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san4es73 [151]
3 years ago
11

Which part(s) of the electromagnetic spectrum are visible to humans?

Physics
1 answer:
jarptica [38.1K]3 years ago
4 0
Color aka the visible light spectrum
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Maximum current problem. If the current on your power supply exceeds 500 mA it can damage the supply. Suppose the supply is set
VikaD [51]

To solve this problem we will apply Ohm's law. The law establishes that the potential difference V that we apply between the ends of a given conductor is proportional to the intensity of the current I flowing through the said conductor. Ohm completed the law by introducing the notion of electrical resistance R. Mathematically it can be described as

V = IR \rightarrow R = \frac{V}{I}

Our values are

I = 500mA = 0.5A

V = 37V

Replacing,

R = \frac{V}{I}

R = \frac{37}{0.5}

R = 74 \Omega

Therefore the smallest resistance you can measure is 74 \Omega

8 0
3 years ago
How long does it take an automobile traveling 66.7 km/h to become even with a car that is traveling in another lane at 52.7 km/h
tresset_1 [31]

Answer:

The  time taken is  t =  32.5 \  s

Explanation:

From the question we are told that

   The  speed  of  first car is  v_1  =  66.7 \ km/h  =  18.3 \  m/s

    The  speed  of  second car is v_2  =  52.7 \ km/h  =  14.64 \  m/s

   The  initial distance of separation is  d =  119 \ m

The distance covered by first car is mathematically represented as

     d_t =  d_i  +  d_f

Here  d_i is the initial distance which is  0 m/s

  and  d_f  is the final distance covered which is  evaluated as d_f  =  v_1 * t

So

     d_t =  0 \  m/s  +  (v_1 * t )

     d_t =  0 \  m/s  +  (18.3 * t )

The distance covered by second  car is mathematically represented as

     d_t =  d_i  +  d_f

Here  d_i is the initial distance which is  119 m

  and  d_f  is the final distance covered which is  evaluated as d_f  =  v_2* t

       d_t =  119  + 14.64 *  t

Given that the two car are now in the same position we have that

    119  + 14.64 *  t  =   0   +  (18.3 * t )

   t =  32.5 \  s

6 0
3 years ago
The density of atmosphere (measured in kilograms/meter3) on a certain planet is found to decrease as altitude increases (as meas
alexgriva [62]

Answer:

B.  inverse plot, 0.51 kilograms/meter3

Explanation:

First of all, we note that the relationship between the altitude and the atmospheric density is an inverse relationship. In fact, an inverse relationship is a relationship between the x-variable and the y-variable of the form

y \propto \frac{1}{x}

Therefore, as the x increases, the y decreases, and as the x decreases, they increases. This is exactly what occurs with the altitude and the atmospheric density in this plot: as the altitude increases, the density decreases, and vice-versa.

Moreover, we can infer the value of the atmospheric density at an altitude of 1,291 km. This point is located between point A (2550 km) and point B(1000 km), so the density must have a value between 0.30 kg/m^3 and 0.54 kg/m^3, so the correct choice is

B.  inverse plot, 0.51 kilograms/meter3


5 0
3 years ago
1. Two forces act on a box as follows: F= 100 N at 02 = 170. and F2= 75 N at
Setler [38]

Answer:

yfv.

Explanation:

7 0
3 years ago
Will Mark Brainliest if Correct PLZ!!!!! A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on
qaws [65]

Answer:

≅50°

Explanation:

We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:

Δx=V₀t+at²/2

And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

  • Δx=760 m
  • V₀=87 m/s
  • t=13.6 s
  • a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!

With that we can plug the values in to get:

760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}

760=(1183.2)(cos\theta)

cos\theta=\frac{760}{1183.2}

\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}

3 0
3 years ago
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