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blsea [12.9K]
3 years ago
13

 15cm3 block of gold weighs  2.8N it is carefully submerged in a tank of mercury. one cm3 of mercury weight 0.!3n a will the mer

cury be displaced by the gold?
Physics
1 answer:
NNADVOKAT [17]3 years ago
3 0
You have said that 15cm³ of gold weighs 2.8N.  So I may infer that each cm³
of gold weighs about 0.19N.  When I compare that figure with the 0.13N per cm³
of mercury, it becomes immediately apparent that the gold is more dense than
mercury. Therefore, the sample of gold, no matter what its size or weight, will
displace its total volume of mercury, and will go on to sink entirely beneath the
waves in the mercury.
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Answer:

F=-88Ib

Explanation:

From the question we are told that:

Force P=88Ib

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Generally the equation for Frictional force F is mathematically given by

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4 0
3 years ago
One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe
Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

\mu = \frac{m}{l}

\mu = \frac{8.79*10^{-3}}{70*10^{-2}}

\mu = 0.01255kg/m

The expression for the wavelength of the standing wave for the second overtone is

\lambda = \frac{2}{3} l

Replacing we have

\lambda = \frac{2}{3} (70*10^{-2})

\lambda = 0.466m

The frequency of the sound wave is

f_s = \frac{v}{\lambda_s}

f_s = \frac{344}{0.768}

f_s = 448Hz

Now the velocity of the wave would be

v = f_s \lambda

v = (448)(0.466)

v = 208.768m/s

The expression that relates the velocity of the wave, tension on the string and linear mass density is

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

v^2 = \frac{T}{\mu}

T= \mu v^2

T = (0.01255kg/m)(208.768m/s)^2

T = 547N

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the n^{th} harmonic frequency is

f_n = nf_1

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

n=3

Then,

f_3 = 3f_1

Rearranging to find the fundamental frequency

f_1 = \frac{f_3}{3}

f_1 = \frac{448Hz}{3}

f_1 = 149.9Hz

7 0
3 years ago
A 10.0-kg box starts at rest and slides 3.5 m down a ramp inclined at an angle of 10° with the horizontal. If there is no fricti
Alenkinab [10]

Answer:

v_f = 3.45 m/s

Explanation:

As we know that box was initially at rest

so here work done by all forces on the box = change in its kinetic energy

so we will have

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now we have

m = 10 kg

\theta = 10^0

L = 3.5 m

so we will have

10(9.81)sin10 \times 3.5 = \frac{1}{2}(10)(v_f^2 - 0)

2(9.81) sin10 \times 3.5 = v_f^2

so final speed is given as

v_f = 3.45 m/s

6 0
3 years ago
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