Answer:
the first one is D
Explanation:
so if the others u put are right the the second would be c
Answer: A negatively charged particle that is found in atoms.
Answer: This is true
Explanation: v2=800/25= 32m/s
Answer:
ExplanaThis is a way of measuring how much gravity there is. The formula is: weight/mass = gravitational field strength.
Gravitational field strength = Weight/mass unit is N/kg
Weight = mass x gravitational field strength unit is N
On Earth the gravitational field strength is 10 N/kg. Other planets have different gravitational field strengths. The Moon has a gravitational field strength of 1.6 N/kg. You might have seen films of astronauts leaping high on the moon.
Here on Earth, if I jump I am pulled back to ground by gravity. What is my weight? My mass is 80kg and if we multiply by gravitational field strength (10N/kg) - my weight is 800N. Now if I go to the moon, my mass will be the same, 80kg. We multiply that by the moon's gravitational field strength, which is 1.6 N/ kg. That means my weight on the moon is 128N. So I have different weights on the Earth and on the Moon. That's why astronauts can jump high into the air on the moon - they're lighter up there.
Jupiter is a very large planet with strong gravitational field strength of 25 N/ kg. My body is 80kg. If I go to Jupiter my weight is going to be 25 x 80 = 2,000 N. That means I wouldn't be able to get off the ground or stand up straight! I would probably be lying down all the time there. So weight varies depending on which planet you are on. You can find out more yourself by looking up tables of weight on different planets.tion:
pls brainlieste
Answer:
Explanation:
Given:
mass of first particle, 
mass of second particle, 
mass of third particle, 
coordinate position of first particle in meters, 
coordinate position of second particle in meters, 
coordinate position of third particle in meters, 
<u>Now, gravitational force on particle 3 due to particle 1:</u>
towards positive Y axis.
<u>gravitational force on particle 3 due to particle 2:</u>
towards positive X axis.
<u>Now the net force</u>
<em>For angle in counterclockwise direction from the +x-axis</em>
