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3 years ago

In longitudinal waves the places where the coils are bunched together are called *

Physics

1 answer:

Firdavs [7]3 years ago

3 0

In longitudinal waves the places where the coils are bunched together are called *
Compressions

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A 2 kg block is pushed against a spring (k = 400 N/m), compressing it 0.3 m. When the block is released, it moves along a fricti

Kitty [74]

Answer:

$2.29 \mathrm{m} \text { the block slide if the } \mathrm{u}_{\mathrm{s}}=0.4$

$4.58 \mathrm{m} \text { the block slide if the } \mathrm{u}_{\mathrm{k}}=0.2$

Explanation:

Given values

Mass (m) = 2kg

K = 400 N/M

Compressing it 0.3 m

<u>The law of conservation of energy</u>:

$\frac{m v^{2}}{2}+\frac{k x^{2}}{2}=\text { constant }$

$\text { Where, } \frac{m v^{2}}{2} \text { is kinetic energy of the block. }$

$\frac{k \Delta l^{2}}{2}$ Energy of the spring deformation.

M mass of the block

x spring deformation

Therefore, if block left the spring (x = 0)

$\frac{m v^{2}}{2}+0=0+\frac{k \Delta l^{2}}{2}$

Where, Δl is initial spring deformation

$\frac{m v^{2}}{2}=\frac{k \Delta l^{2}}{2}$

$\mathrm{v}^{2}=\frac{k \Delta l^{2}}{m}$

$\mathrm{v}=\sqrt{\frac{k}{m} \times \Delta l^{2}}$

$v=\sqrt{\frac{400}{2} \times(0.3)^{2}}$

$\mathrm{v}=\sqrt{200 \times 0.09}$

<u>The law of conservation of energy</u>:

$\frac{m v^{2}}{2}+m g h=\text { constant }$

Where h is height

$\frac{m v^{2}}{2}+0=0+m g h$

$\frac{m v^{2}}{2}=m g h$

Cancel mass "m" each side

$\mathrm{h}=\frac{v^{2}}{2 g}$

Distance along incline equals

$\begin{array}{ll}{\text { For friction us }} & {\left(L=\frac{h}{u_{s}}\right)} \\ {\text { For friction } u_{k}} & {\left(L=\frac{h}{u_{k}}\right)}\end{array}$

$\begin{array}{l}{\mathrm{u}_{\mathrm{s}}=0.4} \\ {\mathrm{U}_{\mathrm{k}}=0.2} \\ {\text { For friction } \mathrm{u}_{\mathrm{s}}}\end{array}$

$\begin{array}{l}{\mathrm{h}=\frac{v^{2}}{2 g u_{s}}} \\ {\mathrm{L}=\frac{4.24^{2}}{2 \times 9.8 \times 0.4}} \\ {\mathrm{L}=\frac{17.9776}{784}}\end{array}$

$\begin{array}{l}{L=2.29 \mathrm{m}} \\ {2.29 \mathrm{m} \text { the block slide if the } \mathrm{u}_{5}=0.4} \\ {\text { For friction } \mathrm{u}_{\mathrm{k}}} \\ {\mathrm{L}=\frac{4.24^{2}}{2 \times 9.8 \times 0.2}}\end{array}$

$\begin{array}{l}{L=\frac{17.9776}{3.92}} \\ {L=4.58 \mathrm{m}} \\ {4.58 \mathrm{m} \text { the block slide if the } \mathrm{u}_{5}=0.4}\end{array}$

8 0

3 years ago

If Earth were the size of a grape, how big would the Moon be? How far away would the Moon be from the Earth? How large would the

dangina [55]

Answer:

The moon would be the size of a raspberry. It would be as far a like 4 or 5 inches. The sun would be as big as an cantaloupe.The sun would be as far as 10 or 11 inches.

Explanation:

3 0

3 years ago

A motor does a total of 480 joules of work in 5.0 seconds to lift a 12-kilogram block to the top of a rampThe average power deve

Flauer [41]

Answer:

there it is fella

we neglect the mass data

3 0

3 years ago

an apple in a tree has a gravitational potential energy of 175J and a mass of 0.36g . how high from the ground is the apple

Neko [114]

The equation for potential energy is denoted as;

Pe = mgh,

where m = the mass, g = acceleration due to gravity, and h = vertical height of the apple. We are given the units for everything but height, which is also what we are solving for. We can then algebraically rearrange our initial equation to solve for h;

h = (Pe)/(mg)

Plug in your given units, and solve!

Post-check:

h = Pe/mg

h = 175J/(0.36g)(-9.81m/s^2)

h = appr. 49.5 meters

Note: Potential energy is a vector quantity; the displacement of the apple will be a negative number, but the distance itself, a scalar quantity, will be the absolute value of that.

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3 years ago

Please help asap I need this rn

den301095 [7]

It’s the third one .

6 0

3 years ago