amount of work done is 5880 J
Given:
mass of object = 50kg
Final height = 20m
initial height = 8m
To Find:
amount of work done
Solution:
work is done when a force acts upon an object to cause a displacement. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.
The work done by gravity is given by the formula,
W = mgh
W = 50 x 9.8 x ( 20-8)
= 5880 J
So the work done is 5880 J
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Answer:
Explanation:
This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.
If an increase in the distance of charges (let's denote with "d") causes the electric field strength (let's denote with"E") to increase, then the mathematical representation can be illustrated as d ∝ E (meaning distance of charge is directly proportional to electric field strength).
But if an increase in the distance of the charges causes the electric field strength to decrease, then the mathematical representation can be illustrated as d ∝ 1/E (meaning distance of charge is inversely proportional to electric field strength).
A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".
Larger. How big depends on the size of the vector. But think about pythagoras theorem to solve for the vector.
Answer:
m(P4) = 46.175 (grams)
m (KClO3) = 149 (grams)
Explanation:
1) n(P4) = n(P4O10);
m(P4)/M(P4) = m(P4O10)/M(P4O10);
m(P4) = M(P4)*m(P4O10)/M(P4O10)
= 123.90*105.8/283.89
= 46.175 (grams)
2) Analogously, 10n(P4O10) = 3n(KClO3)
m (KClO3) = 10M(KClO3)*m(P4O10)/3M(P4O10)
= 10*122.55*105.8/283.89/3
= 149 (grams).
The correct answer is "an attractive force" between the wires.
Let's see why. Assume we have wire A on the left and wire B on the right, and that the current in both wires go upward. First, let's find the direction of the magnetic field produced by wire A at wire B: by using the right-hand rule, we see that since the current (the thumb) goes upward, the magnetic field (given by the other fingers) at wire B is directed inside the paper.
Then we can apply again the right-hand rule to see what is the force on wire B. The index gives the direction of the current (upward), the middle finger the direction of the magnetic field (inside the paper), and the thumb gives the direction of the force: to the left, so toward wire A. This means the force is attractive. (you can re-do the procedure on wire A, and you will find the force on wire A is directed toward wire B)
