(a) Determine the circumference of the Earth through the equation,
C = 2πr
Substituting the known values,
C = 2π(1.50 x 10¹¹ m)
C = 9.424 x 10¹¹ m
Then, divide the answer by time which is given to a year which is equal to 31536000 s.
orbital speed = (9.424 x 10¹¹ m)/31536000 s
orbital speed = 29883.307 m/s
Hence, the orbital speed of the Earth is ~29883.307 m/s.
(b) The mass of the sun is ~1.9891 x 10³⁰ kg.
Less because the ramp is letting off force but i does depend on the way you are going on the ramp
Answer:
a = 120 m/s²
Explanation:
We apply Newton's second law in the x direction:
∑Fₓ = m*a Formula (1)
Known data
Where:
∑Fₓ: Algebraic sum of forces in the x direction
F: Force in Newtons (N)
m: mass (kg)
a: acceleration of the block (m/s²)
F = 1200N
m = 10 kg
Problem development
We replace the known data in formula (1)
1200 = 10*a
a = 1200/10
a = 120 m/s²
Here's the equation you use: Density = mass/volume
1) 5.2g/cm^3 = m/3.7cm^3
2) m = 5.2g/cm^3 x 3.7cm^3
3) m = 19.24g
You can check the answer by plugging it in
19.24g/3.7cm^3
= 5.2g/cm^3
Single replacement would be represented by a single element being replaced.
This is shown in answer choice B
Where the positions of A and B are swapped