At the north pole:
9.832 m/s²
32.258 ft/s²
At the equator:
9.780 m/s²
32.088 ft/s²
It's about 0.532% greater at the pole than at the equator.
A large person who weighs 200 pounds on the equator
would weigh 1 pound 1 ounce more at the pole.
(Even more if he was properly dressed for it !)
If the net force acting on a moving object causes no change in its velocity, the object's momentum will stay the same.
<h3>What is momentum?</h3>
Momentum of a body in motion refers to the tendency of a body to maintain its inertial motion.
The momentum is the product of its mass and velocity.
This suggests that if the net force acting on a moving object causes no change in its velocity, the momentum of the object will remain the same.
Therefore, if the net force acting on a moving object causes no change in its velocity, the object's momentum will stay the same.
Learn more about momentum at: brainly.com/question/13554527
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The atomic number tells you the number of protons in an atom. This value never changes because the number of protons in the nucleus always remains constant. The mass number tells you the number of protons and neutrons (or nucleons) together: protons + neutrons = mass number. Since the number of neutrons in the nucleus varies, you can have different amounts of neutrons in the same type of atom. These varied types are called isotopes.
Hope this helps.
Answer:
d. 6.0 m
Explanation:
Given;
initial velocity of the car, u = 7.0 m/s
distance traveled by the car, d = 1.5 m
Assuming the car to be decelerating at a constant rate when the brakes were applied;
v² = u² + 2(-a)s
v² = u² - 2as
where;
v is the final velocity of the car when it stops
0 = u² - 2as
2as = u²
a = u² / 2s
a = (7)² / (2 x 1.5)
a = 16.333 m/s
When the velocity is 14 m/s
v² = u² - 2as
0 = u² - 2as
2as = u²
s = u² / 2a
s = (14)² / (2 x 16.333)
s = 6.0 m
Therefore, If the car had been moving at 14 m/s, it would have traveled 6.0 m before stopping.
The correct option is d
Answer:
See the explanation.
Explanation:
The speeds of sound depend upon the temperature of the medium. As the temperature increased in the afternoon, the speed of sound increases. Then the time taken by reflected sound will be less than 0.1 sec in the afternoon. And to hear an echo the time gap between an original sound and reflected sound must be at least 0.1 sec. That is why the student could not hear the echo at all in the afternoon.
