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svet-max [94.6K]
4 years ago
11

Please help! What are 3 things that make Longitudinal and Transverse waves the same?

Physics
1 answer:
Flura [38]4 years ago
4 0
"<span>All waves have frequency, wavelength, speed and amplitude." </span>
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SUB TO YFBG JAY ON YT RN ASAP
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Oki said he would not hood him to go on a bike but i said he would be ok if you don’t have a ride home with you and green one for a
6 0
3 years ago
Difference between freefall and weightlessness.​
Margaret [11]

Answer:

Differences between freefall and weightlessness are as follows:

<h3><u>Freefall</u></h3>
  • When a body falls only under the influence of gravity, it is called free fall.
  • Freefall is not possible in absence of gravity.
  • A body falling in a vacuum is an example of free fall.

<h3><u>Weightlessness</u></h3>
  • Weightlessness is a condition at which the apparent weight of body becomes zero.
  • Weightlessness is possible in absence of gravity.
  • A man in a free falling lift is an example of weightlessness.

Hope this helps....

Good luck on your assignment....

6 0
3 years ago
Jen pushed a box for a distance of 80m with 20 N of force. How much work did she do?
Drupady [299]
The answer is 1,600 J.

A work (W) can be expressed as a product of a force (F) and a distance (d):
W = F · d<span>

We have:
W = ?
F = 20 N = 20 kg*m/s</span>²
d = 80 m
_____
W = 20 kg*m/s² * 80 m
W = 20 * 80 kg*m/s² * m
W = 1600 kg*m²/s²
W = 1600 J
8 0
3 years ago
When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam
Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s

b) the change in the combined kinetic energy of the two-car system during this collision

\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))

substitute the value in the equation above

=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J

Hence, the change in combine kinetic energy is -2534.78J

8 0
3 years ago
Twenty is the _________________ of potassium
Mumz [18]
Twenty is the atomic number of potassium.
7 0
4 years ago
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