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3 years ago

15. What is a thermograph?

Physics

2 answers:

Genrish500 [490]3 years ago

8 0

Answer: A device that uses infrared sensors.

Explanation:

Anni [7]3 years ago

8 0

Answer:

a device that uses infrared sensors to create pictures that show variations in temperature

Explanation:

Gradpoint

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Compare the forces in a small nucleus to the forces in a large nucleus

pochemuha

The comparison of the forces in a small nucleus to the forces of a large one is the fact that they are capable of holding the protons and neutrons which made it no matter what their size may be. Therefore, as long as there is a nucleus, their forces can both hold together the two atoms tight.

6 0

3 years ago

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

7 0

3 years ago

Compare physical changes and chemical changes

JulijaS [17]

Answer:

The difference between a physical reaction and a chemical reaction is composition. In a chemical reaction, there is a change in the composition of the substances in question; in a physical change there is a difference in the appearance, smell, or simple display of a sample of matter without a change in composition. Although we call them physical "reactions," no reaction is actually occurring. In order for a reaction to take place, there must be a change in the elemental composition of the substance in question. Thus, we shall simply refer to physical "reactions" as physical changes from now on.

Explanation:

Physical changes are limited to changes that result in a difference in display without changing the composition. Some common changes (but not limited to) are:

Texture

Color

Temperature

Shape

Change of State (Boiling Point and Melting Point are significant factors in determining this change.)

Physical properties include many other aspects of a substance. The following are (but not limited to) physical properties.

Luster

Malleability

Ability to be drawn into a thin wire

Density

Viscosity

Solubility

Mass

Volume

4 0

3 years ago

URGANT!!!!!!! NEED ANSWERS TODAY!!!!

natima [27]

2H2 + O2 = 2H2O + Energy

5 0

2 years ago

A diverging lens (f1 = -12.0cm) is located 50.0 cm to the left of a converging lens (f2 = 34.0 cm). A 2.0 cm-tall object stands

pishuonlain [190]

Answer:

The final image relative to the converging lens is 34 cm.

Explanation:

Given that,

Focal length of diverging lens = -12.0 cm

Focal length of converging lens = 34.0 cm

Height of object = 2.0 cm

Distance of object = 12 cm

Because object at focal point

We need to calculate the image distance of diverging lens

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-12}$

$v=\infty$

The rays are parallel to the principle axis after passing from the diverging lens.

We need to calculate the image distance of converging lens

Now, object distance is ∞

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{34}-\dfrac{1}{\infty}$

$v=34$

The image distance is 34 cm right to the converging lens.

Hence, The final image relative to the converging lens is 34 cm.

5 0

3 years ago