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3 years ago

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the density of gold is 19.3 g/cm3. suppose a certain gold wedding ring deplaced 0.55mL of liquid when dropped in a glass of spar

kling cider. how much did this wedding ring cost

1 answer:

adell [148]3 years ago

8 0

Answer:

19.3

Explanation:

Assuming we have to find Specific gravity of gold.

As we know that specific gravity is defined as the ratio of weight of the object and weight of the water displaced by the object

so it is given by

specific gravity = weight of the object/weight of the water displaced

now we have

weight of the object = (density)(volume)g

weight of object = (19.3)(0.55)g

now weight of the liquid displaced is given by

weight of water displaced = (1 g/cm^3)(0.55ml)g

now we have

specific gravity = (19.3×0.55)/(1×0.55)

specific gravity= 19.3

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Mechanical advantage is a quantity that measures how much a machine multiplies force or distance.

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A 0.8 kg bead slides on a curved wire, starting

TEA [102]

Answer:

vB = 12 m/s

W = 36 J

Explanation:

Potential energy converts to kinetic energy

½mv² = mgh

v = √(2gh)

vB = √(2(9.8)(7.1 - 0) = 11.796609... m/s

W = mg(hA) - mg(hB) = 0.8(9.8)(7.1 - 2.5) = 36.064 J

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3 years ago

On a cold day, the speed of sound in air is 330 m/s. A note with a frequency of 1,320 Hz is played on an instrument. What is the

solmaris [256]

As we know that speed of sound is

$v = 330 m/s$

frequency of sound is

$f= 1320 Hz$

now we know that frequency of sound and speed of sound is related to each other by the following formula

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$v = \lambda * f$

$330 = 1320 * \lambda$

$\lambda = 0.25 m$

<em>so wavelength of sound is 25 cm or 0.25 m</em>

9 0

3 years ago

Read 2 more answers

a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions doe

kramer

Complete Question

a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions does each tire make before the car comes to a stop? Assume that the car does not skid and that each tire has a radius of 0.307 m. Answer in units of rev.

Answer:

The value is $N = 109 \ rev$

Explanation:

From the question we are told that

The speed of the car is $u = 28.4 \ m/s$

The constant deceleration experienced is $a = 1.92 \ m/s^2$

The radius of the tire is $r = 0.307 \ m$

Generally from kinematic equation we have that

$v^2 = u^2 + 2as$

Here v is the final velocity which is 0 m/s

So

$0^2 = 28.4^2 + 2 * 1.92 * s$

=> $s = 210.04 \ m$

Generally the circumference of the tire is mathematically represented as

$C = 2 \pi r$

=> $C = 2 * 3.142 * 0.307$

=> $C = 1.929 \ m$

Generally the number of revolution is mathematically represented as

$N = \frac{ s}{C}$

=> $N = \frac{210.04}{1.929}$

=> $N = 109 \ rev$

5 0

3 years ago

Two vectors are given by a = 8.6i + 5.1 j and b = 931 + 9.5.

skelet666 [1.2K]

Answer:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>

a) a×b = 34.27k

b) a·b = 128.43

c) (a + b)·b = 305.17

d) The component of a along the direction of b = 9.66

Explanation:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:

a) The vectorial product, a×b is:

$a \times b = (8.6*9.5 - 5.1*9.3)k = 34.27k$

b) The escalar product a·b is:

$a\cdot b = (8.6*9.3) + (5.1*9.5) = 128.43$

c) <u>Asumming (a</u><u> </u><u>+ b)·b</u> <em>instead a+b·b</em> we have:

$(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17$

d) The component of a along the direction of b is:

$a*cos(\theta) = \frac{a\cdot b}{|b|} = \frac{128.43}{\sqrt{9.3^{2} + 9.5^{2}}} = 9.66$

I hope it helps you!

5 0

4 years ago