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3 years ago

5

An electric iron is used to remove wrinkles from clothing. The electric iron is made of metal. Which property of this substance

is most useful for this application?
A.
boiling point
B.
density
C.
electrical conductivity
D.
thermal conductivity

1 answer:

Hoochie [10]3 years ago

7 0

Answer:

d

Explanation:

the heat can move through the metal and onto the clothing

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The density of ice is 0.93 g/cm3. what is the volume, in cm3, of a block of ice whose mass is 5.00 kg? remember to select an ans

serious [3.7K]

The Volume of the ice block is 5376.344 cm^3.

The density of a material is define as the mass per unit volume.

Here, the density of ice given is 0.93 g/cm^3

Mass of the ice block given is 5 kg or 5000 g

Now calculate the volume of the ice block

density=mass/volume

0.93=5000/Volume

Volume =5376.344 cm^3

Therefore the volume of ice block is 5376.344 cm^3

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A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m3 , undergoes a constant-pressure expansion at

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Answer:

Work: 4.0 kJ, heat: 4.25 kJ

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For a gas transformation at constant pressure, the work done by the gas is given by

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$p = 2 bar = 2\cdot 10^5 Pa$ is the pressure

$V_i = 0.1 m^3$ is the initial volume

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Substituting,

$W=(2\cdot 10^5)(0.12-0.10)=4000 J = 4.0 kJ$

The 1st law of thermodynamics also states that

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Q is the heat absorbed by the gas

Here we know that

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Therefore we can re-arrange the equation to find the heat absorbed by the gas:

$Q=\Delta U + W = 0.25 kJ + 4.0 kJ = 4.25 kJ$

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3 years ago

What distance does electromagnetic radiation travel in 55.0 μs ?

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Electromagnetic radiation is an energy that is known as light. so electromagnetic radiation will have the same speed as the speed of light which is 3 x 10^8 m/s. so the distance it travel at 55 x 10^-6 s is:
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What is the atomic mass of an element?

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3 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0

3 years ago