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Ratling [72]
3 years ago
8

The first simple, one-celled organisms that developed during the Precambrian era are called A. Trilobites B. Prokaryotes C. Stro

matolites
Physics
2 answers:
Ulleksa [173]3 years ago
7 0
B Prokaryotes, which are single celled organisms
sergeinik [125]3 years ago
3 0

Answer: B. Prokaryotes

On the basis of the fossil evidences found the prokaryotes including the bacteria and archea were the first single celled organisms, which were expected to be present in the 3.5 to 3.8 billion years ago typically in the Precambrian Period. On the basis of the fossil evidences obtained these organisms were abundantly present.

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9. If the musician hit the drum on a stage, how would the sound wave behave differently if he hit it the drum if the drum were s
egoroff_w [7]
I think it would be yes because the drum is submerged in water and the water would slow the sound waves, making the sound softer. Right?
4 0
3 years ago
A wooden block meauring 40cm x 10cm x 5cm has a mass 850gm . find the density of wood?
Lelechka [254]

Answer:

Explanation:

Density = Mass / Volume = 850 / 40*10*5 = 0.425 g /cm^3

5 0
3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
3 years ago
What momentum does a 70 kg person running 10m/s (a fast sprint) have ?
gogolik [260]

Momentum = (mass) x (speed)

Momentum = (70 kg) x (10 m/s)

<em>Momentum = 700 kg-m/s</em>

8 0
3 years ago
;-; please help me....​
sineoko [7]
The answer is 24N. Since the body is moving with constant velocity all the forces must balance (equal & opposite)
5 0
3 years ago
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