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Elena-2011 [213]
3 years ago
5

Help me for a physics project please

Physics
1 answer:
lesantik [10]3 years ago
4 0

-cant travel through space since there's no molecules to travel through

-sound travels 4.3 times faster in water than air

-sounds are waves that pass through our ears via vibrations and travel by vibrations of molecules

-different types of sound like audible, inaudible, infrasonic, ultrasonic,

-sounds waves are either longitudinal, mechanical and pressure waves

-sound travels at 767 miles per hour

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If you increase the length ? of a pendulum by a factor of 9, how will the period t increase?
garik1379 [7]
T = 2 * pie √(L/g)

so, if length is increased by 9

then time period is increased by √9 = 3

hope it helped :)
3 0
3 years ago
Read 2 more answers
Which statement represents a difference between horizontal and vertical relationships?
marishachu [46]
D.) Vertical relationships involve unequal status, while horizontal relationships represent equal status.
HOPE THIS HELPS!
6 0
3 years ago
What wavelength of light (in nm) is associated with a frequency of 8.01E15 Hz?
AlekseyPX

Answer: 37.5 nm

Explanation: speed of light c= 3.00·10^8 m/s.

I use same accuracy to speed of light as it's for frequency.

Frequency f= 8.01·10^15 1/s

Speed c = wavelength · frequency

Wavelength = c/f = 3.745·10^-8 m

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
At what point does the external energy enter the system?
Phoenix [80]
The correct answer as the first one above !
8 0
3 years ago
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