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lana [24]
2 years ago
14

Sunspots appear dark because they are cooler than the surrounding solar surface. True or false

Physics
1 answer:
DiKsa [7]2 years ago
6 0

Answer:

False, Sunspots appear dark (in visible light) due to their low temperature(cooler) than rest of the sun

Explanation:

Sunspots appear dark because they are much cooler( have low temperature than the rest of the surface contained by Sun. As they appear dark, but still they have very temperature that's why so hot. Sunspots have temperatures ranges  3,500 Celsius (3773 kelvin) and the surrounding surface of the sun has a temperature much higher of about 5,500 Celsius(5773 Kelvin). Even if we see a sunspot alone in space,  it will glow so brightly.

Learn more about sunspots :

brainly.com/question/27774496

#SPJ4

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An 8.00- W resistor is dissipating 100 watts. What are the current through it and the difference of potential across it?
hjlf

Answer:

I= 3.5 amps

Explanation:

Step one:

given data

rating of resistor R= 8 ohms

power P= 100W

Required

The current I

Step two

Yet this power is also given by

P = I^2R

make I subject of the formula we have

I= \sqrt{\frac{P}{R} }

substitute

I= \sqrt{\frac{100}{8} }\\\\I=\sqrt{12.5}\\\\I= 3.5 amps

8 0
3 years ago
Not sure if it went through last time. Please help asap!
Olin [163]
The equation for force is F=ma. Because we have the value of mass (0.42 kg) and the acceleration (14.8 m/s^2), simply plug them into the equation for force to get
0.42 \times 14.8 = 6.22
The answer is 6.22 N because newtons are the unit used to measure force.
8 0
3 years ago
Read 2 more answers
A block is resting on a platform that is rotating at an angular speed of 2.4 rad/s. The coefficient of static friction between t
Sloan [31]

Answer:

r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.

Explanation:

From a sum of forces:

Ff = m*a   where Ff = μ * N    and a = \frac{V^2}{r}=\omega^2*r

N - m*g = 0   So, N = m*g.   Replacing everything on the original equation:

\mu*m*g = m*\omega^2*r   (eq2)

Solving for r:

r = \frac{\mu*g}{\omega^2}=1.41m

If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.

4 0
3 years ago
A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre
Brut [27]

Answer:

The time taken is  t =  0.356 \ s

Explanation:

From the question we are told that

  The length of steel the wire is  l_1  = 31.0 \ m

   The  length of the  copper wire is  l_2  = 17.0 \ m

    The  diameter of the wire is  d =  1.00 \ m  =  1.0 *10^{-3} \ m

     The  tension is  T  =  122 \ N

     

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

              t  =  t_s  +  t_c

Where  t_s is the time taken to transverse the steel wire which is mathematically represented as

         t_s  = l_1 *  [ \sqrt{ \frac{\rho * \pi *  d^2 }{ 4 *  T} } ]

here  \rho_s is the density of steel with a value  \rho_s  =  8920 \ kg/m^3

   So

      t_s  = 31 *  [ \sqrt{ \frac{8920 * 3.142*  (1*10^{-3})^2 }{ 4 *  122} } ]

      t_s  = 0.235 \ s

 And

        t_c is the time taken to transverse the copper wire which is mathematically represented as

      t_c  = l_2 *  [ \sqrt{ \frac{\rho_c * \pi *  d^2 }{ 4 *  T} } ]

here  \rho_c is the density of steel with a value  \rho_s  =  7860 \ kg/m^3

 So

      t_c  = 17 *  [ \sqrt{ \frac{7860 * 3.142*  (1*10^{-3})^2 }{ 4 *  122} } ]

      t_c  =0.121

So  

   t  = t_c  + t_s

    t =  0.121 + 0.235

    t =  0.356 \ s

4 0
3 years ago
Rainwater draining from a neighborhood street initially travels at 4 ft/s through a pipe with a cross-sectional area of 15.7 ft2
Fudgin [204]

Answer:

The  velocity  is v_2  =  0.96 \ ft/s

Explanation:

From the question we are told that

   The initial speed is  v_1  =  4 \ ft/s

   The  cross -sectional area of the first pipe is  A_1  =  15.7 \ ft

   The  cross -sectional area of the second pipe is A_2 =  65.4 \  ft

Generally from continuity equation we have that

     A_1 * v_1 =  A_2  * v_2

So  

     v_2  =  \frac{A_1 *  v_1  }{A_2 }

=>   v_2  =  \frac{15.7  *  4  }{65.4 }

=>   v_2  =  0.96 \ ft/s

6 0
4 years ago
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