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3 years ago

WHAT IS THE UNBALANCED FORCE THAT SLOWS DOWN SLIDING DESKS AND MOVING CARS

Physics

2 answers:

gogolik [260]3 years ago

6 0

Friction is a forces that slows things down.

umka2103 [35]3 years ago

3 0

Friction slows it down

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A magnetic field is created by ____.

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3 years ago

An aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 30mm in diameter and i

ASHA 777 [7]

Answer:

$\delta = 0.385\,m$ (Compression)

Explanation:

The aluminium bar is experimenting a compression due to an axial force, that is, a force exerted on the bar in its axial direction. (See attachment for further details) Under the assumption of small strain, the deformation experimented by the bar is equal to:

$\delta = \frac{P\cdot L}{A \cdot E}$

Where:

$P$ - Load experimented by the bar, measured in newtons.

$L$ - Length of the bar, measured in meters.

$A$ - Cross section area of the bar, measured in square meters.

$E$ - Elasticity module, also known as Young's Module, measured in pascals, that is, newtons per square meter.

The cross section area of the bar is now computed: ( $D_{o} = 0.04\,m$ , $D_{i} = 0.03\,m$ )

$A = \frac{\pi}{4}\cdot (D_{o}^{2}-D_{i}^{2})$

Where:

$D_{o}$ - Outer diameter, measured in meters.

$D_{i}$ - Inner diameter, measured in meters.

$A = \frac{\pi}{4}\cdot [(0.04\,m)^{2}-(0.03\,m)^{2}]$

$A = 5.498 \times 10^{-4}\,m^{2}$

The total contraction of the bar due to compresive load is: ( $P = -180\times 10^{3}\,N$ , $L = 0.1\,m$ , $E = 85\times 10^{9}\,Pa$ , $A = 5.498 \times 10^{-4}\,m^{2}$ ) (Note: The negative sign in the load input means the existence of compressive load)

$\delta = \frac{(-180\times 10^{3}\,N)\cdot (0.1\,m)}{(5.498\times 10^{-4}\,m^{2})\cdot (85\times 10^{9}\,Pa)}$

$\delta = -3.852\times 10^{-4}\,m$

$\delta = -0.385\,mm$

$\delta = 0.385\,m$ (Compression)

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3 years ago

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4 years ago

A body accelerates uniformly from rest at 2m/s square. calculate it's velocity after traveling 9m.

Talja [164]

Answer:

6m/s

Explanation:

Given parameters:

Initial velocity = 0m/s

Acceleration = 2m/s²

Distance = 9m

Unknown:

Final velocity = ?

Solution:

To solve this problem, we use the expression below:

v² = u² + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

v² = 0² + (2 x 2 x 9) = 36

v = 6m/s

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3 years ago