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aksik [14]
3 years ago
12

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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1. In a(n) ______________________ circuit, all parts are connected one after another along one path.
poizon [28]

Answer:

1. A <em>series circuit </em>is a closed circuit which has all loads connected in a row and there is only one path for the current to flow.

2. The <em>Ohm's Law </em>state that a current flow through a resistor is directly proportional to the voltage across it R = \frac{V}{I}

3. A <em>parallel circuit </em>is a closed circuit divided into branches that it has two o more paths for the current to flow and the loads are parallel to each other which mean the voltage across them is the same for all loads.

3 0
3 years ago
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A bike travels 4 miles in half an hour at what speed
Sloan [31]
8 miles per hour
because if it is moving at 4 miles every half hour that means you have to multiply it by two to get it in miles per hour and we all know that 4 times 2 is 8 so it would be 8 miles per hour =)
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3 years ago
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3. A sprinter leaves the starting blocks with an acceleration of 4.5 m/s2. What is the
UkoKoshka [18]

Hi there! :)

\large\boxed{v_{f} = 18 m/s}

Use the following kinematic equation to solve for the final velocity:

v_{f} = v_{i} + at

In this instance, the runner started from rest, so the initial velocity is 0 m/s. We can rewrite the equation as:

v_{f} = at

Plug in the given acceleration and time:

v_{f} = 4.5 * 4 = 18 m/s

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3 years ago
Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?
shtirl [24]

Answer:

L=6.21m

Explanation:

For the simple pendulum problem we need to remember that:

\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0,

where \theta is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

\omega^{2}=\frac{g}{L},

where \omega is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

\omega=\frac{2\pi}{T},

where T is the oscillation period. Now, we can easily solve for L:

(\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m

3 0
3 years ago
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