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3 years ago

How is desert pavement formed?

1 answer:

mihalych1998 [28]3 years ago

5 0

Desert pavement or commonly called stone or stony pavement is a type of desert surface that is covered with pebbles and stone tight and intertwined. Many of these layers form a type of pavement known as mosaic. When the particles begin to be stacked between different natural agents such as rain, gravity and animals compact the layers and generates a layer that allows the pavement not to erode. Basically runoff and deflation eliminate silt and clay, leaving thicker particles behind.

The correct answer is A.

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How many turns should a 10-cm long solenoid have if it is to generate a 1.5 x 10-3 t magnetic field on 1.0 a of current?

stepan [7]

We can answer this problem using Ampere’s Law:

Bh = μoNI

Where:

B = Magnetic Field

h = coil length

μo = permeability =4π*10^-7 T·m/A

N = number of turns

I = current

It is given that B=0.0015T, I=1.0A, h=10 cm = 0.1m

Use Ampere's law to find # turns:
Which can be rewritten as:
N = Bh/μoI 
N = (0.0015)(0.1)/(4π*10^-7)(1.0)
N = 119.4

Answer: 119.4 turns

3 0

3 years ago

Consider a steel guitar string of initial length L=1.00L=1.00 meter and cross-sectional area A=0.500A=0.500 square millimeters.

erma4kov [3.2K]

Answer:

The extent to which it would stretch is $\Delta L = 0.015 \ m$

Explanation:

From the question we are told that

The initial length is $L = 1.00m$

The area is $A = 0.500 mm^2 = \frac{0.500}{1 *10^6} = 0.500*10^6 \ m^2$

The Young modulus of the steel is $Y = 2.0*10^{11} Pa$

The tension is $T =1500 N$

The Young modulus is mathematically represented as

$Y = \frac{\sigma}{e}$

Where $\sigma$ is the stress which is mathematically represented as

$\sigma = \frac{F}{A}$

Substituting values

$\sigma = \frac{1500}{0.500*10^{-6}}$

$\sigma = 3.0*10^9 N/m^2$

And e is the strain which is mathematically represented as

$e = \frac{\Delta L}{L }$

Where $\Delta L$ The extension of the steel string

Substituting these into the equation above

$Y = \frac{3.0*10^9}{\frac{\Delta L}{L} }$

Substituting values

$2.0 *10^{11} = \frac{3.0*10^9}{\frac{\Delta L}{L} }$

$\Delta L = \frac{3.0*10^9 * 1}{2.0 *10^{11}}$

$\Delta L = 0.015 \ m$

7 0

3 years ago

A dumbbell-shaped object is composed by two equal masses, m, connected by a rod of negligible mass and length r. If I1 is the mo

Sati [7]

Answer:

The correct answer to the following question will be Option A (I1 > I2).

Explanation:

Method for moment of inertia because of it's viewpoint including object at a mean distance "r" from the axis is,

⇒ mr²

For Case 1:

Let the length of a rod be "r".

The axis passes via the middle of that same rod so that the range from either the axis within each dumbbell becomes " $\frac{r}{2}$ ".

Now,

Now total moment of inertia = sum of inertial moment due to all of the dumbbell

⇒ $l1=m(\frac{r}{2})^2+m(\frac{r}{2})^2$

⇒ $\frac{mr^2}{2}$

For Case 2:

Axis moves via one dumbbell because its range from either the axis becomes zero (0) and its impact is zero only at inertia as well as other dumbbell seems to be at a range "r" from either the axis

Now,

Total moment of inertia = moment of inertia of dumbbell at distance "r".

$l2=mr^2$