, the magnitude of ship's momentum is 6.33\times 10^8\ kg\cdot m/s.

6 0
2 years ago
A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of
koban [17]

<u>Answer:</u>

 Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

  Speed of truck = 25 m/s north = 25 j m/s

  Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

                          = (1.43 i + 1.00 j) m/s

    Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

    Magnitude of velocity = 26.04 m/s

    Angle from positive horizontal axis = 86.85⁰

 So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.

4 0
2 years ago
A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu
melomori [17]

Answer:

please read the answer below

Explanation:

The angular momentum is given by

|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)

By taking into account the angles between the vectors r and v in each case we obtain:

a)

v=(2,0)

r=(0,1)

angle = 90°

L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

b)

r=(0,-1)

angle = 90°

L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

c)

r=(1,0)

angle = 0°

r and v are parallel

L = 0kgm/s

d)

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

e)

r=(1,1)

angle = 45°

L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}

f)

r=(-1,1)

angle = 45°

the same as e):

L = 5kgm/s

g)

r=(-1,-1)

angle = 135°

L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}

h)

r=(1,-1)

angle = 135°

the same as g):

L = 5kgm/s

hope this helps!!

4 0
2 years ago
As the drawing illustrates, a siren can be made by blowing a jet of air through 20 equally spaced holes in a rotating disk. The
Aneli [31]

Answer:

ω = 630.2663 = 630[rad/s]

Explanation:

Solution:

- We can tackle this question by simple direct proportion relation between angular speed for the disk to rotate a cycle that constitutes 20 holes. We will use direct relation with number of holes per cycle to compute the revolution per seconds i.e frequency of speed f.

                                  1rev(20 hole) -> 20(cycle)/rev  

                                        2006.2(cycle) -> f ?  

                              f = 2006.2/20 = 100.31rev at second  

- The relation between angular frequency and angular speed is given by:

                                 ω = 2πf

                                 ω = 2*3.14*100.31

                                 ω = 630.2663 = 630[rad/s]

4 0
2 years ago
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