Darker, cooler areas on the sun's surface are called sunspots.
Hope that helped =)
Acceleration is the rate of change of velocity, so can be calculated by:
a = v2 - v1 / t where v2 and v1 are the final and initial velocities of the object respectively and t is the time taken. Therefore the acceleration is:
a = 30 m/s - 20 m/s / 2.0 s = 10 m/s / 2.0 s = 5 m/(s^2)
Hope this helps!
Answer:
Accretion
Explanation:
Accretion is the process by which there is an accumulation of particles into a bigger object by attracting more mass by gravitational force into an accretion disk. This is one of the first steps in the formation of our solar system. There was a collapse of a gas cloud which resulted in most of the mass collecting in the center leading to the formation of the sun and the rest spread out forming the planets.
Lower mass: 1.20 kg, upper mass: 1.28 kg
Explanation:
In order to solve the problem, we consider the forces acting on the upper mass only first.
The upper mass is acted upon three forces:
- The applied force
, upward - The weight of the mass itself,
, where
is the upper mass and
is the acceleration of gravity, downward - The tension in the string,
, downward
Therefore, the equation of the forces for the upper mass is:

where
is the acceleration (upward)
Solving for
,

Now we can find the lower mass by considering the forces acting on it:
- The tension in the string, T = 16 N, upward
- The weight of the mass itself,
, where
is the lower mass, downward
So the equation of the forces is

And solving for the mass,

Learn more about acceleration and forces:
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Answer:
The current in the second loop will stay constant
Explanation:
Since the induced emf in the second coil, ε due to the changing current i₁ in the first wire loop ε = -Mdi₁/dt where M = mutual inductance of the coils and di₁/dt = rate of change of current in the first coil = + 1 A/s (positive since it is clockwise)
Now ε = i₂R where i₂ = current in second wire loop and R = resistance of second wire loop.
So, i₂R = -Mdi₁/dt
i₂ = -Mdi₁/dt/R
Since di₁/dt = + 1 A/s,
i₂ = -Mdi₁/dt/R
i₂ = -M × + 1 A/s/R
i₂ = -M/R
Since M and R are constant, this implies that i₂ = constant
<u>So, the current in the second wire loop will stay constant.</u>