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1 year ago

Most scientists, inventors and engineers do not come up with their ideas all on their own but rather

Physics

1 answer:

Natalka [10]1 year ago

6 0

Answer:

Benjamin Franklin, Thomas Edison

and Robert Anderson, Thomas Davenport

Explanation:

Edison was Franklin's idol in studying electricity (Edison's electric light bulb, and Franklin's metal key string kite static experiment)

Anderson and Davenport for the first electric car

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A man is standing on the shore of a beach, up to his knees in water. Every second two waves hit him. What is the frequency of th

laila [671]

Answer:

f = 2 Hz

Explanation:

The frequency of a wave is defined as the no. of waves passing per unit of time. Therefore, the frequency of a wave can be calculated by the following formula:

$f = \frac{n}{t}$

where,

f = frequency of the wave = ?

t = time passed = 1 s

n = no. of waves passing in time t = 2

Therefore,

$f = \frac{2}{1\ s}$

3 0

2 years ago

Pwease help me with thesse three

DiKsa [7]

Answer:

1. Spring

2. 23.5

3. Summer

7 0

2 years ago

What current is needed in the solenoid's wires?

marta [7]

Using Ampere's Law, the magnetic field produced inside this solenoid is given by
B = uo N I / h
where uo is the vacuum permeability, N is the number of turns in the solenoid and h is the length of the solenoid. Earth's magnetic field is around 50 microteslas in North America thus the current needed in the solenoid is
I = B h / (uo N) = (50 E-6 ) (4) / ((4 pi E-7)(6000) ) = 0.026 A
I = 26 mA
So you need a current of around 26 mA.

5 0

2 years ago

A solid disk of mass 2 kg and radius 2 m is given a horizontal push of 20N at a point .3 m above its center. a. What is the mini

Margaret [11]

Answer:

$\mu_s=1.0205$

Explanation:

Given:

mass of solid disk, $m=2\ kg$

radius of disk, $r=2\ m$

force of push applied to disk, $F=20\ N$

distance of application of force from the center, $s=0.3\ m$

<em>For the condition of no slip the force of static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

$\therefore F$

where:

$f_s$ = static frictional force

$\Rightarrow 20$

$\mu_s>1.0204$

7 0

2 years ago

F = 2.0*10^20 N,

natka813 [3]

$F=\dfrac{Gm_1m_2}{r^2}$

With the given values of $F,G,m_1,r$ , we have

$2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}$