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Maru [420]
3 years ago
8

The airplane is flying with a constant velocity. Which force acting on the airplane below represents the friction from air resis

tance?
A


B


C


D

Physics
2 answers:
tankabanditka [31]3 years ago
8 0

The correct answer to the question is  C i.e C represents the friction from air resistance.

EXPLANATION:

Before coming into any conclusion, first we have to understand friction.

The friction is the opposing force which acts tangentially between two bodies in contact when there is a relative motion between them.

The air resistance is that frictional force which is provided by the air to the moving body through it. Hence, the friction from air resistance will be directed opposite to the motion of the body.

In the given diagram, the airplane is going horizontally. The force A acts in forward direction while force C acts in backward direction. The forces B and D are acting vertically.  There is no motion in vertical direction. Hence, the net force of A and C will cause the airplane to move.

As the plane is moving along the direction of A, the frictional force must act along the direction of C.

trasher [3.6K]3 years ago
8 0

C, since friction is always going towards the opposite direction the object/person is traveling.

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5 0
3 years ago
What is a physical form in which a substance can exist?
Rama09 [41]
The three physical forms are:

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8 0
2 years ago
A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A
Y_Kistochka [10]

Answer:

The tensions in T_{BC} is approximately 4,934.2 lb and the tension in T_{BD} is approximately  6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = T_{BC}

The tension in rope segment BD = T_{BD}

The tension in rope segment AB = T_{AB} = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

T_{BC} × cos(26.0°) + T_{BD} × cos(21.0°) = T_{AB} = 1200 lb

T_{BC} × cos(26.0°) + T_{BD} × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

T_{BC} × sin(26.0°) + T_{BD} × sin(21.0°) = 0...........................(2)

Which gives;

T_{BC} × sin(26.0°) = - T_{BD} × sin(21.0°)

T_{BC} = - T_{BD} × sin(21.0°)/(sin(26.0°))  ≈ - T_{BD} × 0.8175

Substituting the value of, T_{BC}, in equation (1), gives;

- T_{BD} × 0.8175 × cos(26.0°) + T_{BD} × cos(21.0°) = 1200 lb

- T_{BD} × 0.7348  + T_{BD} ×0.9336 = 1200 lb

T_{BD} ×0.1988 = 1200 lb

T_{BD} ≈ 1200 lb/0.1988 = 6,035.6938 lb

T_{BD} ≈ 6,035.6938 lb

T_{BC} ≈ - T_{BD} × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

T_{BC} ≈ -4934.1733 lb

From which we have;

The tensions in T_{BC} ≈ -4934.2 lb and  T_{BD} ≈ 6,035.7 lb.

8 0
3 years ago
A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a
Zolol [24]
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R_{eq}=R+10 \\ R_{eq}=14+10 \\ \boxed {R_{eq}=24-Ohm}

If you notice any mistake in my english, please let me know, because i am not native.

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Answer:

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Explanation:

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