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crimeas [40]
3 years ago
6

A certain wave has a wavelength of 45.0 meters and a frequency of 9.00 Hz. What is the speed of the wave? 405 m/s 5.00 m/s 0.200

m/s It depends on the length of the rope.
Physics
2 answers:
kirill115 [55]3 years ago
5 0

Wave speed  =  (wavelength)  x  (frequency)

                   =   (45 meters)   x   (9 per second)

                   =         405 meters per second . 
Tpy6a [65]3 years ago
4 0

Answer:

405 m/s

Explanation:

A certain wave has a wavelength of 45.0 meters and a frequency of 9.00 Hz. What is the speed of the wave?

405 m/s

5.00 m/s

0.200 m/s

It depends on the length of the rope.

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2 years ago
Identify the independent, dependent, and constant variables for different experiments.
diamong [38]

ANSWER:

IV, Type of dish detergent. DV, height of foam. CV, type of container, amount of water in container, temperature of water, time the container is agitated.

Explanation:

Independent variable(IV)- what you change during the experiment.

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3 0
3 years ago
The massless spring of a spring gun has a force constant k=12~\text{N/cm}k=12 N/cm. When the gun is aimed vertically, a 15-g pro
ASHA 777 [7]

Answer:

0.011 m.

Explanation:

Energy stored in the spring = Energy of the projectile.

1/2ke² = mgh ................ Equation 1

Where k = spring constant, e = extension or compression, m = mass of the projectile, g = acceleration due to gravity, h = height.

make e the subject of the equation

e = √(2mgh/k)............................. Equation 2

Given: k = 12 N/cm = 1200 N/m, m = 15 g = 0.015 kg, h = 5.0 m

Constant: g = 9.8 m/s²

Substitute into equation 2

e = √(2×0.015×5/1200)

e = √(0.15/1200)

e = √(0.000125)

e = 0.011 m.

4 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
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