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ivann1987 [24]
3 years ago
10

a man pushes on a wall with a force of 50N. what are the size and direction of the force that the wall exert on the man?

Physics
1 answer:
Ira Lisetskai [31]3 years ago
5 0
According to Newton's 3rd law of motion, every action has an equal and opposite reaction. So Wall must apply a force of 50 Newton but in opposite direction. Hope you got it :-D
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You are given two vectors vector A = 4.9 at 31o vector B = 6 at 156o Angles are measured counterclockwise from the x-axis. What
Ket [755]

Answer:

   C_{y} = 4.96  and     θ' = 104,5º

Explanation:

To add several vectors we can decompose each one of them, perform the sum on each axis, to find the components of the resultant and then find the module and direction.

Let's start by decomposing the two vectors.

Vector A

             sin θ = A_{y} / A

             cos θ = Aₓ / A

             A_{y} = A sin  θ

             Ax = A cos θ

             A_{y} = 4.9 sin 31 = 2.52

             Ax = 4.9 cos 31 = 4.20

Vector B

           B_{y} = B sin θ

           Bx = B cos θ

           B_{y} = 6 sin 156 = 2.44

           Bx = 6 cos 156 = -5.48

The components of the resulting vector are

X axis

         Cx = Ax + B x

         Cx = 4.20 -5.48

         Cx = -1.28

Axis y

         C_{y} = Ay + By

         C_{y} = 2.52 + 2.44  

         C_{y} = 4.96

Let's use the Pythagorean theorem to find modulo

         C = √ (Cₙ²x2 + Cy2)

         C = Ra (1.28 2 + 4.96 2)

         C = 5.12

We use trigonemetry to find the angle

         tan θ = C_{y} / Cₓ

          θ’ = tan⁻¹ (4.96 / (1.28))

           θ’ = 75.5

como el valor de Cy es positivo y Cx es negativo el angulo este en el segundo cuadrante, por lo cual el angulo medido respecto de eje x positivo es

       θ’ = 180 – tes

        θ‘= 180 – 75,5

        θ' = 104,5º

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Dima020 [189]
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Red light is bent the least of all colors as it passes through a prism. What does this tell you about red light? It has a short
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Answer:

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Explanation:

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The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is
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To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

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g = 32.17405ft/s^2

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For the air the defined properties would be

\rho_a = 0.0075lb/ft^3

g = 32.17405ft/s^2

h_2 = ?

We have for equilibrium that

\text{Pressure change in Air}=\text{Pressure change in Mercury}

\rho_m g h_1 = \rho_a g h_2

Replacing,

(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)

Rearranging to find h_2

h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}

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Therefore the elevation of the mountain top is 9400ft

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3 years ago
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