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rewona [7]
3 years ago
8

Explains this: a light source can emit more than one type of light

Physics
1 answer:
Radda [10]3 years ago
6 0

Answer:

The electromagnetic spectrum comprise a lot of waves length. Usually, different waves length are called as different lights, and a light source can emit in more than a different wave length, as the sun does, for example. The sun emit the visible light, UV light, infrared, etc.

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What is the danger to spacecraft and astronauts from micrometeoroids?
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Answer:

Suppose the micrometeoroid weighed 1 g = .001 kg

Suppose also the spacecraft were moving at 18,000 mph (1.5 hrs per rev)

Usually, the smaller particle would be moving but for simplicity suppose that it were stationary wrt the ground

v = 18000 miles / hr * 1500 m/mile / 3600 sec/hr = 7500 m/s

KE = 1/2 * .001 kg * (7500 m)^2 = 28,125 Joules

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(NEED TO HELP NOWW ). While Bob is studying waves in science and he learns that the energy of a wave is directly proportional to
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Which model could represent a neutral atom of oxygen?<br> A. 2<br> B. 4<br> C. 3<br> D. 1
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Conduction is greater in solids .explain in your own words Why?
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2 years ago
A race car starting from rest accelerates uniformly at a rate of 4.90 meters per squared. What is the cars speed after it has tr
shusha [124]
Ok, we need to find a relation for the speed as it relates to the acceleration.  This is given by the integral of acceleration:

v= \int\limits^{t}_{0} {a} \, dt' =at

Where we have the initial velocity is 0m/s and a will be 4.90m/s².

But we see there is an issue now... We know the velocity as a function of time, but we don't know how long the car has been accelerating!  We need to calculate this time by now finding the position function as a function of time.  This way we can solve for the time, t, that it takes to go 200m accelerating this way and then substitute that time into our velocity equation and get the velocity. 
Position is just the integral of velocity:

s= \int\limits^{}_{} {at} \, dt = \frac{1}{2}at^2

Where the initial velocity and initial position are both zero.

Now we set this position function equal to 200m and find the time, t, it took to get there

\frac{1}{2}(a \frac{m}{s^2} )t^2=200m \\  \\ \frac{1}{2}4.90 \frac{m}{s^2} t^2=200m \\  \\ t^2= \frac{400m}{4.90 \frac{m}{s^2}}=81.63s^2 \\  \\ t= \sqrt{81.63s^2 }  =9.04s

Now let's put t=9.04s into our velocity equation:

v =at=4.9\frac{m}{s^2} \times 9.04s=44.3 \frac{m}{s}


8 0
3 years ago
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