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V125BC [204]
2 years ago
8

An air craft heads north at 320 km/hr relative to the wind. the wind velocity is 80km/hr from the north. find the relative veloc

ity of the air craft ralative to the ground
​
Physics
1 answer:
Gnoma [55]2 years ago
7 0

Answer:

Relative to the ground, the velocity of the aircraft is 240 km/hr

Explanation:

Relative velocity is different from normal velocity;

When 2 objects are moving in opposite directions towards each other, they will appear to be faster than they actually are;

This is known as the relative velocity;

The information tells us we have the aircraft moving 320 km/hr northwards relative to the wind;

The wind is in the opposite direction at 80 km/hr;

R = relative velocity of the aircraft

v = actual velocity of the aircraft

w = velocity of the wind

R = v + w

Note: if the wind was moving in the same direction, the formula would be R = v - w

320 = v + 80

v = 320 - 80

v = 240

The velocity relative to the ground is simply the actual velocity as the ground doesn't move;

So, relative to the ground, the velocity of the aircraft is simply 240 km/hr

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We know that the surface area and volume of the sphere is given by:

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Therefore, the ratio between the surface area and the volume for the sphere will be:

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Equating the volume to the constant c, we will find the value of r.

V=c=\frac{4}{3} \pi r^{3}\\r= (\frac{3c}{4\pi} )^{\frac{1}{3} }

Substituting the value of r in the ration between surface area and volume, we get:

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\frac{4.83598}{c^{\frac{1}{3} } }

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1 year ago
Which is true?The photoelectric effect occurs only for frequencies below the cutoff frequency, regardless of the intensity.The p
Crank

Answer:

The photoelectric effect occurs only for frequencies above the cutoff frequency, regardless of the intensity.

Explanation:

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2 years ago
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You push a box with a force of 80 n. if the net force on the box is 50 n, what is the force on the box due to sliding friction?
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2 years ago
Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the
Gnesinka [82]

Answer:

F = 233.52 N,  θ' = 351.41º

Explanation:

In this exercise we must find the net force applied on the donkey.

For this we use Newton's second law, where we create a reference frame with the horizontal x axis

let's decompose the forces

Jack

        = 80.5 N

Jill

       cos 45 = F_{2x} / F₂2

       sin 45 = F_{2y} / F₂2

       F_{2x} = F₂ cos 45

       F_{2y} = F₂ sin 45

       F_{2x} = 81.7 cos 45 = 57.77 N

       F_{2y} = 81.7 sin 45 = 57.77 N

Jane

      cos (270 + 45) = F_{3x} / F₃3

      sin 315 = F_{3y} / F₃

      F_{3x} = 131 cos 315 = 92.63 N

      F_{3y} = 131 sin 315 = -92.63 N

the force can be found in each axis

X axis

         F_{x} = F_{1x} + F_{2x} + F_{3x}

         F_{x} = 80.5 +57.77 + 92.63

         F_{x} = 230.9 N

Axis y

         F_{y} = F_{1y} + F_{2y} + F_{3y}

         F_{y} = 0 + 57.77 -92.63

         F_{y} = -34.86 N

we can give the result in two ways

a) F = (230.9 i ^ - 34.86 j ^) N

b) in the form of module and angle

we use the Pythagorean theorem

         F = √(Fₓ² + F_{y}²

        F = √(230.9² + 34.86²)

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        tan θ = \frac{F_y}{F_x} }

        θ = tan⁻¹ (\frac{F_y}{F_x} })

        θ = tan⁻¹ (-34.86 / 230.9)

        θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

          θ' = 360 -θ

          θ‘= 360- 8.59

          θ' = 351.41º

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Force= mass x acceleration

79=mass(3)

m=79/3

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