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Anarel [89]
3 years ago
6

If 2.2 lbs = 1.0 kg, and Megan Progress weighs 130 lbs, what is her weight in newtons?

Physics
1 answer:
Dafna11 [192]3 years ago
8 0

Answer:

Force = 2802.8 Newton

Explanation:

Given the following data;

Mass = 130 lbs

But, 2.2 lbs = 1.0 kg

130 lbs to kg = 130 * 2.2 = 286 kg

Acceleration due to gravity = 9.8m/s²

To find the force in Newton;

Force = mass * acceleration due to gravity

Force = 286 * 9.8

Force = 2802.8 Newton

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Energy can not be created or destroyed but can change from one form to another.

example: as a roller coaster cart loses height the more speed it gains, the potential energy is transferred into kenetic energy
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3 years ago
Steel is made of atoms of iron and carbon. Would iron and carbon form metallic bonds? Explain your answer choice.
Hatshy [7]

Sample Response: "No, steel and carbon would not form metallic bonds because metallic bonds only form between metals. Iron is a metal, but carbon is not."

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3 years ago
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With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward. If the force exerted on the book by
lara31 [8.8K]

According to the net force, the acceleration of the book is 16.47 m/s².

We need to know about force to solve this problem. According to second Newton's Law, the force applied to an object will be proportional to mass and acceleration. Hence, it can be written as

∑F = m . a

where F is force, m is mass and a is acceleration

From the question above, we know that

m = 3 kg

g = 9.8 m/s²

F1 = 20 N

Find the net force

∑F = F1 + W

∑F = 20 + m . g

∑F = 20 + 3 . 9.8

∑F = 20 + 29.4

∑F = 49.4 N

Find the acceleration

∑F = m . a

49.4 = 3 . a

a = 16.47 m/s²

Find more on force at: brainly.com/question/25239010

#SPJ4

7 0
1 year ago
Starting from rest, a 6.79 kg block slides 2.82 m down a rough 20.7 ◦ incline. The coefficient of kinetic friction between the b
Veronika [31]

Answer:

23.52092 J

Explanation:

m = Mass of block = 6.79 kg

s = Sliding distance = 2.82 m

\theta = Angle of slide = 20.7°

\mu = Coefficient of kinetic friction = 0.425

g = Acceleration due to gravity = 9.8 m/s²

Work done by the force of gravity is given by

W=mgsin\theta\\\Rightarrow W=6.79\times 9.8\times sin20.7\\\Rightarrow W=23.52092\ J

The work done by the force of gravity is 23.52092 J

8 0
3 years ago
A steel bar of rectangular cross section (1.5 in. 2 3 .0 in.) carries a tensile load P (see fig- ure). The allowable stresses in
lukranit [14]

Explanation:

Value of the cross-sectional area is as follows.

        A = 1.5 \times 2.30

           = 3.45 in^{2}

The given data is as follows.

          Allowable stress = 14,500 psi

          Shear stress = 7100 psi

Now, we will calculate maximum load from allowable stress as follows.

           P_{max} = \sigma_{a}A

                       = 14500 \times 3.45

                       = 50025 lb

Now, maximum load from shear stress is as follows.

           P_{max} = 2 \times \tau_{a} \times A

                      = 2 \times 7100 \times 3.45

                      = 48990 lb

Hence, P_{max} will be calculated as follows.

       P_{max} = min((P_{max})_{\sigma}, (P_{max})_{\tau})

                  = 48990 lb

Thus, we can conclude that the maximum permissible load P_{max} is 48990 lb.

4 0
4 years ago
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