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Elan Coil [88]
2 years ago
10

Running water has materials such as dirt, sand, and dead plants and animals in it. When this water ends up in a lake, the materi

als it was carrying fall to the bottom of the lake and form layers. A layer is thicker when more water enters the lake. For example, thick layers form during times of heavy rain, and thin layers form during times of little rain. Sometimes lakes dry up. The bottoms of dry lakes can change into rock. This rock will still have layers. A geologist studied one of these rocks made from the bottom of a lake. Which rock layer formed during the wettest season? Layer
Physics
1 answer:
defon2 years ago
7 0

Answer:

The outside layer is the wettest.

Explanation:

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Bruno the bat flies at a speed of 0.5 m/s in circle of radius 1 m. What is his acceleration?
beks73 [17]

Answer:

Acceleration is 0.25m/s^2

Explanation:

Given the following :

Speed = 0.5m/s

Radius(r) of circle = 1m

Acceleration round a circular path is given as :

a = v^2 / r

Where

a = acceleration of the body

v = speed / velocity

r = radius

Therefore,

a = v^2 / r

a = (0.5)^2 / 1

a = 0.25m/s^2

7 0
2 years ago
Maslow's hierarchy of needs assumes
dem82 [27]

Answer:

That people are motivated by a series of five universal needs.

Explanation:

6 0
2 years ago
From the center of the Earth to the moon, what should the orbital radius of such satellite be in order to stay over the same poi
yulyashka [42]

In order to have a period that matches the Earth's rotation, a satellite must be in a circular orbit, and 42,164 km from the center of the Earth.

But that's not quite enough to make sure that it always stays over the same point on the Earth's surface (and appears motionless in the sky). For that to happen, the satellite's orbit has to be directly over the Equator.

The Moon has nothing to do with any of this.

3 0
3 years ago
5.3 two 30 kg children in a 20 kg cart are stationary at the top of a hill. They start rolling down the 80 m tall hill and they
Makovka662 [10]

Answer:

<em>60008.4 J</em>

<em></em>

Explanation:

The mass of each kid = 30 kg

mass of the cart = 20 kg

The speed of the cart down the hill = 30 km/hr = 30 x 1000/3600 = 8.33 m/s

The height of the hill = 80 m

The potential energy of the boys at the top of the hill = mgh

where

m is the total mass of the kids and the cart = (30 x 2) + 20 = 80 kg

g is the acceleration due to gravity = 9.81 m/s^2

h is their height above the ground = 80 m (on the top of the hill)

substituting, we have

potential energy PE = 80 x 9.81 x 80 = 62784 J

At an instance at the bottom of the hill

their kinetic energy = \frac{1}{2} mv^2

where

v is their velocity = 8.33 m/s

m is their total mass = 80 kg

substituting, we have

kinetic energy KE = \frac{1}{2}*80*8.33^2 = 2775.6 J

Total work done on the cart is equal to the energy lost by the cart when it reached the bottom of the hill

work done by friction = PE - KE = 62784 - 2775.6 = <em>60008.4 J</em>

5 0
3 years ago
An object is dropped from a bridge. A second object is thrown downward 1.48 s later. They both reach the water 48.1 m below at t
pashok25 [27]

To solve this problem we will apply the linear motion kinematic equations. With the data provided we will calculate the time of the first object to fall. Later we will get the time difference between the two. This difference will allow us to find the free fall distance. Through the distance we will find the initial velocity, that is,

x = v_0 t +\frac{1}{2}at^2

48.1 = 0*t + \frac{1}{2} (9.8)t^2

t = 3.13s

The second object is thrown downward at one second later and it meets the first object at the water is

t' = 3.13 -1.48

t' = 1.65s

The distance of the object will travel due to free fall acceleration is

x = v_0 t+\frac{1}{2} at^2

x = 0*(1.65) +\frac{1}{2}(9.8)(1.65)^2

x = 13.34m

The distance of the object will travel due to its initial velocity is

v_0 = \frac{d_0}{t}

d_0 = v_0 t

48.1-13.34 = v_0 (1.65)

v_0 = 21.06m/s

Therefore the initial speed of the second object is 21.06m/s

8 0
3 years ago
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