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3 years ago

Name the type of reproduction process as shown in Fig 1 and Fig 2. State one point of difference between the two

Physics

1 answer:

likoan [24]3 years ago

6 0

Figure 1= binary fission in amoeba

figure 2= budding in yeast

difference

1.Parent divides to form two daughter cells and itself gets disappeared in binary fission but in budding , a bud gets matured and detaches from the parent

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Have you heard of any particle, which moves faster than light? Explore the recent researches and find about it.

Lisa [10]

Answer:

There is nothing faster then light

Explanation:

4 0

3 years ago

What is the momentum of a 0.000015 kg mosquito flying straight at you with a velocity of 5.20 m/s?

Setler [38]

Answer:

Momentum, $p=7.8\times 10^{-5}\ N-m$

Explanation:

Given that,

Mass of a mosquito, m = 0.000015 kg

Velocity of the mosquito, v = 5.2 m/s

We need to find the momentum of the mosquito. The momentum of an object is given by :

p = mv

Put all the values in the above formula.

$p=0.000015\ kg\times 5.2\ m/s\\\\p=0.000078\ N-m\\\\\text{or}\\\\p=7.8\times 10^{-5}\ N-m$

So, the momentum of the mosquito is $7.8\times 10^{-5}\ N-m$ .

3 0

3 years ago

A projectile is launched from the ground at an angle of 60o above the horizontal. At what point in its trajectory does it have t

Elodia [21]

Answer:

It's constant everywhere in its trajectory.

Explanation:

the projectile was launched with an initial velocity, the only acceleration that is affecting the projectile's velocity is gravity.

The acceleration of gravity is practically equal everywhere on earth, so during its trajectory, we have to take into consideration only the acceleration because of gravity.

This is only correct because the projectile was launched with an initial velocity and it's not accelerating from rest and then falls.

5 0

3 years ago

In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s

Damm [24]

Answer:

another ball velocity = 3.92 m/s and with 30° clockwise from initial direction

Explanation:

given data

mass m1 = 0.165 kg

mass m2 = 0.155 kg

before collision velocity v1 = 5.8 m/s

before collision velocity v2 = 0

angle = 35.0° from initial direction

after collision 1st ball velocity v3 = 3.2 m/s

to find out

after collision another ball velocity v4

solution

we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

so from conservation of momentum we say

m1v1 = m1v3cos35 + m2v4cosθ with x axis .............1

m1v3sin35 = m2v4sinθ with y axis .............2

so from 1 equation

0.165 × 5.8 = 0.165(3.2)cos35 + 0.155(v4)cosθ

v4 cosθ = 3.38 .................3

form 2 equation

0.165(3.2)sin35 = 0.155(v4)sinθ

v4 sinθ = 1.95 ......................4

so magnitude of another ball velocity is square and adding equation 3 and 4

another ball velocity = √(3.39²+1.96²)

another ball velocity = 3.92 m/s

and direction is tanθ = 1.96/3.39

θ = 30° clockwise from initial direction

3 0

3 years ago

Please help me with this

masha68 [24]

Answer:

>400N is needed to balance that lever

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2 years ago