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3 years ago

A meter rule balance on a knife headge at the 55cm mark wheb a mass of 40g from 995cm .find the mass of the meter rule.

Physics

1 answer:

umka2103 [35]3 years ago

7 0

Answer:

Let the weight of the ruler be X.

Now X acts in the middle, so we have 55-50 = 50cm

Since left hand movement= right hand movement

X(55-50)=90(95-55)

5X= 90×40

X=320cm

Hence the weight of the ruler is 320g

Explanation:

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1. The Moon's mass is 7.34 x 1022 kg, and it is 3.8 x 105 km away from Earth. Earth's

sineoko [7]

Answer:

2.03 x 10²⁴N

Explanation:

Given parameters:

Mass of moon = 7.34 x 10²²kg

Mass of the earth = 5.97 x 10²⁴kg

Distance = 3.8 x 10⁵km

Unknown:

Gravitational force of attraction = ?

Solution:

To find the gravitational force of attraction between the masses, we use the expression below;

F = $\frac{Gm_{1} m_{2} }{r^{2} }$

G is the universal gravitation constant

m is the mass

1 and 2 represents moon and earth

r is the distance

F = $\frac{6.67 x 10^{-11} x 7.34 x 10^{22} x 5.97 x 10^{24} }{(3.8 x 10^{5})^{2} }$

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8 0

2 years ago

An old-fashioned LP record rotates at 33 1/3 RPM

Ksenya-84 [330]

Answer:

Part a) $\frac{5}{9}\ \frac{rev}{sec}$

Part b) $\frac{9}{5}\ \frac{sec}{rev}$

Explanation:

Part a) what is its frequency, in rev/s

we have that

An old-fashioned LP record rotates at 33 1/3 RPM

$33\frac{1}{3}\ \frac{rev}{min}$

Convert mixed number to an improper fraction

$33\frac{1}{3}\ \frac{rev}{min}=\frac{33*3+1}{3}=\frac{100}{3}\ \frac{rev}{min}$

Remember that

$1\ min=60\ sec$

Convert rev/min to rev/sec

$\frac{100}{3}\ \frac{rev}{min}=\frac{100}{3}(\frac{1}{60})=\frac{100}{180}\ \frac{rev}{sec}$

Simplify

$\frac{5}{9}\ \frac{rev}{sec}$

Part b) what is it period, in seconds

we know that

The period is the reciprocal of the frequency

therefore

the frequency is

$\frac{9}{5}\ \frac{sec}{rev}$

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3 years ago

do constructive inference occur when the compression of one wave meets up with the compression of a second wave

Ugo [173]

Answer:

Yes

Explanation:

There are two types of interference possible when two waves meet at the same point:

- Constructive interference: this occurs when the two waves meet in phase, i.e. the crest (or the compression, in case of a longitudinale wave) meets with the crest (compression) of the other wave. In such a case, the amplitude of the resultant wave is twice that of the original wave.

- Destructive interferece: this occurs when the two waves meet in anti-phase, i.e. the crest (or the compression, in case of a longitudinal wave) meets with the trough (rarefaction) of the other wave. In this case, the amplitude of the resultant wave is zero, since the amplitudes of the two waves cancel out.

In this problem, we have a situation where the compression of one wave meets with the compression of the second wave, so we have constructive interference.

6 0

3 years ago

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BaLLatris [955]

Answer:

v = 20.31 m/s

Explanation:

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3 years ago

A charge q of magnitude 6.4 × 10^-19 coulombs moves from point A to point B in an electric field of 6.5 × 10^4 newtons/coulomb.

lana [24]