Answer:
Explanation:
The weight of an object on Earth is given by 
, so we can calculate its mass by doing 
, which for our values is:
<em>Nothing is being asked</em> about Io but if one wanted to know the weight <em>W'</em> of the watermelon there one just have to do:
In order for Greg to safely drain the water out of the noodles, he should use potholders or any thing that is does not conduct heat or transfer heat. Some pots are also equipped with handles that are made of plastics for safely transferring of its content to another container.
Answer:
The moon is 1,079.4 mi.
Mars is 2,106.1 mi
Multiply your weight by the moon's gravity relative to earth's, which is 0.165. Solve the equation. In the example, you would obtain the product 22.28 lbs. So a person weighing 135 pounds on Earth would weigh just over 22 pounds on the moon
Being that Mars has a gravitational force of 3.711m/s2, we multiply the object's mass by this quanitity to calculate an object's weight on mars. So an object or person on Mars would weigh 37.83% its weight on earth.
Explanation:
~Hope this helps
Answer:
A. The applied force should be the same size as the friction force
Explanation:
Whenever we apply a force to an object it moves if the force applied to that object is unbalanced and there is no force or a lesser force to counter it. According to Newton's Second Law of motion, when an unbalanced force is applied to an object it produces an acceleration in the object in its own direction. So, the two forces acting on this box are the frictional force and the applied force in horizontal direction. In order to move the box at constant speed, the applied force must first, overcome the frictional force, so the object can start its motion. Since, the motion has constant velocity, it means no acceleration. So, the force must be balanced in order to avoid acceleration as a consequence of Newton's Second Law of motion. Therefore, the correction in this case will be:
<u>A. The applied force should be the same size as the friction force</u>
The equation we use is mλ=dsinθ for intensity maximas. We are given at the first maximum (m=1), it occurs at 17.8 degrees. Thus we can solve for d by substituting known values into our equation.
(1) (632.8*10^-9m)=dsin(17.8) => d = 2.07*10^-6m
Next we want to find the angle at the second maximum (m=2) so we need to solve for θ.
(2) (632.8*10^-9m) = (2.07*10^-6m)sinθ
θ=37.69 degrees
Hopes this helps!
P.S. I hope this is right. If not sorry in advance.
