Answer:
MOMENTUM
Explanation:
another way of saying getting the smallest force possible is the word " MOMENTUM".
momentum is the ability to keep maintaining,incresing or itself developing to move at constant speed or to increase the speed.
A bond with elements from B.
The vector force on the unit positive charge placed at any location in the field defines the strength of the electric field at that point. The charge used to determine field intensity (field strength) is known as the test charge. Now, a field line is defined as a line to which the previously mentioned field strength vectors are tangents at the relevant places. When we study positive charge field lines, the field strength vectors point away from the positive charge. If there is a negative charge anywhere in the vicinity, the field lines that began from the positive charge will all terminate at the negative charge if the value of the negative charge is the same as the value of the positive charge. Remember that the number of field lines originating from positive charge is proportional to the charge's value, and similarly, the number of field lines terminating at negative charge is proportionate to the charge's value. As a result, if all charges are equivalent, all lines originating from the positive charge terminate at the negative charge. If the value of the positive charge is greater than the value of the negative charge, the number of lines ending at the negative charge will be proportionally less than the number of lines beginning at the positive charge. The remaining lines that do not end at the negative charge will go to infinity. If the positive charge is less, all lines from it terminate at a negative charge, and any other reasonable number of ines terminate at a negative charge from infinity. We should also keep in mind that the number of lines that run perpendicular to the field direction across a surface of unit area is proportional to the field strength at that location. As a result, lines are dense in the strong field zone and sparse in the low intensity region.
Answer:
v = 34.128 km/hr
Explanation:
Given that,
The initial speed of a truck, u = 0
Acceleration of the truck, a = 0.3 m/s²
Distance moved, d = 150 m
Let the final speed of the truck is v. Using third equation of motion i.e.
Put all the values,
or
v = 34.128 km/h
So, the final speed of the truck is equal to 34.128 km/h.
