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alina1380 [7]
3 years ago
13

Based on the passage, how do the boys resolve their conflict?

Physics
2 answers:
scZoUnD [109]3 years ago
7 0

the answer is B

i jus took the assement

Naya [18.7K]3 years ago
6 0

Answer:

B

Explanation:

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The distance between two corresponding parts of a wave is its
oksano4ka [1.4K]

Answer:

wavelenght

Explanation:

The wavelength is the spatial period of a wave, analogous to the temporal period, it is the distance between two consecutive points with maximum amplitude that are repeated in space . In the waves of the sea, the wavelength is easily observed in the separation between two consecutive ridges.

4 0
3 years ago
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A bird flies 10 miles south, turns and flies 15 miles east. What distance did it fly.
vfiekz [6]

Answer:

wouldn't it be 25 miles?? yeah

Explanation:

4 0
3 years ago
What is the mass of an object that experiences a gravitational force of 510 N near Earth's surface?
sineoko [7]

Answer:The mass of an object is 52 kg.

Explanation:

Gravitational force on the object ,F=510 N

Acceleration due to gravity = g = 9.8 m/s^2

Mass of the object = m

Force = mass × acceleration

510 N=m\times 9.8 m/s^2

m=52.04 kg\approx 52 kg

The mass of an object is 52 kg.

8 0
3 years ago
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Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
3 years ago
A certain shade of blue has a frequency of 7.15 × 1014 hz. what is the energy of exactly one photon of this light?
Ket [755]
The energy carried by a single photon of frequency f is given by:
E=hf
where h=6.6 \cdot 10^{-34} m^2 kg s^{-1} is the Planck constant. In our problem, the frequency of the photon is f=7.15 \cdot 10^{14}Hz, and by using these numbers we can find the energy of the photon:
E=(6.6\cdot 10^{-34}m^2 kg s^{-1})(7.15 \cdot 10^{14}Hz)=4.7 \cdot 10^{-19}J
4 0
3 years ago
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