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ohaa [14]
3 years ago
10

A pilot heads his jet due east. The jet has a speed of 475 mi/h relative to the air. The wind is blowing due north with a speed

of 30 mi/h. (Assume that the i vector points east, and the j vector points north.)
a. Express the velocity of the wind as a vector in component form.
b. Express the velocity of the jet relative to the air as a vector in component form.
c. Find the true velocity of the jet as a vector.
d. Find the true speed of the jet. (Round your answer to the nearest integer.)
e. Find the direction of the jet.
Physics
1 answer:
maria [59]3 years ago
4 0

Answer:

a) V = 30j

b) U = 475i

c) W = 475i + 30j

d) true speed = 475.95mi/h

e) Direction = 3.62°

Explanation:

The speed of the jet is 475mi/h relative to air.

The wind speed is 30mi/h to the north.

a) The velocity of the wind represented by a ve tor v in component form is given by:

V = 30j

b) The velocity of the jet relative to the air represented by a vector U in component form is given by:

U = 475i

c) The true velocity of the jet as a vector W is represented as W is given by:

W = U + V = 475i + 30j

d) The true speed of the jet is given by:

W = Sqrt(475^2 +30^2)

W = Sqrt(225625 + 900)

W = Sqrt(226525)

W = 475.95mi/h

e) The direction of the jet is given by :

Theta = Tan^-1 (30/475)

Theta = Tan^-1(0.06316)

Theta = 3.62°

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Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

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Side = 0.2 m

Force F_{1}=18\ N

Force F_{2}=26\ N

Force F_{3}=14\ N

We need to calculate the torque due to force F₁

Using formula of torque

\tau_{1}=-F_{1}d_{1}

\tau_{1}=-F_{1}\times\dfrac{a}{2}

Put the value into the formula

\tau_{1}=-18\times\dfrac{0.2}{2}

\tau_{1}=-1.8\ N-m

We need to calculate the torque due to force F₂

Using formula of torque

\tau_{2}=F_{2}d_{2}

\tau_{2}=F_{2}\times\dfrac{a}{2}

Put the value into the formula

\tau_{2}=26\times\dfrac{0.2}{2}

\tau_{2}=2.6\ N-m

We need to calculate the torque due to force F₃

Using formula of torque

\tau_{3}=F_{3}d_{3}

\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}

Put the value into the formula

\tau_{3}=0.1(14\sin45+14\cos45)

\tau_{3}=1.92\ N-m

We need to calculate the net torque on the square plate

\tau=\tau_{1}+\tau_{2}+\tau_{3}

\tau=-1.8+2.6+1.92

\tau=2.72\ N-m

Hence, The net torque on the square plate is 2.72 N-m.

3 0
3 years ago
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