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Tcecarenko [31]
3 years ago
6

During which month is the bacteria population just under 1 million?

Physics
1 answer:
Thepotemich [5.8K]3 years ago
5 0
D. March because it is just below the 1 million marker on the graph and it is the only one that low.
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While passing through a human cell, an x-ray photon interacts with and inactivates the cell’s master molecule. What is the conse
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Read 2 more answers
The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion.
lesya [120]

Answer:

(a) m = 1.6 x 10²¹ kg

(b) K.E = 2.536 x 10¹¹ J

(c) v = 7.12 x 10⁵ m/s

Explanation:

(a)

First we find the volume of the continent:

V = L*W*H

where,

V = Volume  of Slab = ?

L = Length of Slab = 4450 km = 4.45 x 10⁶ m

W = Width of Slab = 4450 km = 4.45 x 10⁶ m

H = Height of Slab = 31 km = 3.1 x 10⁴ m

Therefore,

V = (4.45 x 10⁶ m)(4.45 x 10⁶ m)(3.1 x 10⁴ m)

V = 6.138 x 10¹⁷ m³

Now, we find the mass:

m = density*V

m = (2620 kg/m³)(6.138 x 10¹⁷ m³)

<u>m = 1.6 x 10²¹ kg</u>

<u></u>

(b)

The kinetic energy will be:

K.E = (1/2)mv²

where,

v = speed = (1 cm/year)(0.01 m/1 cm)(1 year/365 days)(1 day/24 h)(1 h/3600 s)

v = 3.17 x 10⁻¹⁰ m/s

Therefore,

K.E = (1/2)(1.6 x 10²¹ kg)(3.17 x 10⁻¹⁰ m/s)²

<u>K.E = 2.536 x 10¹¹ J</u>

<u></u>

(c)

For the same kinetic energy but mass = 77 kg:

K.E = (1/2)mv²

2.536 x 10¹¹ J = (1/2)(77 kg)v²

v = √(2)(2.536 x 10¹¹ J)

<u>v = 7.12 x 10⁵ m/s</u>

7 0
3 years ago
A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and
monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

\mu mg=\dfrac{mv^2}{r}

v is the speed of cat, v=\dfrac{2\pi r}{t}

\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0
3 years ago
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