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3 years ago

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Archimedes supposedly was asked to determine whether a crown made for the king consisted of puregold. According to legend, he so

lved this problem by weighing the crown first in air and then in water. Suppose the scale read 7.84 N with the crown in air and 6.84 N when it was in water. What should Archimedes have told the king

1 answer:

DENIUS [597]3 years ago

4 0

Answer:

the crown was not made of pure gold

Explanation:

Mass of gold = weight in air/ g = 7.84N/10ms-2= 0.784 Kg or 0.8Kg

From Archimedes principle:

Upthrust= weight in air- weight in a fluid

Upthrust= volume × density × g

Note density of water = 1000kgm-3

7.84-6.84= V × 1000kgm-3×10ms-2

V= 1/10000= 1×10-4 m^3

Density = mass/ volume= 0.8/1×10-4

= 8×10^3 Kgm-3

But we know the density of gold to be 19.3 ×10^3 kgm-3

Hence the crown was not made of pure gold

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A ski gondola is connected to the top of a hill by a steel cable oflength 620 m and diameter 1.5 cm. As thegondola comes to the

klemol [59]

Answer:

Explanation:

Given

Length of cable $L=620 m$

Diameter of cable $d=1.5 cm$

time taken to return to original position $T=14 s$

time taken to cover distance L

$t=\frac{T}{2}=7 s$

velocity

$v=\frac{L}{t}=\frac{620}{7}=88.57 m/s$

(b)Relation between velocity of wave Tension is

$v=\sqrt{\frac{T}{\mu }} , where \mu =$ mass per unit Length

$T=v^2\cdot \mu$

$T=(88.57)^2\cdot \frac{m}{L}$

$T=(88.57)^2\cdot \frac{\rho AL}{L}$

where $\rho =density\ of\ steel =7850 kg/m^3$

$A=area\ of\ cross-section=\frac{\pi }{4}d^2=1.76\times 10^{-4} cm^2$

$T=(88.57)^2\cdot 7850\times 1.76\times 10^{-4}$

$T=10,883 N$

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3 years ago

An astronaut in a satellite 1600km above the Earth experience a gravitational force magnitude 700N on earth.The Earth's radius i

Brilliant_brown [7]

Answer:

The magnitude of the gravitational force which the astronaut experience in satellite $= 448$ Newton

Explanation:

As we know

Gravitational force F = $\frac{GMm}{r^2}$

Where G is the gravitational constant $6.67 * 10^{-11}$

M is the mass of earth $= 6* 10^{24}$

m is not given

r is the radius of earth $6 400*10^3$

Substituting the given values, we get -

$F = \frac{6.67*10^{−11}*6*10^{24}* m}{(6400*10^3)^2}$

$m = 71.6$ Kg

Gravitational force in the satellite

$F = \frac{6.67*10^{−11}*6*10^{24}* m}{(6400*10^3 + 1600 *10^3)^2}\\F = 448$ N

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3 years ago

a missile is moving 1810 m/s at a 20.0. It needs to hit a target 29,500m away in a 65.0 magnitude in 9.20s. What acceleration mu

Vesna [10]

Answer:

323 m/s²

Explanation:

Given:

x₀ = 0 m

y₀ = 0 m

x = 29500 cos 65°

y = 29500 sin 65°

v₀x = 1810 cos 20°

v₀y = 1810 sin 20°

t = 9.20

Find:

ax, ay, θ

First, in the x direction:

x = x₀ + v₀ t + ½ at²

29500 cos 32° = 0 + (1810 cos 20°) (9.20) + ½ ax (9.20)²

25017 = 15648 + 42.32 ax

ax ≈ 221.4

And in the y direction:

y = y₀ + v₀ t + ½ at²

29500 sin 32° = 0 + (1810 sin 20°) (9.20) + ½ ay (9.20)²

15633 = 5695 + 42.32 ay

ay ≈ 234.8

Therefore, the magnitude of the acceleration is:

a² = ax² + ay²

a² = (221.4)² + (234.8)²

a ≈ 322.7

Rounded to 3 significant figures, the magnitude of the acceleration is approximately 323 m/s².

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3 years ago

A race car starting from rest accelerates uniformly at a rate of 3.90 m/s^2 what is the cars speed after it has traveled 200 m.

snow_tiger [21]

The formula for accelerational displacement is at^2/2, so we know that 3.9t^2/2 = 200, or 3.9t^2 = 400. t = $\sqrt{400/3.9} \approx 10.12739367$ , at = v, so $3.9 * 10.12739367 \approx 39.5 m/s$

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3 years ago

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A speaker generates a continuous tone of 440 Hz. In the drawing, sound travels into a tube that splits into two segments, one lo

chubhunter [2.5K]

Answer:

The minimum difference between the lengths of the two tubes should be 0.385 meters.

Explanation:

As we known that for any two waves to arrive in phase at any point the difference in the path traveled by the waves should be an integral multiple of the wavelength of the wave.

Mathematically we can write:

$\Delta x=n\frac{\lambda }{2}$

For the given wave we have

$\lambda =\frac{v}{\nu }$

Applying values we get

$\lambda =\frac{339}{440 }=0.77m$

Thus the minimum difference in the lengths of the tubes can be obtained by putting the value of n = 1

$\therefore \Delta x=1\times \frac{0.77}{2}=0.385m$

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4 years ago