Answer:
The car strikes the tree with a final speed of 4.165 m/s
The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m
Explanation:
First we need to calculate the initial speed 
Once we have the initial speed, we can isolate the final speed from following equation:
Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.
To do that, we replace 62 m in the first formula, as follows:
Answer:
B) Pressure on the scale, not registered as weight.
Explanation:
This is because energy (derived from weight) becomes compiled on the tips of your toes, and therefore does not increase your weight, but simply the pressure at a smaller point
Answer:
The crate was being lifted by a height of 1.48 meters.
Explanation:
In an attempt o move a crate;
Force applied = 2470 N
Work done by the force = 3650 J
We know that the work done is defined as the force used to move an object to a distance.
Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.
Work done is defined as:
Work = Force*distance covered in the direction of the force
3650 = 2470*distance
distance = 3650/2470
distance = 1.48 meters
Answer:
B) Diphosphorus pentoxide
Explanation:
"<em>F = dP/dt. </em> The net force acting on an object is equal to the rate at which its momentum changes."
These days, we break up "the rate at which momentum changes" into its units, and then re-combine them in a slightly different way. So the way WE express and use the 2nd law of motion is
"<em>F = m·A.</em> The net force on an object is equal to the product of the object's mass and its acceleration."
The two statements say exactly the same thing. You can take either one and work out the other one from it, just by working with the units.
