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3 years ago

احسبي محصلة أزاحة طالب يتحرك مسافة 40m شمالا ثم يتحرك 20m جنوبا ما محصلة ازاحته وفي اي اتجاه ؟

Physics

2 answers:

Phantasy [73]3 years ago

6 0

Answer:

احسبي محصلة أزاحة طالب يتحرك مسافة 40m شمالا ثم يتحرك 20m جنوبا ما محصلة ازاحته وفي اي اتجاه ؟

Serggg [28]3 years ago

4 0

No idea sorry for that

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Why is a standard system of measurement important?

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Which of the following properties of a sound wave determines its loudness (or intensity): wavelength, speed, amplitude, or frequ

lara31 [8.8K]

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Explanation: The intensity or loudness of a sound depends upon the extent to which the sounding body vibrates (i.e the amplitude of vibration).

Loudness is measured in units called decibels and amplitude in metre

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3 years ago

The mass of the parts is always ___ the mass of the whole when looking at nuclear masses.

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4 years ago

determine the pressure exerted on the surface of a submarine cruising 175 ft below the free surface of the sea. assume that the

Alenkasestr [34]

Answer:

92.81 psia.

Explanation:

The density of water by multiplying its specific gravity by the density of sea water.

SG = density of sea water/density of water

ρ = SG x ρw

1 kg/m3 = 62.4 lbm/ft^3

= 1.03 * 62.4

= 64.27lbm/ft^3.

The absolute pressure at 175 ft below sea level as this is the location of the submarine.

P = Patm +ρgh

= 14.7 + 64.27 * 32.2 * 175

Converting to pound force square inch,

= 14.7 + 64.27 * (32.2ft/s^2) * (175ft) * (1lbf/32.2lbm⋅ft/s^2) * (1ft^2/144in^2 )

= 14.7 + 78.11 psia

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3 years ago

A wet bar of soap slides down a ramp 9.2 m long inclined at 8.0∘ .

xenn [34]

Answer:

The time taken by the bar to reach the bottom t=4.886s

Given:

Displacement of the bar S=9.2m

Angle of inclination $\theta=8.0^{\circ}$

Coefficient of friction factor $\mu k=0.056$

To find:

How long it takes to reach the bottom ‘t’

Step by Step Explanation:

Solution:

We know that the formula for weight of the soap bar is given as

$F_{g}=m g \sin \theta$

The frictional force acting on this soap bar is determined by

$F_{f}=\mu m g \cos \theta$

To determine the constant acceleration of the bar, we derive as

$F=m a$

Here $F=F_{g}-F_{f}$ and thus

$F_{g}-F_{f}=m a$ $m g \sin \theta-\mu m g \cos \theta=m a$

$a=g \sin \theta-\mu g \cos \theta$

Where $F_{g}$ =Force imparted due to weight

$F_{f}$ =Frictional Force

m=Mass of the bar

g=Acceleration due to gravity

a=Acceleration of the bar

$\sin \theta$ and $\cos \theta$ are the angles involved in the system

If the bar starts from the rest

Equations of motion involved in calculating the displacement of the bar is given as

$s=\frac{1}{2} a t^{2}$ , From this

$a t^{2}=2 s$

$t^{2}=\frac{2 s}{a}$

$t=\sqrt{\frac{2 s}{a}}$

Where s= displacement or length moved by the bar

a=Acceleration of the bar

t=Time taken to reach bottom

Substitute all the known values in the above equation we get

$t=\sqrt{\frac{2 \times 9.2}{a}}$ and we know that

$a=g \sin \theta-\mu g \cos \theta$

$=9.8 \times \sin 8.0-0.056 \times 9.8 \times \cos 8.0$

$=1.364-0.543$

$a=0.821$

$t=\sqrt{\frac{2 \times 9.2}{0.821}}$

$t=\sqrt{\frac{19.6}{0.821}}$

$t=\sqrt{23.87332}$

t=4.886s

Result:

Thus the time taken by the bar to reach the bottom is t=4.886s

3 0

3 years ago