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4 years ago

How do sex cells differ from other human cells?

Physics

1 answer:

VLD [36.1K]4 years ago

7 0

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

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Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 23.5 m above water wit

Elodia [21]

As stated in the statement, we will apply energy conservation to solve this problem.

From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

$\Delta KE = \Delta PE$

$\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1$

Where,

m = mass

$v_{f,i}$ = initial and final velocity

g = Gravity

h = height

As the mass is tHe same and the final height is zero we have that the expression is now:

$\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2$

$\frac{1}{2} (v_f^2-v_0^2) = gh_2$

$(v_f^2-v_0^2) = 2gh_2$

$v_f = \sqrt{2gh_2+v_0^2}$

$v_f = \sqrt{2(9.8)(23.5)+13.6^2}$

$v_f = 25.4m/s$

7 0

3 years ago

What type of force causes your car to move?

amid [387]

Answer:

frictional force

Explanation:

3 0

3 years ago

The process of recording the strength of the contraction of a muscle when it is stimulated by an electrical current is known

Naddik [55]

i dont know the answer of your question correctly

3 0

3 years ago

If the Hubble Space Telescope is 598 m above the surface of the Earth and is traveling at 7.56 x 103 m/s, how long does it take

viva [34]

Answer:

5069.04 seconds

Explanation:

The parameter we are looking for is called the Orbital period of the Hubble Space Telescope.

It is given as:

$T = \sqrt{\frac{4\pi^2r^3 }{GM} }$

where r = radius of orbit of Hubble Space Telescope

G = gravitational constant = $6.67408 * 10^{-11} m^3 kg^{-1} s^{-2}$

M = Mass of earth

We are given that:

r = radius of the earth + distance of HST from earth

r = $6.38 * 10^6 + 598 = 6380598 m$

M = $5.98 * 10^{24} kg$

Therefore, T will be:

$T = \sqrt{\frac{4*\pi^2 * 6380598^3 }{6.67408 * 10^{-11} * 5.98 * 10^{24}}}$

$T = 5069.04 secs$

The orbital period of the Hubble Space Telescope is 5069.04 seconds.

7 0

3 years ago

Read 2 more answers

A student lifts a box of books 2 meters with a force of 45 N. He then carries the box 10 meters to the living room. What is the

Alisiya [41]

Answer:

90J

Explanation:

The only time work is being done is when he lifts the box off the ground. Therefore, using the work formula, 2 x 45, you get 90J. Hope this helps someone.

5 0

3 years ago