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solmaris [256]
4 years ago
12

Use the drop-down menus to determine which physical properties are being described. Copper wire is used to transfer electricity.

There are 19.3 grams of gold in one cubic centimeter. Iron turns to liquid at 1,534.85°C. Silver is very easy to bend. Fluorite is a green crystal.
Physics
2 answers:
Natalka [10]4 years ago
6 0

Answer:

Copper wire is used to transfer electricity. CONDUCTIVITY

There are 19.3 grams of gold in one cubic centimeter.  DENSITY

Iron turns to liquid at 1,534.85°C. MEALTING POINT

Silver is very easy to bend. HARDNESS

Fluorite is a green crystal.   COLOR

Explanation:

Just took the edgu. test

xz_007 [3.2K]4 years ago
4 0

Following are the physical properties of the given conditions:

  1. Copper wire is used to transfer electricity. This property is called Electrical Conductivity. It is defined as the property of material to conduct electricity. Copper is considered as the best conductive material.
  2. There are 19.3 grams of gold in one cubic centimeter. This statement describes the Density of gold. (Density= mass/volume) Density of material represent mass of material per unit of its volume.
  3. Iron turns to liquids at 1,534.85°C. That is the given temperature is Melting Point of iron. Melting point is the temperature where a solid starts turning into liquid.
  4. Silver is very easy to bend. This is called Flexibility of material. Flexibility is the ability of material to deform.
  5. Fluorite is a green crystal. This shows the color and structure of Fluorite. Green shows the color of material. Whereas crystal is the type of structure the material possess.
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3 years ago
1.60 kg frictionless block is attached to an ideal spring with force constant 315 N/m . Initially the spring is neither stretche
Tatiana [17]

Answer:

(a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

Explanation:

Given that,

Mass of block =1.60 kg

Force constant = 315 N/m

Speed = 13.0 m/s

(a). We need to calculate the amplitude of the motion

Using conservation of energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2

A^2=\dfrac{mv^2}{k}

Put the value into the relation

A^2=\dfrac{1.60\times13.0^2}{315}

A=\sqrt{0.858}

A=0.926\ m

(b). We need to calculate the block’s maximum acceleration

Using formula of acceleration

a=A\omega^2

a=A\times\dfrac{k}{m}

Put the value into the formula

a=0.926\times\dfrac{315}{1.60}

a=182.31\ m/s^2

(c). We need to calculate the maximum force the spring exerts on the block

Using formula of force

F=ma

Put the value into the formula

F= 1.60\times182.31

F=291.69\ N

Hence, (a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

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3 years ago
Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
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Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

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Answer:

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