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Vadim26 [7]
3 years ago
7

Pls pls help me out AHH, what is not true about MEIOSIS?

Physics
1 answer:
agasfer [191]3 years ago
4 0

Answer:

B.

Explanation:

This is not true as the number of chromosomes in the daughter cells are half the number in the parent's cells

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Explain a compression and a refraction of a sound wave traveling though the air
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Hmm this is difficult you need to go to answers.con and it come out cya :D
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4 years ago
A flute player hears four beats per second when she compares her note to an 880 Hz tuning fork (note A). She can match the frequ
ludmilkaskok [199]

Answer:

884Hz

Explanation:

Beats is the absolute difference between two frequencies therefore

Beats = f1-f2

4=f1-880

F1=880+4

F1=884Hz

7 0
3 years ago
When the spacecraft is at the halfway point, how does the strength of the gravitional force on the spaceprobe by Earth compre wi
mixer [17]

Solution :

When the spacecraft is at halfway point, the distance from the Earth as well as Mars are same. We have to account the masses of the planets. The gravitational force that is exerted by the Earth is greater because of its combined mass with the space probe.

The mass of Earth is greater than the mass of Mars. Therefore, the force of Earth is more than Mars.

5 0
3 years ago
A baseball has mass 0.147 kg. If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55
Anit [1.1K]

Explanation:

We have,

Mass of a baseball is 0.147 kg

Initial velocity of the baseball is 44.5 m/s

The ball is moved in the opposite direction with a velocity of 55.5 m/s

It is required to find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Change in momentum,

\Delta p=mv-mu\\\\\Delta p=m(v-u)\\\\\Delta p=0.147\times ((-55.5)-44.5)\\\\\Delta p=-14.7\ kg-m/s\\\\|\Delta p|=14.7\ kg-m/s

Impulse = 14.7 kg-m/s

Therefore, the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat is 14.7 kg-m/s

4 0
3 years ago
A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an ini
Volgvan

Answer:

The horizontal range will be 2.55\times 10^5m

Explanation:

We have given initial speed of the shell u = 1.6\times 10^3m/sec

Angle of projection = 51°

Acceleration due to gravity g=9.8m/sec^2

We have to find maximum range

Horizontal range in projectile motion is given by

R=\frac{u^2sin2\Theta }{g}=\frac{(1.60\times 10^3)^2sin(2\times 51^{\circ})}{9.81}=2.55\times 10^5m

So the horizontal range will be 2.55\times 10^5m

6 0
3 years ago
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