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3 years ago

What causes the granular structure on the sun's photosphere?

Physics

1 answer:

Charra [1.4K]3 years ago

4 0

Something to do with how the suns magnetic field interacts with the surface plasmas I think.

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What is the kinetic energy of a bicycle with a mass of 16 kg traveling at a velocity of 5 m/s

fenix001 [56]

Answer:

200J

Explanation:

K.E=½mv²

K.E=½×16kg×5m/s²

K.E=8kg×25

K.E=200J

4 0

2 years ago

A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bum

AleksandrR [38]

Answer:

Hits per second=199 hit/s

Explanation:

#Given the angular velocity, $\omega=33\frac{1}{3} rev/min$ , radius of the record $r=0.1m$ and the distance between any two successive bumps on the groove as $d=1.75mm$ .

The linear speed of the record in meters per second is:

$v=\omega r=33\frac{1}{3}\times\frac{2\pi}{60}\times 10\times 10^{_2}\\\\=0.3843m/s\\$

#From $v$ above, if the bumps are uniformly separated by 1m, then the rate at which they hit the stylus is:

$Hits/second=v/d \ \ \ \ d=1.75mm\\\\=0.3483/0.000175\\\\=199.0385714\approx 199$

Hence the bumps hit the stylus at around 199hit/s

8 0

2 years ago

A certain light truck can go around a flat curve having a radius of 150 m with a maximum spee dof 32.0 m/s. With what maximum sp

Dvinal [7]

Answer

Maximum speed at 75 m radius will be 22.625 m /sec

Explanation:

We have given radius of the curve r = 150 m

Maximum speed $v_{max}=32m/sec$

Coefficient of friction $\mu =\frac{v_{max}^2}{rg}=\frac{32^2}{150\times 9.8}=0.6965$

Now new radius r = 75 m

So maximum speed at new radius $v_{max}=\sqrt{\mu rg}=\sqrt{0.6965\times 75\times 9.8}=22.625m/sec$

7 0

2 years ago

A space probe is coasting through space at a steady speed of 100p feet per second. the booster rocket fires for 1/2 seconds so t

aleksandr82 [10.1K]

Answer:

Acceleration: $9800 ft/s^2$

Explanation:

The acceleration of an object is equal to the rate of change of velocity:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

For the space probe in this problem, we have:

u = 100 ft/s (initial velocity)

v = 5000 ft/s (final velocity)

t = 0.5 s (time taken)

Therefore, the acceleration is

$a=\frac{5000-100}{0.5}=9800 ft/s^2$

6 0

2 years ago

The length (L) of a simple pendulum is directly proportional to the square of the time period (T). Which type of graph can best

iris [78.8K]

L=T^2 so that would be a parabola

8 0

3 years ago

Read 2 more answers