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3 years ago

Does a light bulb at a temperature of 2500K produce as white a light as the sun at 6000K

Physics

1 answer:

sattari [20]3 years ago

7 0

Answer:

No. ... UV light causes sunburn, whereas visible light does not.

Explanation:

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What must happen to the temperature of a material for thermal expansion to occur.

Aloiza [94]

In most cases the temperature must increase for thermal expansion to occur. Most substances expand as temperature increases because the atoms or molecules vibrate faster as temperature increases and experience greater separation.

8 0

3 years ago

The Cassegrain design provides more compact (shorter) telescopes. Why? (Examine figures 2.4.2 and 2.4.3). The shorter design is

Ipatiy [6.2K]

Answer:

Because the light reflects multiple times until it gets to the Cassegrain focus.

Explanation:

The Cassegrain design can be seen in a reflecting telescope. In this type of design the light is collected by a concave mirror, and then intercepted by a secondary convex mirror, and sends it down to a central opening in the primary mirror (concave mirror), in which a detector is placed (Cassegrain focus)

Since, the light is reflected many times due to Cassegrain design, that leads to shorter telescopes.

5 0

3 years ago

A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).

just olya [345]

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles ( $\vec r_{cm}$ ), measured in meters, is defined by this weighted average:

$\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}}$ (1)

Where:

$m_{i}$ - Mass of the i-th particle, measured in kilograms.

$\vec r_{i}$ - Location of the i-th particle with respect to origin, measured in meters.

If we know that $\vec r_{cm} = (-0.500\,m,-0.700\,m)$ , $m_{1} = 1\,kg$ , $\vec r_{1} = (-1.20\,m, 0.500\,m)$ , $m_{2} = 4.50\,kg$ , $\vec r_{2} = (0.600\,m, -0.750\,m)$ and $m_{3} = 4\,kg$ , then the coordinates of the third particle are:

$(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}$

$(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})$

$(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)$

$(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)$

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

5 0

2 years ago

In which test did the air parcel rise the highest? Is there a pattern in the relationship between starting air temperature and p

wel

Answer:

xitxitx

Explanation:

ufxigxigxitcutcocyocoycitcotcitcohcohxitxigcigxigxkgxigxigxigxigxigxigxitx8y

5 0

3 years ago

Object a travels in the +x-direction before hitting a stationary object

Leto [7]

The object’s resultant angle of motion with the +x-axis after the collision is 47°

<span>From object A:

1) x-momentum is 5.7 × 10^4 kilogram meters/second,
2) y-momentum is 6.2 × 10^4 kilogram meters/second.

Now, we know, tan</span>Ф = $\frac{y}{x}$

⇒tanФ = $\frac{6.2 × 10^4 }{5.7 × 10^4}$

⇒tanФ = 1.088

⇒ Ф = $tan^{-1}$ 1.088
= 47.4 ≈ 47

8 0

3 years ago

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