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2 years ago

A wave created by shaking a rope up and down is called a.

1 answer:

5 0

A wave created by shaking a rope up and down is called a:
(According to me and certified experts) Transverse wave.

If this helps, then please give Brainliest!

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Which structures allow for the removal of wastes from the developing fetus?

castortr0y [4]

The fetus relies upon its mother as it develops. These are some of the things it needs:

protection against knock and bumps, and temperature changes oxygen for respiration nutrients (food and water)

The developing fetus also needs its waste substances removing.

The fetus is protected by the uterus and the amniotic fluid, a liquid contained in a bag called the amnion.

5 0

4 years ago

A boy is looking into a plane mirror .His nose is 30 cm in front of the mirror .How far is his nose from where its image appears

natima [27]

Answer:

60cm

Explanation:

30+30=60

8 0

3 years ago

Examples of stored energy

jeyben [28]

Answer:

Biomass, petroleum, natural gas, and propane are examples of stored chemical energy. Energy is energy stored in objects by the application of a force

3 0

4 years ago

How do sex cells differ from other human cells?

VLD [36.1K]

Sex cells have half as many chromosomes as other cells. Sex cells are therefore called haploid.

7 0

4 years ago

A tank whose bottom is a mirror is filled with water to a depth of 20.0 cm. A small fish floats motionless 7.0 cm under the surf

liberstina [14]

Answer:

The apparent depth of (a) the fish is 5.3 cm and (b) the image of the fish is 24.8 cm.

Explanation:

According to the following equation:

$\frac{n_{w} }{s} +\frac{n_{a} }{s'} = \frac{n_{a}- n_{w}}{R_{c} } \\$

where nw and na is the refractive indices of water (1.33) and air (1.00) respectively; s is the depth of the fish below the surface of the water; s' is the apparent depth of the fish from normal incidence and Rc is the radius of curvature of the mirror at the bottom of the tank.

Note that the bottom of the tank is assumed to be a flat mirror, therefore the radius of curvature is very large (R⇒∞).

Therefore, the above equation can be expressed as:

$\frac{n_{w}}{s} +\frac{n_{a}}{s'}=0$

Now we can solve for the apparent depth of the fish.

(a) $s'=-(\frac{n_{a}}{n_{w}})x s$ (Make s' subject of the formula from the above equation)

$s'=(\frac{1.00}{1.33} )x7cm$

∴ $s'=5.3 cm.$

(b) The motionless fish floats 13 cm above the mirror, therefore the image of the fish will be situated at 13 + 20 =33 cm away from the real fish.

Therefore, s = 33 cm

$s'=-(\frac{n_{a}}{n_{w}})x s$

$s'=(\frac{1.00}{1.33} )x 33 cm$

$s'=24.8 cm.$

NB: Here, it is assumed that the water is pure, as impurities may alter the refractive index of water.

8 0

4 years ago