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Alona [7]
2 years ago
15

How mesuremerment of gravity using simple pendulum ​

Physics
1 answer:
ale4655 [162]2 years ago
6 0

Answer:

See below

Explanation:

Set up your pendulum

  measure its length  and time the period ( you could time 100 of them and divide the time result by 100 to get the period, T)

    then use  

             T = 2 pi  sqrt (L/g)        T = period    L = length   g = gravity

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A _________________ is when the courts allow people who have all been harmed to pool grievances and sue for damages on behalf of
Sliva [168]

A class action law suit is one in which the courts allow several
people who all claim to have been harmed to pool grievances
and sue for damages on behalf of the group.

4 0
4 years ago
Read 2 more answers
A car is on cruise control at v=30m/s. If the wind resistance is equal to. F=0.42v² what energy will the car expend to drive 1km
weeeeeb [17]

Answer:

D)378kJ

Explanation:

Applying,

E = F×d................ Equation 1

Where E = Energy expanded by the car, F = Wind resistance, d = distance

From the question,

F = 0.42v²............ Equation 2

Susbtitute equation 2 into equation 1

E = 0.42v²d.............. Equation 3

Given: v = 30 m/s, d = 1 km = 1000 m

Substitute these values into equation 3

E = 0.42(30²)(1000)

E = 378000

E = 378 kJ

Hence the right option is D)378kJ

6 0
3 years ago
The Sloan Digital Sky Survey includes many thousands of galaxies in its spectroscopic catalogs. How are the distances to most of
Katarina [22]

The distances to the most of these galaxies estimated from their red shifts and the application of Hubble's Law.

<h3>What is Hubble's Law?</h3>

The discovery in physical cosmology that galaxies are travelling away from Earth at rates proportionate to their distance is known as Hubble's law, often referred to as the Hubble-Lemaître law or Lemaître's law. In other words, they are travelling away from Earth more quickly the more away they are. The redshift of the galaxies—a shift in the light they produce toward the red end of the visible spectrum—has been used to calculate their velocities.

The Sloan Digital Sky Survey includes many thousands of galaxies in its spectroscopic catalogs. The distances to the most of these galaxies estimated

from their red shifts and the application of Hubble's Lawh

to learn more about Hubble's law go to - brainly.com/question/3050512

#SPJ4

3 0
2 years ago
The apparent visual magnitude of a star is 7.3. This tells us that the star is _______.
fredd [130]

Answer:

Option (B)

Explanation:

In the field of astronomy, the magnitude scale of star is usually defined as a logarithmic scale that measures the brightness of the star, and this directly the relative distance from the earth. This magnitude value is dependent on the two main factors, namely the its brightness and the distance from a fixed point.

The given visual magnitude of a star is 7.3, and is comparatively very less brighter. This shows a relatively very large distance from the earth and due to this large distance and low luminousity, this star is not visible to our naked eye.

Thus, the correct answer is option (B).

8 0
4 years ago
A flat, circular, copper loop of radius r is at rest in a uniform magnetic field of magnitude B that extends far beyond the edge
GarryVolchara [31]

Answer:

     i =- \frac{r \ A'}{2 \ rho} ,  i =  0.92   A

Explanation:

This exercise asks for the electromotive force, which can be calculated with Faraday's law

          fem = - \frac{d \Phi_B }{dt}

where the magnetic flux

          Ф = B. A

bold letters indicate vectors. We can write this equation

          Ф = B A cos θ

In this case the magnetic field is perpendicular to the page and the normal to the loop of the loop is also parallel to the page, therefore the angle is zero and the cosine is 1

the loop is

          A = π r²

we substitute in the first equation

          fem = - π r² \frac{dB}{dt}

we substitute the values

          fem = -π r² 1

          fem = - π r²

to calculate the current let's use ohm's law

           V = i R

            R = ρ L / A'

where A 'is the area of ​​the wire and L is the length of the loop

            L = 2π r

           

            V = i (ρ 2π r / A ')

            I = \frac{V \ A'}{2\pi  \ r  \ rho}

         

In this case

           V = fem

 

           I = fem / R            

           i =- \frac{r \ A'}{2 \ rho}

         

In order to complete the calculation, you need the radius of the loop and / or the wire cutter.

if we assume that the loop has a radius of r = 1 cm = 0.01 m and an area of ​​the wire   A'= π 10⁻⁶ m²   a radius of the wire 1 mm

           i = - 10⁻² π 10⁻⁶ / ( 2 1.7 10-8)

           i =  0.92   A

5 0
3 years ago
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