The simplest way to do this is to set up equivalent fractions, like this-
=
Solve for x by using cross multiplication.
40*2.2= 88
1*x=88
x=88
Therefore, the boy weighs 88lbs.
Answer:
Displacement: 2.230 km Average velocity: 1.274
Explanation:
Let's represent displacement by the letter S and the displacement in direction 49.7° as A. Displaement is a vector, so we need to decompose all the bird's displacement into their X-Y compoments. Let's go one by one:
- 0.916 km due east is an horizontal direction and cane be seen as direction towards the negative side of X-axis.
- 0.928 km due south is a vertical direction and can be seen as a direction towards the negative side of Y-axis.
- 3.52 km in a direction of 49.7° has components on X and Y axes. It is necessary to break it down using trigonometry,
First of all. We need to sum all the X components and all the Y componets.
∑
⇒ ∑![Sx = [tex]3.52cos(49.7) - 0.916](https://tex.z-dn.net/?f=Sx%20%3D%20%5Btex%5D3.52cos%2849.7%29%20-%200.916)
∑
∑
⇒ ∑
∑
The total displacement is calculated using Pythagoeran therorem:
⇒
With displacement calculated, we can find the average speed as follows:
⇒ 
Answer: Your answer is<u> 1.36.</u>
Hope this helps!
Answer:
The current in wire resistance 2Ω
a). 8696 A
b). fraction power 15.1% a 115kV
Explanation:
Resistance
Ω/Km*40km
R=2Ω
P=1000 MW
a).
Using law ohm
b).
%
Answer;
D. The car would begin to move in the direction it was headed in a straight line.
Explanation;
-Centripetal force is any net force causing uniform circular motion. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration.
-The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.
-Therefore,If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular
path were reduced then the car would begin to move in the direction it was headed in a straight line.
