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adoni [48]
3 years ago
12

A student ties down one end of a length of rope, then moves the other end back and forth rapidly. The wave that results in the r

ope is an example of a _____.
A.electromagnetic wave


B.longitudinal wave


C.transverse wave


D.mechanical wave of zero amplitude
Physics
2 answers:
Romashka-Z-Leto [24]3 years ago
6 0
I would say the answer is B
Alexus [3.1K]3 years ago
3 0

The answer is C, transverse wave. A student ties down one end of a length of rope, then moves the other end back and forth rapidly. The wave that results in the rope is an example of a transverse wave.

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21. light changes its direction when travelling through a new medium because denser mediums have a higher angle of refraction.
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1.
ale4655 [162]

Answer:

5.0 atm

Explanation:

P₁V₁=P₂V₂

P₁V₁/V₂=P₂

(1)(2.5)/(0•50)=P₂

P₂=5

Pressure is now 5.0 atm

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3 years ago
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A student in Denver (altitude = 1 mile = 1609 m above sea level) brings a physics book of mass 1.3 kg to the top of a ten story
Nesterboy [21]

Answer:

Explanation:

Potential energy is the energy of a body due to is virtue of rest.

Potential energy is given as mgh

g is a constant and it is 9.81m/s²

And also the mass of the body is given as 1.3kg

Now the height of the body is

He took a book to a storey building of height 26m

He still holds the book 151 cm (1.51m) above the house.

The house is on an altitude of 1609m from the sea level.

Total Ug with out the sea level is

Ug=mgh

Ug=1.3 × 9.81 ×(26+1.51)

Ug=350.84J

Then, the potential energy due to the sea level is given as

Ug=mgh

Where g = 1/6371 m/s²

Therefore

Ug=mgh

Ug=1.3 × 1/6371 ×1609

Ug=0.328J

Total energy = 0.328+350.84

Ug=351.17J

8 0
3 years ago
One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.​
Zepler [3.9K]

Answer:

The answer is "\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}"

Explanation:

Given:

\to v=1\ liter= 10^{-3} \ m^3\\\\\to  w= 9.6 \ N\\

calculation:

Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978

4 0
3 years ago
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