walk on the moon
orbit the earth
do a spacewalk
Answer: orbit the earth
Explanation:
In 1962, with three orbits around the Earth, John Glenn is the first American Astronaut to do so.
John Glen was selected in NASA in 1959. He flew Friendship 7 mission on 20th February, 1962. He was the first American to orbit the Earth. He took three rounds about the Earth.
B) the distance between the charged particles is decreased.
C) the magnitude of the charge on both the particles is decreased.
D) the magnitude of the charge on one of the particles is decreased.
The force of repulsion between two like-charged particles will increase if the distance between the charged particles is decreased.
According to Coulomb's law: The magnitude of each of the electric forces with which two point-at-rest charges interact is directly proportional to the product of the magnitude of both charges
And inversely proportional to the square of the distance that separates them and has the direction of the line that joins them.
The formula will be given as:
We can see that the force is inversely proportional to the distance of the two charges.
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Answer:
B). the distance between the charged particles is decreased.
Explanation:
Answer:
their distance and displacment are both 400 meters. and No distance and displacement are not equal for a person who runs 2.5 times a circular track.
Explanation:
if you run in a ircle you are coming back to the starting poing ad remember distane is the total length travlled while displacement is the final- initial length
The object be going at the bottom of the hill with velocity 7.75 m/s.
When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.
Given in the question a 400 kg object is sitting at rest at the top of a hill that is 30.0 m high and 80.0 m long measured along the hill . If there is no friction, velocity with which object be going at the bottom of the hill,
θ =sin^-1(30/80)
θ =22
F = 400 g sin(22) = 150 N
F = ma
150 = 400 a
a = 0.375 m/s²
v² - u²= 2 a s
v² - 0 = 2 x 0.375 x 80
v = 7.75 m/s
The object be going at the bottom of the hill with velocity 7.75 m/s.
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