We learned that We are in the disk of the Galaxy, about 5/8 of the way from the center.
<h3>What is the work of Harlow Shapley?</h3>
Shapley, who was headquartered in Boulder, Colorado, used Cepheid variable stars to estimate the size of the Milky Way Galaxy and its position relative to the Sun. In 1953, he published his "liquid water belt" theory, today known as the concept of a livable zone.
There are many stars, grains of dust, and gas in the Milky Way. It is known as a spiral galaxy because, from the top or bottom, it would appear to be whirling like a pinwheel. About 25,000 light-years from the galaxy's nucleus, the Sun is situated on one of the spiral arms.
Approximately 5/8 of the way from the galaxy's nucleus, we are in the disc. William Herschel believed that the Sun and Earth were about in the middle of the vast cluster of stars known as the Milky Way.
To learn more about Harlow Shapley's original estimate go to - brainly.com/question/28145909
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Answer:
B = 4.059 x 10¹⁵ T
Explanation:
Given,
Number of loop, N = 400
radius of loop, r = 0.65 x 10⁻¹⁵ m
Current, I = 1.05 x 10⁴ A
Magnetic field at the center of the loop
B = 4.059 x 10¹⁵ T
The direction is the same as the direction the ball is moving in. Because of the rolling of the ball, the direction of movement of the surface of the ball is opposite the overall direction of the ball. Since friction will oppose the direction of movement of the surface of the ball, it is in the same direction as the net direction of movement of the ball.
Answer:
112 m/s², 79.1°
Explanation:
In the x direction, given:
x₀ = 0 m
x = 19,500 cos 32.0° m
v₀ = 1810 cos 20.0° m/s
t = 9.20 s
Find: a
x = x₀ + v₀ t + ½ at²
19,500 cos 32.0° = 0 + (1810 cos 20.0°) (9.20) + ½ a (9.20)²
a = 21.01 m/s²
In the y direction, given:
y₀ = 0 m
y = 19,500 sin 32.0° m
v₀ = 1810 sin 20.0° m/s
t = 9.20 s
Find: a
y = y₀ + v₀ t + ½ at²
19,500 sin 32.0° = 0 + (1810 sin 20.0°) (9.20) + ½ a (9.20)²
a = 109.6 m/s²
The magnitude of the acceleration is:
a² = ax² + ay²
a² = (21.01)² + (109.6)²
a = 112 m/s²
And the direction is:
θ = atan(ay / ax)
θ = atan(109.6 / 21.01)
θ = 79.1°
M = mass of the whale = 1000 kg
m = mass of the seal = 200 kg
V = initial velocity of whale before collision with the seal = 6.0 m/s
v = initial velocity of the seal before collision with the whale = 0 m/s
V' = final velocity of two sea creatures after collision = ?
Using conservation of momentum
M V + m v = (M + m) V'
inserting the above values in the equation
(1000 kg) (6.0 m/s) + (200 kg) (0 m/s ) = (1000 kg + 200 kg) V'
6000 kgm/s + 0 kgm/s = (1200 kg) V'
V' = (6000 kgm/s ) /(1200 kg)
V' = 5 m/s
