density = mass / volume.
277/38 approx 280/40 approx 7
tin or zinc. use calculator
None of the above are the name of an air mass. This is probably a trick question.
Answer:
a) For an object with mass M, in a region with a gravitational acceleration g, is:
W = M*g
Then the weight is g times the mass of the object, this means that the weight is not always equal to the mass of the object, this first statement is false.
For example, in Earth the gravitational acceleration is 9.8m/s^2
Then there is no object in Earth with a weight equal to its mass.
b) The force of tension can be the force in a piece of string that is holding an object with mass M.
The force of tension will be always pointing in the direction to the "center" of the string, then it does not push, the tension force "pulls"
The statement is false.
c) we know that the weight is:
W = M*g
This is a force, that for an object that is in the air, will pull the object back to the ground.
Suppose that we also have that object attached to a string, and the object is in the air, now the object starts to fall due to its weight and we also pull down with the string, then the total force pulling down will be the weight plus the tension of the string, then we will have a force larger (in magnitude) than the weight, which means that the statement is false.
Answer
The force on the left across the lab table.
Explanation
The Newton's third law of motion states that; <em>action and reaction are equal but a ct in opposite direction. </em>
When the block of is pulled on the right with a force of X Newtons then there is a force -X Newtons applying and equal force on the left. For every action there must be a reaction with is equal and applying in the opposite direction.
So, if the block is pulled on the right by a force of 8 N there is another equal force applying on the left.
Answer:
1. B. Increase
2. A. Decrease
Explanation:
To understand this issue, we need to put some values and using the ohm's law we can corroborate the two situations.
Ohm's law:
Now for the voltage we will use V = 110 [V], for resistance R = 10 [ohms]
Replacing the values we have:
Now let's double the voltage 220 [V]:
Therefore the current will be increased.
Let's do the same for the resistance if originally we have R = 10 [ohms]
Now let's double the resistance 20 [ohms]:
Therefore the current will be decreased.