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snow_tiger [21]
3 years ago
6

In which condition the acceleration of a moving vehicle become zero​

Physics
1 answer:
Alekssandra [29.7K]3 years ago
4 0

Explanation:

When,the vehicle has uniform velocity, it's acceleration becomes zero

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What of the following does NOT influence resistance? *
Marina86 [1]

Answer:

D. Number of components

Explanation:

4 0
3 years ago
Any body moving with simple harmonic motion is being acted on by a force that is:__________.
Stella [2.4K]

Answer:

B

Explanation:

Because this oscillations occur when the restoring force is directly proportional to displacement, given as

F=-kx

Where k= force constant

X= displacement

6 0
3 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 uniform solid cylindrical log begins rolling without slipping down a ramp that rises 28.0° above the horizontal. After it has
Novay_Z [31]

Answer:

3.07 m/s

Explanation:

6 0
3 years ago
Solve:40KMEN
Oksanka [162]

The spider will cross the driveway <u>25.71 in seconds</u>.

Why?

It's a conversion problem, so, in order to solve it, we need to convert the units of the given information.

Let's convert the given speed (in cm/s) to m/s.

We know that:

1m=100cm

So, converting we have:

14\frac{cm}{s}*\frac{1m}{100cm}=0.14\frac{m}{s}

Then, calculating the time, we have:

distance=speed*time\\\\3.6m=0.14\frac{m}{s}*time\\\\time=\frac{3.6m}{0.14\frac{m}{s}}=25.71s

We have that the spider will cross the driveway in 25.71 seconds.

Have a nice day!

8 0
3 years ago
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