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slamgirl [31]
3 years ago
10

Fraternity hazing is acceptable because it is an initational rite to the brotherhood

Physics
1 answer:
Fantom [35]3 years ago
8 0

Answer:

that is the right answer

Explanation:

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The question is on the picture
Firlakuza [10]

Answer:

<em> think 2 also if not im so sorry  but i think it is :)</em>

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3 years ago
A magnet can attract or repel another magnet from a distance true or false?
Anuta_ua [19.1K]
False the strength off the magnet lessens the farther you get from it
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3 years ago
Read 2 more answers
A very long string (linear density 0.7 kg/m ) is stretched with a tension of 70 N . One end of the string oscillates up and down
rewona [7]

To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by

V = \sqrt{\frac{T}{\rho}}

Where,

T = Tension

\rho = Linear density

Our data are given by

Tension , T = 70 N

Linear density , \rho = 0.7 kg/m

Amplitude , A = 7 cm = 0.07 m

Period , t = 0.35 s

Replacing our values,

V = \sqrt{\frac{T}{\rho}}

V = \sqrt{\frac{70}{0.7}

V = 10m/s

Speed can also be expressed as

V = \lambda f

Re-arrange to find \lambda

\lambda = \frac{V}{f}

Where,

f = Frequency,

Which is also described in function of the Period as,

f = \frac{1}{T}

f = \frac{1}{0.35}

f = 2.86 Hz

Therefore replacing to find \lambda

\lambda = \frac{10}{2.86}

\lambda = 3.49m

Therefore the wavelength of the waves created in the string is 3.49m

3 0
4 years ago
Lucy is cruising through space in her new spaceship. As she coasts along, a spacebug drifts into her path and bounces off the wi
Lera25 [3.4K]
The answer is A, because it’s the first one
5 0
3 years ago
A particle was moving in a straight line at 172.8 km/hr. If it decelerated over 120 meters to come to rest, find the time taken
Bezzdna [24]

Answer:

v=s/t

s=vt

t=s/v

t=(120×10‐³)/172.8

(the distance meters has been changed to kilometres)

t=1/1440 hrs

Given ,

8 0
3 years ago
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