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Naya [18.7K]
3 years ago
14

Which principle(s) of the cell theory shows that all organisms are somehow similar?

Physics
1 answer:
Stolb23 [73]3 years ago
7 0

Explanation:

The unified cell theory states that: all living things are composed of one or more cells; the cell is the basic unit of life; and new cells arise from existing cells.

hope it helps

plz mark as brainliest

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The curved line PQR is the velocity-time graph for a car starting from rest.
Elden [556K]
His. Curbs I b h bs. H b u b
3 0
3 years ago
A bottle rocket takes off with a = 34.5 m/s2. It
sergiy2304 [10]

Answer:3.4 seconds

Explanation:

Initial velocity(u)=0

acceleration=34.5m/s^2

Height(h)=200m

Time =t

h=u x t - (gxt^2)/2

200=0xt+(34.5xt^2)/2

200=34.5t^2/2

Cross multiply

200x2=34.5t^2

400=34.5t^2

Divide both sides by 34.5

400/34.5=34.5t^2/34.5

11.59=t^2

t^2=11.59

Take them square root of both sides

t=√(11.59)

t=3.4 seconds

8 0
3 years ago
A boy drags a box of toys.It experiences a frictional force of 10N .calculate acceleration of the system
hodyreva [135]

Answer:

20 N

Explanation:

i believe that is right

7 0
3 years ago
The maximum tension the wire can withstand without breaking is 300 N . A 0.800 kg projectile traveling horizontally hits and emb
topjm [15]

Answer:

Let's investigate the case where the cable breaks.

Conservation of angular momentum can be used to find the speed.

\vec{L}_1 = \vec{L}_2\\\vec{L}_1 = m\vec{v_0} \\\vec{L}_2 = I\vec{\omega}\\

The projectile embeds itself to the ball, so they can be treated as a combined object. <u>The moment of inertia of the combined object is equal to the sum of the moment of inertia of both objects. </u>

I = I_{projectile} + I_{ball}\\I = mr^2 + mr^2\\I = 2mr^2

where r is the length of the cable.

<u>After the collision, the ball and the projectile makes a circular motion because of the cable.</u> So, the force (tension) in circular motion is

F = \frac{mv^2}{r}

The relation between linear velocity and the angular velocity is

v = \omega r

So,

F = \frac{m(\omega r)^2}{r} = m\omega^2 r = 300\\mv_0r  =I\omega\\\\\omega = mv_0r/I\\300 = m(\frac{mv_0r}{I})^2r = m(\frac{mv_0r}{2mr^2})^2r = m(\frac{v_0}{2r})^2r = \frac{mv_0^2r}{4r^2} = \frac{mv_0^2}{4r}\\300 = \frac{0.8v_0^2}{4r}\\1500 = v_0^2/r\\v_0 = \sqrt{1500r}

As can be seen, the maximum velocity for the projectile without breaking the cable is \sqrt{1500r}, where r is the length of the cable.

6 0
3 years ago
Bonus: We have looked at scenarios under many ideal conditions. In reality when an object falls for long enough it reaches termi
MrMuchimi

Answer:

The answer is A.

Explanation:

4 0
3 years ago
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