Infrared thermometers are commonly called laser thermometers true or false?

Physics

1 answer:

crimeas [40]2 years ago

6 0

Answer:

<h2>False </h2>

Explanation:

They are sometimes called laser thermometers as a laser is used to help aim the thermometer, or non-contact thermometers or temperature guns, to describe the device's ability to measure temperature from a distance. Infrared thermometers are a subset of devices known as "thermal radiation thermometers".

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How does the gravitational pull of the blue supergiants impact the direction of your star?

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Answer:

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Explanation:

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2 years ago

An object with little mass won't require a lot of force to move.

Anna35 [415]

Answer:

True

Explanation:

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A light rope is attached to a block with mass 3.60 kg that rests on a frictionless, horizontal surface. The horizontal rope pass

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Answer and Explanation:

(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope ( $F_{t}$ ), vertical gravitational force ( $F_{g}$ ) and vertical normal force ( $F_{n}$ ), due to the surface. Since there is no vertical movement, $F_{g}$ and $F_{n}$ cancels it out. So, for this block, net force is horizontal due to the rope $F_{t}$ .

The block of mass m is hanging from the pulley, so there is the force of the rope ( $F_{t}$ ) and the gravitational force ( $F_{g}$ ). Both are vertical, because there is no surface "holding" block m.

(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:

$F_{r}=m.a$

$a=\frac{F_{r}}{m}$

$a=\frac{18.8}{3.6}$

a = 5.22

The acceleration of either block is 5.22 m/s².

(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².

Suppose positive referential is going up. To determine mass:

$F_{r}=m.a$

$F_{t}-F_{g}=m.a$