<span># of protons + # of neutrons = atomic mass</span>
Answer:
I thinl it is I'm not sure though
Answer:
gravitational field strength (g) is measured in newtons per kilogram (N/kg)
When we jump from the truck and accelerate towards the earth surface, the earth also accelerates towards us but it's acceleration is very negligible.
To find the answer, we need to know about the acceleration of earth due to the gravitational attraction.
<h3>What's the gravitational force between the earth and a person?</h3>
- Gravitational attraction force is GMm/r² between the earth and a person.
- M= mass of the earth
m= mass of the person
r= separation between them.
<h3>What's the acceleration of the earth towards the person when he jumps from a truck?</h3>
- According to Newton's second law, Force = M×acceleration
- Acceleration= Force / M
- Here, Force = GMm/r²,
so acceleration of earth= Gm/r²
- As this acceleration is very small, so we can't notice it.
Thus, we can conclude that the earth also accelerates towards us.
Learn more about the gravitational force here:
brainly.com/question/72250
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Answer:
a) w = 7.27 * 10^-5 rad/s
b) v1 = 463.1 m/s
c) v1 = 440.433 m/s
Explanation:
Given:-
- The radius of the earth, R = 6.37 * 10 ^6 m
- The time period for 1 revolution T = 24 hrs
Find:
What is the earth's angular speed?
What is the speed of a point on the equator?
What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?
Solution:
- The angular speed w of the earth can be related with the Time period T of the earth revolution by:
w = 2π / T
w = 2π / 24*3600
w = 7.27 * 10^-5 rad/s
- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.
v1 = R*w
v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)
v1 = 463.1 m/s
- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
π/2 ........... s
x ............ 1/5 s
x = π/2*5 = 18°
- The radius of the earth R' at point where θ = 18° from the equator is:
R' = R*cos(18)
R' = (6.37 * 10 ^6)*cos(18)
R' = 6058230.0088 m
- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.
v2 = R'*w
v2 = (6058230.0088)*(7.27 * 10^-5)
v2 = 440.433 m/s