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3 years ago

Can you give me some project ideas

Physics

1 answer:

astraxan [27]3 years ago

3 0

project ideas for what exactly?

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Who was the first person to describe the earth as a magnet

denis-greek [22]

Answer:

William Gilbert

Explanation:

first described the Earth as a giant dipole magnet 400 years ago. But, as Rod Wilson recounts, he did far more than this.

5 0

2 years ago

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A bus starts from rest.if the acceleration is 2m/s square, find

MrMuchimi

Answer:

The velocity after 2 seconds can be found through:

V = u +a*t

Where V is final velocity, u is initial velocity, a is acceleration and t is time.

V = 0 + 2* 2= 4 meters/second

The distance (s) can be found through:

V^2= u^2 +2*a* s

Where V is final velocity, u is initial velocity, a is acceleration.

4^2= 0^2 + 2 *2*s

16= 0 + 4s

s= 4 meters

Distance (s) can also be found through:

s= ut + 1/2 at^2

s= 0+ 1/2 *2*2^2= 1 *2*2

s= 4 meters

Explanation:

3 0

1 year ago

A 100-m long transmission cable is suspended between two towers. If the mass density is 2.01 kg/m and the tension in the cable i

Inga [223]

Answer:

The speed is $v =122.2 \ m/s$

Explanation:

From the question we are told that

The length of the wire is $L = 100 \ m$

The mass density is $\mu = 2.01 \ kg/m$

The tension is $T = 3.00 *10^{4} \ N$

Generally the speed of the transverse cable is mathematically represented as

$v = \sqrt{\frac{T}{\mu} }$

substituting values

$v = \sqrt{\frac{3.0 *10^{4}}{2.01} }$

$v =122.2 \ m/s$

3 0

2 years ago

If it requires 4.5 J of work to stretch a particular spring by 2.3 cm from its equilibrium length, how much more work will be re

saveliy_v [14]

Answer:

$\Delta W=24.1162\ J$

Explanation:

Given:

work done to stretch the spring, $W=4.5\ J$

length through which the spring is stretched beyond equilibrium, $\Delta x=2.3\ cm=0.023\ m$

additional stretch in the spring length, $\delta x=3.5\ cm=0.035\ m$

<u>We know the work done in stretching the spring is given as:</u>

$W=\frac{1}{2} \times k.\Delta x^2$

where:

k = stiffness constant

$4.5=0.5\times k\times 0.023^2$

$k=17013.2325\ N.m^{-1}$

Now the work done in stretching the spring from equilibrium to ( $\Delta x+\delta x$ ):

$W'=0.5\times k.(\Delta x+\delta x)^2$

$W'=0.5\times 17013.2325\times 0.058^2$

$W'=28.6162\ J$

So, the amount of extra work done:

$\Delta W=W'-W$

$\Delta W=28.6162-4.5$

$\Delta W=24.1162\ J$

4 0

2 years ago

Convert 636 g/cm to lbs per foot

Gemiola [76]

636 g/cm would be 42.7372 lb/ft

8 0

3 years ago

Read 2 more answers