Answer:
1.029
Explanation:
1.0090 can also be looked at as "1.009"
0.02 can also be looked at as "0.020"
I think of it as 20+9 which is 29. There for your answer should be 1.029. There are no measurement rules applying to this equation since they are both in centimeters. So you don't have to convert anything.
W-APE. For example, work W done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative APE. There must be a minus sign in front of APE to make W positive. PE can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.
( The capital A’s in the words are supposed to be triangles ! I also hoped this helped ! Please mark me as brainliest !! )
To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.
Planet gravitational force
Distance between planet and star
Gravitational force is
Applying the new distance,
Replacing with the previous force,
Replacing our values
Therefore the magnitude of the force on the star due to the planet is 
Answer:
1923 N
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 65 Kg
Radius (r) = 2.5 m
Velocity (v) = 8.6 m/s
Centripetal force (F) =?
The centripetal force, F, can be obtained by using the following formula:
F = mv²/r
F = 65 × 8.6² / 2.5
F = 65 × 73.96 / 2.5
F = 4807.4 / 2.5
F = 1922.96 ≈ 1923 N
Thus, the magnitude of the centripetal's force acting on the student is approximately 1923 N
Answer:
Option C
Explanation:
Answer C is the correct option. water can be written as H₂O, which means that there are 2 Hydrogen atoms for every oxygen atom, therefore it will occupy more space than oxygen and push more. there is also one more possibility, if the splitting takes place in Hoffman's Voltameter then the Hydrogen will be close to the cathode as hydrogen is positive. Otherwise, option C is correct answer. Hope this Helps you!
