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marin [14]
3 years ago
15

A spring scale calibrated in kilograms is used to determine the density of a rock specimen. The reading on the spring scale is 0

.45 kg when the specimen is suspended in air and 0.36 kg when the specimen is fully submerged in water. If the density of water is 1000 kg/m3, the density of the rock specimen is_________.
Physics
1 answer:
denpristay [2]3 years ago
4 0

Answer:

5000 kg/m^3

Explanation:

Here. we are asked to calculate the density of the rock specimen.

we proceed as follows;

mass of water displaced is calculated by finding the difference between the actual and apparent masses

This has a value of 0.45kg - 0.36kg = 0.09kg

The rock and water that is displaced have exactly the same volume and thus their densities is the same. This makes the ratio of their masses to be the same

Ratio of masses is

0.45 / 0.09 = 5.0

Here we can see that the mass of the rock is five times the mass of the water so it must be five times denser

Thus, since the density of water is 1000 kg/m^3 , the density of rock is 5000 kg/m^3

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A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant
mixas84 [53]

Answer:

1.92 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 200 Kg

Spring constant (K) = 10⁶ N/m

Workdone =?

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass (m) = 200 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = m × g

F = 200 × 9.8

F = 1960 N

Next we shall determine the extent to which the spring stretches. This can be obtained as follow:

Spring constant (K) = 10⁶ N/m

Force (F) = 1960 N

Extention (e) =?

F = Ke

1960 = 10⁶ × e

Divide both side by 10⁶

e = 1960 / 10⁶

e = 0.00196 m

Finally, we shall determine energy (Workdone) on the spring as follow:

Spring constant (K) = 10⁶ N/m

Extention (e) = 0.00196 m

Energy (E) =?

E = ½Ke²

E = ½ × 10⁶ × (0.00196)²

E = 1.92 J

Therefore, the Workdone on the spring is 1.92 J

3 0
2 years ago
The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?
Harrizon [31]

We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours.  So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes.  So this will be easy.

-- After 1 half-life . . .

. . . . . 50% decays, 50% is still there.

-- After the 2nd half-life . . .

. . . . . (half of the leftover 50%) = another 25% decays, 25% is left.

-- After the 3rd half-life . . .

. . . . . (half of the leftover 25%) = another 12.5% decays, 12.5% is left.

12.5% of 4,000g = (0.125 x 4,000g) = <em>500 g</em> .

============================================

<u>Another way</u>:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

After 3 half-lifes, 1/8 is left.

1/8 of 4,000g = (4,000g/8) = <em>500 g </em>.

5 0
3 years ago
Which will produce a magnetic field?
Yanka [14]

Answer:

Answer:

a magnet

3 0
3 years ago
Modern nuclear bomb tests have created an extra high level of14C in our atmosphere. When future archaeologistsdate samples from
Ipatiy [6.2K]

Answer:

Too old(Ex. if real time is 1000 then they estimate >1000)

Explanation:

This is because with time our planet may have a definite function which describes temperature.(Because of all the factors and global warming except nuclear bomb testing)

Now nuclear test on planet have significant effect on temperature rise.

Also 14°C rise in temperature is good one because of this.

If future archaeologists only consider that  uniform function as above mentioned then they estimate more time then the real one.

Thus too old is right answer.

8 0
3 years ago
A person walks 9 meters to the right and 8 meters to the left what's the distance and displacement​
nordsb [41]
The distance is 17 and the displacement is 1
5 0
2 years ago
Read 2 more answers
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