My answer -
the corona,
the sun's outer layer, reaches temperatures of up to 2 million degrees
Fahrenheit (1.1 million Celsius). At this level, the sun's gravity can't
hold on to the rapidly moving particles, and it streams away from the
star.
The sun's activity shifts over the course of its 11-year cycle, with
sun spot numbers, radiation levels, and ejected material changing over
time. These alterations affect the properties of the solar wind,
including its magnetic field properties, velocity, temperature and
density. The wind also differs based on where on the sun it comes from
and how quickly that portion is rotating.
The velocity of the solar wind
is higher over coronal holes, reaching speeds of up to 500 miles (800
kilometers) per second. The temperature and density over coronal holes
are low, and the magnetic field is weak, so the field lines are open to
space. These holes occur at the poles and low latitudes, and reach their
largest when activity on the sun is at its minimum. Temperatures in the
fast wind can reach up to 1 million degrees F (800,000 C).
At the coronal streamer belt around the equator, the solar wind travels
more slowly, at around 200 miles (300 km) per second. Temperatures in
the slow wind reach up to 2.9 million F (1.6 million C).
p.s
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Assume the snow is uniform, and horizontal.
Given:
coefficient of kinetic friction = 0.10 = muK
weight of sled = 48 N
weight of rider = 660 N
normal force on of sled with rider = 48+660 N = 708 N = N
Force required to maintain a uniform speed
= coefficient of kinetic friction * normal force
= muK * N
= 0.10 * 708 N
=70.8 N
Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.
The force of gravity is the only force that keeps a pendulum in motion. both the force increases the speed of the pendulum on the downswing and decreases it's speed on its upswing.
Answer:
16.4287
Explanation:
The force and displacement are related by Hooke's law:
F = kΔx
The period of oscillation of a spring/mass system is:
T = 2π√(m/k)
First, find the value of k:
F = kΔx
78 N = k (98 m)
k = 0.796 N/m
Next, find the mass of the unknown weight.
F = kΔx
m (9.8 m/s²) = (0.796 N/m) (67 m)
m = 5.44 kg
Finally, find the period.
T = 2π√(m/k)
T = 2π√(5.44 kg / 0.796 N/m)
T = 16.4287 s