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
Glad to help you and if you need anything else on brainly let me know so I can elp you again have an AWESOME!!! :^)
Answer:
W = 0 J
Explanation:
The amount of work done by gas at constant pressure is given by the following formula:
where,
W = Work done by the gas
P = Pressure of the gas
ΔV = Change in the volume of the gas
Since the volume of the gas is constant. Therefore, there is no change in the volume of the gas:
<u>W = 0 J</u>
Answer: 810 J
Explanation: work W = F·s = 101 N · 8.0 m = 808 J
Force F = u mg = 101 N in which u is friction constant. Also mass is included in force.
Answer:
1069.38 gallons
Explanation:
Let V₀ = 1.07 × 10³ be the initial volume of the gasoline at temperature θ₁ = 52 °F. Let V₁ be the volume at θ₂ = 97 °F.
V₁ = V₀(1 + βΔθ) β = coefficient of volume expansion for gasoline = 9.6 × 10⁻⁴ °C⁻¹
Δθ = (5/9)(97°F -52°F) °C = 25 °C.
Let V₂ be its final volume when it cools to 52°F in the tank is
V₂ = V₁(1 - βΔθ) = V₀(1 + βΔθ)(1 - βΔθ) = V₀(1 - [βΔθ]²)
= 1.07 × 10³(1 - [9.6 × 10⁻⁴ °C⁻¹ × 25 °C]²)
= 1.07 × 10³(1 - [0.024]²)
= 1.07 × 10³(1 - 0.000576)
= 1.07 × 10³(0.999424)
= 1069.38 gallons
bonsoir, je peux vous aider avec les maths laissez-moi comprendre merci
