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san4es73 [151]
3 years ago
5

A standing wave has a frequency of 471 Hz and a wavelength of 1.9. What is the speed of the

Physics
1 answer:
Brums [2.3K]3 years ago
6 0

Answer:

c = 894.90 m/s

Explanation:

Given data:

Frequency of wave = 471 Hz

Wavelength of wave = 1.9 m

Speed of wave = ?

Solution:

Formula:

Speed of wave = frequency × wavelength

c = f×λ

c = 471 Hz × 1.9 m

  Hz = s⁻¹

c = 471s⁻¹ × 1.9 m

c = 894.90 m/s

The speed of wave is 894.90 m/s.

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Can someone help me with this real quick?
ch4aika [34]

Answer:

7,546 J

Explanation:

recall that Potential energy is given by

P.E = mgΔh

where m = 70kg (given)

g = 9.8 m/s² (acceleration due to gravity)

Δh = change in height

= distance from top of building to top of car

= height of building - height of car

= (5+8) - 2

= 11m

substituting all these into the equation:

P.E = mgΔh

= 70 x 9.8 x 11

= 7,546 J

4 0
3 years ago
A race-car driver achieved a velocity of 53 m/s in 14 seconds after taking off from rest at the starting line. What was the aver
Shtirlitz [24]
53/14 meters per second.
Or
About 3.78571 meters per second
Hope this helps
3 0
3 years ago
Paleontologists are experts on the rock cycle true or false
kow [346]

Answer:

I cannot do the laws of physics cuz I hate science but I'm just dance

Explanation:

<u>it's</u><u> it's true</u>

8 0
3 years ago
Two particles are fixed to an x axis: particle 1 of charge −8.00 ✕ 10⁻⁷ C at x = 6.00 cm, and particle 2 of charge +8.00 ✕ 10⁻⁷
victus00 [196]

Answer:

0 N/C

Explanation:

Parameters given:

q_1 = -8.00 * 10^{-7} C

x_1 = 6.00 cm

q_2 = +8.00 * 10^{-7} C

x_2 = 21 cm

The distance between q_1 and q_2 is

21 - 6 = 15cm

Electric field is given as

E = \frac{kq}{r^2}

r = 15/2 = 7.5cm = 0.075m

The electric field at their midpoint due to q_1 is:

E = \frac{9 * 10^9 * -8.0 * 10^{-7}}{0.075^2}

E_1 = -1.28 * 10^6 N/C

The electric field at the midpoint due to q_2 is:

E = \frac{9 * 10^9 * 8.0 * 10^{-7}}{0.075^2}

E_2 = 1.28 * 10^6 N/C

The net electric field will be:

E = E_1 + E_2

E = -1.28 * 10^6 + 1.28 * 10^6

E = 0 N/C

7 0
4 years ago
Technician A says test lights are great for quick tests on non-computerized circuits. Technician B says you can use a test light
Tpy6a [65]

Answer:

that technician A is right

Explanation:

The test lights are generally small bulbs that are turned on by the voltage and current flowing through the circuit in analog circuits, these two values ​​are high and can light the bulb. In digital circuits the current is very small in the order of milliamps, so there is not enough power to turn on these lights.

From the above it is seen that technician A is right

4 0
4 years ago
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