Answer:
1.99 parsecs.
Explanation:
We have been given that the most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-years.
We know that one light year equals to 0.306601 parsecs. To convert 6.5 light-years to parsecs, we will multiply 0.306601 by 6.5.
Therefore, Luhman 16 is approximately 1.99 parsecs away from the Earth.
We know, weight = mass * gravity
10 = m * 9.8
m = 10/9.8 = 1.02 Kg
Now, Let, the gravity of that planet = g'
g' = m/r² [m,r = mass & radius of that planet ]
g' = M/10 / (1/2R)² [M, R = mass & radius of Earth ]
g' = 4M / 10R²
g' = 2/5 * M/R²
g' = 2/5 * g
g' = 2/5 * 9.8
g' = 3.92
Weight on that planet = planet's gravity * mass
W' = 3.92 * 1.02
W' = 4 N
In short, Your Answer would be 4 Newtons
Hope this helps!
17
What would the scale read? zero
18 In free fall you are being pulled by a gravity. "Truly" weightless presumably happens in deep space where there is nothing to pull you.
19 coasters accelerate down to simulate weight loss/zeroised. As do NASA planes,
Roller coasters are for fun seekers. NASA is for science
Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
█ Question <span>█
</span><span>In an electronic transition, an atom cannot emit what?
</span>█ Answer █
When an electronic transition is occurring, an atom cannot emit ultra-violet light.
<span>Hope that helps! ★ <span>If you have further questions about this question or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia</span></span>
