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Wittaler [7]
3 years ago
7

The type of exercise that you choose will determine what _____will be worked

Physics
2 answers:
OlgaM077 [116]3 years ago
6 0
The correct answer is "muscles." Different exercises work different muscles.
-BARSIC- [3]3 years ago
4 0
Does muscle go in the blank?
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How does a Freebody diagram tell you about the net force an object?
Sloan [31]
So you subtract the numbers that are on the same axis. So if your gravitational force is 10 and your normal force is 5 you do 5-10 to get -5 since gravity acts downward
6 0
3 years ago
Draw a stationary wave and show the position of node and antinode​
Aneli [31]

I hope that the attachment helps you..

3 0
2 years ago
Statement A: Area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.
pochemuha

Explanation:

Formula for calculating the area of a  rectangle A = Length *width

For statement A;

Given area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.

Area of the rectangle = 2.536mm * 1.4mm

Area of the rectangle = 3.5504mm²

The rule of significant figures states that we should always convert the answer to the least number of significant figure amount the given value in question. Since 1.4mm has 2 significant figure, hence we will convert our answer to 2 significant figure.

Area of the rectangle = 3.6mm² (to 2sf)

For statement B;

Given area of a rectangle with measured length = 2.536 mm and width = 1.41 mm.

Area of the rectangle = 2.536mm * 1.41mm

Area of the rectangle = 3.57576mm²

Similarly, Since 1.41mm has 3 significant figure compare to 2.536 that has 4sf, hence we will convert our answer to 3 significant figure.

Area of the rectangle = 3.58mm² (to 3sf)

Based on the conversion, it can be seen that 3.6mm²  is greater than 3.58mm², hence the area of rectangle in statement A is greater than the area of the rectangle in statement B.

7 0
3 years ago
Whitch two options are forms of kinetic energy?
valentina_108 [34]

Answer:the witch has nothing to do with the problem

Explanation:

7 0
2 years ago
The aircraft wing from problem 6 experiences temperature extremes that span 210 degrees Celsius. The component for the wing will
ExtremeBDS [4]

Answer:

α = 2,857 10⁻⁵ ºC⁻¹

Explanation:

The thermal expansion of materials is described by the expression

          ΔL = α Lo ΔT

          α = \frac{\Delta L}{L_o \ \Delta T}

in the case of the bar the expansion is

         ΔL = L_f - L₀

         ΔL=   1.002 -1

         ΔL = 0.002 m

the temperature variation is

         ΔT = 100 - 30

         ΔT = 70º C

we calculate

         α = 0.002 / 1 70

         α = 2,857 10⁻⁵ ºC⁻¹

5 0
3 years ago
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