From the Water in each of the next move

An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in

valkas [14]

Answer:

Volume of the sample: approximately $\rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}$ .

Average density of the sample: approximately $\rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}$ .

Assumption:

$\rm g = 9.81\; N \cdot kg^{-1}$ .
$\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ .
Volume of the cord is negligible.

Explanation:

<h3>Total volume of the sample</h3>

The size of the buoyant force is equal to $\rm 17.50 - 11.20 = 6.30\; N$ .

That's also equal to the weight (weight, $m \cdot g$ ) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with $g$ .

$\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg$ .

Assume that the density of water is $\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ . To the volume of water displaced from its mass, divide mass with density $\rho(\text{water})$ .

$\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}$ .

Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.

$V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}$ .

<h3>Average Density of the sample</h3>

Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with $g$ .

$\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg$ .

The volume of the sample is found in the previous part.

Divide mass with volume to find the average density.

$\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}$ .

3 0

3 years ago

A question is scientific if you can answer it using which of the following?

pav-90 [236]

C. Observations and measurements

7 0

2 years ago

Two sources of light of wavelength 700 nm are 9 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for

Lelechka [254]

Answer:

The distance is $D = 0.000712 \ m$

Explanation:

From the question we are told that

The wavelength of the light source is $\lambda = 700 \ nm = 700 *10^{-9} \ m$

The distance from a pin hole is $x = 9\ m$

The diameter of the pin hole is $d = 1.2 \ mm = 0.0012 \ m$

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

$D = \frac{1.22 \lambda }{d }$

substituting values

$D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }$

$D = 0.000712 \ m$

5 0

2 years ago

Operators must slow to no wake speed when within what distance of a u.s. naval vessel?

Alex Ar [27]

Recreational boaters have a role in keeping the waterways safe and secure. Violators of the restrictions can expect a quick and severe response. The operator should not approach within yards and slow to minimum speed within 500 yards of any of united states naval vessel. If the vessel need to pass within 100 yards of a united states naval vessel for safe passage, must contact the united states naval vessel or the united states coast guard escort vessel on VHF-FM channel 16. In addition,

• The operator must perceive and avoid all security areas.

• Avoid commercial port operation areas particularly those that include military cruise line or petroleum amenities.

• The operator must perceive and avoid other controlled areas near dams, power plants, etc.

• The operator of the vessel must not discontinue or anchor underneath bridges or in the channel.

6 0

2 years ago

Read 2 more answers

A missile is launched vertically from a missile silo, it will explode after 32 s. It's launch speed was 145 m/s, and it doesn't

Karolina [17]

Answer:

Landed before it explodes

Explanation:

vf = vi + at,

0 = 145 - (9.8)t,

t = 14.79 s (Time to reach highest point)

14.79 x 2 = 29.59 s (Time to land on the ground)

It will have landed before it explodes because both the time to reach the highest point and the time to land on the ground are less than 32 seconds.

4 0

2 years ago