Answer:1 trip around the earth is an angular displacement of 2*pi
3.6525*10^2 days
I
Explanation:24 h/1 day * 3.600*10^3 s/1h = 3.156*10^7 s
Angular speed = angular displacement / time
Angular speed = 2*pi rads / 3.156*10^7 s = 1.9910*10^-7 rad/s
Answer:
3.42N
Explanation:
*not too sure bc i left my physics notes at school so it might not be 100% accurate :p*
Use the equation: F = (GMm)/(r^2)
F = force of gravity
G = gravitational constant (6.7x10^-11)
M = mass1 (2.5x10^30kg)
m = mass2 (1kg)
r = radius (7000m)
Plug it in: F = ((6.7x10^-11)(2.5x10^30)(1)) / (7000^2)
F = (1.675x10^20) / (4.9x10^7)
F = 3.4183673x10^12
F = 3.42N
The mass of this bag of cement in S.I. units (kg) is equal to 0.062 kilograms.
<u>Given the following data:</u>
- Mass of cement = 62 grams.
To calculate the mass of this bag of cement in S.I. units (kg):
<h3>How to convert to
S.I. units.</h3>
In Science, kilograms (kg) is the standard unit of measurement or S.I. units of the mass of a physical object. Thus, we would convert the value of the mass of this bag of cement in grams to kilograms (kg) as follows:
<u>Conversion:</u>
1000 grams = 1 kilograms.
62 grams = X kilograms.
Cross-multiplying, we have:
X = 
X = 0.062 kilograms.
Read more on mass here: brainly.com/question/13833323
Answer: there is zero kinetic energy but there is Gravitational Potential Energy (GPE) and GPE = 8826.3 J
Explanation:
Answer:
See the answers below.
Explanation:
to solve this problem we must make a free body diagram, with the forces acting on the metal rod.
i)
The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.
We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.
For the summation of forces we will take the forces upwards as positive and the negative forces downwards.
ΣF = 0

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.
ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.
ΣM = 0
![(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]](https://tex.z-dn.net/?f=%2815%2A9%29%20-%20%2818%2AW%29%20%3D%200%5C%5C135%20%3D%2018%2AW%5C%5CW%20%3D%207.5%20%5BN%5D)