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3 years ago

Who and when was mechanical energy discovered? help me please!

Physics

2 answers:

Greeley [361]3 years ago

8 0

Answer:

Explanation:

Thomas Young (1773 − 1829) first introduced the word “energy” to the field of physics in 1800, but the word did not gain popularity. Thomas Young later established the wave nature of light through interference experiments.

Gala2k [10]3 years ago

3 0

Answer:

June 1849 - James Prescott Joule

Explanation:

James Joule is the physicist who discovered the relationship between the loss of mechanical energy and the gain of heat. If an object contains mechanical energy, then it has the ability to do work.

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A 30 kg child rides a 20 kg bicycle. Together, the child and the bicycle have a momentum of 110 kg-m/s. What is the velocity of

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Answer:

The velocity of the boy and the bicycle is 2.2 m/s.

Explanation:

We have,

Mass of child is 30 kg and the mass of bicycle is 20 kg. The combined momentum of the child and the bicycle is 110 kg-m/s.

It is required to find the velocity of the boy and the bicycle. The momentum of an object is given in terms of mass and its velocity. So,

$p=Mv$

M is combined mass of child and bicycle

$v=\dfrac{p}{M}\\\\v=\dfrac{110}{30+20}\\\\v=2.2\ m/s$

So, the velocity of the boy and the bicycle is 2.2 m/s.

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3 years ago

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Answer:

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Explanation:

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2 years ago

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3 years ago

A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

$y(x, t) = Acos(kx -wt)$

The relationship between velocity and propagation constant is

$v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\$

v = 15.87m/s

Time taken, $t = \frac{\lambda}{v}$

$= \frac{1.3}{15.87}\\\\=0.0819 sec$

t = 0.0819s

The velocity of transverse wave is given by

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

$v = \sqrt{\frac{W}{\frac{m}{\lambda}}}$

mass of string is calculated thus

mg = 0.0125N

$m = \frac{0.0125N}{9.8N/s}$

m = 0.00128kg

$\omega = \frac{v^2m}{\lambda}$

$\omega = \frac{(15.87^2)(0.00128)}{1.30}$

$\omega =$ 0.25N

The propagation constant k is

$k=\frac{2\pi}{\lambda}$

hence

$\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}$

$\lambda =$ 0.036 m

No of wavelengths, n is

$n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\$

n = 36

The equation of wave travelling down the string is

$y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]$

$without, unit\\\\y(x , t)= Acos[172x + 2730t]$

7 0

3 years ago

A small current element carrying a current of I = 1.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic

xxTIMURxx [149]

Answer:

the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)

the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)

Explanation:

Use Biot, Savart, the magnetic field

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

Given that,

i = 1.00A

d → l = 4.00 m m ^ j

r = 2.5m

Displacement vector is

$\bar{r}=x\hat i+y\hat j+z \hat k\\$

$\bar{r}= (2.5m) \hat i +(0m)^2 + (0m)^2$

=2.5m

on the axis of x at x = 2.5

$r = \sqrt{(2.5)^2 + (0)^2 + (0)^2}$

r = 2.5m

And unit vector

$\hat r =\frac{\bar{r}}{r}$

$= \frac{2.5 \hat i}{2.5}\\\\= 1\hat i$

Therefore, the magnetic field is as follow

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

$d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(2.50)^2} \\\\d\bar{B} = -6.4\times10^{-11} T$

(Along z direction)

B)r = 5.00m

Displacement vector is

$\bar{r}=x\hat i+y\hat j+z \hat k\\$

$\bar{r}= (5.00m) \hat i +(0m)^2 + (0m)^2$

=5.00m

on the axis of x at x = 5.0

$r = \sqrt{(5.00)^2 + (0)^2 + (0)^2}$

r = 5.00m

And unit vector

$\hat r =\frac{\bar{r}}{r}$

$= \frac{5.00 \hat i}{5.00}\\\\= 1\hat i\\$

Therefore, the magnetic field is as follow

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

$d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(5.00)^2} \\\\d\bar{B} = 1.6\times10^{-11} T$

(Along x direction)

7 0

3 years ago

Who and when was mechanical energy discovered? help me please!​

Who and when was mechanical energy discovered? help me please!