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gregori [183]
3 years ago
5

In 2-3 complete sentences, explain the relationship between speed, frequency, and wavelength.

Physics
1 answer:
Ahat [919]3 years ago
7 0
The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves: vw = fλ, where vw is the speed of sound, f is its frequency, and λ is its wavelength. ... The frequency is the same as that of the source and is the number of waves that pass a point per unit time.
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A person run 21.0 km west then runs 10.0 km east what is the persons distance
VashaNatasha [74]

Answer:

31.0 km

Explanation:

21.0+10.0=31.0 km

6 0
3 years ago
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Average velocity is different than average speed because calculating average velocity involves
Artemon [7]

Answer:

calculating displacement.

Explanation:

It's not true that displacement and distance would be the same always. Displacement is always smaller than or equal to distance as it is the smallest path between the initial and final point whereas distance is the measure of the total path covered.

8 0
3 years ago
Rahim is watching his favorite football team on television. In order to work, the television must be plugged into an electrical
Rasek [7]
I think the answer is B
8 0
3 years ago
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At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to (a) 2 times that of the Celsius and (b) 1/4 ti
castortr0y [4]

Answer:

(a) F = 320

(b) = F = -5.1625

Explanation:

The formula that converts degree Celsius (C) to degree Fahrenheit (F) is:

F = 1.8C + 32

Solving (a): F = 2C

Substitute 2C for F in the above equation

F = 1.8C + 32

2C = 1.8C + 32

Collect like terms

2C - 1.8C = 32

0.2C = 32

Multiply both sides by 5

5 * 0.2C = 32 * 5

C = 160

Recall that F = 2C

F = 2 * 160

F = 320

Solving (b): F = ¼C

Substitute ¼C for F in the above formula

F = 1.8C + 32

¼C = 1.8C + 32

Convert fraction to decimal

0.25C = 1.8C + 32

Collect like terms

0.25C - 1.8C = 32

-1.55C = 32

Divide both sides by -1.55

C = 32/(-1.55)

C = -32/1.55

C = -20.65

Recall that: F = ¼C

F = -¼ * 20.65

F = -5.1625

5 0
3 years ago
A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)
wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m           [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = 75\times 10^{-6}\ \mu C    [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}

Plug in the given values for each point and solve.

(a)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C

(b)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C

(c)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C

(d)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C

7 0
3 years ago
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