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goldenfox [79]
3 years ago
6

Please answer. Thanks

Physics
1 answer:
Brilliant_brown [7]3 years ago
8 0
I believe #1. is the answer " Mitosis and specialization", because all the information needed for growth, development, and reproduction is present in a Zygote
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A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa
DIA [1.3K]

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

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4 years ago
Which diagram shows diffraction as light passes through an opening?
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3 years ago
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the gravitational force between two objects is 1600 and what will be the gravitational force between the objects if the distance
Xelga [282]

I believe this is what you have to do:

The force between a mass M and a point mass m is represented by

F = G\frac{Mm}{r^{2} }

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁

So F₁ = G(Mm/r^2)

Now the distance has doubled so lets account for this in F₂:

F₂ = G(Mm/(2r)^2)

Now square the 2 that gives you four and we can pull that out in front to give

F₂ = \frac{1}{4} G(Mm/r^2)

Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations

now we see that:

F₂ = \frac{1}{4} F₁

So the second force will be 0.25 (1/4) x 1600 or 400 N.



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3 years ago
An Olympic high jumper, with a mass of 82 kg, has a
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Explanation:

I really wanted to help you, but then I realized i didnt know how to

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A child pulls a sled up a snow covered hill. If the child does 504J of work on the sled while pulling the sled 23m up the hill t
zhenek [66]

Explanation:

ans is equal to 504j* 23 m* 10 ms

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