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TiliK225 [7]
2 years ago
14

What do echolocation and ultrasounds have in common?

Physics
1 answer:
Natali5045456 [20]2 years ago
8 0
They both are mechanical waves.

Explanation:

Echolocation and ultrasounds are both mechanical waves that require a medium to trace through. They both process and transfer information through waves; the difference is that we humans cannot hear ultrasound waves.
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An electric dipole consisting of charges of magnitude 1.70 nC separated by 6.80 μm is in an electric field of strength 1160 N/C.
bazaltina [42]

Answer:

p = 1.16 10⁻¹⁴ C m     and  ΔU = 2.7 10 -11 J

Explanation:

The dipole moment of a dipole is the product of charges by distance

                        p = 2 a q

With 2a the distance between the charges and the magnitude of the charges

                        p = 1.7 10⁻⁹ 6.8 10⁻⁶

                        p = 1.16 10⁻¹⁴ C m

 

The potential energie dipole  is described by the expression

                       U = - p E cos θ

Where θ is the angle between the dipole and the electric field, the zero value of the potential energy is located for when the dipole is perpendicular to the electric field line

Orientation parallel to the field

                      θ = 0º

                      U = 1.16 10⁻¹⁴ 1160 cos 0

                      U1 = 1.35 10⁻¹¹ J

Antiparallel orientation

                       θ = 180º

                      cos 180 = -1

                      U2 = -1.35 10⁻¹¹ J

The difference in energy between these two configurations is the subtraction of the energies

                         ΔU = | U1 -U2 |

                         ΔU = 1.35 10-11 - (-1.35 10-11)

                         ΔU = 2.7 10 -11 J

6 0
3 years ago
A 50.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her and to the bridge. The unstretched length of the
lina2011 [118]

Answer:

The the analysis for the free fall part should be done under the constant acceleration.

Explanation:

In the given problem, the jumper is falling under the free fall. Since, no external force is acting on the body therefore, the fall will be under the action gravity only. also, the acceleration due to gravity is always constant.

Therefore, the the analysis for the free fall part should be done under the constant acceleration.

8 0
3 years ago
____________ is a medium-sized flat-topped hill with cliff-face sides that is taller than it is wide.
Luda [366]

Answer:

mesa

Explanation:



A mesa is a flat-topped mountain or hill. It is a wide, flat, elevated landform with steep sides. ... Spanish explorers of the American southwest, where many mesas are found, used the word because the tops of mesas look like the tops of tables.

5 0
3 years ago
What is the ideal banking angle for a gentle turn of 1.20-km radius on a highway with a 105 km/h speed limit (about 65 mi/h), as
Mnenie [13.5K]

Answer:

4.14°

Explanation:

given:

r = 1.2 km

v = 105 km/h

1) <em>convert your given </em>

a) r = 1.2 km to m = 1200m

b) v = 105 km/h  to m/s = 29.2 m/s

2) <em>plug into your ideal banking angle equation</em>

tan^-1(\frac{v^2}{rg}) = \frac{29.2^2}{(1200)(9.8)} = 4.14°

8 0
2 years ago
In an oil drop experiment, a drop with a weight of 1.9 x 10-14 N was suspended when the potential difference between 2 plates th
AlekseyPX

Answer:

Study More You will get it :L

5 0
2 years ago
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