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Natalka [10]
3 years ago
8

A toy car runs off the edge of a table that is 2.225 m high. If the car lands 0.400 m away from the base of the table, how fast

was the car going when it fell off?
Physics
1 answer:
ruslelena [56]3 years ago
8 0

Answer:

0.594 m/s

Explanation:

First, find the time it takes to land.

Given, in the y direction:

Δy = 2.225 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(2.225 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.674 s

Next, find the horizontal velocity.

Given, in the x direction:

Δx = 0.400 m

a = 0 m/s²

t = 0.674 s

Find: v₀

Δx = v₀ t + ½ at²

(0.400 m) = v₀ (0.674 s) + ½ (0 m/s²) (0.674 s)²

v₀ = 0.594 m/s

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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
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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.

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We transform this formula to get a:
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a=\frac{2}{10}\frac{N}{kg}=0.2\frac{m}{s^{2}}
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