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igomit [66]
3 years ago
10

The freezer compartment in a conventional refrigerator can be modeled as a rectangular cavity 0.3 m high and 0.25 m wide with a

depth of 0.5 m. Determine the thickness of styrofoam insulation (k = 0.30 W/m⋅K) needed to limit the heat loss to 400 W if the inner and outer surface temperatures are −10 and 33°C, respectively.
Physics
1 answer:
Anon25 [30]3 years ago
8 0

Answer:

Thickness of Styrofoam insulation is 0.02741 m.

Explanation:

Given that,

Height = 0.25 m

Depth = 0.5 m

Power = 400 W

Temperature = 33°C

We need to calculate the area of Styrofoam

Using formula of area

A=2(lb+bh+hl)

Put the value into the formula

A=2(0.3\times0.5+0.25\times0.5+0.5\times0.3)

A=0.85\ m^2

Inner surface temperature of freezer

T_{i}=-10°C=263\ K

Outer surface temperature of freezer

T_{o}=33+273=306\ K

We need to calculate the thickness of Styrofoam insulation

Using Fourier law,

q=\dfrac{kA}{L}(T_{o}-T_{i})

L=\dfrac{kA}{q}(T_{o}-T_{i})

Put the value into the formula

L=\dfrac{0.30\times0.85}{400}(306-263)

L=0.02741\ m

Hence, Thickness of Styrofoam insulation is 0.02741 m.

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In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude
melamori03 [73]
Answers:
(a) B_o  = 0.3466μT
(b) \lambda = 0.4488μm
(c) f = 6.68 * 10^{14}Hz

Explanation:
Given electric field(in y direction) equation:
E_y = 104sin(1.40 * 10^7 x -\omega t)

(a) The amplitude of electric field is E_o = 104. Hence

The amplitude of magnetic field oscillations is B_o =  \frac{E_o}{c}
Where c = speed of light

Therefore,
B_o =  \frac{104}{3*10^8} = 0.3466μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
\frac{2 \pi }{\lambda} = 1.40 * 10^7
\lambda =  \frac{2 \pi}{1.40} * 10^{-7}
\lambda =  0.4488μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = 6.68 * 10^{14}Hz


6 0
3 years ago
According to Newton’s first law of motion, when will an object at rest begin to move?
Natasha_Volkova [10]

Newton's first law of motion says something like "An object remains
in constant, uniform motion until acted on by an external force".

Constant uniform motion means no change in speed or direction.
If an object changes from rest to motion, that's definitely a change
of speed.  So it doesn't remain in the state of constant uniform
motion (none) that it had when it was at rest, and that tells us
that an external force must have acted on it.

8 0
4 years ago
Read 2 more answers
if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu
RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

F = \frac{k|q_1||q_2|}{r^2}

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N

Therefore, the mutual force between the two point charges is 319.64 N

4 0
3 years ago
A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force resisting the carts moti
-BARSIC- [3]

The force applied by the man is 60 N

Explanation:

We can solve this problem by applying Newton's second law, which states that:

\sum F = ma (1)

where

\sum F is the net force acting on the child+cart

m is the mass of the child+cart system

a is their acceleration

In this problem, we have:

m = 30.0 kg is the mass

a=1.50 m/s^2

And there are two forces acting on the child+cart system:

  • The forward force of pushing, F
  • The force resisting the cart motion, R = 15.0 N

Therefore we can write the net force as

\sum F = F -R

where R is negative since its direction is opposite to the motion

So eq.(1) can be rewritten as

F-R=ma

And solving for F,

F=ma+R=(30.0)(1.50)+15.0=60 N

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

4 0
4 years ago
Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box o
Lina20 [59]

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

\Sigma F = F - F' = M\cdot a

Box with mass 2M

\Sigma F = F' - F'' = 2\cdot M \cdot a

Box with mass 3M

\Sigma F = F'' = 3\cdot M \cdot a

On the third equation, acceleration can be modelled in terms of F'':

a = \frac{F''}{3\cdot M}

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

F' = 2\cdot M \cdot a + F''

F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''

F' = \frac{5}{3}\cdot F''

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)

F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''

F = 2\cdot F''

F'' = \frac{1}{2}\cdot F

Afterwards, F' as function of the external force can be obtained by direct substitution:

F' = \frac{5}{6}\cdot F

The net forces of each block are now calculated:

Box with mass M

M\cdot a = F - \frac{5}{6}\cdot F

M\cdot a = \frac{1}{6}\cdot F

Box with mass 2M

2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F

2\cdot M \cdot a = \frac{1}{3}\cdot F

Box with mass 3M

3\cdot M \cdot a = \frac{1}{2}\cdot F

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

8 0
4 years ago
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