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Anna35 [415]
3 years ago
15

In our usual coordinate system( +x to the right, +y up (away from the center of the Earth), +z out of the page toward you), what

is the vector gravitational force on a 31 kg object sitting on the ground? (From the momentum principle you can conclude that a force of the same magnitude is exerted upward by the ground on the object if the momentum isn't changing. In Chapter 4 we'll see more about the origin of this force.)
Physics
1 answer:
skad [1K]3 years ago
5 0

Answer:

The vector gravitational force is  W = 303.8\  N( -y)

Explanation:

From the question we are told that

      The mass of the object is  m  = 31 \ kg

      The acceleration due to gravity experienced by the object is g =  9.8 m/s^2

Generally the vector gravitational force on the object is mathematically represented as

                       W = 31 *  9.8

                       W = 303.8\  N (-y)

The  -y indicates that the direction of the force is towards the center of the earth

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Calculate the ratio of the drag force on a jet flying at 950 km/h at an altitude of 10 km to the drag force on a prop-driven tra
Black_prince [1.1K]

Answer:

\frac{F_1}{F_2}=3.55

Explanation:

F = Force

C = Drag coefficient equal for both aircrafts

ρ = Density of air

A = Surface area equal for both aircrafts

v = Velocity

v_2=\frac{2}{5}v_1

F_1=\frac{1}{2}\rho_1 CAv_1^2

F_2=\frac{1}{2}\rho_2 CAv_2^2\\\Rightarrow F_2=\frac{1}{2}\rho_2 CA\left(\frac{2}{5}v_1\right)^2

Dividing the above two equations we get

\frac{F_1}{F_2}=\frac{\frac{1}{2}\rho_1 CAv_1^2}{\frac{1}{2}\rho_2 CA\left(\frac{2}{5}v_1\right)^2}\\\Rightarrow \frac{F_1}{F_2}=\frac{\rho_1}{\rho_2\frac{4}{25}}\\\Rightarrow \frac{F_1}{F_2}=\frac{0.38}{0.67\frac{4}{25}}\\\Rightarrow \frac{F_1}{F_2}=3.55

The ratio of the drag forces is \mathbf{\frac{F_1}{F_2}}=\mathbf{3.55}

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3 years ago
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