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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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A gravitational blank exist between you and every object in the universe
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I think it might be pull
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3 years ago
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a spiral spring of natural length 10.0cm has a scale pan hanging freely in its tower ends . When an object of mass of 20g is pla
weqwewe [10]

Answer:

Explanation:

There seems to be a typo in the problem statement.  It says the spring stretches to a shorter length after more mass is added.  Please check the problem statement.  I'm going to do the calculations assuming that the first length should be 11.80 cm and the second length should be 12.05 cm.

Hooke's law states that the force needed to compress or extend a linear spring is:

F = kΔx, where k is the stiffness and Δx is the displacement.

When a 20g object is placed in the pan, the spring stretches to a length of 11.80 cm.  The force of the spring is counteracting the weight of both the pan and the object. Therefore:

(m + 0.020) g = k (0.1180 - 0.100)

And when another 30g object is placed in the pan, the spring stretches to a length of 12.05 cm.

(m + 0.020 + 0.030) g = k (0.1205 - 0.100)

We now have two equations and two variables.  If we divide the second equation by the first equation:

(m + 0.050) / (m + 0.020) = (0.1205 - 0.100) / (0.1180 - 0.100)

(m + 0.050) / (m + 0.020) = 0.02050 / 0.0180

0.0180 (m + 0.050) = 0.02050 (m + 0.020)

0.0180 m + 0.0009 = 0.02050 m + 0.00041

0.00049 = 0.0025 m

m = 0.196

The pan has a mass of 0.196 kg, or 196 g.

4 0
3 years ago
Two light bulbs are 2.0 m apart. From what distance can these light bulbs be marginally resolved by a small telescope with a 4.5
andrezito [222]

Answer:

R = 1.2295 10⁵  m

Explanation:

After reading your problem they give us the diameter of the lens d = 4.50 cm = 0.0450 m, therefore if we use the Rayleigh criterion for the resolution in the diffraction phenomenon, we have that the minimum separation occurs in the first minimum of diffraction of one of the bodies m = 1 coincides with the central maximum of the other body

            θ = 1.22 λ / D

where the constant 1.22 leaves the resolution in polar coordinates and D is the lens aperture

             

how angles are measured in radians

          θ = y / R

where y is the separation of the two bodies (bulbs) y = 2 m and R the distance from the bulbs to the lens

            \frac{y}{R} = 1.22 \frac{ \lambda}{D}

            R = \frac{ y \ D}{1.22 \lambda}

let's calculate

            R = \frac{ 2 \ 0.045}{ 1.22 \ 600 \ 10^{-9}}

            R = 1.2295 10⁵  m

3 0
2 years ago
How much work is done? A Net Force of 9.0 N acts through a distance of 3.0 m in a time of 3.0 s. The answers are 3.0 J, 9.0 J, 2
Vadim26 [7]
If the force and the motion are along the same direction (like it is here) then work is force*distance.  The time doesn't come into play until you want the power used.  So here
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5 0
3 years ago
When a certain air-filled parallel-plate capacitor is connected across a battery, it acquires a charge of magnitude 172 μC on ea
crimeas [40]

Answer:

k = 2.279

Explanation:

Given:

Magnitude of charge on each plate, Q = 172 μC

Now,

the capacitance, C of a capacitor is given as:

C = Q/V

where,

V is the potential difference

Thus, the capacitance due to the charge of 172 μC will be

C = \frac{(172\ \mu C)}{V}

Now, when the when the additional charge is accumulated

the capacitance (C') will be

C' = \frac{(172+220)\ \mu C)}{V}

or

C' = \frac{(392)\ \mu C)}{V}

now the dielectric constant (k) is given as:

k=\frac{C'}{C}

substituting the values, we get

k=\frac{\frac{(392\ \mu C)}{V}}{\frac{(172)\ \mu C)}{V}}

or

k = 2.279

6 0
3 years ago
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