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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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If a 400-mm diameter pipe with a pipe roughness coefficient of 100 flows full of pressurized water with a head loss of 0.4 ft pe
RoseWind [281]

Answer:

Q = 913.9 gpm

Explanation:

The Hazen Williams equation can be written as follows:

P = \frac{4.52\ Q^{1.85}}{C^{1.85}d^{4.87}}

where,

P = Friction Loss per foot of pipe = \frac{0.4}{1000\ ft} = 4 x 10⁻⁴

Q = Flow Rate in gallon/min (gpm) = ?

d = pipe diameter in inches = (400 mm)(0.0393701 in/1 mm) = 15.75 in

C = roughness coefficient = 100

Therefore,

4\ x \ 10^{-4} = \frac{4.52\ Q^{1.85}}{(100)^{1.85}(15.75)^{4.87}}\\\\Q^{1.85} = \frac{4\ x \ 10^{-4}}{1.33\ x\ 10^{-9}} \\\\Q = (300384.75)^\frac{1}{1.85}

<u>Q = 913.9 gpm</u>

5 0
3 years ago
A ball traveling at a speed ν0 rolls off a desk and lands at a horizontal distance x0 away from the desk, as shown in the figure
klasskru [66]

Answer:

3x_0

Explanation:

The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by

d=v_x t

where

v_x is the horizontal velocity

t is the time of flight

The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.

In the first situation, the horizontal distance covered is

d=v_0 t = x_0

in the second case, the horizontal velocity is increased to

v_x' = 3v_0

And so the new distance travelled will be

d' = v_x' t = 3 v_0 t = 3 x_0

So, the distance increases linearly with the horizontal velocity.

5 0
3 years ago
The blades in a blender rotate at a rate of 6100 rpm. When the motor is turned off during operation, the blades slow to rest in
MissTica

Answer:

<em>155.80rad/s</em>

Explanation:

Using the equation of motion to find the angular acceleration:

\omega_f = \omega_i + \alpha t

\omega_f is the final angular velocity in rad/s

\omega_i  is the initial angular velocity in rad/s

\alpha is the angular acceleration

t is the time taken

Given the following

\omega_f = 6100rpm

Time = 4.1secs

Convert the angular velocity to rad/s

1rpm = 0.10472rad/s

6100rpm = x

x = 6100 * 0.10472

x  = 638.792rad/s

\omega_f = 638.792rad/s\\

Get the angular acceleration:

Recall that:

\omega_f = \omega_i + \alpha t

638.792 = 0 + ∝(4.1)

4.1∝ = 638.792

∝ = 638.792/4.1

∝ = 155.80rad/s

<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>

5 0
2 years ago
You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh
lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

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

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}

And the force on the charge due to the charge on the north is,

\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of 45^\circ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

brainly.com/question/2037071

#SPJ4

3 0
2 years ago
A physicist found that a force of 1.32 N was measured between two charged spheres. The distance between the spheres was 95 cm. C
labwork [276]

The electric force (and the gravitational force too) is inversely proportional
to the square of the distance between the objects involved.

In this question, the distance is increased by a factor of  (1.25/0.95) .

So the electric force will change by the factor of  (0.95/1.25)² .

The new force is

           (1.32 N) · (0.95/1.25)²  =  0.762... newton   (rounded)
5 0
3 years ago
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