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Ymorist [56]
3 years ago
12

U1=20 m/s turn it to km/h

Physics
1 answer:
notka56 [123]3 years ago
7 0
It is 72 km/h
I hope it helps
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The frequency of a person's pendulum is 0.3204 Hz when at a location where g is known to be exactly 9.800 m/s^2.
wel

Answer:

A) g = 9.751 m/s², B)       h = 2.573 10⁴ m

Explanation:

The angular velocity of a pendulum is

           w = √ g / L

Angular velocity and frequency are related.

          w = 2π f

          f = 1 / 2π √ g / L

A) with the initial data we can look for the pendulum length

        L = 1 /4π²   g / f²

        L = 1 /4π²   9,800 / 0.3204²

        L = 2.4181 m

The length of the pendulum does not change, let's look for the value of g for the new location

        g = 4π² f²  L

        g = 4π² 0.3196²  2.4181

        g = 9.75096 m / s²

         g = 9.751 m/s²

B) The value of the acceleration of gravity can be found with the law of universal gravitation

           F = G m M / R_{e}²

And Newton's second law

           W = m g

           W = F

         G m M / R_{e}² = mg

         g = G M / R_{e}²

         R_{e}² = G M / g

Let's calculate

          R_{e}² = 6.67 10⁻¹¹ 5.98 10²⁴ /9.75096

          R = √ 4.0905 10¹³ = √ 40.9053 10¹²

          R = 6.395726 10⁶ m

The height above sea level is

         h = R - [tex]R_{e}[/tex

         h = (6.395726 -6.37) 10⁶

         h = 0.0257256 106

         h = 2.573 10⁴ m

7 0
4 years ago
Suppose the microphone on an AM radio is not working properly. Which best explains how this will affect the process of amplitude
Anna007 [38]
You can have it your own definition of
3 0
3 years ago
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The greater the mass is in an object, the higher resistance to a change in movement the object will have. Please select the best
Fofino [41]
This statement is true. The greater the mass is in an object, it is indeed the higher resistance to a change in movement the object will have. That only mean that the mass of an object and its resistance to change of movement is directly proportional.
3 0
3 years ago
Select all the correct answers.
sdas [7]
I believe there are two correct answers, and those answers are A and D
5 0
3 years ago
Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po
gregori [183]

Answer:

E = 0    r <R₁

Explanation:

If we use Gauss's law

      Ф = ∫ E. dA = q_{int} / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

           E = 0    r <R₁

6 0
3 years ago
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