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3 years ago

A submarine is observed to rise from a real depth of 80 cm to 60 cm in water calculate the change in apparent depth

Physics

1 answer:

Bezzdna [24]3 years ago

6 0

Answer:

15m

Explanation:

change in real depth = 80 - 60 = 20m

Let us assume that the index is 4/3

Refractive index = change in real depth ÷ change in apparent depth

change in apparent depth = change in real depth÷ refractive index

= 20÷(4/3) = 20×(3/4) = 5×3 = 15m

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4. Which of the following statements about the state of matter is correct? A. Molecules in a solid are disorderly arranged B. So

Zolol [24]

Answer:

B. Solids are highly compressible

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3 years ago

What is the EM used for?

Phantasy [73]

Answer:

Applications of electromagnetic waves

Explanation:

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3 years ago

What represents the impulse of the force in a graph of force versus time?

vovikov84 [41]

In the graph of the force vs the time:
Force is the gradient of the momentum versus the time.

If we get the area under the curve in this graph:
it will be calculated as : force * time
which gives the change in momentum or the impulse.

Therefore, the area under the curve represents the impulse of the force in a graph of force versus time

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4 years ago

The block in the drawing has dimensions L0×2L0×3L0,where L0 =0.5 m. The block has a thermal conductivity of 200 J/(s·m·C˚). In d

Dennis_Churaev [7]

Answer:

$Q_p=18000\ J$

$Q_o=8000\ J$

$Q_g=72000\ J$

Explanation:

Given:

Length of block, $l=3L_o=3\times 0.5=1.5\ m$

Breadth of block, $b=L_o=0.5\ m$

height of block, $h=2L_o=2\times 0.5=1\ m$

Thermal conductivity of the block, $k=200\ W.m.^{\circ}C$

Temperature on the hotter side, $T_H=37^{\circ}C$

temperature on the cooler side, $T_L=7^{\circ}C$

time for which the heat flows, $t=4\ s$

REFER THE ATTACHED IMAGE FOR THE REFERENCE

The rate of heat flow using Fourier's law of conduction is given as:

$\frac{Q}{t}=k.A.\frac{dT}{dx}$

Now the amount heat flow perpendicular to the pink surface:

$\frac{Q_p}{4}=200\times (0.5\times 1.5).\frac{30}{1}$

$Q_p=18000\ J$

Now the amount heat flow perpendicular to the orange surface:

$\frac{Q_o}{4}=200\times (0.5\times 1).\frac{30}{1.5}$

$Q_o=8000\ J$

Now the amount heat flow perpendicular to the green surface:

$\frac{Q_g}{4}=200\times (1.5\times 1).\frac{30}{0.5}$

$Q_g=72000\ J$

6 0

3 years ago

A copper wire has a diameter of 4.00 x 10-2 inches and is originally 10.0 ft long. What is the greatest load that can be support

jek_recluse [69]

Complete question is;

A copper wire has a diameter of 4.00 × 10^(-2) inches and is originally 10.0 ft long. What is the greatest load that can be supported by this wire without exceeding its elastic limit? Use the value of 2.30 × 10⁴ lb/in² for the elastic limit of copper.

Answer:

F_max = 28.9 lbf

Explanation:

Elastic limit is simply the maximum amount of stress that can be applied to the wire before it permanently deform.

Thus;

Elastic limit = Max stress

Formula for max stress is;

Max stress = F_max/A

Thus;

Elastic limit = F_max/A

F_max is maximum load

A is area = πr²

We have diameter; d = 4 × 10^(-2) inches = 0.04 in

Radius; r = d/2 = 0.04/2 = 0.02

Plugging in the relevant values into the elastic limit equation, we have;

2.30 × 10⁴ = F_max/(π × 0.02²)

F_max = 2.30 × 10⁴ × (π × 0.02²)

F_max = 28.9 lbf

5 0

3 years ago